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Transcript
```A Course in Modal Logic
（2007）
LI Xiaowu
Institute of Logic and Cognition
of Sun Yat-sen University
1
Abstract
This book is an introductory text in Modal logic, the logic of necessity
and possibility. More formally speaking, Modal logic is the study of
validities of formalized inferences (inference forms) expressed by the
modal language containing the operators of necessity and possibility.
It is the foundation of philosophical logics, such as temporal logic
(tense logic, time logic), deonic logic, epistemic logic (doxastic logic),
dynamic epistemic logic (dynamic doxastic logic), conditional logic,
intuitionist logic, nonmonotonic modal logic, update logic, and so on.
More generally, it is the foundation of logics with operators which are
not seen as truth connectives, such as dynamic logic, logic of
provability, logic of language, hybrid logic, arrow logic, logic of
game, and so on. As a course, it is one of the foundations of logic
students to learn. We will introduce the basic content of modal logic
at the sentence (proposition) level.
2
Contents
Preface ................................................................... 5
Chapter 1 Axiomatic Systems ............................. 7
§1 Axiomatic Systems,
Harmony and
Coherence ..................................................................... 8
§2 Elementary Systems............................................... 33
§3 Basic Systems ........................................................ 56
§4 Degenerate Systems ............................................... 92
§5 Other Important Systems ..................................... 104
Chapter 2 Strong Derivation, Consistency and
Maximal Consistency ....................................... 111
§1 Strong Derivation................................................. 112
§2 S-Consistency and Maximality ............................ 116
Chapter 3 Relation Semantics and Soundness
Theorem ............................................................ 139
§1 Relation Semantics............................................... 140
§2 Correspondence Theorem .................................... 153
§3 Soundness Theorem ............................................. 164
§4 Countermodel Method ......................................... 178
Chapter 4 Canonical Model and Completeness
3
Theorem ............................................................ 187
§1 Completeness Definition...................................... 189
§2 Canonical Model and Elementary Theorem ........ 192
§3 Completeness Theorem........................................ 197
§4 Characterization Theorem.................................... 204
§5 Characterization Theorem of KG mn
jk .................... 223
Chapter 5 Finite Model Property
and
Decidablilty ....................................................... 233
§1 Finite Model Property
and Finite
Frame Property.......................................................... 235
§2 Filtration and Finite Model Property　............... 238
§3 Minimal Canonical Model
and Finite
Model Property ......................................................... 254
§4 Finite Model Property and Decidablilty .............. 266
§5 A System without
the Finite
Model Property ......................................................... 271
Chapter 6 Neighborhood Semantics............... 278
§1 Neighborhood Semantics ..................................... 280
§2 Characterization Theorem.................................... 297
§3 Filtration and Finite Model Property ................... 308
§4 Neighborhood Semantics
vs
Relation Semantics.................................................... 313
Literature .......................................................... 320
4
Postscript........................................................... 324
Preface
Modal logic is the study of validities of formalized inferences
(inference forms) expressed by the modal language containing the
operators of modality: necessity and possibility.
We first will intuitively discuss modalities and their types and
properties. So called modalities we mean states of things. For example,　(1) Earth necessarily moves around the sun,　(2) There possibly is water in Mars,　(3) You ought to learn harder than that.
The above ‘necessity’, ‘possibility’ and ‘oughtness’ are all
modalities. If we consider the modalities at sentence level only, then
the previous propositions can be transformed into the following
propositions: (1*) It is necessary that Earth moves around the sun.　(2*) It is possible that there is water in Mars.
(3*) It is ought that you learn harder than that.
This means that the modalities are acted on the sentences, rather
than diving into the sentences as (1) - (3). The words expressing modalities are called modality words, such as
5
the above ‘is necessary’, ‘is possible’ and ‘is ought’. In this course, we study two modalilies only: necessity and
possibility, at sentence level. In our thinking, there are many different necessities, such as the
mathematical necessity, the philosophical (especially, the
epistemological) necessity, the biological necessity, and the
sociological necessity. There are also many meanings of the modality, possibility, the reader
can give out some examples. From the point of view of logical semantics, modal logics is
essentially the study of the concepts of necessary truth and possible
truth. tautologousness and validity in the classical logics (the classical
propositional calculus and the first-order logic) are two concepts of
necessary truth, whereas satisfiability in it is a concept of possible
truth. In our thinking, the modalities can be also iterated. For example,　(1) if a sentence A is necessary, then the sentence ‘A is necessary’ is
also necessary, (2) if a sentence A is possible, then the sentence ‘A is possible’ is
also necessary.
In the following we will study the modal logics that allow iterated
modalities. 6
Chapter 1 Axiomatic Systems
In§1 of this chapter, we will present modal languages and modal
formulas, discuss their syntactic characteristics, introduce the axioms
and inference rules characterizing modality, and then define the
axiomatic systems (axiomatization systems) concerned in this course
and the concepts of formal proof and derivation in modal systems.
Finally, we prove that these system are harmonious and coherent.
From § 2 to the end of this chapter, we will introduce the
elementary systems, the basic systems, the degenerate systems and
other important systems. We will consider the properties of the proof
theory of these systems respectively, paying special attention to
proving some important theorems and derived rules of these systems,
as well as some important interrelations of them. We will also prove
the metatheorems for some important systems, such as Equivalent
Replacement Theorem, Dual Formula Theorem, Dual Symbol String
Theorem, Deduction Theorem, Reduction Theorem, PostCompleteness Theorem and Modal Conjunction Normal Form
Existence Theorem. Most of them are the natural generalizations of
the related theorems for the classical propositional calculus.
7
§1 Axiomatic Systems,
Harmony and Coherence
In this section, we will present modal languages and modal formulas,
discuss their syntactic characteristics, introduce the axioms and
inference rules characterizing modality, and then define the axiomatic
systems (axiomatization systems) concerned in this course and the
concepts of formal proof and derivation in modal systems. Finally, we
prove that these system are harmonious and coherent.
A language is called an object language if it is a language
describing directly objects we want to study. A language describing
an object language is called metalanguage. For simplicity, in this
course, in the sense of a metalanguage, we use ⇔ as ‘… if and only if
…’, ⇒ as ‘if …, then …’, ~ as ‘not’, ∀ as ‘for all’, ∃ as ‘there are
(is) ’, respectively.
In this course we use ω as the set of all natural numbers, and use n
< ω as n ∈ ω.
Definition 1.1.1
(1) L is a (sentence) modal similarity type ⇔ L is constituted by the
following different types of primitive symbols:
① A set of sentence symbols: At := {p0, …, pn (, …)} for n < ω; ①
② Logical symbols: ¬, ∧, □;
①
‘:= ’ is read as ‘is defined as’.
8
③ Technique symbols: ( , ).
(2) Let L be a modal similarity type. ML(L) is called the modal
language formed by L ⇔ ML(L) is inductively defined (recursively
defined) as the smallest set closed under (satisfying) the following
formation rules:
① At ⊆ ML(L); (induction base)
② A, B ∈ ML(L) ⇒ ¬A, (A ∧ B), □A ∈ ML(L). (induction step)
①
┤
Remark. (Ⅰ) By (1), a modal similarity type L has three disjoint
sorts:
¬, ∧, □;
( , ).
p0, …, pn (, …);
By ① in (1), the nonempty set At is countable, namely, At is finite
or countable infinite. Usually, the elements in At are also called
propositional variables or sentential variables or atomic formulas. If
not especially mention henceforth, we always let At be countable
infinite.
We often use ML as ML(L) if no confusion will arise. The elements
in ML are called formulas or sentences.
Simply, as in the literature, the language ML consists of a set of
formulas A, given by the following rules:
A := pn | ¬B | (B ∧ C) | □B for all pn ∈ At.
¬, ∧ and □ are signs of negation, conjunction and necessity,
respectively. ¬ and ∧ are connective symbols. The intended
interpretations of ¬A, (A ∧ B) and □A are ‘it is not the case that A’,
‘A and B’ and ‘it is necessary that A’, respectively.
¬ and ∧ usually are called truth connective symbols or proposition
Simply speaking,
a modal language = a modal similarity type + a set of modal formation rules.
①
9
connective symbols, and □ a modal symbol or (unary) modal
operator. By the previous definition, □ is also a unary connective
symbol as it connects a formula like ¬, but in this course we do not
generally interpret it as a truth connective symbol.
In this course, intuitively interpreted, we see □ as a modal
concept we will study. This concept is usually understood as some
form of ‘necessity’. Actually, people have various kinds of modal
concepts in the everyday thinking, their intuitive meanings are quite
different, but in this course, we do not generally discuss the
differences. In this course, we will mostly study the technique part of
modal logic, even if, from semantic aspect, also mostly study two
types of formal semantics. Of course, different formal semantics,
especially different axiomatic systems, characterize different modal
concepts, whereas these differences are actually supported by
different intuitive semantics (even philosophical background). The
reader who is interested in this can refer to the related literature.
(Ⅱ) The cardinality of At is finite or countable infinite, but, in fact,
most of results given in this course all hold for the language formed
by any sentence symbol set.
(Ⅲ) For any formula of the form □A, A is called the scope of □.
(Ⅳ) It is easy to see that ML is an object language. If not especially
mention henceforth, we always use metavariables p, q, r, … (with or
without subscripts) as formulas in At, where p, q, r especially denote
p0, p1, p2, respectively; use metavariables A, B, C, D, … (with or
without subscripts) as formulas in ML, and use Γ, Φ, Ψ, Θ, Ξ, …
(with or without the subscript) as formula sets, namely, subsets of ML.
The metavariables are also called metalogical variables. They are not
among the symbols of the language, but are used in talking about the
language. Hence p ∈ At, A ∈ ML and Φ ⊆ ML mean respectively that
the formula p denotes is in At, the formula A denotes is in ML, and
10
the formula set Φ denotes is contained within ML as a subset.
(Ⅴ) The formulas in ML obtained by (2) have unique readability
and enumerability, as the symbols mentioned in ① - ③ of (1)
above are different each other. Assume that a modal similarity type L
is countable infinite, namely, At is countable infinite, since every
formula in ML constituted by L is a symbol string constituted by finite
symbols, by a basic fact in the set theory, the cardinal number of (the
number of) all finite subsets of a countable infinite set is still
countable infinite, and thus we can enumerate all formulas in ML as
follows:
A1, …, An, ….
For the strict expression and proof of this fact, the reader can refer to
0.2.11 and 2.1.15 in  of LI Xiaowu and  of Enderton.
(Ⅵ) ‘ ┤’ is called end symbol. It shall be used in ends of the
following definitions, lemmas, theorems, corollaries, and so on.
Definition 1.1.2 (Abbreviation Definition)
(Df∨) (A ∨ B) := ¬(¬A ∧ ¬B),
(Df→) (A → B) := ¬(A ∧ ¬B),
(Df↔) (A ↔ B) := ((A → B) ∧ (B → A)),
(Df◇) ◇A := ¬□¬A,
(Df T) T := (p ∨ ¬ p) (= (p0 ∨ ¬ p0)),
(Df⊥) ⊥ := ¬T. ┤
Remark. (Ⅰ) Df∨, Df→, … in front of the above definitions are the
abbreviations of the related definitions, respectively. It is very
convenient to use these during proving the properties of the proof
theory of the following systems.
( Ⅱ ) ∨, →, ↔ and ◇ are signs of disjunction, implication
(conditionality), equivalence (biconditionality) and possibility,
11
respectively. ∨, → and ↔ are connective symbols. The intended
interpretations of (A ∨ B), (A → B), (A ↔ B) and ◇A are ‘A or B’, ‘A
implies B’, ‘A and B are equivalent’, and ‘it is possible that A’,
respectively.
(Ⅲ) By Df◇, it is easy to see that ◇ is the dual of □. As □,
◇ is also called modal symbol or modal operator. In the following
□ and ◇ are simply called modalities. Note that abstractly
speaking, □ and ◇ are read as box and diamond, respectively.
Convention 1.1.3
(1) In this course we allow ourselves to omit the outermost brackets
(parentheses) round any complete formula. We also usually omit the
inside parentheses subject to the convention that
¬, □, ◇, ∧, ∨, →, ↔.
are taken in this order of priority.
(2) Same connective symbol satisfies rightward associative law.
For example, we use
A1 → A2 →…→ An − 1 → An
as
A1 → (A2 →…→ (An − 1 → An) …).
(3) If Φ ≠ ∅(∅ denotes the empty set) is a finite set, then we
denote by ∧Φ and ∨Φ a conjunction and a disjunction of all
elements of Φ in some fixed order, respectively.
If Φ = {A}, then ∧Φ := A, ∨Φ := A.
(4) If Φ = ∅, then ∧Φ := T, ∨Φ := ⊥. ┤
Remark. Consider ∧∅ := T in (4): as far as I can see, the sufficient
and necessary condition that ∧Φ is true is
12
(☆) if A ∈ Φ, then A is true.
But when Φ = ∅, that A ∈ Φ does not hold, so (☆) vacuously holds,
so∧∅ is defined as T.
Similarly, ∨∅ is defined as⊥.
Definition 1.1.4
(1) The subformula set Sub(A) of A is the smallest set Ψ such that
① A ∈ Ψ;
② ¬B ∈ Ψ ⇒ B ∈ Ψ;
③ B ∧ C ∈ Ψ ⇒ B, C ∈ Ψ;
④ □B ∈ Ψ ⇒ B ∈ Ψ.
[This is another type of inductive definition by descending
complexity.]
(2) We use Sub(Φ) as the subformula set of Φ:
Sub(Φ) := ∪{Sub(A) | A ∈ Φ}.
(3) Φ is closed under subformula ⇔ Sub(Φ) = Φ.
(4) If A is ¬B or □B, then B is called the proper main subformula
of A. ¬ and □ here are called the main connective symbols of A,
respectively. If A is B ∧ C, then B and C are called the proper main
subformulas of A. ∧ here is called the main connective symbol of A. A
proper main subformula is simply called a main subformula. ┤
Definition 1.1.5
(1) (Substitution Definition) σ is a substitution mapping ① ⇔ σ is
a mapping from At into ML.
Given any formula A, Aσ is a substitution instance of A ⇔ Aσ is
inductively defined as follows:
①
A mapping is equal to a function defined in the set theory.
13
① pnσ = σ(pn) for all n < ω,
② (¬B)σ = ¬(Bσ),
③ (B ∧ C)σ = (Bσ) ∧ (Cσ),
④ (□B)σ = □(Bσ).
Given A, we usually use A(pn/B) as: there is a substitution mapping
σ such that A(pn/B) = Aσ, σ(pn) = B and σ(pm) = pm for all m ≠ n.
(2) (Replacement Definition) A(B∥C) is a formula obtained via
replacing some appointed subformula B in A by a formula C ⇔ A(B∥
C) is inductively defined as follows:
① if A is the appointed B, then A(B∥C) = C;
② if A = ¬D and the appointed B is in D (namely, this is not the
case of ①), then
A(B∥C) = ¬(D(B∥C));
③ if A = D ∧ E and the appointed B is in D, then
A(B∥C) = (D(B∥C)) ∧ E;
④ if A = D ∧ E and the appointed B is in E, then
A(B∥C) = D ∧ (E(B∥C));
⑤ if A = □D and the appointed B is in D, then
A(B∥C) = □(D(B∥C)). ┤
Remark. Intuitively, A(pn/B) is obtained from A by uniformly
(simultaneously) substituting B for pn, namely, A(pn/B) is the formula
obtained by using B substituting all occurrences of pn in A, whereas
A(B∥C) is obtained from A by replacing the special B by C, namely,
A(B∥C) is the formula obtained by using C substituting the special
occurrence of B in A.
Definition 1.1.6 We use PC as the system consisting of the following
14
axioms and inference rules:
(Ax1) p → q → p,
(Ax2) (p → q → r) → (p → q) → p → r,
(Ax3) (¬p → q) → (¬p → ¬q) → p,
(MP) A, A → B∕B,
(US) A∕Aσ where σ is any substitution mapping. ┤
Remark. (Ⅰ) MP is called modus ponens (or Rule of Modus Ponens
or Detachment Rule or Rule of Detachment Rule).
US is called (Uniform) Substitution Rule or Rule of (Uniform)
Substitution.
The above inference rules are also called Transformation Rules.
The inference rules can also be defined in the terms of the set
theory. Generally speaking, an inference rule can formally be defined
as a subset of the Cartesian product ML × … × ML. For instance, MP
can be defined as the following subset of the Cartesian product ML ×
ML × ML:
{<A, A → B, B> | A, B ∈ ML}.
US is defined as the following subset of the Cartesian product ML ×
ML:
{<A, Aσ> | A ∈ ML and σ is any substitution mapping}.
(Ⅱ) In the following we see PC as the smallest (vacuous) modal
system.
Before constructing various kinds of (non-vacuous) modal systems,
we first consider the following axioms and inference rules (except
RER) apparently denoting modalities. They are often used below:
Definition 1.1.7
15
(1) Axioms:
(M) □(p ∧ q) → □p ∧ □q,
(C) □p ∧ □q → □(p ∧ q),
(N) □T,
(K) □(p → q) → □p → □q,
(D) □p → ◇p,
(T) □p → p,
(4) □p → □□p,
(B) p → □◇p,
(5) ◇p → □◇p,
(G) ◇□p → □◇p,
(McK) □◇p → ◇□p,
(H) □(p ∧ □p → q) ∨ □(q ∧ □q → p),
(W) □(□p → p) → □p,
(Lem) □(□p → q) ∨ □(□q → p),
(Dum) □(□(p → □p) → p) → ◇□p → p.
(2) Inference Rules (Transformation Rules)
(RN) A∕□A,
(RM) A → B∕□A → □B,
(RE) A ↔ B∕□A ↔ □B,
(RR) A ∧ B → C∕□A ∧ □B → □C,
(RK) A1 ∧…∧ An → A∕□A1 ∧…∧ □An → □A for all n < ω,
(RER) A ↔ B∕C ↔ C(A∥B). ┤
Remark. (Ⅰ) The capital letter strings of the left-hand side of the
above axioms and rules denote their names (usually are the
abbreviations of their English names).
(Ⅱ) When n = 0, namely {A1, …, An} = ∅, by 1.1.3(4), RK is T →
16
A∕T → □A, so, in essence, RK is RN when n = 0.
(Ⅲ) In the following we simply show the names of part of the
above axioms and rules:
K is named in honour of Kripke who was the founder of modern
modal logic.
D is the abbreviation of deontic.
T is first mentioned and named by R.Feys and sometimes called
Axiom of Necessity.
5 is also denoted as E, whereas the latter is the abbreviation of
Euclidean.
B is the abbreviation of Brouwer who was the founder of
intuitionistic logic, as the system T + B mentioned below associates
with the intuitionist logic. Thus B is called the Brouwersche axiom
for the curious reason that when it is stated equivalently as
p → ¬◇¬◇p
and the modality ¬◇ is replaced by the intuitionistic negation sign, ∼,
the result is
p → ∼ ∼ p,
the intuitionistically valid version of the law of double negation.
McK is the abbreviation of McKinsey, also denoted as .1.
G is the abbreviation of Geach, also denoted as .2.
Lem is the abbreviation of Lemmon, also denoted as .3.
H is the abbreviation of Hintikka.
Dum is the abbreviation of Dummett.
W is the abbreviation of (anti-) wellordered. In the literature, W is
also denoted as G or GL where L is named in honour of M. H. Löb
who gave out the system KW mentioned below.
RN is the abbreviation of Rule of Necessitation (Necessitation
Rule).
RM is the abbreviation of Rule of Monotonicity (Monotonicity
17
Rule).
RE is the abbreviation of Rule of Equivalence (Provability
Equivalence Rule or Congruence Rule or Rule of Congruence).
RR is the abbreviation of Rule of Regular (Regular Rule).
RK is the abbreviation of Rule of Kripke (Normality Rule).
RER is the abbreviation of Rule of Equivalence Replacement
(Provability Equivalence Replacement Rule).
(Ⅳ) The rules mentioned above can also be defined in the terms of
the set theory. For instance, RN is defined as the following subset of
the Cartesian product ML × ML:
{<A, □A> | A ∈ ML}.
Definition and Convention 1.1.8 Let Γ is an axiom set (in essence, Γ
is a set of formulas), R is a set of inference rules (refer to 1.1.11(5)
below).
(1) S = <Γ; R> is a (general) modal system ⇔ S is a system
consisting of as follows:
① the axioms of S are the axioms of PC and Γ,
② the inference rules of S are MP, US and R.
(2) Γ here is called the character axiom set of S (w.r.t PC), R is
called the character (inference) rule set of S (w.r.t PC). MP, US and
all rules of R are called the primitive inference rules of S. ┤
Remark. (Ⅰ) For convenience, when Γ or R is the empty set, we
use S = <R> or S = <Γ> as S = <∅; R> or S = <Γ; ∅>, respectively.
Note that when Γ and R are both the empty set, S = <∅; ∅> = PC.
(Ⅱ) Given any modal system S = <Γ; R>. Let Γ1 be a new axiom
set and R1 a new inference rule set. For convenience, we also use S +
Γ1 + R1 as <Γ ∪ Γ1; R ∪ R1>.
Let A be a new axiom, we also simply use SA as S + A.
18
We still use S − A, S − Γ and S − R as the systems obtained by
eliminating axiom A, axiom set Γ and rule set R from S, respectively.
The modal systems given in the following definition are central
studying objects in this course.
Definition 1.1.9
E := <RE>,
M := EM,
R := MC,
K := <K; RN>,
D := KD,
T := KT,
B := TB,
S4 := T4,
S5 := T5,
K4.3 := K4Lem,
S4.1 := S4McK,
S4.2 := S4G,
S4.3 := S4Lem. ┤
In the following we introduce the concepts of formal proof,
theorem and derivation, respectively:
Definition 1.1.10 Let S be any modal system.
(1) A formula sequence A1, …, An is a formal proof in S ⇔ for all i
such that 1 ≤ i ≤ n, one of the following conditions at least holds:
19
① Ai is a substitution instance of an axiom of S ① ;
② there are some j, k < i such that Ai is obtained from Aj and Ak
by MP, namely, there are some j, k < i such that Aj = Ak → Ai;
③ there is some j < i such that Ai is obtained from Aj by a
character rule (modal rule) of S.
(2) A has a formal proof in S ⇔ there is a formal proof A1, …, An in
S such that An = A.
(3) A is a theorem of S, denoted as ⊢ S A, ⇔ A has a formal proof in
S.
If A is a theorem of S, then we also say that S has the theorem A.
We use Th(S) as the set of all theorems of S.
(4) If A ∉ Th(S), then we also use ⊬ S A as A ∉ Th(S). A is called a
non-theorem of S.
If A ∉ Th(S), then we also say that S has no theorem of the form A.
(5) A and B are (provably) equivalent in S ⇔ ⊢ S A ↔ B. ┤
Remark. (Ⅰ) The above conditions ② - ③ can be represented
simply as:
Ai is obtained from some formulas in front of it by a rule of S.
In other words，
Ai is derived from one or more formulas occurring earlier in the
sequence, by one of inference rules of S.
(Ⅱ) Consider the following definition:
1.1.10 (1*) A formula sequence A1, …, An is a formal proof in S ⇔
for all i such that 1 ≤ i ≤ n, one of the following conditions at least
holds:
In other words, there is an axiom A of S and a substitution function σ
such that Ai = Aσ. in other words, Ai is an axiom schema.
①
20
① Ai is an axiom of S;
② there are some j, k < i such that Ai is obtained from Aj and Ak by
MP;
③ there is some j < i such that Ai is obtained from Aj by US,
namely, there is some j < i and a substitution mapping σ such that Ai =
Ajσ;
④ there is some j < i such that Ai is obtained from Aj by a character
rule of S.
It is provable that A has a formal proof in S defined by 1.1.10 (1)
⇔ A has a formal proof in S defined by 1.1.10 (1*). We leave the
details of the equivalence proof as an exercise.
(Ⅲ) The theorems of S are those formulas which can be derived
from its axioms by applying its inference rules.
(Ⅳ) It is easy to see that any axiom of S itself is also a theorem of S.
(Ⅴ) Let A1, …, An be a formal proof in S. Then for all i such that 1
≤ i ≤ n, the sequence A1, …, Ai is also a formal proof in S, and every
Ai is a theorem of S.
Definition 1.1.11
(1) A formula sequence A1, …, An is a derivation from Φ in S ⇔
for all 1 ≤ i ≤ n, one of the following conditions holds:
① Ai is a substitution instance of an axiom of S;
② Ai ∈ Φ;
③ there are some j, k < i such that Ai is obtained from Aj and Ak
by MP;
④ there is some j < i such that Ai is obtained from Aj by a
character rule of S.
Here Φ is called the hypothesis set of the derivation A1, …, An.
(2) A has a derivation from Φ in S or A is derived from Φ in S or A
21
is inferred from Φ in S or A is deduced from Φ in S, denoted as Φ ⊢ S
A, ⇔ there is a derivation A1, …, An from Φ in S such that An = A.
The sequence A1, …, An is also called a derivation of from Φ to A
in S or a derivation of A from Φ in S .
For convenience, below we use Φ ⊢ SΨ as
Φ ⊢ S A for all A ∈ Ψ.
We also use CnS(Φ) as {A ∈ ML | Φ ⊢ S A}, and CnS(Φ) is called
the provable consequence set of Φ in S.
(3) If A ∉ CnS(Φ), namely, A have no derivation from Φ in S, then
we also use Φ ⊬ S A as A ∉ CnS(Φ).
(4) If Φ = {A1, …, An}, then we use A1, …, An ⊢ S A as Φ ⊢ S A, and
A1, …, An ⊬ S A as Φ ⊬ S A.
(5) R is a derived rule of S ⇔ R can be derived from the theorems
and rules of S. In other words, given any the premise(s) of R, the
conclusion of R can be derived from the theorems and rules of S.
(6) A rule R = A1, …, An∕A of S is strong, denoted as R ∈ Rule(S),
⇔ A1, …, An ⊢ S A.
In the terms of the set theory, a strong rule can formally be defined
as:
n+1
{< A1, …, An, A> ∈ ML
| A1, …, An ⊢ S A}.
A strong rule of S is also called a rule of S if no confusion will
arise.
(7) A rule R = A1, …, An∕A of S is weak, denoted as R ∈
WRule(S), ⇔ if ⊢ S A1, …, and ⊢ S An, then ⊢ S A.
(8) A1, …, An∕A is a virtual rule of S ⇔ there is some 1 ≤ i ≤ n
such that S has no theorem of the form Ai.
Otherwise, A1, …, An∕A is an actual rule of S. ┤
22
Remark. (Ⅰ) We often omit the subscript ‘S’ if no confusion will
arise.
(Ⅱ) Let A1, …, An be a derivation from Φ in S. Then for all i such
that 1 ≤ i ≤ n, the sequence A1, …, Ai is also a derivation from Φ in S.
(Ⅲ) It is clear that CnS(∅) = Th(S).
(Ⅳ) According to (6) and (1), MP is a strong rule of S, so is a
character rule of S, and so is a rule which can be derived from such
rules and the theorems of S. In other words, a strong rule is a rule
which can be applied to the elements in Φ of a derivation from Φ in S
or to the formulas which can be derived from the elements in Φ
directly or indirectly.
According to (7), a weak inference rule is a rule which can be only
applied to the theorems of S, generally not to the elements in Φ of a
derivation from Φ in S or to the formulas which can be derived from
the elements in Φ directly or indirectly. For example, in the previous
definition, Substitution Rule can be only applied to the axioms of S
(in the following we shall prove that Substitution Rule can be only
applied to the theorems of S). Otherwise, for example, we have
p ⊢ ¬ p or p ⊢ ⊥,
whereas these derivations are not what we want.
(Ⅴ) In the following we use RPC as (any rule of) Rule(PC).
(Ⅵ) If R ∈ (W)Rule(S), then we say that S has R; otherwise, we
say that S does not have R.
(Ⅶ) The rules we concern except MP are all single premise rules.
For a single premise rule A∕B, we have
A∕B is a virtual rule of S ⇔ S has no theorem of the form A.
So a virtual rule is always a (vacuous) weak rule.
For convenience, we divide the above modal systems into the
23
following several classes:
Definition 1.1.12 Let S be a modal system.
(1) S is a congruence system ⇔ S has Rule RE.
(2) S is a monotonic system ⇔ S has Rule RM.
(3) S is a regular system ⇔ S has Rule RR.
(4) S is a normal system ⇔ S has Rule RK.
(5) E, M and R are called the elementary systems,
K, D, T, B, S4 and S5 are called the basic systems. ┤
Definition 1.1.13 Let S and S1 be two modal systems.
(1) S is a subsystem of S1 or S1 is an extension system of S, denoted
as S ⊆ S1, ⇔ Th(S) ⊆ Th(S1).
(2) S is a proper subsystem of S1 or S1 is a proper extension system
of S, denoted as S ⊂ S1, ⇔ Th(S) ⊂ Th(S1).
(3) S and S1 are (deductively) equivalent, denoted as S = S1, ⇔
Th(S) = Th(S1). ┤
Remark. In order to prove S ⊆ S1, it suffices to show that the
axioms and primitive rules of S are the theorems and (derived) rules
of S1.
Definition 1.1.14 Let S be a modal system.
(1) S is harmonious ⇔ there is not a formula A such that A and ¬A
are both the theorem of S.
(2) S is coherent ⇔ there is a formula A such that A is not a
theorem of S. ┤
First, we will show that the systems defined in 1.1.9 are
harmonious and coherent. For this, we first give out a definition:
24
Definition 1.1.15 Given any formula A, the PC-transformation (A)
of A is inductively defined as follows:
－□
= pn for all n < ω,
(1) (pn)
－□
－□
(2) (¬B)
= ¬(B) ,
－□
－□
－□
(3) (B ∧ C)
= (B)
∧ (C) ,
－□
－□
(4) (□B)
= (B) . ┤
－□
－□
is the formula obtained by deleting each
Remark. (Ⅰ) (A)
－□
occurrence of □ in A where ◇ is seen as ¬□¬. We use A as
－□
(A)
if no confusion will arise.
－□
(Ⅱ) (…)
is actually an interpretation mapping from a modal
language ML into a classical sentence language.
Theorem ① 1.1.16 Let S = <Γ; R> be a system defined in 1.1.9.
(1) S is harmonious.
(2) S is coherent.
Proof. (1) Examine the axioms one by one in Γ, it is easy to see that
－□
① If A is a substitution instance of an axiom in Γ, then ⊢ PC A . ②
Examine the rules in R one by one, it is easy to see that
－□
－□
－□
② If B1, …, Bn∕C ∈ R, then B1 , …, Bn ∕C ∈ RPC.
We will show that
－□
③ ⊢ S A ⇒ ⊢ PC A .
Assume that ⊢ S A. Then A has a formal proof A1, …, An in S. By
induction on 1 ≤ i ≤ n, we will show that
Don’t confuse Theorem here and a theorem of a system!
Among the axioms introduced heretofore, there is just W such that W－□ is
no tautology.
①
②
25
④ ⊢ PC Ai .
If Ai is a substitution instance of an axiom of PC, then we get easily
④.
If Ai is a substitution instance of an axiom in Γ, then by ① we get
easily ④.
If Ai is obtained from the formulas in front of it by MP, then we get
easily ④ by the induction hypotheses.
If Ai is obtained from the formula in front of it by a rule in R, then
we get easily ④ by the induction hypothesis and ②.
Thus we get ③.
Hypothesize that S is not harmonious, then there is some A such
that ⊢ S A and ⊢ S ¬A. By ③, we have
⊢ PC A－□ and ⊢ PC ¬ A－□,
(2) By (1). ┤
－□
Remark. (Ⅰ) There are two types of methods which are usually
use to prove that S is harmonious. One is just the above method –
Deleting Modality Method. ① According to the above ③ , by
deleting the modal operators of formulas in Th(S), we can reduce the
harmony of S to the harmony of PC.
Other method shall be introduced below.
(Ⅱ) In the following we will still introduce other systems. Most of
them can be proved harmonious and coherent by using this method.
In this course, there are a lot of exercises. By taking the exercises,
This method is usually used in the predicate logics. Of course, where the
things we delect are quantifiers.
①
26
the reader can consolidate own knowledge and methods learnt. The
best method of enhancing logical skills is doing exercises as usual as
possible.
Logical exercises usually be divided into two class: calculation
exercises and proof exercises.
Here, by calculation we mean that given a certain formula A (and
formula set Φ), find out a certain formal proof of A (or a certain
derivation of A from Φ ).
Usually, there are not definite laws to do calculation exercises. Art
comes from skill. The more you try, the more better you are.
There are generally the following several steps to do calculation
exercises:
Step 1: Compare the result we will prove with the results we have
proven, look which of the latter is the most approximative to the
former by their formal structures?
Step 2: Find out whether there is a lack of which middle result to
obtain the result we will prove.
Step 3: First consider the middle result. If we can prove it then first
prove it, and hence the result we will prove . Otherwise, enter into
Step 4: Find out whether there is a lack of which second-level of
middle result to obtain the first-level of middle result. If we can prove
the second-level of middle result then first prove it, and hence again
the first-level of middle result. Otherwise, continue this process.
In general case, we can always trace to the result we have proven,
even axiom.
It is usually more difficult to do proof exercises than calculation
exercises. There are generally the following several methods to do
proof exercises:
(1) Definition proof Method. In order to do proof exercises is to
make the reader understood the exact meanings of the basic concepts
27
given via definition from the different angles. For such exercises, the
reader can prove them as long as carefully analyses the definitions.
(2) Induction proof Method. There are some exercises that are more
abstract or more general than others. For such exercises, the reader
should first prove the induction case, and hence prove the induction
step by the induction hypothesis (hypotheses). The induction methods
we usually use are Formula Structure Induction (Formula Complexity
Induction, Formula Length Induction) and Sequence Length Induction.
The former is an induction on the structure (complexity, length) of a
formula; the latter is an induction on the length of a sequence
(specially, a formal proof or a derivation). Both such methods can be
boiled down to an induction on natural numbers, namely, consider n
running over all natural numbers or all natural numbers in a range.
The key to use an induction method is to grasp the induction
hypothesis (hypotheses), namely, to correctly understand what is (are)
the associated induction hypothesis (hypotheses).
(3) Reduction to Absurdity (reductionto absurdity). In order to
prove a proposition ϕ, first assume that ϕ does not hold, and hence by
the other assumptions given and the results grasped before if there are,
try to educe a contradiction, and thus indirectly prove ϕ.
(4) Proof Method via Contraposition. Many results we will prove
are the conditionals of the form:
ϕ ⇒ ψ where ϕ and ψ are two propositions
Sometimes, it is not easy to prove that ψ holds after assuming that ϕ
holds. Here it suffices to assume that ψ does not hold, then try to
prove that ϕ does not hold. The method is also called Proof Method
(5) Proof Method via Exhaustion Cases. Sometimes, there are
several possiblites for a proposition ϕ we will prove. We have to
prove that ϕ holds in each case.
(6) Constructivity Proof Method. In order to prove the result we
28
will prove, we will construct a thing (usually is a set). This method is
usually applied to existence proofs.
(7) Falsification Method via Giving an Counterexample. Give out
an counterexample to falsify a general proposition (such a
proposition can be represented apparently or impliedly as a
universally quantified proposition). Such a counterexample is
required to be simple to the best of one's abilities. This method is
general proposition holds. If you feel that it does not hold, it is a
quite good choice to use Falsification Method via Giving an
Counterexample. For example, in the fourth section of Chapter 3, we
will use the countermodel method to determine the proper inclusion
relations between the familiar systems.
(8) Reduction Method. Via the method, we can reduce unknown
things to known things, or complex things to simple things. For
example, in 1.1.16, we reduced the harmony of some modal system S
to the harmony of PC. The method we used in 1.3.17 – 18 and etc.
below is also a kind of reduction method.
Noteworthily, in an actual proof, sometimes we use one of the
above methods independently, sometimes simultaneously use the
above several methods.
Finally, there is also a point we have to show: to learn logic well,
the reader still need to bring up own logical imaginary power.
Sometimes, it is also a kind of good method to bring up own logical
imagination to extend a concrete example to more complex cases.
Exercises 1.1
29
1.1.17 Prove Unique Readability (Decomposition) Theorem: For
each A ∈ ML, there is and there is only one of the following
conditions is satisfied:
(1) A ∈ At,
(2) There are unique formula B ∈ ML and unique symbol O ∈ {¬,
□} such that A = OB,
(3) There are unique formula tuple <B, C> ∈ ML × ML such that A
= B ∧ C.
[Hint: refer to 2.1.15 in  of LI Xiaowu and  of
Enderton. ] ┤
1.1.18 (Induction Principle) Let ϕ be a property. Assume that
(1) Each sentence symbol in At has ϕ;
(2) If A has ϕ, then ¬A and □A have ϕ;
(3) If A and B have ϕ, then A ∧ B has ϕ.
Prove that each formula in ML has ϕ.
[Hint: refer to  of Enderton. ] ┤
1.1.19 Given any formula A. Prove that a substitution instance of a
substitution instance of A is also a substitution instance of A. ┤
1.1.20 Let S be any modal system. Prove that S has Weak Rule US:
⊢ S A ⇒ ⊢ S Aσ for any substitution mapping σ,
where ‘⊢ S A’ means that A has a formal proof in S defined by 1.1.10
(1). ┤
[The exercise show that Substitution Rule US can be applied to the
theorems of S. ]
1.1.21(Problem) Let S1 be an extension system of S. Does the
30
following implication relation hold?
R ∈ Rule(S) ⇒ R ∈ Rule(S1). ┤
1.1.22 Let S1 be an extension system of S such that S1 and S have the
same primitive inference rules, and let Φ ⊆ Ψ. Prove that
Φ ⊢ S A ⇒ Ψ ⊢ S1 A. ┤
1.1.23
(1) Let S and S1 be two modal systems such that S ⊆ S1. Prove that
(a) If S1 is harmonious, then S is harmonious.
(b) If S1 is coherent, then S is coherent.
(2) Let S be a modal system. Prove that
S is harmonious ⇔ S is coherent. ┤
1.1.24 Prove that A has a formal proof in S defined by 1.1.10 (1) ⇔ A
has a formal proof in S defined by 1.1.10 (1*). ┤
1.1.25(Problem) Let S be a modal system. Do the following
propositions hold?
(1) Th(S + A) ⊆ CnS(A).
(2) CnS(A) ⊆ Th(S + A) .
(3) Th(S + Φ) ⊆ CnS(Φ).
(4) CnS(Φ) ⊆ Th(S + Φ) .┤
1.1.26 Let S and S1 be two modal systems. Prove that all the axioms
and primitive rules of S are the theorems and (derived) rules of S1 ⇒
S ⊆ S1. ┤
1.1.27
(1) Let At given in 1.1.1 be countable infinite. Prove that the
31
cardinal number of the language ML formed by At is also countable
infinite.
(2) Let At = {p0} given in 1.1.1. How many formulas are there in
the language ML formed by At? ┤
1.1. 28(Problem) Let S be a modal system.
(1) Is A ∧ ¬A∕B a weak rule of S?
(2) Is A ∨ ¬A∕B a weak rule of S?
32
§2 Elementary Systems
From this section to the end of this chapter, we will introduce the
elementary systems, the basic systems, the degenerate systems and
other important systems. We will consider the properties of the proof
theory of these systems respectively, paying special attention to
proving some important theorems and derived rules of these systems,
as well as some important interrelations of them. We will also prove
the metatheorems for some important systems, such as Equivalent
Replacement Theorem, Dual Formula Theorem, Dual Symbol String
Theorem, Deduction Theorem, Reduction Theorem, PostCompleteness Theorem and Modal Conjunction Normal Form
Existence Theorem. Most of them are the natural generalizations of
the related theorems for the classical propositional calculus.
1. Congruence Systems
In this section, we will consider some properties of proof theory of the
congruence systems. In the following we will see, all non-vacuous
systems mentioned in this course are all congruence systems, so the
results that will be proved below hold for all non-vacuous modal
systems.
We first will prove that these systems have some important
theorems and derived rules which will be constantly used henceforth.
For convenience, we will use S(1), …, S(n) as the appointed theorems
and derived rules of the modal system S.
33
Definition 1.2.1 Given any n < ω.
(1) Inductively define □nA as follows:
n+
□0A = A, …, □ 1A = □□nA.
(2) Inductively define ◇nA as follows:
n+
◇0A = A, …, ◇ 1A = ◇◇nA. ┤
Remark. □nA and ◇nA mean that there are n □s and ◇s in
front of A, respectively.
Theorem 1.2.2
(1) The following are the derived rules and theorems of E:
E(1) A ↔ B∕◇A ↔ ◇B (= RE◇).
E(2) RER, and thus we have A ↔ B, C∕C(A∥B). ①
E(3) □p ↔ ¬◇¬ p, ¬□p ↔ ◇¬ p.
E(4) ◇p ↔ ¬□¬ p, ¬◇p ↔ □¬ p.
E(5) □n p ↔ ¬◇n¬ p for all n < ω.
E(6) ◇n p ↔ ¬□n¬ p for all n < ω.
(2) E = <RER>.
(3) ((Provability) Equivalent Replacement Theorem)
⊢ A ↔ B ⇒ ⊢ C ↔ C(A∥B),
⊢ A ↔ B and ⊢ C ⇒ ⊢ C(A∥B).
Proof. (1) Prove E(1):
① A ↔ B,
Assumption
①, RPC
② ¬A ↔ ¬B,
③ □¬A ↔ □¬B,
②, RE
④ ¬□¬A ↔ ¬□¬B,
③, RPC
In the following we call A ↔ B, C∕C(A∥B) RER too, although, strictly
speaking, it is weaker than RER w.r.t E.
①
34
⑤ ◇A ↔ ◇B.
④, Df◇
[Remark. (Ⅰ) This is a usual expression of a derivation: this
derivation can be divided into three vertical columns: the first vertical
column is composed of the reference numbers ①, ②, …which
denote the steps of the derivation; the second is the derivation itself;
the third is composed of the justifications (bases) of the derivation
(the starting point of the derivation (Assumption), axioms, inference
rules (include derived rules), the theorems proved before, the
abbreviation definitions of the form Df◇). For example, ‘①, RPC’ in
the above third vertical column means that a rule of RPC is applied to
A ↔ B mentioned in ①, ….
For simplicity, below, in the third vertical column of a formal proof
or a derivation, we often omit the reference numbers as follows if no
confusion will arise: if a formula is directly derived from the formula
denoted by the above line, then we will not write out the reference
number denoting the above line in this line. For example, all ①, ②,
③ and ④ can be omitted in the above third vertical column. Note
that for a formal proof or a derivation, the second vertical column is
just essential.
For simplicity, below we also omit the punctuation in the second
vertical column of a formal proof or a derivation.
(Ⅱ) In this course, suppose the reader is familiar with the firstorder logic, especially the most important formal proofs and
derivations in the classical propositional calculus, so, usually, we only
give out the key parts of a formal proof or a derivation below.
Sometimes, several steps are abbreviated to one step. The reader
should be able to supply the omitted parts and bases of a formal proof
or a derivation. ]
Prove E(2): Assume that in E, ⊢ A ↔ B. We will show that ⊢ C ↔
C(A∥B). Consider five cases of Replacement Definition:
35
Case 1 C is the appointed A: Then C(A∥B) = B, so, by the
assumption and Replacement Definition we have ⊢ C ↔ C(A∥B).
Case 2 C = ¬D and the appointed A is in D: By the induction
hypothesis, we have ⊢ D ↔ D(A∥B), so ⊢ ¬D ↔ ¬D(A∥B) by RPC,
hence ⊢ ¬D ↔ (¬D)(A∥B) by Replacement Definition.
Case 3 C = D ∧ E and the appointed A is in D: By the induction
hypothesis, we have ⊢ D ↔ D(A∥B), so ⊢ D ∧ E ↔ (D(A∥B)) ∧ E
by RPC ((Provability) Equivalent Replacement Theorem of PC),
hence ⊢ D ∧ E ↔ (D ∧ E)(A∥B) by Replacement Definition.
Case 4 C = D ∧ E and the appointed A is in E: By the induction
hypothesis, we have ⊢ E ↔ E(A∥B), so ⊢ D ∧ E ↔ D ∧ (E(A∥B))
by RPC, hence ⊢ D ∧ E ↔ (D ∧ E)(A∥B) by Replacement Definition.
Case 5 C = □D and the appointed A is in D: By the induction
hypothesis, we have ⊢ D ↔ D(A∥B), so ⊢ □D ↔ □(D(A∥B)) by
RE, hence ⊢ □D ↔ (□D)(A∥B) by Replacement Definition.
Prove E(3):
PC ①
① □p ↔ ¬¬□p
② □p ↔ ¬¬□¬¬ p
RER
③ □p ↔ ¬◇¬ p
Df◇
Prove E(4):
PC
① ¬□¬ p ↔ ¬□¬ p
② ◇p ↔ ¬□¬ p
Df◇
Prove E(5): if n = 0, then it holds clearly; if n = 1, then by E(3).
Assume that the result we will prove holds for n = k. Next we prove
From now on, PC in the third vertical column of a formal proof or a
derivation means that the formula in the line is a substitution instance of some
tautology.
①
36
that the result we will prove holds for n = k + 1:
Induction Hypothesis
① □kp ↔ ¬◇k¬ p
+
RE
② □k 1p ↔ □¬◇k¬ p
+
③ □k 1p ↔ ¬¬□¬◇k¬ p
RPC (or RER)
+
+
Df◇
④ □k 1p ↔ ¬◇k 1¬ p
Prove E(6):
E(5), US
① ¬◇n¬¬ p ↔ □n¬ p
② ◇n¬¬ p ↔ ¬□n¬ p
RPC
n
n
③ ◇ p ↔ ¬□ ¬ p
RER
(2) By E(2), it suffices to show that E ⊆<RER>, namely, <RER>
has RE:
① A↔B
Assumption
② □A ↔ □B
RER
(3) By E(2). ┤
Remark. By (2), we see System E is exactly the system obtained by
applying Equivalence Replacement Theorem of PC to the formula of
the form □ A, so all the congruence systems (namely, all the
extension systems of E) have Equivalence Replacement Theorem (for
ML).
Definition 1.2.3 (Dual Formula Definition) Assume that A contains
no → and ↔. ① (A)d is the dual formula of A ⇔ the following
conditions are satisfied:
(1) (pn) d = ¬pn for all n < ω,
(2) (¬B) d = ¬(B) d,
(3) (B ∧ C) d = (B)d ∨ (C)d,
①
Here we assume that A can contain ∨ and ◇.
37
(4) (B ∨ C) d = (B)d ∧ (C)d,
(5) (□B) d = ◇(B) d,
(6) (◇B) d = □(B)d. ┤
Remark. We use A d as (A)d. By the previous definition if no
confusion will arise, and say that ∧ and ∨ are dual each other, and □
and ◇ are dual each other, too. (4) and (6) in the previous definition
are not essential, adding them is just for simplicity. For example, if
not use (4), then we have
(B ∨ C) d = (¬(¬B ∧ ¬C)) d
= ¬(¬B d ∨ ¬C d) by (2) and (3),
whereas in PC, and thus in E, we have the theorem
¬(¬B d ∨ ¬C d) ↔ B d ∧ C d.
Example 1.2.4 Let A = ◇¬□◇□p, then Ad is □¬◇□◇¬ p.
┤
Remark. Ad can be intuitively understood as: first we add ¬ to the
front of A, and then move ¬ rightward, transform □ (or ◇) into
◇ (or □) when ¬ crosses over □ (or ◇), and transform ∧ (or ∨ )
into ∨ (or ∧) when ¬ crosses over ∧ (or ∨), as far as ¬ cannot be
moved rightward (namely, ¬ shall be finally moved to the front of
sentence symbols).
Theorem 1.2.5 (Dual Formula Theorem) Let S be a congruence
system. Then, in S, ⊢ A ↔ ¬Ad.
Proof. By induction on (the structure or length of) A. ①
①
Here we assume that A contains ∨ and ◇.
38
Case 1 A = pn. since ⊢ pn ↔ ¬¬ pn, it follows that ⊢ pn ↔ ¬ pnd
by Dual Formula Definition.
Case 2 A = ¬B. Then
Induction Hypothesis
① B ↔ ¬B d
② ¬B ↔ ¬¬B d
RPC
d
③ ¬B ↔ ¬(¬B)
Dual Formula Definition
Case 3 A = B ∧ C. Then
Induction Hypotheses
① B ↔ ¬B d, C ↔ ¬C d
② B ∧ C ↔ ¬B d ∧ ¬C d
RPC
③ B ∧ C ↔ ¬(B d ∨ C d)
RPC (de Morgan law)
④ B ∧ C ↔ ¬(B ∧ C) d
Dual Formula Definition
Case 4 A = B ∨ C. This proof resembles the proof of Case 3.
Case 5 A = □B. Then
Induction Hypothesis
① B ↔ ¬B d
② □B ↔ □¬B d
RE
③ □B ↔ ¬¬□¬B d RPC
④ □B ↔ ¬◇B d
Df◇
⑤ □B ↔ ¬(□B) d
Dual Formula Definition
Case 6 A = ◇B. This proof resembles the proof of Case 5. ┤
Remark. (Ⅰ) Dual Formula Theorem shows that any formula and
the negation of its dual form are equivalent.
(Ⅱ) If we see ∨ and ◇ as the abbreviation definition, without
considering Case 4 and Case 6 in the proof of the previous theorem,
we only need induction on the proper main subformulas of A when A
has no sentence symbols in itself.
Definition 1.2.6 A symbol string s of L is a modal symbol string ⇔
s is φ or a finite string consisting of some symbols from {¬, □, ◇},
39
where φ denotes the empty string (the empty, or null, modality). In
other words, s is a modal symbol string ⇔ s ∈ MSS such that MSS is
inductively defined as the smallest set closed under the following
formation rules:
① φ, ¬, □, ◇ ∈ MSS;
② s ∈ MSS and s ≠ φ ⇒ ¬ s, □ s, ◇s ∈ MSS. ┤
Remark. (Ⅰ) φ and ∅ are different, the latter denotes the empty set.
(Ⅱ) Strictly speaking, all modal symbol strings except φ are
composed of unary logic symbols. Note that ◇ is not essential.
(Ⅲ) By the previous definition, the following are all modal symbol
strings:
φ, ¬, □, ◇, □◇¬, ◇□¬□◇.
Corollary 1.2.7 The following are the theorems or derived rules of E:
E(9) A∕A(B∥¬B d), A∕A(¬B∥B d).
E(10) A ↔ (A d) d.
E(13) A → B∕B d → A d.
E(14) A ↔ B∕A d ↔ B d.
E(15) A∕A′ where A′ is obtained from A by using the following
principles some times: if sB is a subformula of A such that s contains
¬. If we move ¬ in s rightward (or leftward), then transform □ (or
◇) into ◇ (or □) when ¬ crosses over □ (or ◇), and delete ¬¬
when ¬ encounters ¬.
Proof. We exemplify E(10):
40
Dual Formula Theorem
② A ↔ (¬A) d
Dual Formula Definition
③ A ↔ (A d) d
E(8), RER
We exemplify E(15): by E(3), E(4), US and the theorems of PC,
we have in E:
(☆) ⊢ ¬□B ↔ ◇¬B, ⊢ ¬◇B ↔ □¬B, ⊢ ¬¬B ↔ B.
So by using RER some times, we easily obtain A′′ from A. ┤
Remark. (Ⅰ) In the following we shall use LMC as E(15).
(Ⅱ) Four important instances of LMC are:
① A∕A(¬□B∥◇¬B),
② A∕A(¬◇B∥□¬B),
③ A∕A(◇¬B∥¬□B),
④ A∕A(□¬B∥¬◇B).
(Ⅲ) LMC can be seen as an instance of RER.
(Ⅳ) By the proof of E(15), even we have ⊢ E A ↔ A′.
Definition 1.2.8 Let s be a modal symbol string.
(1) s is positive ⇔ ¬ occurs even times (include 0) in s.
(2) s is negative ⇔ ¬ occurs oddly times in s.
(3) s is iterative ⇔ □ or ◇ occurs n ≥ 2 times in s.
(4) s is standard ⇔ s contains no ¬ or the unique ¬ occurs only
in the most front of s.
(5) s is pure ⇔ s contains no ¬. ┤
Definition 1.2.9 (Dual Symbol String Definition) Let s be a modal
symbol string. (s)d is the dual symbol string of s ⇔ (s)d is obtained
from s by uniformly (simultaneously) substituting the each occurrence
41
of □ and ◇ in s for ◇ and □. That is, (s)d is inductively defined
as follows:
(1) (φ) d = φ,
(2) (¬t)d = ¬(t)d,
(3) (□t)d = ◇(t)d,
(4) (◇t)d = □(t)d. ┤
Remark. We use sd as (s)d if no confusion will arise.
Theorem 1.2.10 Let s and t be modal symbol strings and S a
congruence system. Then in S:
(1) ⊢ sp ↔ ¬ sd ¬ p.
(2) ⊢ ¬ s¬ p ↔ sd p.
(3) ⊢ sp ⇔ ⊢ ¬ sd ¬ p,
⊢ ¬s¬ p ⇔ ⊢ sd p.
(4) (Converse Duality Theorem) Given
(A) sp → t p,
(Ad) td p → s d p.
Then in S:
⊢ A ⇔ ⊢ Ad. ①
(5) S + A = S + Ad.
(6) (Equivalence Duality Theorem)
⊢ sp ↔ t p ⇔ ⊢ sd p ↔ t d p.
Proof. The proof of (1):
Dual Formula Theorem
① sp ↔ ¬(sp) d
② sp ↔ ¬ sd p d
Dual Formula Definition, Dual Symbol String Definition
①
Ad is called as the converse dual formula of A.
42
③ sp ↔ ¬ sd ¬ p
Dual Formula Definition
The proof of (2):
(1)
① sp ↔ ¬ sd ¬ p
d
② ¬ s¬ p ↔ s ¬¬ p
US, RPC
③ ¬ s¬ p ↔ sd p
RER
The proof of (3): by (1) and (2).
The proof of (4). ‘⇒’:
① sp → t p
A
② ¬t p → ¬ sp
RPC (Hypothetical-transposition Rule)
US
③ ¬t¬ p → ¬ s¬ p
④ t dp → sd p
(2)
‘⇐’:
① t d p → sd p
② ¬ sd p → ¬t d p
RPC
③ ¬ sd ¬ p → ¬t d ¬p
US
④ sp → tp
(1)
We get easily (5) and (6) by (4). ┤
Remark. By (5), in the congruence systems, using ◇p → □◇p as
an axiom and using ◇□p → □p as an axiom are equivalent.
Definition 1.2.11 Let S be any modal system.
(1) Let s and t be two modal symbol strings.
s and t are equivalent in S ⇔ ⊢ S sp ↔ tp.
We here say that s (or t ) is equivalent to t (or s) in S.
(2) In the following we use Modal(S) as the cardinal number of the
modal symbol strings that are not equivalent to each other in S.
We here say that E has Modal(S) modal symbol strings that are not
equivalent to each other. ┤
43
In 1.2.18 below, we will show that E has countable infinite modal
symbol strings that are not equivalent to each other.
Corollary 1.2.12
(1) ¬◇¬□ and □□ are equivalent in E.
(2) A positive modal symbol string is equivalent to a pure modal
symbol string in E.
(3) A modal symbol string is equivalent to a standard modal symbol
string in E. ┤
2. Monotonic Systems Theorem 1.2.13
(1) The following are the theorems or derived rules of M:
M(1) RM,
M(2) A → B∕◇A → ◇B (= RM◇),
M(3) ◇(p ∧ q) → ◇p ∧ ◇q,
M(4) ◇p ∨ ◇q → ◇(p ∨ q) (= C◇),
M(5) □p ∨ □q → □(p ∨ q) (= Fc),
M(6) (◇p → □q) → □(p → q).
(2) E ⊆ M.
Proof. (1) Prove M(1):
① A→B
Assumption
② A↔A∧B
RPC
③ □A ↔ □(A ∧ B)
RE
④ □(A ∧ B) → □A ∧ □B
M, US
44
⑤ □A → □B
③, ④, RPC
Prove M(2):
Assumption
① A→B
② ¬B → ¬A
RPC
③ □¬B → □¬A
RM
④ ¬□¬A → ¬□¬B
RPC
⑤ ◇A → ◇B
Df◇
By M(2), PC and RPC, we get easily M(3) and M(4).
Prove M(5):
PC, RM
① □p → □(p ∨ q), □q → □(p ∨ q)
② □p ∨ □q → □(p ∨ q)
RPC
Prove M(6):
PC
① (◇p → □q) → ¬◇p ∨ □q
② (◇p → □q) → □¬ p ∨ □q
LMC
③ (◇p → □q) → □(¬ p ∨ q)
M(5)
④ (◇p → □q) → □(p → q)
RER
(2) By 1.1.13, it suffices to show that Th(E) ⊆ Th(M), but this
holds clearly. ┤
Theorem 1.2.14 (Dual Symbol String Theorem) Let S be a
monotonic system, s and t positive modal symbol strings. Given
(A) sp → t p,
(Ad) t d p → sd p,
(RA) A → B∕sA → tB, (Positive String Monotonicity Rule) ①
(RAd) A → B∕t d A → sdB.
Then
(1) ⊢ S A ⇔ RA ∈ Rule(S),
①
It is easy to see that it is a generalization of RM and RM◇.
45
Proof. (1) ‘⇒’:
① A→B
Assumption
RM, RM◇, Hypothetical-transposition Rule ①
② t A → tB
③ s A → tA
A, US
④ s A → tB
③, ②, RPC (Syllogism Rule)
‘⇐’:
PC
① p→p
② sp → tp
RA
(2) ‘⇒’:
Assumption
① A→B
② tdA → t dB
RM, RM◇, Hypothetical-transposition Rule
d
d
③ t B→sB
④ t d A → sdB
②, ③, RPC
‘⇐’:
PC
① p→p
② t d p → sd p
Remark. The assumption of this theorem about positive modal
symbol string is essential, because ¬p → ¬q cannot be derived from p
→ q.
Corollary 1.2.15 Let S be a monotonic system. Then
S + A = S + Ad = S + RA = S + RAd.
Proof. Just apply 1.2.10(4) and the previous theorem. ┤
To prove it strictly also needs induction on the length of s. Since there are
even ¬s in s, we can be by induction for even ¬s.
①
46
Remark. According to the previous corollary, in a monotonic
system, we can use any above axioms or use any rules equivalently.
For example,
M = E + M = E + RM.
3. Regular Systems
Theorem 1.2.16
(1) The following are the theorems or derived rules of R:
R(1) □(p ∧ q) ↔ □p ∧ □q (= R),
R(2) ◇(p ∨ q) ↔ ◇p ∨ ◇q (= R◇),
R(3) ◇(p ∨ q) → ◇p ∨ ◇q (= M◇),
R(4) RR,
R(5) K,
R(6) A1 ∧…∧ An → A∕□A1 ∧…∧ □An → □A (n ≥ 1),
R(7) A → B ∨ C∕◇A → ◇B ∨ ◇C (= RR◇),
R(8) □p ∧ ◇q → ◇(p ∧ q),
R(9) ◇(p → q) ↔ □p → ◇q.
(2) M ⊆ R.
(3) R = <RR> = <RR◇> Df□ = <M◇; RM◇> Df□
= <C◇, M◇; RE◇> Df□ = <K; RM>,
where the subscript ‘Df□’ means that the primitive modal symbol of
the modal similarity type of the language describing the current
system is ◇, and the abbreviation definition is
(Df□) □A := ¬◇¬A.
Proof. (1) By Axioms C and M, it is clear that R(1) holds.
Prove R(2):
R, US
① □(¬ p ∧ ¬ q) ↔ □¬ p ∧ □¬ q
② □¬(p ∨ q) ↔ □¬ p ∧ □¬ q
RER
47
③ ¬□¬(p ∨ q) ↔ ¬□¬ p ∨ ¬□¬ q RPC
④ ◇(p ∨ q) ↔ ◇p ∨ ◇q
Df◇
Prove R(4):
① A∧B→C
Assumption
RM (1.2.13 M(1))
② □(A ∧ B) → □C
③ □A ∧ □B → □(A ∧ B)
C, US
④ □A ∧ □B → □C
③, ②, RPC
Prove R(5):
PC
① (p → q) ∧ p → q
② □(p → q) ∧ □p → □q
RR
③ □(p → q) → □p → □q
RPC
Prove R(6):
Assumption
① A1 ∧…∧ An → A
② □(A1 ∧…∧ An) → □A
RM
③ □A1 ∧…∧ □An → □A
R, RER × n − 1
Prove R(7):
① A→B∨C
Assumption
② ¬B ∧ ¬C → ¬A
RPC
RR
③ □¬B ∧ □¬C → □¬A
④ ◇A → ◇B ∨ ◇C
RPC, Df◇
Prove R(8):
K, US
① □(p → ¬ q) → □p → □¬ q
② □¬(p ∧ q) → □p → □¬ q
RER
③ ¬◇(p ∧ q) → □p → ¬◇q
LMC
④ □p ∧ ◇q → ◇(p ∧ q)
RPC
Prove R(9):
R◇, US
① ◇(¬ p ∨ q) ↔ ◇¬ p ∨ ◇q
② ◇(¬ p ∨ q) ↔ ¬□p ∨ ◇q
LMC
48
③ ◇(p → q) ↔ □p → ◇q
RER
(2) It is clear that (2) holds.
(3) By the following examples, we will prove circularly:
<RR> = <RR◇>Df□ = <M◇; RM◇>Df□.
First we will show that <RR> ⊆ <RR◇>Df□:
① A∧B→C
Assumption
② ¬C → ¬A ∨ ¬B
RPC
③ ◇¬C → ◇¬A ∨ ◇¬B
RR◇
④ ¬◇¬A ∧ ¬◇¬B → ¬◇¬C
RPC
⑤ □A ∧ □B → □C
Df□
Next we will show that <RR◇>Df□ ⊆ <M◇; RM◇>Df□:
① A→B∨C
Assumption
② ◇A → ◇(B ∨ C)
RM◇
③ ◇A → ◇B ∨ ◇C
M◇, RPC
Finally we will show that <M◇; RM◇>Df□ ⊆ <RR>.
First prove M◇:
① □¬p ∧ □¬q → □¬(p ∨ q)
PC, RR
② ¬□¬(p ∨ q) → ¬□¬ p ∨ ¬□¬ q
RPC
③ ◇(p ∨ q) → ◇p ∨ ◇q
Df◇
Next prove RM◇:
① A→B
Assumption
② ¬B ∧ ¬B → ¬A
RPC
③ □¬B ∧ □¬B → □¬A
RR
④ ¬□¬A → ¬□¬B
RPC
⑤ ◇A → ◇B
Df◇
The rest unproved are left as exercises. ┤
Remark. (Ⅰ) The restrictive condition ‘n ≥ 1’ in R(6) is essential
49
and cannot be altered as n ≥ 0. That is, R(6) ≠ RK.
(Ⅱ) <RR> and <RR◇>Df□ can be called (pairwise) dual systems. In
the following we can see that there are a lot of such dual systems.
(Ⅲ) In order to prove the equivalences of n systems, by using the
proof method of pairwise inclusion, we need to give out 2(n − 1)
proofs of inclusion relations, but by using the proof method of
circular inclusion, we need only give out n proofs of inclusion
relations.
Theorem 1.2.17 (Chellas,  (p. 243)) Let S = E + Φ such that Φ
⊆ {M, C, N}. Then S has the following weak rules (refer to 1.1.11(6)):
(1) □A∕A,
(2) ◇A∕A,
(3) □A → □B∕A → B,
(4) ◇A → ◇B∕A → B,
(5) □A ↔ □B∕A ↔ B,
(6) ◇A ↔ ◇B∕A ↔ B.
Proof. First we will define the one step deleting modality mapping δ
from ML into ML as follows:
δ(pn) = pn for all n < ω,
δ(¬A) = ¬δ(A),
δ(A ∧ B) = δ(A) ∧ δ(B),
δ(□A) = A.
It is easy to prove that
δ(A O B) = δ(A) O δ(B) where O ∈ {∨, →, ↔}.
－□
[Note that (Ⅰ) δ(◇A) = δ(¬□¬A) = ¬¬A. (Ⅱ) δ(A) and A defined
in 1.1.15 are different. For example, it is easy to see that
δ(□p → □□p) = p → □p, but
－□
(□p → □□p)
= p → p. ]
50
First we will show that S has the following weak rule:
① ⊢ A ⇒ ⊢ δ(A).
Assume that ⊢ A, then there is a formal proof A1, …, An = A in S. By
induction on the length of this proof, we will show that
② ⊢ δ(Ai) for all 1 ≤ i ≤ n.
Case 1 Ai is a substitution instance of an axiom of S: it suffices to
consider the modal axioms.
If Ai = □(B ∧ C) → □B ∧ □C is a substitution instance of Axiom
M, then δ(Ai) = B ∧ C → B ∧ C is a theorem of PC.
If Ai = □B ∧ □C → □(B ∧ C) is a substitution instance of Axiom
C, then δ(Ai) = B ∧ C → B ∧ C is a theorem of PC.
If Ai = □T is Axiom N, then δ(Ai) = T is a theorem of PC. ①
Case 2 Ai is obtained from some formulas Aj and Aj → Ai in front
of it by MP: by the induction hypotheses, we have
⊢ δ(Aj), ⊢ δ(Aj → Ai).
By the latter, we have ⊢ δ(Aj) → δ(Ai), so ⊢ δ(Ai) by MP.
Case 3 Ai is obtained from some Aj = B ↔ C in front of it by RE:
Since each of A1, …, An is a theorem, it follows that
③ ⊢ Aj .
Since δ(Ai) = δ(□B ↔ □C) = B ↔ C = Aj, it follows that ⊢ δ(Ai) by
③. ②
Thus we get ②, and thus prove that δ(A1), …, δ(An) = δ(A) is a
formal proof of δ(A) in S, so ① holds.
By ① it is easy to prove that (1) - (6). ┤
①
②
In this case we have actually proved a stronger result: ⊢ PC δ(Ai).
Note that in this case we don’t need the induction hypothesis.
51
Remark. (Ⅰ) In the previous proof, we mostly use a kind of
deleting modality method that is different to the method used in
1.1.15. In the following we shall still introduce another deleting
modality method.
(Ⅱ) In the following Exercise 6.1.27, we require the reader to
prove that E, M and R have no theorem of the form □A, so □A∕A
is also a virtual rule of these three systems.
In the following we consider the cardinal number of the modal
symbol strings that are not equivalent to each other in the above
systems.
Theorem 1.2.18 Let S = E + Φ such that Φ ⊆ {M, C, N}. Then S has
countable infinite modal symbol strings that are not equivalent to each
other.
Proof. Define another deleting modality mapping τ from ML into ML
as follows:
τ(pn) = pn for all n < ω,
τ(¬A) = ¬τ(A),
τ(A ∧ B) = τ(A) ∧ τ(B),
τ(□A) = T.
As the proof of above theorem, it is easy to prove that (Case 3 there is
simpler here), in S, we have
① ⊢ A ⇒ ⊢ τ(A).
Given any 1 ≤ m < ω, then τ(p ↔ □mp) = p ↔ T. We will show that
② ⊬ p ↔ □m p.
Hypothesize that the result doesn’t hold, then ⊢ p ↔ □m p, so, by ①,
we have ⊢ p ↔ T, so ⊢ p, hence, by US, ⊢ A for any A. So S is not
harmonious, and thus we have a contradiction.
52
By ② and 1.2.17(5), we have
③ ⊬ □n p ↔ □n + m p for all n < ω.
By ③, it is easy to get the result we will prove . That is, we have
Modal(S) = ℵ0. ┤
Corollary 1.2.19 The elementary systems have countable infinite
modal symbol strings that are not equivalent to each other. ┤
Exercises 1.2
1.2.20 Prove that E = <RE◇>Df□. ┤
1.2.21 Prove that M = <RM> = <RM◇>Df□. ┤
1.2.22
(Ⅰ) Prove that the following are the theorems or derived rules of
M:
(1) ◇(p → q) ∨ □(q → p),
(2) □p1 ∨…∨ □pn → □(p1 ∨…∨ pn) for 1 ≤ n < ω,
(3) A∕◇B → ◇(A ∧ B),
(4) ◇m(A1 ∧…∧ An) → ◇mA1 ∧…∧ ◇mAn for m, n < ω,
(5) A → B∕ sA → sB where s is a positive modal symbol string.
(Ⅱ) □A → B∕□A → □B is a derived rule of M4.
(Ⅲ) Prove that □(p → ◇p) is a theorem of M5. ┤
1.2.23
(Ⅰ) Prove that the following are the theorems or derived rules of R:
(1) □(p → q) ∧ □(q → r) → □(p → r),
53
□(p → q) ∧ ◇(p ∧ r) → ◇(q ∧ r),
◇(p → q ∧ r) → (□p → ◇q) ∧ (□p → ◇r),
□(p → q) → ◇p → ◇q,
□(p → q) ∧ □p → ◇(q → r) → ◇(p ∧ r),
◇T ↔ (□p → ◇p),
□(p ↔ q) → (□p ↔ □q),
□(p ∨ q) → ◇p ∨ □q,
□(p1 ∨…∨ pn + 1) → ◇p1∨…∨ ◇pn ∨ □pn + 1
where 1 ≤ n < ω,
(10) □(p1 ∧…∧ pn) ↔ □p1 ∧…∧ □pn where 1 ≤ n < ω,
(11) □p1 ∧…∧ □pn ∧ ◇pn + 1 → ◇(p1 ∧…∧ pn + 1)
where 1 ≤ n < ω,
(12) □□p ∧ ◇(q ∧ ◇q1 ∧…∧ ◇qn)
→ ◇(q ∧ ◇(q1 ∧ p) ∧ …∧ ◇(qn ∧ p)) where 1 ≤ n < ω,
(13) A → B1 ∨…∨ Bn∕◇mA → ◇mB1 ∨…∨ ◇mBn
where m < ω and 1 ≤ n < ω,
(14) A → B1 ∨…∨ Bn∕◇A → ◇B1 ∨…∨ ◇Bn where 1 ≤ n < ω.
(Ⅱ) Prove the unproved part of 1.2.16(3).
(Ⅲ) Prove that R + ◇(p → p) = R + D.
(Ⅳ) (Deduction Theorem for R4)
Φ ∪ {□A} ⊢ R4 B ⇒ Φ ⊢ R4 □A → B.
(Ⅴ) Prove that □p → □◇p is a theorem of R5.┤
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
From now on we mark the superscript * after the reference number
of difficult exercises:
1.2.24* Prove that □◇□◇p ↔ □◇p is a theorem of R4. ┤
[Remark. By this exercise, □◇□◇ and □◇ are equivalent in
54
R4. ]
1.2.25(Problem) Let S and S1be modal systems. Does the following
proposition hold?
S ⊆ S1 ⇔ Modal(S1) ≤ Modal(S).
55
§3 Basic Systems
In this section we will study the basic systems K, D, T, B, S4, S5 that
are six most familiar normal systems. They are also the emphasis we
will study in this course.
Theorem 1.3.1
(1) The following are the theorems and derived rules of K:
K(1) □(p → p),
K(2) RM,
K(3) RE, (and thus) RER,
K(4) □(p ∨ q) → □p ∨ ◇q,
K(5) □p ∧ □q → □(p ∧ q) (= C),
K(6) □(p ∧ q) → □p ∧ □q (= M),
K(7) □(p ∧ q) ↔ □p ∧ □q (= R).
(2) Equivalent Replacement Theorem:
⊢ A ↔ B ⇒ ⊢ C ↔ C(A∥B),
⊢ A ↔ B and ⊢ C ⇒ ⊢ C(A∥B).
(3) R ⊆ K.
(4) K = <N, R; RE> = <C, K(1); RM>
= <N; RR> = <RK>.
(5) K = EMCN = MCN = RN.
Proof. (1) Prove K(1): by RN.
Prove K(2):
① A→B
Assumption
56
② □(A → B)
RN
③ □(A → B) → □A → □B
K, US
④ □A → □B
②, ③, MP
Prove K(3): By K(2), we have RE. Like the proof of E(2) in 1.2.2,
we get easily RER.
Prove K(4):
① □(¬ q → p) → □¬ q → □p
K, US
② □(¬ q → p) → ¬□¬ q ∨ □p
RPC
③ □(p ∨ q) → □p ∨ ◇q
RER, Df◇
Prove K(5):
PC
① p→q→p∧q
② □p → □(q → p ∧ q)
RM
③ □(q → p ∧ q) → □q → □(p ∧ q)
K, US
④ □p ∧ □q → □(p ∧ q)
②, ③, RPC
Prove K(6):
□(p ∧ q) → □q
PC, RM
① □(p ∧ q) → □p,
② □(p ∧ q) → □p ∧ □q
RPC
Prove K(7): By K(5) and K(6).
(2) By 1.2.2 (3).
(3) holds clearly.
(4) We will prove it circularly. First prove K ⊆<N, R; RE>.
Prove RN:
Assumption
① A
② T↔A
RPC
③ □T ↔ □A
RE
④ □A
③, N, RPC
By the proof of M(1) in 1.2.13, we get RM easily from M and RE.
We will show that K:
PC, RM
① □((p → q) ∧ p) → □q
57
② □(p → q) ∧ □p → □q
R, RPC
③ □(p → q) → □p → □q
RPC
Next we will show that <N, R; RE> ⊆ <C, K(1); RM>: we get
easily RE by RM, get N by K(1) and RM, get M by RM (refer to the
proof of K(6)), and get R by C and M.
Next we will show that <C, K(1); RM> ⊆ <N; RR>.
Prove RM:
① A→B
Assumption
② A∧A→B
RPC
③ □A ∧ □A → □B
RR
④ □A → □B
RPC
We get K(1) by N and RM and get C by RR easily.
Next we will show that <N; RR> ⊆ <RK>. It is easy to see that RR
is an instance of RK (when n = 2). We will prove N:
① T→T
PC
RK (when n = 0, RK is T → A∕T → □A.)
② T → □T
③ □T
RPC
Finally prove <RK> ⊆ K: By induction on n.
When n = 0, RK is T → A∕T → □A, so
① T→A
Assumption
② A
RPC
③ □A
RN
④ T → □A
RPC
When n = 1, RK is RM (= K(2)).
Inductively hypothesize that when n = k, the result we will prove
holds. Then
① A1 ∧…∧ A k + 1 → B
Assumption
② A1 ∧…∧ A k → A k + 1 → B
RPC
③ □A1 ∧…∧ □A k → □(A k + 1 → B)
Induction Hypothesis
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④ □A1 ∧…∧ □A k → □A k + 1 → □B
K, RPC
⑤ □A1 ∧…∧ □A k + 1 → □B
RPC
(5) holds clearly by (4). We leave the details of the proof as an
exercise. ┤
Remark. By (4), a system having RK has K and RN, so
a normal system is always an extension system of K.
Theorem 1.3.2 The following are weak rules of K and D:
(1) □A∕A,
(2) ◇A∕A, ①
(3) □A → □B∕A → B,
(4) ◇A → ◇B∕A → B.
Proof. This proof resembles the proof of 1.2.17. So below we supply
the differences only:
…
Case 1 Ai is a substitution instance of some axiom: it suffices to
consider the modal axioms.
If Ai = □(B → C) → □B → □C is a substitution instance of K,
then δ(Ai) = (B → C) → B → C is a theorem of K.
If Ai = □B → ◇B is a substitution instance of D, then δ(Ai) = B →
¬¬B is a theorem of D.
…
Case 3 Ai is obtained from some Aj in front of it by RN: since ⊢
Aj and δ(Ai) = δ(□Aj) = Aj, it follows that ⊢ δ(Ai).
…. ┤
①
Refer to the below exercise 3.4.10 (1)①.
59
Remark. For the other basic systems, the previous theorem does not
hold. The reason is that Case 1 above cannot be passed through:
(1) □◇p → ◇p is a substitution instance of Axiom T, but below,
by using the countermodel method, we can prove that it is not
necessary that δ(□◇p → ◇p) = ◇p → ¬¬ p is a theorem of a
system containing T.
(2) □ p → □ □ p is Axiom 4, but below, by using the
countermodel method, we can prove that it is not necessary that δ(□p
→ □□p) = p → □p is a theorem of a system containing 4.
(3) ◇ p → □ ◇ p is Axiom 5, but below, by using the
countermodel method, we can prove that it is not necessary that δ(◇p
→ □◇p) = ¬¬ p → ◇p is a theorem of a system containing 5.
Here and in 1.2.17 we can see that (the substitution instances of )
Axioms C, M, N, K and D have a kind of symmetry. There is not such
symmetry for the other axioms except Axiom W mentioned in 1.1.7.
In the following we will consider Deduction Theorem for K.
Definition 1.3.3 Let A1, …, An be a derivation of from Φ to A in K.
For each 1 ≤ i, m ≤ n, Ai depends on Am in the derivation ⇔ one of the
following conditions is satisfied:
(1) Ai = Am;
(2) There are j, k < i such that Ai is obtained from Aj and Ak by MP,
and there is at least one of Aj and Ak that depends on Am in the
derivation;
(3) There is some j < i such that Ai is obtained from Aj by RN, and
Aj depends on Am in the derivation. ┤
Remark. The intuitive meaning of ‘Ai depends on Am in the
derivation’ is: Ai is Am or Ai is obtained from Am (as a premise,
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possibly still from other premises) by some steps of derivations.
Lemma 1.3.4 Assume that B does not depend on A in some derivation
of Φ ∪ {A} ⊢ K B. Then Φ ⊢ K B.
Proof. Let B1, …, Bn is a derivation from Φ ∪ {A} to B in K such that
B does not depend on A in the derivation. By induction on 1 ≤ i ≤ n, it
suffices to show that
(#) If Bi does not depend on A in the derivation, then Φ ⊢ Bi.
Assume that Bi does not depend on A in the derivation. To prove
(#), there are three cases to consider:
Case 1 Bi is a substitution instance of some axiom of K or Bi ∈ Φ
①
: then we have Φ ⊢ Bi by the definition of derivation.
Case 2 There are some j, k < n such that Bi is obtained from Bj
and Bk by MP: Bi does not depend on A in the derivation, so do Bj and
Bk. By the induction hypotheses, Φ ⊢ Bj and Φ ⊢ Bk, so, by MP, it is
easy to see that Φ ⊢ Bi.
Case 3 There is some j < n such that Bi is obtained from Bj by RN:
Bi does not depend on A in the derivation, neither does Bj. By the
induction hypothesis, Φ ⊢ Bj, so, by RN, it is easy to see that Φ ⊢ Bi.
┤
Lemma 1.3.5 Assume that Φ ∪ {A} ⊢ K B, and there is a derivation
from Φ ∪ {A} to B in K such that RN is applied to the formulas
depending on A m < ω times. Then
Φ ⊢ K □0A ∧…∧ □mA → B.
Proof. Assume that B1, …, Bn is a derivation from Φ ∪ {A} to B in K
Since Bi does not depend on A in the derivation, we need not to consider
the case of Bi = A.
①
61
such that RN is applied to the formulas depending on A m < ω times.
By induction on 1 ≤ i ≤ n, we will show that in K,
(☆) Φ ⊢ □0A ∧…∧ □hA → Bi,
where h is the number of formulas depending on A to which RN is
applied in the derivation B1, …, Bi. So h = m when i = n.
Case 1 Bi is a substitution instance of an axiom of K: Then it is
easy to see that Φ ⊢ Bi, by RPC (Implication Introduce Rule), it is
easy to prove that Φ ⊢ A → Bi. Since A = □0A, it follows that
Φ ⊢ □ 0 A → Bi .
We do not use RN, so (☆) holds.
Case 2 Bi ∈ Φ ∪ {A}: If Bi ∈ Φ, then Φ ⊢ Bi. The rest of this
subcase is as the proof of Case 1. If Bi = A, then Φ ⊢ A → Bi. The rest
of this subcase is also as the proof of Case 1.
Case 3 there are some j, k < i such that Bi is obtained from Bj and
Bk = Bj → Bi by MP, and RN is applied to the formulas depending on
A h1 times in the derivation B1, …, Bj, RN is applied to the formulas
depending on A h2 times in the derivation B1, …, Bk: By the induction
hypotheses,
h
Φ ⊢ □0A ∧…∧ □ 1A → Bj, and
h
Φ ⊢ □0A ∧…∧ □ 2A → Bj → Bi.
Since h1, h2 ≤ h, it is easy to prove that
h
Φ ⊢ □0A ∧…∧ □hA → □0A ∧…∧ □ 1A, and
h
Φ ⊢ □0A ∧…∧ □hA → □0A ∧…∧ □ 2A,
by Syllogism Rule (RPC), it is easy to prove that
Φ ⊢ □0A ∧…∧ □hA → Bj, and
Φ ⊢ □0A ∧…∧ □hA → Bj → Bi.
By Implication Distribution Rule (RPC), it is easy to see that (☆)
62
holds.
Case 4 there is some j < i such that Bi is obtained from Bj by RN.
Subcase 1 Bj does not depend on A: Then by the definition of ‘to
depend on’, B1, …, Bj is a derivation from Φ ∪ {A} to Bj in K such
that Bj does not depend on A in this derivation. By the previous
Lemma, Φ ⊢ Bj, so, by RN, Φ ⊢ Bi, and thus Φ ⊢ A → Bi, and hence
it is easy to see that (☆) holds.
Subcase 2 Bj depends on A, and RN is applied to the formulas
depending on A h1 times in the derivation B1, …, Bj: By the induction
hypothesis,
h
Φ ⊢ □0A ∧…∧ □ 1A → Bj,
by RK,
h +1
Φ ⊢ □1A ∧…∧ □ 1 A → □Bj.
Since h1 < h, it follows that h1 + 1 ≤ h, and thus
h +1
Φ ⊢ □0A ∧…∧ □hA → □1A ∧…∧ □ 1 A,
by Syllogism Rule again,
h
Φ ⊢ □0A ∧…∧ □ A → □Bj,
so we have (☆). ┤
By the previous lemma, we have
Theorem 1.3.6 (Deduction Theorem for K)
(1) Assume that Φ ∪ {A} ⊢ K B, and there is a derivation from Φ
∪ {A} to B in K such that RN is not applied to the formulas
depending on A. Then Φ ⊢ K A → B.
(2) Assume that A ⊢ K B, and there is a derivation from A to B in K
such that RN is not applied to the formulas depending on A. Then
⊢ K A → B. ┤
63
In the following we will consider the system D.
Theorem 1.3.7
(1) K ⊆ D.
(2) The following are theorems or derived rule of D:
D(1) ◇(p → p),
D(2) A∕◇A (= RN◇),
D(3) ◇p ∨ ◇¬ p.
(3) D = K + ◇(p → p).
(4) D = K + A∕◇A.
Proof. (1) holds clearly.
(2) Prove D(1):
① □(p → p) → ◇(p → p)
D, US
② □(p → p)
(1), K(1)
③ ◇(p → p)
①, ②, MP
Prove D(2):
Assumption
① A
② □A
RN
③ □A → ◇A
D, US
④ ◇A
②, ③, MP
Prove D(3):
PC, D(2)
① ◇(p ∨ ¬ p)
② ◇p ∨ ◇¬ p
R(2) (since R ⊆ K ⊆ D)
(3) By D(1), it suffices to show that D can be inferred from the
system K + ◇(p → p):
① ◇(p → p)
Axiom
② ◇(p → p) ↔ □p → ◇p
R(9)
③ □p → ◇p
①, ②, MP
64
(4) By (3) and D(2), it suffices to show that ◇(p → p) can be
inferred from the system K + A∕◇A:
① p→p
PC
② ◇(p → p)
A∕◇A ┤
Definition 1.3.8 A is a constant formula ⇔ A is a formula
constructed by using T and ¬, ∧ or □: A is a formula constructed by
using T as atomic formula. In other words, A is a constant formula ⇔
A ∈ CML such that CML is inductively defined as the smallest set
closed under the following formation rules:
① T ∈ CML;
② B, C ∈ CML ⇒ ¬B, (B ∧ C), □B ∈ CML. ┤
By a contradictory formula we mean a formula that is always false
in the sense of the classical two-value semantics. ① From the point of
view of PC, A is called a contradictory formula ⇔ ⊢ PC A ↔ ⊥.
Theorem 1.3.9 Let A be a constant formula.
－□
is a tautology, then A is a theorem of D.
(1) If A
－□
(2) If A
is not a tautology, then ¬A is a theorem of D.
Proof. Let ⊢ = ⊢ D. By induction on A.
－□
= T is a tautology, so ⊢ A.
Case 1 A = T: then A
Case 2 A = ¬B: There are two subcases to consider:
－□
－□
－□
is a tautology: then ¬A
= ¬¬B
is not a
Subcase 1 A
－□
－□
tautology (in fact, ¬A
is a contradictory formula), so B
is not a
tautology, and hence ⊢ ¬B by the induction hypothesis about (2), and
①
The subformulas of the form □B in A are all seen as sentence symbols.
65
thus ⊢ A.
－□
Subcase 2 A
is not a tautology: first we will prove:
(☆) a formula D constructed by using T and ¬ or ∧ is a tautology
By induction on D. If D is T, then D is a tautology.
If D = ¬E. Then E is a tautology or a contradictory formula by the
a tautology.
If D = E ∧ F. Then each of E and F is a tautology or a contradictory
formula by the induction hypotheses about (☆).
Assume that E and F are both tautologies: then so is D.
Assume that E or F is not a tautology: then E or F is a contradictory
formula by the induction hypothesis about (☆), so is D.
Hence (☆) holds. Now we return the thesis.
－□
－□
is not a tautology, it follows that A
Since A
－□
is a tautology, and thus ⊢ B by the induction
formula by (☆), so B
hypothesis about (1), and hence ⊢ ¬¬B, so ⊢ ¬A.
Case 3 A = B ∧ C: There are two subcases to consider:
－□
－□
－□
－□
is a tautology: Since A
= B
∧ C , it
Subcase 1 A
－□
－□
and C
are both tautologies. So ⊢ B and ⊢ C by
follows that B
the induction hypotheses about (1), and thus ⊢ A.
－□
－□
Subcase 2 A
is not a tautology: Then it is easy to see that B
－□
is not a tautology. So ⊢ ¬B or ⊢ ¬C by the induction
or C
hypothesis about (2), and hence we always have ⊢ ¬B ∨ ¬C, and thus
⊢ ¬(B ∧ C), so ⊢ ¬A.
①
(☆) is actually a result of the classical propositional calculus.
66
Case 4 A = □B: There are two subcases to consider:
－□
－□
－□
Subcase 1 A
is a tautology: since A
= B , it follows that B
－□
is a tautology, and hence ⊢ B by the induction hypothesis about
(1), and thus ⊢ □B by RN, so ⊢ A.
－□
－□
Subcase 2 A
is not a tautology: then B
is not a tautology,
⊢ ¬B by the induction hypothesis about (2), so ⊢ □¬B by RN. By
Axiom D, we have ⊢ ◇¬B, so ⊢ ¬□B by LMC, and thus ⊢ ¬A. ┤
Remark. (Ⅰ) This theorem shows the maximality of Th(D) w.r.t.
CML: for all A ∈ CML, A ∈ Th(D) or ¬A ∈ Th(D). (Refer to 2.2.1(2)
below.)
(Ⅱ) The induction used in the previous theorem can be called a
simultaneous induction.
In the following we will consider the system T.
Theorem 1.3.10
(1) The following are the theorems or derived rule of T:
T(1) p → ◇p (= T◇),
T(2) A∕◇A (= RN◇),
T(3) D,
T(4) ◇(p → □p).
T(5) □¬(p → □p) → □¬ p.
(2) D ⊆ T.
(3) T = K + T1 where
(T1) (□p → p) ∨ (q → ◇q).
Proof. Prove T(1):
T, RPC
① ¬ p → ¬□p
67
② ¬¬ p → ¬□¬ p
US
③ p → ◇p
RER, Df◇
Note that T(1) is a converse dual formula of T, so T(1) can also be
obtained directly from T and Converse Duality Theorem 1.2.10(4).
Prove T(2): by T(1).
Prove T(3): by T and T(1).
Prove T(4):
① □p → ◇□p
T(1), US
② ◇(p → □p) ↔ (□p → ◇□p)
R(9), US
③ ◇(p → □p)
①, ②, RPC
Prove T(5):
T(4)
① ◇(p → □p)
② ¬□¬(p → □p)
Df◇
③ □¬(p → □p) → □¬ p
RPC
(2) By T(3).
(3) It is clear to prove T1 from T. So below it suffices to prove T
from K + T1:
T1, US
① (□p → p) ∨ (¬p → ◇¬ p)
② (□p → p) ∨ (¬◇¬p → p)
RER
③ (□p → p) ∨ (□p → p)
LMC
④ □p → p
RPC ┤
In the following we will consider the system B.
Theorem 1.3.11
(1) T ⊆ B.
(2) The following are the derived rule or theorems of B:
B(1) ◇A → B∕A → □B (= RB),
B(2) ◇□p → p (= B◇),
68
B(3) ◇□p → □◇p (= G),
B(4) ◇□p ∧ ◇□q → □◇(p ∧ q).
(3) B = T + RB.
Proof. (1) holds clearly.
(2) Prove B(1):
① ◇A → B
Assumption
② □◇A → □B
RM
③ A → □◇A
B, US
④ A → □B
③, ②, RPC
Prove B(2): by B and Converse Duality Theorem 1.2.10(4).
Prove B(3): by B(2) and B.
Prove B(4):
① p ∧ q → □◇(p ∧ q)
B, US
② ◇□p → p
B(2)
③ ◇□q → q
US
④ ◇□p ∧ ◇□q → p ∧ q
②, ③, RPC
⑤ ◇□p ∧ ◇□q → □◇(p ∧ q)
④, ①, RPC
(3) By B(1), it suffices to prove B from T + RB:
PC
① ◇p → ◇p
② p → □◇p
RB ┤
In the following we will consider the S4, this system has a lot of
interesting properties.
By 1.1.20, any modal system S has Weak Rule US, so if A contains
exactly a sentence symbol pn and ⊢ S A(pn), then for all B, we have
⊢ S A(pn/B).
So we can give out the following convention:
Convention 1.3.12 Let A contain exactly a sentence symbol pn, and
69
let A
－ pn
be the formula obtained by deleting each occurrence of pn in
A. We use ⊢ S A
－ pn
as ⊢ S A for any modal system S. ┤
Remark. We can see more distinctly the relations between the
modalities and logical symbols in a theorem of a modal system by
using the previous convention.
Theorem 1.3.13
(1) T ⊆ S4.
(2) The following are the theorems of S4:
S4(1) □ ↔ □□,
S4(2) ◇ ↔ ◇◇,
S4(3) ◇□◇ → ◇,
S4(4) □◇ → □◇□◇,
S4(5) □◇ ↔ □◇□◇,
S4(6) ◇□ ↔ ◇□◇□,
S4(7) □ → □◇□,
S4(8) □(□p → q) → □p → □q.
(3) (Deduction Theorem for S4)
Φ ∪ {A} ⊢ S4 B ⇒ Φ ⊢ S4 □A → B.
Proof. (1) holds clearly.
(2) Prove S4(1):
4
① □ → □□
② □□ → □
T, US
③ □ ↔ □□
①, ②, RPC
Prove S4(2): by S4(1) and Equivalence Duality Theorem 1.2.10(6).
Prove S4(3):
T, US
① □◇ → ◇
② ◇□◇ → ◇◇
RM◇
70
③ ◇□◇ → ◇
S4(2), RER
Prove S4(4):
T◇, US
① □◇ → ◇□◇
② □□◇ → □◇□◇
RM
③ □◇ → □◇□◇
S4(1), RER
Prove S4(5):
S4(3), RM
① □◇□◇ → □◇
② □◇ ↔ □◇□◇
S4(4), RPC
Prove S4(6): by S4(5) and Equivalence Duality Theorem 1.2.10(6).
Prove S4(7): by S4(3) and Converse Duality Theorem 1.2.10 (4).
Prove S4(8):
① □(□p → q) → □□p → □q
K
② □(□p → q) → □p → □q
S4(1), RER
(3) Assume that Φ ∪ {A} ⊢ S4 B. Then there is a derivation B1, …,
Bn from Φ ∪ {A} to B in S4. It suffices to show that in S4,
(☆) Φ ⊢ □A → Bi for all 1 ≤ i ≤ n.
By induction on 1 ≤ i ≤ n. For simplicity, we consider the untrivial
case only:
there is some j < i such that Bi is obtained from Bj by RN.
By the induction hypothesis,
Φ ⊢ □A → Bj.
So Φ ⊢ □□A → □Bj by RM, and thus Φ ⊢ □A → □Bj by 4, and
hence Φ ⊢ □A → Bi.
Please supply the rest of the cases. ┤
By (2) of the previous theorem, we get easily:
Corollary 1.3.14 The modal symbol strings in following sequences
71
are both equivalent in S4:
<□□, □>,
<□◇□◇, □◇>,
<◇◇, ◇>;
<◇□◇□, ◇□>. ┤
Definition 1.3.15
(1) Modal-degree Deg(A) of A is defined as follows:
① Deg(pn) = 0 for all n < ω,
② Deg(¬B) = Deg(B),
③ Deg(B ∧ C) = max{Deg(B), Deg(C)},
④ Deg(□B) = Deg(B) + 1.
(2) A is a n-degree formula ⇔ Deg(A) = n.
(3) Modal-degree Deg(s) of a modal symbol string s is defined as
follows:
① Deg(φ) = 0,
② Deg(¬ s) = Deg(s),
③ Deg(□s) = Deg(s) + 1.
(4) s is a n-degree modal symbol string ⇔ Deg(s) = n. ┤
Remark. It is easy to see that the modal-degree of a pure modal
symbol string is its length, and
Deg(◇s) = Deg(s) + 1.
Definition 1.3.16
(1) Let s and t be modal symbol strings such that Deg(s) > Deg(t).
s is reducible to t in S, denoted as s ➞S t, ⇔ ⊢ S s ↔ t.
Otherwise, s is unreducible to t in S.
(2) s is reducible in S ⇔ there is a modal symbol string t such that
Deg(s) > Deg(t) and s ➞S t.
Otherwise, s is unreducible in S.
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(3) Let s and t be modal symbol strings such that Deg(A) > Deg(B).
A is reducible to B in S ⇔ ⊢ S A ↔ B.
Otherwise, A is unreducible to B in S.
(4) s absorbs t in S ⇔ ⊢ S s ↔ ts. ┤
Remark. (Ⅰ) We often omit the subscript ‘S’ if no confusion will
arise.
(Ⅱ) By the previous definition, Modal(S) means the cardinal
number of the standard modal symbol strings which are unreducible
in S. By 1.3.14, we have the following theorem:
Theorem 1.3.17 (Reduction Theorem for S4) We have the
following reduction relation in S4:
◇◇ ➞ ◇,
□□ ➞ □,
□◇□◇ ➞ □◇, ◇□◇□ ➞ ◇□.
And thus, in S4, □ absorbs □, ◇ absorbs ◇, □◇ absorbs □◇,
◇□ absorbs ◇□, respectively. ┤
Theorem 1.3.18 Modal(S4) ≤ 14.
Proof. By 1.2.12(3), it suffices to show that S4 has at most the
following unreducible modal symbol strings:
φ
□
◇
□◇
◇□
□◇□
◇□◇
¬ ¬□ ¬◇ ¬□◇ ¬◇□ ¬□◇□ ¬◇□◇.
First we consider the above first line. First we number them as follows:
(1) φ,
(2) □,
(3) ◇,
(4) □◇,
(5) ◇□,
(6) □◇□,
(7) ◇□◇.
It is easy to see that (1) is a unique 0-degree modal symbol string, (2)
73
and (3) are two 1-degree modal symbol strings. Adding a modal
symbol to (2) and (3) (it can be added to the front of (2) and (3) or the
back of them), we have at most 4 unreducible 2-degree modal symbol
strings:
(8) □□,
(9) ◇◇,
(10) □◇,
(11) ◇□.
By Reduction Theorem, (8) and (9) are reducible to (2) and (3),
respectively.
Adding a modal symbol to (10) and (11) (it can be added to the
front of or the back of (10) and (11)), we have at most 6 unreducible
3-degree modal symbol strings:
(12) □□◇,
(13) ◇□◇,
(14) □◇□,
(15) □◇◇,
(16) ◇◇□,
(17) ◇□□.
By Reduction Theorem for S4, (12) and (15) are reducible to (4), (16)
and (17) are reducible to (5).
Adding a modal symbol to (13) and (14) (it can be added to the
front of (13) and (14) or the back of them), we have at most 6
unreducible 4-degree modal symbol strings:
(18) □◇□◇,
(19) ◇◇□◇,
(20) ◇□◇□,
(21) ◇□◇◇,
(22) □□◇□,
(23) □◇□□.
By Reduction Theorem for S4, (18) is reducible to (4), (19) and (21)
is reducible to (7), (20) is reducible to (5), (22) and (23) is reducible
to (6).
It is easy to see that there is not a unreducible 4-degree pure modal
symbol string in S4, and thus
(a) S4 has at most 7 unreducible pure modal symbol strings.
Adding ¬ to the front of the above 7 pure modal symbol strings, we
will show that
(b) S4 has no theorem of the form sA ↔ ¬ s A.
Hypothesize that (b) does not hold, then sA → ¬sA and ¬sA → sA are
both the theorems of S4, it is easy to prove that ¬sA and sA are both
74
the theorems of S4, contradicting Harmony of S4 (refer to 1.1.16(1)).
By (a) and (b), it is easy to see that S4 has at most 14 unreducible
modal symbol strings. ┤
Remark. (Ⅰ) In the following we shall prove that Modal(S4) = 14.
Parry originally gave out the proof of Modal(S4) = 14 in 1939, the
reader can refer to  (p.59) of Segerberg.
(Ⅱ) In the final part of the previous proof, we actually prove a
more generalized proposition:
(b*) A harmonious system has no theorem of the form sA ↔ ¬ sA.
(Ⅲ) There is the following relation among the unreducible pure
modal symbol strings in S4: where s → t denotes ⊢ S4 s → t:
□ → □◇□ → ◇□
⎥
↓
↓
⎥
□◇ → ◇□◇
↓
↓
φ ⎯⎯⎯⎯⎯⎯→ ◇
The first → of the first line is by S4(3) and Converse Duality
Theorem 1.2.10(4).
The second → of the first line and the left two ↓’s of the second
line are by T.
The third ↓ of the second line, the → of the third line and the →
of the fifth line are by T◇.
The right ↓ of the fourth line is by S4(3).
In the following we will consider S5, this system is more
interesting than S4.
75
Theorem 1.3.19 The following are the theorems of S5.
S5(1) ◇ ↔ □◇,
S5(2) □ ↔ ◇□, ①
S5(3) 4, (and thus S4 ⊆ S5),
S5(4) □(p ∨ □q) ↔ □p ∨ □q,
S5(5) □(p ∨ ◇q) ↔ □p ∨ ◇q,
S5(6) ◇(p ∧ ◇q) ↔ ◇p ∧ ◇q,
S5(7) ◇(p ∧ □q) ↔ ◇p ∧ □q,
S5(8) B, (and thus B ⊆ S5),
S5(9) G, (and thus S4.2 ⊆ S5).
Proof. Prove S5(1):
① ◇ → □◇
5
② □◇ → ◇
T, US
③ ◇ ↔ □◇
①, ②, RPC
Prove S5(2): by S5(1) and Equivalence Duality Theorem 1.2.10(6).
Prove S5(3):
T◇, US
① □ → ◇□
② ◇□ → □◇□
5, US
③ □ → □◇□
①, ②, RPC
④ □ → □□
S5(2), RER
Prove S5(4):
K(4), US
① □(p ∨ □q) → □p ∨ ◇□q
② □(p ∨ □q) → □p ∨ □q
①, S5(2), RER
③ □p ∨ □□q → □(p ∨ □q)
M(5), US
④ □p ∨ □q → □(p ∨ □q)
③, S4(1), RER
⑤ □(p ∨ □q) ↔ □p ∨ □q
②, ④, RPC
①
Hence, in S5, ◇ absorbs □, and □ absorbs ◇.
76
Prove S5(5):
S5(4), US
① □(p ∨ □◇q) ↔ □p ∨ □◇q
② □(p ∨ ◇q) ↔ □p ∨ ◇q
◇ absorbs □ in S5
Prove S5(6):
S5(4), US
① □(¬p ∨ □¬q) ↔ □¬p ∨ □¬q
② ¬□(¬p ∨ □¬q) ↔ ¬(□¬p ∨ □¬q)
RPC
③ ◇¬(¬p ∨ ¬◇q) ↔ ¬(¬◇p ∨ ¬◇q)
LMC
④ ◇(p ∧ ◇q) ↔ ◇p ∧ ◇q
RER
Prove S5(7):
S5(6), US
① ◇(p ∧ ◇□q) ↔ ◇p ∧ ◇□q
② ◇(p ∧ □q) ↔ ◇p ∧ □q
□ absorbs ◇ in S5
Prove S5(8):
T◇
① p → ◇p
② ◇p → □◇p
5
③ p → □◇p
①, ②, RPC
Prove S5(9):
D
① □→◇
② ◇□ → □◇
S5(2), S5(1), RER (or by absorbability) ┤
Theorem 1.3.20 (Reduction Theorem for S5)
(1) We have the following reduction relation in S5:
(☆) □□ ➞ □, ◇◇ ➞ ◇, ◇□ ➞ □, □◇ ➞ ◇,
and thus there is not a unreducible 2-degree modal symbol string in
S5, so Modal(S5) ≤ 6.
(2) A modal-degree > 1 formula is reducible to a modal-degree ≤ 1
formula in S5 ①.
Proof. (1) By Reduction Theorem for S4 (1.3.17), S5(1) and S5(2), it
①
Refer to 1.3.16 (3).
77
is easy to see that (☆) holds, so there is not a unreducible 2-degree
modal symbol string in S5, and thus there are at most six unreducible
standard modal symbol strings in S5
φ,
□,
◇,
¬,
¬□,
¬◇.
So Modal(S5) ≤ 6.
(2) Let A be a modal-degree > 1 formula, we will reduce the modaldegree of A by using the following method:
① eliminate → and ↔ by using the abbreviation definitions and
RER;
② move ¬ to the front of sentence symbols by using LMC and
etc.;
③ reduce an iterative modal symbol string to a non-iterative modal
symbol string by using the reduction relation given in (1);
④ if the formula A1 obtained by the above approach is still not a 1degree formula, then continue the reduction by the following
approach again:
Assume that A1 contains subformula □B such that Deg(B) > 0.
Case 1 B = C ∧ D: move □ into B by R and RER.
Case 2 B = C ∨ D:
Subcase 1 C or D has the form □E or ◇E: then move □ into
B by S5(4) - S5(5) and RER.
Subcase 2 both of C and D have no the form □E or ◇E: since
Deg(B) > 0, it follows that Deg(C) > 0 or Deg(D) > 0. By the
assumption of the subcase and ②, without loss of generality, we can
assume that C = C1 ∧ C2. By RER and R,
⊢ □((C1 ∧ C2) ∨ D) ↔ □(C1 ∨ D) ∧ □(C2 ∨ D).
Move □ into B by RER again.
If, after moving as above, the modal-degrees of the conjuncts or
78
disjuncts obtained are still large than 1, then continue the reduction by
③ or ④ again. Finally, □ in front of B can always be absorbed by
the modalities in B. ①
Assume that A1 contains subformula ◇B such that Deg(B) > 0.
Case 1 B = C ∨ D: move ◇ into B by R◇ and RER.
Case 2 B = C ∧ D:
Subcase 1 C or D has the form □E or ◇E: then move ◇ into
B by S5(6) - S5(7) and RER.
Subcase 2 both of C and D are not of the form □E or ◇E:
since Deg(B) > 0, it follows that Deg(C) > 0 or Deg(D) > 0. By the
assumption of the subcase and ②, without loss of generality, we can
assume that C = C1 ∨ C2. By RER and R◇,
⊢ ◇((C1 ∨ C2) ∧ D) ↔ ◇(C1 ∧ D) ∨ ◇(C2 ∧ D).
Move ◇ into B by RER again as above.
If, after moving as above, the modal-degrees of the conjuncts or
disjuncts obtained are still large than 1, then continue the reduction by
③ or ④ again. Finally, ◇ in front of B can always be absorbed by
the modalities in B.
Continue the reduction, finally, we can prove (2).
[Note that some of the formulas we encounter in the start, middle or
the end of the reduction process can be directly reduced to modaldegree 0 formulas, such as the substitution instances of tautologies or
Remark. (Ⅰ) By the above (1), for S5, an iterative modal symbol
string cannot give us any thing new. More carefully speaking: if
O1…On is a iterative pure modal symbol string, then O1…On is
①
Refer to (☆) of (1).
79
reducible to On in S5. In other words, On absorbs O1…On − 1.
(Ⅱ) In the following we will show that K, D, T and B do not have
the above Reduction Theorem.
Since both □A → A and A → ◇A are the theorems of S5, it
follows that
Corollary 1.3.21 In S5, the unreducible pure modal symbol strings
can be formed a linear order:
□ → φ → ◇. ┤
Remark. We shall prove that Modal(S5) = 6 later.
In the following we will prove Modal Conjunction Normal Form
Existence Theorem for S5. For this, we first give the following
definition:
Definition 1.3.22
(1) A is a 1-degree modal simple disjunction ⇔ A = A1 ∨…∨ An
such that for all 1 ≤ i ≤ n,
① Deg(Ai) = 0, or
② Deg(Ai) = 1 and Ai has the form □B or ◇B.
Here Ai is also called a disjunct of A.
(2) A is a 1-degree modal conjunction normal form ⇔ A = A1 ∧…∧
An such that for all 1 ≤ i ≤ n, Ai is a 1-degree modal simple disjunction.
Here Ai is also called a conjunct of A. ┤
Remark. It is easy to see that Deg(A) ∈ {0, 1} for all 1-degree
modal conjunction normal form A.
80
Example 1.3.23
(1) The following formulas are 1-degree modal conjunction normal
forms:
p,
□p ∧ (q ∨ r),
(□p ∨ ◇q ∨ r) ∧ (◇p ∨ ◇(q ∨ r)).
(2) The following formulas are not 1-degree modal conjunction
normal forms:
□□p, (□(p ∨ ◇q) ∨ r) ∧ r, □p ∧ (◇r ∨ (q ∧ ◇p)). ┤
Theorem 1.3.24
(1) Given any formula A, there is a formula B = B1 ∧…∧ Bn such
that for all 1 ≤ i ≤ n, Bi is a disjunction constituted by some formulas
of the form
Bi, 1, …, Bi, j, ¬Bi, 1, …, ¬Bi, k,
and ⊢ PC A ↔ B.
(2) (1-degree Modal Conjunction Normal Form Existence
Theorem for S5) Given any formula A, there is a 1-degree modal
conjunction normal form B such that ⊢ S5 A ↔ B.
Proof. (1) Without loss of generality, we can see the subformulas of
the form □ B and ◇ B as atomic formulas, and hence use
Conjunction Normal Form Existence Theorem for PC. We leave the
details of the proof as an exercise.
(2) Given any formula A.
Case 1 Deg(A) = 0: By the previous definition, A itself is a 1degree modal conjunction normal form, so the result we will prove
holds clearly.
Case 2 Deg(A) = 1: By (1), there is a formula B = B1 ∧…∧ Bn
defined as (1) such that ⊢ PC A ↔ B, so ⊢ S5 A ↔ B. If there is a
formula of the form ¬□C or ¬◇C in B, then by LMC, it is replaced
equivalently by a formula of the form ◇¬C or □¬C, so, by
81
Definition 1.3.22(2), B is a conjunction normal form and ⊢ S5 A ↔ B.
Case 3 Deg(A) > 1: By Reduction Theorem for S5 (1.3.20 (2)),
each modal-degree > 1 formula is reducible to a modal-degree ≤ 1
formula in S5. In this way, we return to Case 1 - Case 2 again. ┤
Remark. (Ⅰ) The above proof actually gives out a very applied and
effective program ①.
(Ⅱ) In the above systems we have introduced, no system other
than S5 has 1-degree Modal Conjunction Normal Form Existence
Theorem, since no such theorem other than its Reduction Theorem is
all-around: each modal-degree > 1 formula is reducible to a modaldegree ≤ 1 formula in S5.
Definition 1.3.25
(1) A system S is an n-degree system ⇔ n is a least natural number
such that any modal-degree > n formula is reducible to an m ≤ ndegree formula in S.
(2) S is an infinite-degree system ⇔ there is not a natural number n
such that S is an n-degree system. ┤
Corollary 1.3.26 S5 is a 1-degree system. ┤
We can prove that the elementary systems and the basic systems K,
D, T, B and S4 are all infinite-degree systems.
Note that although we have Reduction Theorem for S4 (1.3.17), a
formula of the form □(A ∨ B) is unnecessarily reducible to a formula
C such that Deg(C) < Deg(□(A ∨ B)) in S4 (this proof refers to
By an effective program we mean that it can be executed in finite steps by
a finite number of rules given in advance.
①
82
[1966a] of Makinson), so S4 is an infinite-degree system.
Finally, we will discuss the virtuality, actuality, strongness and
weakness of inference rules in a system and its extension systems. By
definition, we get immediately:
Corollary 1.3.27
(1) If R is a strong rule of S, then R is a weak rule of S.
(2) If R is a virtual rule of S, then R is a strong rule of S. ┤
Theorem 1.3.28 (the results of negative form)
(1) An actual and weak rule of a system is not necessarily to be an
actual and weak rule of its extension systems.
(2) A virtual and weak rule of a system is not necessarily to be a
weak rule of its extension systems.
(3) A virtual and weak rule of a system is not necessarily to be a
virtual and weak rule of its extension systems.
Proof. (1) It is easy to see that S4 is an extension system of K.
By 1.3.2, □A → □B∕A → B is a weak rule of K.
Since □p → □(p ∨ q) is a theorem of K, it follows that □A →
□B∕A → B is an actual rule of K ①.
On the other hand, by 1.4.16(1) below, □A → □B∕A → B is not
a weak rule of S4. Note that □A → □□A is a theorem of S4.
(2) By 1.3.2, ◇A∕A is a weak rule of K, but by 3.4.10(1) below,
K has no theorem of the form ◇A, so ◇A∕A is a virtual rule of K.
But by 3.4.10(7) below, ◇A∕A is not a weak rule of T.
On the other hand, it is clear that T is an extension system of K.
(3) By (2), ◇A∕A is a virtual and weak rule of K.
①
Refer to 1.1.11(7).
83
Although ◇A∕A is a weak rule of D by 1.3.2, but it is not a
virtual rule of D by D(1). ┤
Remark. By the previous theorem, it is easy to see that it is possible
that a feature of a system is not one of its extensions. For example, by
(2), ◇A∕A is a virtual rule of K because the system is too weak to
have no theorem of the form ◇A, but ◇A∕A is not a virtual rule of
D by (3).
Exercises 1.3
1.3.29
(Ⅰ) Given any n, m < ω, prove that the following are the derived
rules of K:
(1) A1 ∧…∧ An → B∕□mA1 ∧…∧ □mAn → □mB
where 1 ≤ n < ω and m < ω,
m
m
(2) A → B1 ∨…∨ Bn∕◇ A → ◇ B1 ∨…∨ ◇mBn
where 1 ≤ n < ω and m < ω,
(3) □(A ↔ B)∕□A ↔ □B.
(Ⅱ) Prove that K5 = K5V where
(5V) □(□p ∨ q) → □p ∨ □q.
(Ⅲ) Prove that
K = <R◇; RE◇, ¬A∕¬◇A> Df□
= <RR◇, ¬A∕¬◇A> Df□
= <¬◇⊥; RR◇> Df□
= <¬◇⊥, R◇; RE◇> Df□
= <N, C; RM>.
(Ⅳ) Prove that Axiom T ∉ Th(KW), and thus T ∉ Th(K).
84
(Ⅴ) (Modal Conjunction Normal Form Existence Theorem for
K) Prove that any formula, in K, is equivalent to a conjunction of a
formula of the form A ∨ ◇B ∨ □C1 ∨…∨ □Cn where A contains no
modal formula.
(Ⅵ)
(1) Is □A∕A a strong rule of K?
(2) Is □(□p → ◇p) a theorem of K?
(Ⅶ) The version Kb of K is defined as the following system:
① □nA where A is any axiom of PC and n < ω.
② □n (□(p → q) → □p→ □q) where n < ω.
(MP) A, A → B∕B,
(US) A∕Aσ where σ is any substitution mapping.
Similarly define the concept of the theorems of Kb as before.
(1) Prove that RN is a weak rule of Kb.
(2) Does K ⊆ Kb hold?
(3) Does Kb ⊆ K hold?
(4) Given any normal modal system S, analogously define the
version Sb of S. Analogously, do the inclusion relations above hold
yet?
(Ⅷ) Given any normal system S = K + Φ. Can 1.3.3 − 1.3.6 be
generalized to S, namely, whether the result still holds when all K in
1.3.3 − 1.3.6 are replaced by S?
(Ⅸ) Prove that K + Dc = K + Dc* where
(Dc) ◇p → □p,
(Dc*) ◇p ∧ ◇q → ◇(p ∧ q). ┤
1.3.30
(Ⅰ) Prove that the following are the theorems and derived rule of
D:
85
(1) ◇¬ p ∨ ◇¬q ∨ ◇(p ∨ q),
(2) ¬(□p ∧ □¬p),
(3) ¬□(□p ∧ □¬ p),
(4) A ∨ B∕◇A ∨ ◇B,
(5) □n → ◇n (= Dn) where n < ω.
(Ⅱ) Prove that D = K + ◇T.
(Ⅲ)
(1) Is □ → ◇ a theorem of K?
(2) Is A ∨ B∕◇A ∨ ◇B a rule of K? ┤
1.3.31
(Ⅰ) Prove that the following are the theorem and derived rule of T:
(1) □n p → p (= T n) where n < ω.
(2) □p → ◇n p where n < ω.
(3) A ∨ B∕◇A ∨ B.
(Ⅱ) Prove that
(1) T = K + p → ◇p,
(2) T = PC + T + RM + □K where
(□K) □(□(p → q) → □p → □q).
(Ⅲ) Let m, n < ω such that n ≠ m, then □n ↔ □m is not a theorem
of T.
(Ⅳ) Prove that G is a theorem of M + T + Lem.
(Ⅴ) ¬◇p and □◇p are not theorems of T. ┤
Definition 1.3.32 Let s be pure modal symbol string, namely, the
modal symbol string s contains no ¬. Given any n < ω, inductively
define s n A as follows:
s 0A = A, …, s n + 1 A = ss nA. ┤
86
1.3.33
(Ⅰ) Prove that the following are the derived rule and theorems of B:
(1) A → □B∕◇A → B,
(2) □(◇p → q) → p → □q,
(3) ◇(□p ∧ q) → p,
(4) □(p → □q) → ◇p → q,
(5) p → □n◇n p (= B n) where n < ω,
(6) p → (□◇) n p (= B (n)) where n < ω.
(Ⅱ) Prove that B = T + □(◇p → q) → p → □q. ┤
1.3.34
(Ⅰ) Prove that the following are the theorems of S4:
(1) □(p → q) → □(□p → □q) (= K″ ),
(2) □(□p → □q) → □p → □q,
(3) □p ∨ □q ↔ □(□p ∨ □q),
(4) ◇□(p → □◇p) (= ◇□B),
(5) ◇(□p → ◇q) ↔ ◇(p → q),
(6) □(□(p → q) → r) → □(□(p → q) → □r),
(7) □ → □n + 1 (= 4n) where n < ω,
(8) □p ∧ q → □1 p ∧…∧ □n p ∧ q where n < ω,
(9) □(p → q) → □(sp → sq)
where s is a positive modal symbol string.
(Ⅱ) Prove that S4 = T − K + K″. ┤
1.3.35
(Ⅰ) Prove that the following are the theorems of S5:
(1) □(□p → □q) ∨ □(□q → □p),
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(2) □(□p → q) ∨ □(□q → p) (= Lem, and thus S4.3 ⊆ S5),
(3) □(◇p → q) ↔ □(p → □q),
(4) ◇□p → ◇q → □(p ∧ ◇q),
(5) □(□p ∨ q) → □p ∨ □q,
(6) ◇p ∧ ◇q → ◇(◇p → q),
(7) ◇□ ↔ □□,
(8) ◇◇ ↔ □◇,
(9) ◇ → □n◇ (= 5n) where n < ω,
(10) ◇p → □(p → □p) → p.
(Ⅱ) The following systems and S5 are equivalent:
(1) S4 + B,
(2) T + 51 + 52 where
(51) □◇□p → p,
(52) ◇□ → □◇□□.
(3) PC + T + RM2 where
(RM2) A → B∕A → □B, ①
where A has the form □C or ◇C or □C ∧ □D.
(4) T + □(p ∨ □q) → □p ∨ □q.
(5) S4 + p → □□◇p.
(Ⅲ) (Deduction Theorem for S5) Let A have the form □C or ◇
C or □C ∧ □D. Then
Φ ∪ {A} ⊢ S5 B ⇒ Φ ⊢ S5 A → B. ┤
1.3.36 Let MV = K + □p ∨ ◇□p. Prove that MV and the following
systems are equivalent:
①
RM2 corresponds to Consequence Generalization Rule of the first-order
logic.
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(1) K + □q ∨ ◇□p.
(2) K + □⊥ ∨ ◇□⊥. ┤
1.3.37
(Ⅰ) Prove that ◇◇p is not a theorem of K4.
(Ⅱ) Prove that the following are the theorems and derived rules of
K4:
(1) □(p → q) → □(□p → □q) (= K″ ),
(2) □p ∨ □q → □(□p ∨ □q),
(3) □(□(p → q) → r) → □(□(p → q) → □r),
(4) □◇2 ↔ □◇,
(5) ◇□2 ↔ ◇□,
(6) (□◇)2 ↔ □◇,
(7) (◇□)2 ↔ ◇□,
(8) ◇A ∧ ◇□B∕◇(A ∧ □B),
(9) ◇□A ∧ ◇□B∕◇□(A ∧ B). ┤
1.3.38 Prove that the following are the theorems of M5:
(1) □(□p → p) (= U),
(2) □(□□ ↔ □). ┤
1.3.39 Prove that W is not a theorem of S5. ┤
1.3.40 Prove that □(◇ ↔ □◇) is a theorem of K5. ┤
1.3.41 Prove that
(1) Modal(ET5) ≤ 6,
(2) Modal(K5) ≤ 14,
(3) Modal(K45) ≤ 10,
89
(4) Modal(KD5) ≤ 10,
(5) Modal(K4B) ≤ 10,
(6) Modal(KD45) ≤ 6,
(7) Modal(R5) ≤ 10,
(8) Modal(MD5) ≤ 10,
(9) Modal(ED45) ≤ 6. ┤
1.3.42* (Konolige, ) Prove that in K45, any modal-degree > 1
formula is equivalent to a modal-degree ≤ 1 formula, and thus K45 is
a 1-degree system. ┤
1.3.43 (Problem) Do the converse propositions of 1.3.27 hold? ┤
1.3.44* (Problem)
(1) Which systems else are still 1-degree systems? What is a 1degree system included in all 1-degree systems?
(2) Let n ≠ 1 be a natural number. Which system is still n-degree
system? What is an n-degree system included in all n-degree systems?
┤
Definition 1.3.45
(1) S is an n +-degree system ⇔ n is a least natural number such
that any length > n pure modal symbol string is reducible to a length
m ≤ n pure modal symbol string in S.
(2) S is a ∞+-degree system ⇔ there is not a natural number n
such that S is an n +-degree system. ┤
1.3.46 (Problem) Do the following propositions hold?
(1) If S is an n-degree system, then S is an n +-degree system.
(2) If S is an n +-degree system, then S is an n-degree system.
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(3) S4 is a 3 +-degree system.
(4) S5 is a 1 +-degree system.
(5) The elementary systems are ∞+-degree systems.
(6) K, D, T and B are ∞+-degree systems.
(7) If S is an infinite-degree system, then S is a ∞+-degree system.
(8) If S is a ∞+-degree system, then S is an infinite-degree system.
┤
1.3.47* Prove that
(1) E, M and R are infinite-degree systems.
(2) K, D, T, B and S4 are infinite-degree system. ┤
1.3.48 Let S be a coherent normal system. Prove that there is a
formula A such that ◇A ∉ Th(S). ┤
91
§4 Degenerate Systems
In this section we consider 4 degenerate modal systems, they can be
seen as the degenerate systems obtained by extending the modal
systems in 4 different directions.
As far as we know, □ is a unary operator. If not dealing with its
meaning, it is just a unary connective symbol.
Definition 1.4.1 □ is degenerate ⇔ □ degenerates into one of the
(two-value) truth connective symbols. ┤
Remark. As far as we know, there are exactly 4 unary truth
connective symbols, we denoted as: s1, s2, s3, s4, respectively. If we
use 0 and 1 as two truth values: 1 and 0 (truth and falsity), then their
truth- tables are:
p
s1 p
s2 p
s3 p
s4 p
1
1
0
1
0
0
0
1
1
0
□ is seen as one of the truth-value functions by the above truthtables. So there are exactly 4 cases of □ degenerating into truth
connective symbols, thus to describe the above truth-tables (semantics)
by using formulas (syntax) is:
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(1) □p ↔ p,
(2) □p ↔ ¬ p,
(3) □p ↔ T,
(4) □p ↔ ⊥.
The □s in (1) − (4) are also called the identity operator, the
negation operator, the constant-true operator, the constant-false
operator, respectively. In this way, there are exactly the following 4
operator-degenerating modal systems:
Definition 1.4.2 The operator-degenerating modal systems are
defined as follows:
The trivial system Tr := < □p ↔ p>.
The complement system Com := < □p ↔ ¬ p>.
The verum system Ver := < □p ↔ T>.
The falsum system Fal := < □p ↔ ⊥>. ┤
It is easy to see that:
Corollary 1.4.3 Let S be a system defined in 1.4.2, then S is a 0degree system, and thus Modal(S) = 0. ┤
It is easy to prove that
Corollary 1.4.4
(1) Ver = PC + T → □p = PC + □p,
(2) Fal = PC + □p → ⊥ = PC + ¬□p. ┤
Remark. (Ⅰ) By the previous corollary, we shall also see Ver and
Fal as the systems defined as follows:
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Ver := < □p>,
Fal := < ¬□p>.
In fact, Ver and Fal are usually defined in this form in the literature.
(Ⅱ) For convenience, below we use Tr to denote □p ↔ p, Com to
□p ↔ ¬ p, Ver to □p and Fal to ¬□p.
Theorem 1.4.5
(1) Tr and Ver are normal systems.
(2) K is a theorem of the above 4 operator-degenerating systems.
(3) Tr and Ver are incompatible in the following sense: < Tr, Ver >
is not harmonious.
(4) Tr and Ver are harmonious, and each one of them is not a
subsystem of the other, that is:
Tr ⊈ Ver, Ver ⊈ Tr.
Proof. (1) First we will show that Tr is a normal system. We get
easily RN, so we prove K:
① (p → q) → p → q
PC
② □(p → q) → □p → □q
Tr, US, RER
Next we will show that Ver is a normal system. We We get RN
easily, so we prove K:
PC
① □q → □(p → q) → □p → □q
② □q
Ver, US
③ □(p → q) → □p → □q
①, ②
(2) In (1), we proved that K is a theorem of Tr and Ver.
We will show that K is a theorem of Fal:
① ¬□p
Fal
② ¬□p → □(p → q) → □p → □q
PC
③ □(p → q) → □p → □q
①, ②
Finally, we will show that K is a theorem of Com:
① □(p → q) → ¬(p → q)
Com, US
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② □(p → q) → ¬ q
RPC
③ ¬ q → □q
Com
④ □(p → q) → □q
②, ③, RPC
⑤ □(p → q) → □p → □q
RPC
(3) Since □p and □p ↔ p are both the theorems of < Tr, Ver >,
it follows that p is also a theorem of < Tr, Ver >, so, by US, there is a
formula A such that A and ¬A are both the theorems of < Tr, Ver >,
and thus <Tr, Ver > is not harmonious.
(4) By Deleting Modality Method used in the proof of 1.1.16, it is
easy to see that Tr is harmonious. By (3), Ver ⊈ Tr.
On the other hand, define τ(A) as in the proof of 1.2.18. It is easy to
prove that Ver is harmonious. By (3), Tr ⊈ Ver. ┤
Remark. There are some systems such that they are subsystems of
Tr and Ver. For example, K, K4 and K + ◇p → □p.
The following theorem shows that a harmonious extension system
of System D is always a subsystem of Tr. Namely, the harmonious
extension systems starts from D and ends to Tr.
Theorem 1.4.6 Let S ⊇ D be a modal system. If S is harmonious, then
S ⊆ Tr.
Proof. By contraposition, we will show that
(☆) If S ⊈ Tr, then S is not harmonious.
It is easy to see that RE is a derived rule of Tr, and thus RER is a
derived rule of Tr. First we will show that
－□
① A
is a tautology ⇒ ⊢ Tr A.
－□
Assume that A
is a tautology. Since PC ⊆ Tr, it follows that
－□
② ⊢ Tr A .
By Axiom □p ↔ p and US,
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③ ⊢ Tr □B ↔ B for all formula B.
－□
We can always revivify all occurrences of □ deleted in A
by ②,
③ and RER, so ⊢ Tr A. Thus we get ①.
Now we will prove (☆): Assume that S ⊈ Tr, then there is some A
such that
④ A is a theorem of S, and
⑤ A is not a theorem of Tr.
－□
is not a tautology. So
By ⑤ and ①, it is easy to see that A
there is a classical truth mapping v from At into {0, 1} such that the
－□
by 0. We substitute T
classical extension v* of v to ML valuates A
for all sentence symbol pn in A such that v(pn) = 1, and ⊥ for all pn
in A such that v(pn) = 0. Such obtained formula is denoted as A1.
It is clear that A1 is a substitution instance of A and constant
－□
is also not a tautology,
formula (1.3.8). It is easy to prove that (A1)
so, by 1.3.9, ¬A1 is a theorem of D, and thus, by the assumption D ⊆
S, ¬A1 is a theorem of S.
On the other hand, by ④ and US, A1 is a theorem of S, so S is not
harmonious. ┤
Definition 1.4.7 A modal system S is Post-complete ⇔ the following
conditions are satisfied:
(1) S is harmonious, and
(2) Given any A such that A is not a theorem of S. Then S + A is not
harmonious. ┤
Remark. (2) means that the extension system obtained by adding
any non-theorem of S as axiom to S is not harmonious.
Theorem 1.4.8 (Post-Completeness Theorem) Tr and Ver are Post-
96
complete.
Proof. First consider Tr. By 1.4.5(4), Tr is harmonious.
It suffices to show that Tr satisfies 1.4.7(2). For this, we will
introduce the following result we are not going to prove:
① each harmonious normal system is a subsystem of Tr or Ver.
Note that the half proof of ① has proved before (1.4.6)
Given any A such that
② A is not a theorem of Tr.
It suffices to show that
③ Tr + A is not harmonious.
Hypothesize that Tr + A is harmonious. By 1.4.5(1), Tr + A is a
normal system, so, by ①, we have
④ Tr + A ⊆ Tr, or
⑤ Tr + A ⊆ Ver.
But ④ contradicts ② , whereas ⑤ contradicts ‘Tr ⊈ Ver’ in
1.4.5(4). Thus we get ③.
Similarly, we can prove that Ver is Post-complete. ┤
Remark. We can prove that Tr and Ver are precisely the two nonvacuous Post-complete systems w.r.t. the modal logics (refer to
Remark (Ⅱ) below 1.1.6). Of course, PC is also a Post-complete
system w.r.t. the classical propositional calculus.
Definition 1.4.9 Let Φ be any formula set, and Ψ contain no modal
formula. f is a (degenerate) embedding mapping from Φ into Ψ ⇔ f is
a mapping from Φ into Ψ such that:
(1) f(pn) = pn for all n < ω,
(2) f(¬A) = ¬f(A),
(3) f(A ∧ B) = f(A) ∧ f(B),
97
(4) f(□A) = sf(A) where s is a unary truth connective symbol. ┤
Remark. (Ⅰ) By (1) - (3), we have
f(A → B) = f(A) → f(B),
f(A ↔ B) = f(A) ↔ f(B).
(Ⅱ) The function of f is to replace all occurrences of □ in the
current formula by a truth connective symbol, so, by (4), in essence,
f(□A) belongs to {f(A), ¬ f(A), T, ⊥}.
Definition 1.4.10 Let S and S1 be two modal systems.
(1) S is embeddable into S1 ⇔ there is an embedding mapping from
Th(S) into Th(S1) such that
(☆) A ∈ Th(S) ⇔ f(A) ∈ Th(S1) for all A ∈ ML.
(2) S degenerates into S1 or S collapses into S1 ⇔ S1 ⊂ S and S is
embeddable into S1.
(3) S is a degenerate system ⇔ S degenerates into PC. ┤
Remark. ‘⇐’ of (☆) in (1) is essential: Consider {□p} and {p}.
Define f is a mapping from {□p} into {p} such that
(1) f(pn) = pn for all n < ω,
(2) f(¬A) = ¬f(A),
(3) f(A ∧ B) = f(A) ∧ f(B),
(4) f(□A) = f(A).
It is easy to see that f is an embedding mapping from {□p} into {p},
and
p ∉{□p} but f(p) ∈ {p}.
Theorem 1.4.11
(1) Tr is a degenerate system.
(2) Ver is a degenerate system.
98
Proof. (1) Let f be an embedding mapping from Th(Tr) into Th(PC)
such that
f(□A) = f(A) for all A ∈ ML.
By the constitution of Tr, it suffices to show that a substitution
instance of its character axiom p ↔ □p make (☆) of 1.4.10 (1) hold.
As p ↔ □p ∈ Th(Tr), so, by US, it is easy to see that the following
proposition holds trivially:
A ↔ □A ∈ Th(Tr) ⇔ f(A) ↔ f(A) ∈ Th(PC) for all A ∈ ML.
So
A ↔ □A ∈ Th(Tr) ⇔ f(□A ↔ A) ∈ Th(PC) for all A ∈ ML.
It is not difficult to prove by the previous proposition that:
A ∈ Th(Tr) ⇔ f(A) ∈ Th(PC) for all A ∈ ML.
So, (☆) in 1.4.10 (1) holds, it is easy to see that Tr is embeddable
into PC.
It is easy to prove that Tr ∉ Th(PC), so PC ⊂ Tr, and thus Tr is a
degenerate system.
(2) Let f be an embedding mapping from Th(Ver) into Th(PC) such
that
f(□A) = T for all A ∈ ML.
As above, it suffices to consider a substitution instance of the
character axiom of Ver. Since □p ∈ Th(Ver), so, by US, it is easy to
see that the following proposition trivially holds:
□A ∈ Th(Ver) ⇔ f(□A) ∈ Th(PC) for all A ∈ ML.
So, as (1), it is easy to prove that Ver is a degenerate system. ┤
As far as we know, S5 is an extension system of a lot of modal
systems, so, if we can prove that S5 is not a degenerate system, then
that many modal systems are not degenerate systems.
Theorem 1.4.12
99
(1) S5 is not embeddable into PC.
(2) S5 is not a degenerate system.
(3) Any subsystem of S5 is not a degenerate system.
Proof. (1) First we will prove:
① if S5 is embeddable into PC, then ⊢ S5 Tr.
Assume that
② S5 is embeddable into PC.
Then there is an embedding mapping f from Th(S5) into Th(PC) such
that
(☆) A ∈ Th(S5) ⇔ f(A) ∈ Th(PC) for all A ∈ ML.
Hence ⊢ PC f(5), so f(5) is a tautology. By 1.4.9, we have
f(5) = f(◇p) → f(□◇p) = ¬ s ¬ p→ s¬ s ¬ p,
where s is one of s1, s2, s3, s4 in Remark below 1.4.1.
If s = s1, then f(5) = ¬¬ p → ¬¬ p, so f(5) is a tautology.
If s = s2, then f(5) = ¬¬¬ p → ¬¬¬¬ p, so f(5) is not a tautology.
If s = s3, then f(5) = ¬T → T, so f(5) is a tautology.
If s = s4, then f(5) = ¬⊥ → ⊥, so f(5) is not a tautology.
We need not consider the cases of ‘f(5) is not a tautology’, so the
above four cases boil down to the following two cases:
f(◇p) = p,
f(◇p) = ⊥.
Case 1 f(◇p) = ⊥: Since ⊢ PC ¬f(◇p), it follows that
⊢ PC f(¬◇p).
By ② and (☆) of 1.4.10 (1), ⊢ S5 ¬◇p. By ③ in the proof of 1.1.16,
we have ⊢ PC ¬¬¬ p. By RPC and US, it is easy to prove that ⊢ PC⊥,
contradicting the harmony of PC. So Case 1 does not hold.
Case 2 f(◇p) = p: Since ⊢ PC f(◇p) ↔ p, it follows that
⊢ PC f(◇p ↔ p).
By ② and (☆) of 1.4.10 (1), ⊢ S5 ◇p ↔ p, so, by RPC and US,
100
⊢ S5 ¬◇¬ p ↔ ¬¬ p.
By E(3), ⊢ S5 □p ↔ ¬¬ p, it is easy to see that ⊢ S5 Tr.
In this way, we have proved ①.
In the following (3.4.4(7)) we shall show that ⊬ S5 Tr, so, by ①, S5
is not embeddable into PC.
(2) By (1) and 1.4.10(2).
(3) By (2). ┤
Note that by the previous definition, PC is not a degenerate system,
since PC ⊄ PC.
PC is seen as the smallest modal system. From PC, we can give a
lot of extensions of it. By Post-Completeness Theorem, when
extending PC to Tr, we cannot extend Tr to a harmonious proper
extension system of it; when extending PC to Ver from another
direction, we cannot extend Ver to a harmonious proper extension
system of it. But we have proved that Tr and Ver are two degenerate
systems. This means that they are PC in essence. Starting from the
smallest system PC, by adding modal axioms or modal rules, we
extend systems to Tr or Ver step by step, whereas Tr and Ver
degenerate into PC again! This is indeed interesting.
Exercises 1.4
1.4.13 Prove that
(1) Tr = D + p → □p.
(2) Tr = S5 + McK.
(3) Tr = T − K + □(p → q) ↔ (□p → □q).
(4) Ver = KW + Lem.
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(5) S4.1 ⊈ S5. ┤
1.4.14 Prove that Com and Fal are degenerate systems. ┤
1.4.15 Prove that
－□
is a theorem of Tr.
(1) A ↔ A
－□
(2) A ↔ A
is not a theorem of Ver. ┤
1.4.16 Prove that
(1) □A → □B∕A → B is not a weak rule of S4.
(2) □A → □B∕A → B is not a weak rule of T. ┤
1.4.17 Prove that
(1) D is not a theorem of Ver, and thus D ⊈ Ver.
(2) T is not a theorem of Ver, and thus T ⊈ Ver. ┤
1.4.18 (Problem) Except the systems listed in Remark below 1.4.5,
which system is still a subsystem of Tr and Ver? ┤
1.4.19 (Problem)
(1) Which degenerate system is a 0 +-degree system?
(2) Which degenerate system is a 0-degree system? ┤
1.4.20 (Problem) Except the systems mentioned before, is there a
system such that it is simultaneously a proper extension system of S5
and is not a degenerate system? ┤
1.4.21* (Problem)
(1) How many systems describing a binary modal operator are
degenerate or Post-complete?
102
(2) How many systems describing a n-ary modal operator are
degenerate or Post-complete? ┤
103
§5 Other Important Systems
In this section we will study some other important systems, do enough
preparations for the works of the following chapters.
Theorem 1.5.1
(Ⅰ) The following are the theorems or derived rules of S4.1:
(1) □◇p ∧ □◇q → ◇(p ∧ q) (= F).
(2) ◇A, ◇B∕◇(A ∧ B).
(3) ◇A1, …, ◇An∕◇(A1 ∧…∧ An) where 1 ≤ n < ω.
(4) ◇((p1 → □p1) ∧ …∧ (pn → □pn)) where 1 ≤ n < ω.
(Ⅱ) S4.1 and the following systems are equivalent:
(1) S4 + ◇□(p → □p).
(2) S4 + □(p ∨ q) → ◇□p ∨ ◇□q.
(3) S4 + F.
Proof. (Ⅰ) The proof of (1):
McK, RPC
① □◇p ∧ □◇q → □◇p ∧ ◇□q
② □◇p ∧ ◇□q → ◇(□q ∧ ◇p)
R(8), RER
③ □q ∧ ◇p → ◇(p ∧ q)
R(8), RER
④ ◇(□q ∧ ◇p) → ◇◇(p ∧ q)
RM◇
⑤ ◇◇(p ∧ q) → ◇(p ∧ q)
4◇, US
⑥ □◇p ∧ □◇q → ◇(p ∧ q)
①, ②, ④, ⑤
The proof of (2):
① ◇A, ◇B
Assumption
② □◇A, □◇B
RN
104
③ □◇A ∧ □◇B → ◇(A ∧ B)
(1), US
④ ◇(A ∧ B)
②, ③, RPC
The proof of (3): By induction on 1 ≤ n < ω:
The case of n = 1 holds clearly. Inductively hypothesize that the
result we will prove holds when n = k.
① ◇A1, …, ◇Ak + 1
Assumption
② ◇(A1 ∧…∧ Ak)
Induction hypothesis
③ ◇(A1 ∧…∧ Ak + 1)
①, ②, (2)
The proof of (4): By T(4), we have the following theorem:
◇(p1 → □p1), …, ◇(pn → □pn).
By (3), we get easily the result we will prove. ┤
Theorem 1.5.2
(Ⅰ) The following are the theorems or derived rules of GL = KW:
(1) 4,
(2) □(□p → p) ↔ □p,
(3) □(□p ∧ p) ↔ □p,
(4) □A → A∕□A (= RW),
(5) □A ∧ A ∧ □B → B∕□A → □B,
(6) □A1 ∧ A1 ∧…∧ □An ∧ An ∧ □B → B
∕□A1 ∧…∧ □An → □B where n < ω,
(7) □⊥ ↔ □◇p.
(Ⅱ) KW = K4 + W1 where
(W1) □(□p ↔ p) → □p.
Proof.
(Ⅰ) The proof of (1):
PC
① p → □p ∧ □□p → p ∧ □p
② □p → □(□p ∧ □□p → p ∧ □p)
RM
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③ □p → □(□(p ∧ □p) → p ∧ □p)
R, RER
④ □p → □(p ∧ □p)
W
⑤ □p → □□p
R, RPC
The proof of (2):
W
① □(□p → p) → □p
② □p → □(□p → p)
PC, RM
③ □(□p → p) ↔ □p
①, ②
The proof of (3):
PC, RM
① □(□p ∧ p) → □p
② □p → □□p ∧ □p
4, RPC
③ □p → □(□p ∧ p)
R, RPC
④ □(□p ∧ p) ↔ □p
①, ③
The proof of (4):
Assumption
① □A → A
② □(□A → A)
RN
③ □A
W
The proof of (5):
Assumption
① □A ∧ A ∧ □B → B
② □A ∧ A → □B → B
RPC
③ □□A ∧ □A → □(□B → B)
RK
④ □A ∧ □A → □(□B → B)
4
⑤ □A → □B
W, RPC
The proof of (6):
Assumption
① □A1 ∧ A1 ∧…∧ □An ∧ An ∧ □B → B
② □A1 ∧ A1 ∧…∧ □An ∧ An → □B → B
RPC
③ □(□A1 ∧ A1) ∧ …∧ □(□An ∧ An)
→ □(□B → B)
RK
④ □A1 ∧…∧ □An → □(□B → B)
(3), RPC
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⑤ □A1 ∧…∧ □An → □B
W
The proof of (7):
PC, RM
① □⊥→ □◇p
② ◇p → ◇T
PC, RM◇
③ ◇T → □⊥ →⊥
PC, Df◇, Df⊥
④ ◇p → □⊥ → ⊥
②, ③
⑤ □◇p → □(□⊥ → ⊥)
RM
⑥ □◇p → □⊥
W
⑦ □⊥ ↔ □◇p
①, ⑥
(Ⅱ) First we will prove ‘⊆’:
① (q → r) → (q → p) → (r ∧ q ↔ q ∧ p)
② (□p → □□p) → (□p → p)
→ (□□p ∧ □p ↔ □p ∧ p)
③ (□p → p) → (□□p ∧ □p ↔ □p ∧ p)
④ (□p → p) → (□(□p ∧ p) ↔ □p ∧ p)
⑤ □(□p → p) → □(□(□p ∧ p) ↔ □p ∧ p)
⑥ □(□(□p ∧ p) ↔ □p ∧ p) → □(□p ∧ p)
⑦ □(□p → p) → □(□p ∧ p)
⑧ □p ∧ p → p
⑨ □(□p ∧ p) → □p
⑩ □(□p → p) → □p
Next we will prove ‘⊇’: by (Ⅰ) (1), it suffices to
W1:
① (□p ↔ p) → (□p → p)
PC
RM
② □(□p ↔ p) → □(□p → p)
③ □(□p ↔ p) → □p
W ┤
PC
US
4, RPC
R, RER
RM
W1, US
⑤, ⑥
PC
⑧, RM
⑦, ⑨
show that ⊢ KW
Remark. In KW, □ can be intuitively understood as ‘it is provable
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that’ and ◇ as ‘it is consistent that ’, so KW is called the logic of
provability. The reader who is interested in it can refer to  of
Boolos. Where, KW is investigated quite detailedly.
Let Grz be the following formula:
(Grz) □(□(p → □p) → p) → p,
where Grz is the abbreviation of Poland logician Grzegorczyk.
Theorem 1.5.3 The following are the theorems of KGrz:
(1) T,
(2) 4.
Proof. The proof of (1):
① p → □(p → □p) → p
PC
② □p → □(□(p→ □p) → p)
RM
③ □p → p
Grz
The proof of (2): this proof is more complex, so first we let
ϕ = 4 ∧ p,
ψ = □(ϕ → □ϕ).
Then
PC
① (ϕ → □p) → (p → □p)
② ϕ→p
PC
③ □ϕ → □p
RM
④ (ϕ → □ϕ) → (ϕ → □p)
RPC
⑤ (ϕ → □ϕ) → (p → □p)
④, ①, RPC
⑥ □(ϕ → □ϕ) → □p → □□p
RM, K
⑦ ψ→4
Abbreviation
⑧ □(ψ → 4)
RN
⑨ □p → □(ψ → 4)
RPC
⑩ □p → □(ψ → p)
PC, RM
⑪ □(ψ → 4) ∧ □(ψ → p) → □(ψ → 4 ∧ p)
PC, RR
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⑫
⑬
⑭
⑮
⑯
⑰
□p → □(ψ → 4 ∧ p)
□p → □(ψ → ϕ)
□(ψ → ϕ) → ϕ
□p → ϕ
□p → 4
4
⑨, ⑩, ⑪
Abbreviation
Grz, US
⑬, ⑭
RPC
RPC ┤
Exercises 1.5
1.5.4 Prove that
(Ⅰ) The following are the theorems of S4.3:
(1) □(□p → p) (= U),
(2) □(□p → □q) ∨ □(□q → □p),
(3) G (and thus S4.2 ⊆ S4.3).
(Ⅱ) S4.3 = S4 + ◇(□p ∧ q) → □(◇q ∨ p). ┤
1.5.5 Prove that
(1)* KW = K4 + RW where RW refers to 1.5.2(4).
(2) KW is harmonious.
(3) The following two propositions are equivalent in K:
(Ⅰ) ⊢ W.
(Ⅱ) there is A such that
⊢ □A ↔ □p and ⊢ A ↔ (□A → p). ┤
1.5.6 Prove that in S4.2, there are at most 4 pairwise unreducible
nonempty pure modal symbol strings:
□, ◇□, □◇, ◇,
and they can be fomed a linear order by using the implication symbol
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→. ┤
1.5.7 Let S4.3.2 := S4 + f and S4.4 := S4 + w5 where
(f) p ∧ ◇□q → □(◇p ∨ q),
(w5) p → ◇p → □◇p.
Prove that
(1) B ∈ Th(S4.4).
(2) S4.3.2 ⊆ S4.4. ┤
110
Chapter 2 Strong Derivation, Consistency
and Maximal Consistency
In this chapter we will continue studying the proof theory of modal
logics, and thus do enough preparations for the proofs of the
completeness theorems we shall give out in Chapter 4.
In§1 we define the concept of strong derivation in a modal system,
show the relation between the concept and the concept of derivation
defined before, and prove Deduction Theorem for the strong
derivation.
In§2 we define the S-consistency and maximality of formula sets,
and mostly reveal the properties of S-consistent sets and S-maximal
consistent sets.
111
§1 Strong Derivation
In this section we define the concept of strong derivation in a modal
system, show the relation between the concept and the concept of
derivation defined before, and prove Deduction Theorem for the
strong derivation.
Definition 2.1.1
(1) A has a strong derivation from Φ in S or A is strongly inferred
from Φ in S, denoted as Φ ⊢ S A or A ∈ CnS(Φ), ⇔ there is a finite
subset Ψ ⊆ Φ such that ⊢ S ∧Ψ → A.
CnS(Φ) = {A ∈ ML | Φ ⊢ S A} is also called the strong provability
consequence set of Φ in S.
(2) Otherwise, denoted as Φ ⊬ S A. ┤
Remark. (Ⅰ) It is easy to see that the concept of strong derivation
is defined by the concept of theorem (formal proof), it and the concept
of derivation defined before are different. The latter is independently
defined, we even can use it to define the concepts of formal proof and
theorem.
(Ⅱ) Let ℘(ML) be the power set of ML. It is easy to see that the
concept of derivation defined before and the strong derivation defined
here actually define the two provability consequence relations on
Cartesian product ℘(ML) × ML. In other words, ⊢ S and ⊢ S generate
the two provability consequence relations R⊢ and R⊢ S on ℘(ML) ×
S
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ML, respectively:
R⊢ = {<Φ, A > | Φ ⊢ S A},
S
R⊢ S = {<Φ, A > | Φ ⊢ S A}.
Corollary 2.1.2
(1) Φ ⊢ S A ⇒ Φ ⊢ S A, and thus CnS(Φ) ⊆ CnS(Φ).
(2) ∅ ⊢ S A ⇔ ∅ ⊢ S A, and thus CnS(∅) = CnS(∅) = Th(S).
(3) Φ ⊢ PC A ⇔ Φ ⊢ PC A, and thus CnPC(Φ) = CnPC(Φ).
Proof. (1) Assume that Φ ⊢ S A. Then there is a finite subset Ψ ⊆ Φ
such that
⊢ S ∧Ψ → A,
so Ψ ⊢ S ∧Ψ → A. Since Ψ ⊢ S ∧Ψ, it follows that Ψ ⊢ S A by MP
①
, so Φ ⊢ S A.
(2) By (1), it suffices to show that ‘⇐’: Assume that ∅ ⊢ S A, then
⊢ S A, hence ⊢ S T → A, namely, ⊢ S ∧∅ → A, so, by the definition
of strong derivation, ∅ ⊢ S A.
Counterexample 2.1.3 The converse of the above (1) does not hold
generally.
Proof. Assume that S is a normal system such that it does not have the
theorem p → □p. For example, S5.
By RN, we have p ⊢ S □p.
On the other hand, hypothesize that p ⊢ S □p. Then there is a finite
①
Strictly speaking, here MP means Detachment Rule for Derivation.
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subset Ψ ⊆ {p} such that
⊢ S ∧Ψ → □p,
so ⊢ S ∧∅ → □p or ⊢ S ∧{p} → □p. There are two cases to
consider:
Case 1 ⊢ S ∧∅ → □p: then ⊢ S T → □p, so ⊢ S p → □p.
Case 2 ⊢ S ∧{p} → □p: then ⊢ S p → □p.
In the two cases, we have a contradiction. ┤
Theorem 2.1.4 (Deduction Theorem for Strong Derivation)
(1) Φ ∪ {A} ⊢ S B ⇒ Φ ⊢ S A → B.
(2) A ⊢ S B ⇒ ⊢ S A → B.
Proof. (1) Assume that Φ ∪ {A} ⊢ S B, then
① there is a finite subset Ψ ⊆ Φ ∪ {A} such that ⊢ S ∧Ψ → B.
Case 1 A ∉ Ψ: then by ① and RPC,
② Ψ ⊆ Φ and ⊢ S ∧Ψ → A → B.
Case 2 A ∈ Ψ: Let Θ = Ψ − {A}. By ①, ⊢ S ∧Θ ∧ A → B, so
③ Θ ⊆ Φ and ⊢ S ∧Θ → A → B.
By ②, ③ and the definition of strong derivation, Φ ⊢ S A → B.
(2) By (1). ┤
Exercises 2.1
2.1.5 Let S ⊆ S1 be two modal systems. Prove that CnS(Φ) ⊆ CnS1(Φ).
┤
114
2.1.6 Let S be a modal system. Prove that
(1) Φ ⊆ CnS(Φ).
(2) Th(PC) ⊆ CnS(∅).
(3) CnS(∅) ⊆ CnS(Φ).
(4) Th(PC) ⊆ CnS(Φ).
(5) A ∈ Φ ⇒ Φ ⊢ S A.
(6) A ∈ Φ ⇒ Φ ⊢ S A. ┤
2.1.7 Let S be a modal system. Prove that
(☆) Φ ⊢ S A → B and Φ ⊢ S A ⇒ Φ ⊢ S B. ┤
[(☆) can be called Detachment Rule for Strong Derivation.]
2.1.8 Prove that Φ ⊢ K A ⇔ Φ ⊢ Kb A ⇔ Φ ⊢ Kb A. ┤
2.1.9 Let S = K + Ψ and □ωΦ = {□nA | A ∈ Φ and n < ω}.
Prove that
(1) Φ ⊢ S □ωΦ.
(2) Φ ⊢ S A ⇒ □ωΦ ⊢ S A, but its converse does not hold.
(3) (Deduction Theorem for □ωΦ)
□ωΦ ∪ □ω{B} ⊢ S A
⇒ there is some n < ω such that □ωΦ ⊢ S □0B ∧…∧ □nB → A.
(4)* □ωΦ ⊢ S A ⇔ Φ ⊢ S A ⇔ □ωΦ ⊢ S A. ┤
[Remark. The intuitive meaning of □ωΦ is: (the formulas of) Φ is
closed under RN.]
115
§2 S-Consistency and Maximality
In this section we define the S-consistency and maximality of formula
sets, and mostly reveal the properties of S-consistent sets and Smaximal consistent sets.
1. General Properties of Extension Systems of PC
In this subsection we will discuss the general properties of the
extension system S of PC, concretely not deal with the concrete
properties of the character axioms and character rules of S.
Definition 2.2.1
(1) A set Φ is S-consistent, denoted as Φ ∈ CS(S),
⇔ for all finite set Ψ ⊆ Φ, ⊬ S ¬∧Ψ.
Φ is called S-inconsistent ⇔ Φ is not S-consistent.
(2) Φ is maximal (w.r.t. ML) ⇔ A ∈ Φ or ¬A ∈ Φ for all A ∈ ML.
(3) Φ is S-maximal consistent, denoted as Φ ∈ MCS(S),
⇔ Φ is maximal and S-consistent.
(4) A modal system S is consistent ⇔ Th(S) is S-consistent.
S is called inconsistent ⇔ S is not consistent.
(5) We use A ∈ CS(S) as ‘{A} is S-consistent’, and say that A is Sconsistent. ┤
Remark. (Ⅰ) The concept of consistency deals with systems,
116
whereas the concept of maximality deals with languages.
(Ⅱ) CS(S) and MCS(S) denotes the set of all S-consistent sets and
the set of all S-maximal consistent sets, respectively.
(Ⅲ) We can see the rationality of the definition ∧∅ := T from the
concept of consistency, because the empty set ∅ always is a subset of
any set.
( Ⅳ ) There is also an equivalent definition about S-maximal
consistent set:
Φ is S-maximal consistent ⇔ the following conditions are satisfied:
① Φ is S-consistent, and
② A ∉ Φ ⇒ Φ ∪ {A} is S-inconsistent for all formula A. ①
Please prove that the above equivalence (refer to Exercise 2.2.19).
Lemma 2.2.2 (Consistency Lemma)
(1) Φ is S-inconsistent
⇔ there is a finite set Ψ ⊆ Φ such that ⊢ S ¬∧Ψ.
(2) Φ is S-consistent
⇔ there is not a finite set Ψ ⊆ Φ such that ⊢ S ¬∧Ψ.
(3) S is consistent ⇔ ⊬ S ⊥ ⇔ ∅ is S-consistent.
(4) S is inconsistent ⇔ Th(S) = ML.
(5) S is inconsistent ⇔ CS(S) = ∅. ②
(6) A is S-inconsistent ⇔ ⊢ S ¬A.
(7) A is S-consistent ⇔ ⊬ S ¬A.
(8) ¬A is S-consistent ⇔ ⊬ S A.
(9) ¬A is S-inconsistent ⇔　⊢ S A.
② means that Φ has no S-consistent proper extension.
So, below, we always presuppose that S is consistent when consider or
prove that a formula set is S-consistent.
①
②
117
(10) Φ ∪ {A} is S-inconsistent
⇔ there is a finite set Ψ ⊆ Φ such that ⊢ S ∧Ψ → ¬A.
(11) Φ ∪ {¬A} is S-inconsistent
⇔ there is a finite set Ψ ⊆ Φ such that ⊢ S ∧Ψ → A.
(12) Φ ∪ {A} is S-consistent
⇔ for all finite set Ψ ⊆ Φ, ⊬ S ∧Ψ → ¬A.
(13) Assume that Φ ⊆ Th(S).
① If Φ ∪ {A} is S-inconsistent, then ⊢ S ¬A.
② If Φ ∪ {¬A} is S-inconsistent, then ⊢ S A.
Proof. By definition, we get (1) easily.
We get (2) easily by (1).
(3) It suffices to show that
① S is consistent ⇒ ⊬ S⊥ ⇒ ∅ is S-consistent ⇒ S is consistent.
Assume that S is consistent. By 2.2.1(4), Th(S) is S-consistent.
Since ∅ is a finite subset of Th(S), so, by the S-consistency definition,
② ⊬ S ¬∧∅.
Since ¬∧∅ = ¬T = ⊥, it follows that ⊬ S⊥.
Assume that ⊬ S ⊥ , then we have ② , by the S-consistency
definition, it is easy to see that ∅ is S-consistent.
Assume that ∅ is S-consistent. Hypothesize that S is inconsistent,
then by 2.2.1(4), Th(S) is S-inconsistent, so, by (1), there is a finite
subset Ψ ⊆ Th(S) such that ⊢ S ¬∧Ψ. Since Ψ ⊆ Th(S), it follows that
⊢ S ∧Ψ. By ⊢ S ¬A → A → B, it is easy to prove that ⊢ S ⊥. Since ∅
is S-consistent, so, by the S-consistency definition, ⊬ S ⊥, and thus
We get (4) easily by (3).
(5) ‘⇒’: Assume that S is inconsistent. By (3), ⊢ S ⊥, and thus ⊢ S
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¬∧∅. Given any formula set Φ. Since ∅ is a finite subset of Φ, so,
by (1), Φ is S-inconsistent, and thus CS(S) = ∅.
‘⇐’: Assume that CS(S) = ∅. Then ∅ is S-inconsistent, so, by (3),
S is inconsistent.
(6) ‘⇐’: Assume that ⊢ S ¬A, then ⊢ S ¬∧{A}. Since {A} ⊆ {A},
it is easy to see by (1) and 2.2.1(5) that A is S-inconsistent.
‘⇒’: Assume that A is S-inconsistent. By (1) and 2.2.1(5), there is a
finite set Ψ ⊆ {A} such that
(#) ⊢ S ¬∧Ψ.
Case 1 S is consistent: By (3), ⊬ S ⊥. So Ψ ≠ ∅, and thus Ψ =
{A}, and hence ⊢ S ¬A by (#).
Case 2 S is inconsistent: By (4), Th(S) = ML, so ⊢ S ¬A.
(7) By (6).
(8) By (7).
(9) By (8).
(10) ‘⇒’: Assume that Φ ∪ {A} is S-inconsistent. Then by (1),
there is a finite set Θ ⊆ Φ ∪ {A} such that
① ⊢ S ¬∧Θ.
Case 1 A ∉ Θ: Let Ψ = Θ, so ⊢ S ¬∧Ψ by ①. By
⊢ S ¬B → ¬(B ∧ C)
and MP, we have
② ⊢ S ¬(∧Ψ ∧ A).
Case 2 A ∈ Θ: Let Ψ = Θ − {A}, so, by ①, we still have ②.
It is clear that Ψ in the above two cases is a finite subset of Φ.
In this way, by ② and RPC, it is clear that ⊢ S ∧Ψ → ¬A.
‘⇐’: Assume that there is a finite set Ψ ⊆ Φ such that ⊢ S ∧Ψ →
¬A. It is clear that ② holds. Let Θ = Ψ ∪ {A}, then there is a finite
119
set Θ ⊆ Φ ∪ {A} such that ① holds.
By (1), it is easy to see that Φ ∪ {A} is S-inconsistent.
(11) - (12) hold clearly by (10).
(13) ① Assume that Φ ∪ {A} is S-inconsistent. Then by (10),
there is a finite set Ψ ⊆ Φ such that ⊢ S ∧Ψ → ¬A. Since Φ ⊆ Th(S),
it follows that ⊢ S ∧Ψ, and thus by MP, ⊢ S ¬A.
We get ② easily by ①. ┤
Remark. (Ⅰ) In my opinion, (8) is the most important, please fix
this.
(Ⅱ) By (4), the inconsistent system is unique.
(Ⅲ) By 1.1.14, the above (3) or (4), it is easy to see that:
Corollary 2.2.3
(1) Given any modal system S. Then
S is consistent ⇔ S is harmonious ⇔ S is coherent.
(2) Let S be a system defined in 1.1.9. Then S is consistent. ┤
Remark. By the above corollary (1), we can use a property of them
to represent the other properties.
In the following we will consider the relations between the strong
derivation and the S-consistency:
Theorem 2.2.4
(1) Φ ⊢ S A ⇔ Φ ∪ {¬A} is S-inconsistent.
(2) Φ ⊢ S ¬A ⇔ Φ ∪ {A} is S-inconsistent.
(3) Φ ⊬ S A ⇔ Φ ∪ {¬A} is S-consistent.
120
Proof. (1) By the definition of Φ ⊢ S A and Consistency Lemma
2.2.2(11), we get easily the following relation:
Φ ⊢ S A ⇔ there is a finite set Ψ ⊆ Φ such that ⊢ S ∧Ψ → A
⇔ Φ ∪ {¬A} is S-inconsistent.
(2) By the definition of Φ ⊢ S ¬A and 2.2.2(10).
(3) By (1). ┤
Since S is an extension system of PC, it follows that some
important metatheorems for PC can naturally be generalized to the
following metatheorems for S:
Theorem 2.2.5
(1) (Compactness Theorem)
Φ is S-consistent ⇔ all finite subset of Φ are S-consistent.
(2) (Lindenbaum-Lemma)
Φ is S-consistent
⇒ there is a S-maximal consistent set Ψ such that Ψ ⊇ Φ.
Proof. (1) ‘⇒’: Assume that Φ is S-consistent. Then, by definition,
for all finite set Ψ ⊆ Φ,
(☆) ⊬ S ¬∧Ψ.
Hypothesize that Ψ is S-inconsistent, then by Consistency Lemma
2.2.2(1), there is a finite subset Θ ⊆ Ψ such that ⊢ S ¬∧Θ. Since
⊢ S ¬∧Θ → ¬∧Ψ,
it follows that ⊢ S ¬∧Ψ by MP, and thus we have a contradiction with
(☆). So Ψ is S-consistent.
‘⇐’: Assume that Φ is S-inconsistent. Then by 2.2.2(1), there is a
finite subset Ψ ⊆ Φ such that ⊢ S ¬∧Ψ, so there is a finite subset Ψ ⊆
Ψ such that ⊢ S ¬∧Ψ, and thus by 2.2.2(1), Ψ is S-inconsistent.
121
(2) Assume that Φ be S-consistent. We can enumerate all formula
in ML as follows:
A0, A1, …, An, ….
By S-consistency, construct a ⊆-chain of sets as follows:
Φ = Ψ0 ⊆ Ψ1 ⊆…⊆ Ψn ⊆…,
such that for all n < ω,
Ψn ∪ {An} is S-consistent;
Ψn ∪ {An},
Ψn + 1 =
Ψn ∪ {¬An}, Ψn ∪ {An} is S-inconsistent.
By induction on n < ω, we will show that
① Ψn is S-consistent.
By the assumption, Φ = Ψ0 is S-consistent. For any n > 0, inductively
hypothesize that Ψn is S-consistent. We are going to prove that Ψn + 1
is S-consistent. For this, consider the S-consistency of Ψn ∪ {An}.
Case 1 Ψn ∪ {An} is S-consistent: Here Ψn + 1 = Ψn ∪ {An}, and
thus Ψn + 1 is S-consistent.
Case 2 Ψn ∪ {An} is S-inconsistent: Here Ψn + 1 = Ψn ∪ {¬An}.
Hypothesize that Ψn + 1 is S-inconsistent, then by the assumption of
this case and the hypothesis, we have
Ψn ∪ {An} is S-inconsistent, Ψn ∪ {¬An} is S-inconsistent,
so, by Consistency Lemma 2.2.2(10) - (11), there are finite subsets Θ,
Ξ ⊆ Ψn such that
⊢ S ∧Θ → ¬An and ⊢ S ∧Ξ → An,
by RPC, we have
⊢ S ¬(∧Θ ∧ ∧Ξ).
Since Θ ∪ Ξ is a finite subset of Ψn, by 2.2.2(1), Ψn is S-inconsistent,
contradicting the induction hypothesis. So we have proved that Ψn + 1
is S-consistent. Thus we get ①.
Let Ψ = ∪n < ωΨn. Since Ψ0 ⊆ Ψ, it follows that Φ ⊆ Ψ. In the
122
following it suffices to show that
② Ψ is a S-maximal consistent set.
Verify the maximality: Given any A ∈ ML, there always exists n <
ω such that A = An, by the constitution of Ψn + 1, there always exists
one of An or ¬An such that it belongs to Ψn + 1 ⊆ Ψ, so Ψ is maximal.
Verify the consistency: Hypothesize that Ψ is S-inconsistent, then
by Compactness Theorem, there is a finite subset Θ ⊆ Ψ such that Θ
is S-inconsistent. But each finite subset of Ψ is always a subset of
some constructed Ψn ⊆ Ψ, so, by Compactness Theorem, it is easy to
see that Ψn is S-inconsistent, contradicting ①. Thus we get ②. ┤
Remark. Lindenbaum-Lemma means:
every consistent set can be extended to a maximal consistent set.
In the proof of (2), we can also construct Ψn + 1 as follows:
Ψn ∪ {An},
Ψn ∪ {An} is S-consistent;
Ψn + 1 =
Ψn,
Ψn ∪ {An} is S-inconsistent.
Although we can obtain the result above, but its proof herefrom is
different a little. In the proof of above (2), although verifying the
consistency is quite trouble, but verifying the maximality is quite easy,
whereas for Ψn + 1 constructed here, just the reverse is true.
Theorem 2.2.6 (Anti-chain Theorem) Let Φ be a maximal set and Ψ
be a S-consistent set. Assume that Φ ⊆ Ψ, then Φ = Ψ.
Proof. Assume that Φ ⊆ Ψ, and hypothesize that Φ ≠ Ψ, then by the
assumption, Φ ⊂ Ψ. So there is some A such that A ∈ Ψ but A ∉ Φ.
By the latter and Maximality of Φ, ¬A ∈ Φ. Since Φ ⊂ Ψ, it follows
that ¬A ∈ Ψ. On the other hand, since A ∈ Ψ, it follows that there is a
finite set {A, ¬A} ⊆ Ψ such that
123
⊢ S ¬(A ∧ ¬A),
by Consistency Lemma 2.2.2(1), Ψ is S-inconsistent, and thus we
In the following we will consider some important properties of Smaximal consistent sets:
Lemma 2.2.7 (Maximal Consistent Set Lemma) Let Φ be a Smaximal consistent set.
(1) For any A ∈ ML, exactly one of A and ¬A is in Φ, namely,
A ∈ Φ ⇔ ¬A ∉ Φ.
(2)① A ∧ B ∈ Φ ⇔ A ∈ Φ and B ∈ Φ,
② A ∨ B ∈ Φ ⇔ A ∈ Φ or B ∈ Φ. ①
(3) Th(S) ⊆ Φ.
(4) {A, A → B} ⊆ Φ ⇒ B ∈ Φ.
(5) A ∈ Φ and ⊢ S A → B ⇒ B ∈ Φ. ②
(6)① (A ∈ Φ ⇒ B ∈ Φ) ⇔ A → B ∈ Φ,
② (A ∈ Φ ⇔ B ∈ Φ) ⇔ A ↔ B ∈ Φ.
(7) (Anti-Chain Property of S-Maximal Consistent Sets) Let Φ
and Ψ be S-maximal consistent sets. If Φ ⊆ Ψ, then Φ = Ψ.
(8) CnS(Φ) = Φ, namely,
Φ ⊢ S A ⇔ A ∈ Φ.
Proof. (1) By Maximality of Φ, there is at least one of A and ¬A such
that it belongs to Φ. Hypothesize that {A, ¬A} ⊆ Φ. Since
①
Note that the connectives in the sense of an object language correspond to the
connectives in the sense of a metalanguage.
②
By (3), the condition of (5) is strongr than the one of (4), so (5) is weakr than
(4).
124
⊢ S ¬(A ∧ ¬A),
it follows that Φ is S-inconsistent, and thus we have a contradiction.
So there is always one of A and ¬A such that it does not belong to Φ.
(2) First we will prove ①.
‘⇒’: Assume that A ∧ B ∈ Φ. Hypothesize that A or B is not in Φ.
By (1), ¬A or ¬B is in Φ, so {A ∧ B, ¬A} or {A ∧ B, ¬B} is a finite
subset of Φ. Since Φ is S-consistent, by definition, we have
⊬ S ¬(A ∧ B ∧ ¬A) or ⊬ S ¬(A ∧ B ∧ ¬B),
so
⊬ PC ¬(A ∧ B ∧ ¬A) or ⊬ PC ¬(A ∧ B ∧ ¬B);
but this is impossible.
‘⇐’: Assume that A, B ∈ Φ. Hypothesize that A ∧ B ∉ Φ. By (1),
{A, B, ¬(A ∧ B)} ⊆ Φ.
Since Φ is S-consistent, it follows that ⊬ S ¬(A ∧ B ∧ ¬(A ∧ B)), and
thus
⊬ PC ¬(A ∧ B ∧ ¬(A ∧ B));
but this is impossible as well.
② We leave the details of the proof as an exercise.
(3) Assume that ⊢ S A, then ⊢ S ¬¬A. Hypothesize that ¬A ∈ Φ.
Since Φ is S-consistent, it follows that ⊬ S ¬¬A, and thus we have a
contradiction. So ¬A ∉ Φ. By (1), A ∈ Φ.
(4) Assume that {A, A → B} ⊆ Φ. Hypothesize that B ∉ Φ, by (1),
¬B ∈ Φ. Since Φ is S-consistent, it follows that
⊬ S ¬(A ∧ (A → B) ∧ ¬B).
But ¬(A ∧ (A → B) ∧ ¬B) is a substitution instance of a tautology,
so ⊢ S ¬(A ∧ (A → B) ∧ ¬B), and thus we have a contradiction. Hence
B ∈ Φ.
(5) By (3) and (4).
125
(6) Prove ①. ‘⇐’: It holds clearly by (4). ‘⇒’: Assume that
A ∈ Φ ⇒ B ∈ Φ,
then A ∉ Φ or B ∈ Φ, so ¬A ∈ Φ or B ∈ Φ. Since
⊢ S ¬A → A → B and ⊢ S B → A → B,
so, by (5), we always have that A → B ∈ Φ.
② We leave the details of this proof as an exercise.
(7) By anti-chain theorem 2.2.6.
(8) ‘⇒’: Assume that Φ ⊢ S A, then there is a finite set Ψ ⊆ Φ such
that
(☆) ⊢ S ∧Ψ → A.
Since Ψ ⊆ Φ, so, by ① in (2), ∧Ψ ∈ Φ, hence, by (☆) and (5), it
follows that A ∈ Φ.
‘⇐’: By 2.1.6(1). ┤
Remark. (Ⅰ) By (4) above, a S-maximal consistent set is closed
under MP. But we do not prove that a S-maximal consistent set is
closed under RN or the other character rules.
( Ⅱ ) (8) above shows that the set of all strong derivation
consequences of a S-maximal consistent set Φ is exactly Φ itself. In
other words, a S-maximal consistent set is closed under the strong
derivation of using it as a hypothesis set . This is why a modal logic
need introduce the definition of strong derivation.
Definition 2.2.8 The class of all S-maximal consistent sets containing
A is defined as follows:
‖A‖S := {Φ ∈ MCS(S) | A ∈ Φ}.
In other words,
‖A‖S := {Φ | A ∈ Φ and Φ is S-maximal consistent set}. ┤
126
Remark. We often omit the subscript ‘S’ if no confusion will arise.
Theorem 2.2.9
(1) S is consistent ⇔ MCS(S) ≠ ∅.
(2) Assume that Φ is S-consistent. Then
Φ ⊢ S A ⇔ ∀Ψ ∈ MCS(S) (Φ ⊆ Ψ ⇒ A ∈Ψ )
⇔ A belongs to each S-maximal consistent set
containing Φ as a subset.
(3) Assume that S is consistent. Then
⊢ S A ⇔ ∀Ψ∈ MCS(S) (A ∈ Ψ)
⇔ A belongs to each S-maximal consistent set.
Namely,
⊢ S A ⇔ A ∈ ∩MCS(S).
(4) ⊢ S A → B ⇔‖A‖⊆‖B‖.
Proof. (1) ‘⇒’: Assume that S is consistent. By definition, Th(S) is Sconsistent, so, by Lindenbaum-Lemma, get immediately the result we
will prove.
‘⇐’: Assume that S is inconsistent, then by Consistency Lemma
2.2.2(5), CS(S) = ∅. Since MCS(S) ⊆ CS(S), it follows that MCS(S)
= ∅.
(2) ‘⇒’: Assume that Φ ⊢ S A. Given any S-maximal consistent set
Ψ such that Φ ⊆ Ψ. We will show that A ∈ Ψ. Hypothesize that A ∉ Ψ,
then ¬A ∈ Ψ. Since Φ ⊢ S A, it follows that there is a finite set Θ ⊆ Φ
such that ⊢ S ∧Θ → A, and thus ⊢ S ¬(∧Θ ∧ ¬A). Since Φ ⊆ Ψ, it
follows that Θ ∪ {¬A} ⊆ Ψ is a finite set, by Consistency Lemma
2.2.2(1), Ψ is S-inconsistent, and thus we have a contradiction. Hence
A ∈ Ψ.
‘⇐’: Assume that Φ ⊬ S A. By 2.2.4(1), Φ ∪ {¬A} is S-consistent.
127
By Lindenbaum-Lemma, there is a S-maximal consistent set Ψ such
that Φ ∪ {¬A} ⊆ Ψ. So, by Maximal Consistent Set Lemma 2.2.7(1),
A ∉ Ψ.
(3) Since S is consistent, by Consistency Lemma 2.2.2(3), ∅ is Sconsistent, so, by (2) and 2.1.2(2), the result we will prove holds
clearly.
(4) ‘⇒’: Assume that ⊢ S A → B, then by (3) or by 2.2.7(3), for all
S-maximal consistent set Φ, it follows that A → B ∈ Φ. By 2.2.7(6)①,
it follows that
A ∈ Φ ⇒ B ∈ Φ for all S-maximal consistent set Φ.
It is easy to see that‖A‖⊆‖B‖.
‘⇐’: Assume that‖A‖⊆‖B‖. Given any S-maximal consistent
set Φ.
Case 1 Φ ∈‖A‖: Then by the assumption, Φ ∈‖B‖, so B ∈
Φ. Since ⊢ S B → A → B, by 2.2.7(5), it follows that A → B ∈ Φ.
Case 2 Φ ∉‖A‖: Then A ∉ Φ. Since Φ is S-maximal consistent
set, it follows that ¬A ∈ Φ. Since ⊢ S ¬A → A → B, by 2.2.7(5), we
still have that A → B ∈ Φ.
So, by the arbitrariness of Φ and (3), it follows that ⊢ S A → B. ┤
Remark. By ‘⇒’ of (2), it is easy to see that every S-maximal
consistent set is closed under the provability consequence relation ⊢ S
in the following sense: given any S-maximal consistent set Ψ, then
Φ ⊢ S A and Φ ⊆ Ψ ⇒ A ∈ Ψ.
2. Some Important Properties of Modal Systems
In the previous subsection, the results hold for any extension system
128
of PC which we have proved. In other words, the results depend only
on the fact that S is an extension system of PC. S can be any system
that contains no modal axiom and modal inference rule. But the
results in this subsection are not so. For example, some results just
hold for normal systems.
Definition 2.2.10 (Abbreviation Definition) Given formula set Φ.
－
(1) □ Φ := {A | □A ∈ Φ},
－, n
□ Φ := {A | □nA ∈ Φ}.
(2) □+ Φ := {□A | A ∈ Φ}.
(3) □Φ := {□A | □A ∈ Φ}.
(4) ◇+ Φ := {◇A | A ∈ Φ},
+, n
◇ Φ := {◇nA | A ∈ Φ}. ┤
Remark. (Ⅰ) It is easy to see that
+, 0
－, 0
□ Φ = ◇ Φ = Φ,
－, 1
－
□ Φ = □ Φ,
+, 1
◇ Φ = ◇+ Φ.
(Ⅱ) Let Φ = {A → B, ¬□B, ◇C, □D}, then
－
□ Φ = {D},
□Φ = {□D},
□+ Φ = {□(A → B), □¬□B, □◇C, □□D},
◇+ Φ = {◇(A → B), ◇¬□B, ◇◇C, ◇□D}.
Theorem 2.2.11 Let S have the rule RE, and let Φ, Ψ be S-maximal
consistent sets.
－
(1) □ Φ ⊆ Ψ ⇔ ◇+ Ψ ⊆ Φ.
+, n
－, n
(2) □ Φ ⊆ Ψ ⇔ ◇ Ψ ⊆ Φ for n < ω.
129
Proof. (1) ‘⇒’: Assume that □ Φ ⊆ Ψ. Given any ◇A ∈ ◇+ Ψ, we
will show that ◇A ∈ Φ. Since ◇A ∈ ◇+ Ψ, it follows that A ∈ Ψ,
and thus by the S-maximal consistency of Ψ, we have ¬A ∉ Ψ. Since
－
－
□ Φ ⊆ Ψ, it follows that ¬A ∉ □ Φ, and thus □¬A ∉ Φ. By
Maximality of Φ, it follows that ¬□¬A ∈ Φ. By the abbreviation
definition, it follows that ◇A ∈ Φ.
－
‘⇐’: Assume that ◇+ Ψ ⊆ Φ. Given any A ∈ □ Φ, we will show
－
that A ∈ Ψ. Since A ∈ □ Φ, it follows that □A ∈ Φ. On the other
hand, by LMC (or E(3) and weak rule US), we have
⊢ S □A ↔ ¬◇¬A,
so, by Lemma 2.2.7(5), it is easy to see that ¬◇¬A ∈ Φ, and thus we
have ◇¬A ∉ Φ. By ◇+ Ψ ⊆ Φ, it follows that ◇¬A ∉ ◇+ Ψ, hence
¬A ∉ Ψ, and thus A ∈ Ψ.
(2) This proof resembles the proof of (1).
+, n
－, n
‘⇒’: Assume that □
Φ ⊆ Ψ. Given any ◇nA ∈ ◇ Ψ, we
will show that ◇nA ∈ Φ. Since A ∈ Ψ, by Maximality of Ψ, we have
－, n
－, n
¬A ∉ Ψ. Since □ Φ ⊆Ψ, it follows that ¬A ∉ □ Φ, so □n¬A
∉ Φ. By Maximality of Φ, ¬□n¬A ∈ Φ. So ◇nA ∈ Φ by E(6), US
and Lemma 2.2.7(5).
+, n
－, n
‘⇐’: Assume that ◇ Ψ ⊆ Φ. Given any A ∈ □ Φ. We will
show that A ∈ Ψ. Since □nA ∈ Φ, it follows that ¬◇n¬A ∈ Φ by
+, n
E(5), US and 2.2.7(5), so we have ◇n¬A ∉ Φ. By ◇ Ψ ⊆ Φ, it
follows that ¬A ∉ Ψ, so A ∈ Ψ. ┤
－
Theorem 2.2.12 Let S be a normal system.
－
(1) Let ¬□A ∈ Φ. If Φ is S-consistent, then □ Φ ∪ {¬A} is Sconsistent.
(2) Let □A ∉ Φ. If Φ is a S-maximal consistent set, then there is a
130
S-maximal consistent set Ψ such that □ Φ ∪ {¬A} ⊆ Ψ. Thus we
have
－
(#) □ Φ ⊆ Ψ and A ∉ Ψ.
－
Proof. (1) We argue via contraposition. Assume that □ Φ ∪ {¬A}
is S-inconsistent. By Consistency Lemma 2.2.2(11), there is a finite
－
set {A1, …, An} ⊆ □ Φ such that, in S,
⊢ A1 ∧…∧ An → A.
By 1.1.12(4), S has RK, so
⊢ □A1 ∧…∧ □An → □A.
Hence
⊢ ¬(□A1 ∧…∧ □An ∧ ¬□A),
and thus
(☆) ⊢ ¬∧({□A1, …, □An, ¬□A}).
－
Since {A1, …, An} ⊆ □ Φ, by the assumption given, it is easy to see
that {□A1, …, □An, ¬□A} is a finite subset of Φ, so, by (☆) and
Consistency Lemma 2.2.2(1), it is easy to see that Φ is S-inconsistent.
(2) Assume that Φ is a S-maximal consistent set. Since □A ∉ Φ, it
－
follows that ¬□A ∈ Φ, so, by (1) proved before, □ Φ ∪ {¬A} is Sconsistent. By Lindenbaum-Lemma, there is a S-maximal consistent
－
set Ψ such that □ Φ ∪ {¬A} ⊆ Ψ. Herefrom we get (#) easily. ┤
－
Theorem 2.2.13 Let S be a normal system, and let Φ, Ψ be S-maximal
consistent sets. Then for any n < ω,
－, n + 1
Φ ⊆ Ψ ⇔ there is a S-maximal consistent set Θ
□
－
－, n
such that □ Φ ⊆ Θ and □ Θ ⊆ Ψ.
+, n
－, n + 1
－
Proof. ‘⇒’: Assume that □
Φ ⊆ Ψ. Let Ξ = □ Φ ∪ ◇ Ψ.
First we will prove that
(☆) Ξ is S-consistent.
131
Hypothesize that (☆) does not hold, then by Consistency Lemma
2.2.2(1) and RPC, there are two natural numbers m, k < ω such that B1,
+, n
－
…, Bm ∈ □ Φ and ◇nC1, …, ◇nCk ∈ ◇ Ψ and, in S,
⊢ ¬(B1 ∧…∧ Bm ∧ ◇nC1 ∧…∧ ◇nCk).
So, by RPC, we have
① ⊢ B1 ∧…∧ Bm → ¬(◇nC1 ∧…∧ ◇nCk).
[Note that We here actually presuppose that there is at least one of m
and k is not 0. Hypothesize that m = k = 0, then by ①, ⊢ T → ⊥, so
⊢ ⊥. By Consistency Lemma 2.2.2(3), S is inconsistent, and thus by
2.2.2(5), CS(S) = ∅, contradicting the assumption that Φ and Ψ are Smaximal consistent sets. ]
On the other hand, by (4) in 1.2.22 (Ⅰ), we have that
⊢ ◇n(C1 ∧…∧ Ck) → ◇nC1 ∧…∧ ◇nCk.
By RPC,
⊢ ¬(◇nC1 ∧…∧ ◇nCk) → ¬◇n(C1 ∧…∧ Ck).
By LMC,
② ⊢ ¬(◇nC1 ∧…∧ ◇nCk) → □n¬(C1 ∧…∧ Ck).
By ① and ②, we have that
⊢ B1 ∧…∧ Bm → □n¬(C1 ∧…∧ Ck).
By RK,
⊢ □B1 ∧…∧ □Bm → □n + 1¬(C1 ∧…∧ Ck).
Since □B1, …, □Bm ∈ Φ, by Maximal Consistent Set Lemma
2.2.7(2)① and (5), it follows that
n+
□ 1¬(C1 ∧…∧ Ck) ∈ Φ.
n+
Since {A | □ 1A ∈ Φ} ⊆ Ψ, it follows that
③ ¬(C1 ∧…∧ Ck) ∈ Ψ.
On the other hand, since
132
+, n
◇nC1, …, ◇nCk ∈ ◇ Ψ,
it follows that C1, …, Ck ∈ Ψ, and thus by 2.2.7(2)①, C1 ∧…∧ Ck ∈
Ψ, contradicting ③ and 2.2.7(1). Thus we get (☆).
By (☆) and Lindenbaum-Lemma, there is some Θ ∈ MCS(S) such
that Ξ ⊆ Θ.
－
By the constitution of Ξ and Ξ ⊆ Θ, it is easy to see that □ Φ ⊆ Θ
+, n
－, n
and ◇ Ψ ⊆ Θ. By the latter and 2.2.11(2), □ Θ ⊆ Ψ.
‘⇐’: Assume that there is a S-maximal consistent set Θ such that
－
④ □ Φ ⊆ Θ, and
－, n
⑤ □ Θ ⊆ Ψ.
－, n + 1
n+
Given any A ∈ □
Φ. We will show that A ∈ Ψ. Since □ 1A ∈
－
Φ, it follows that □nA ∈ □ Φ, so □nA ∈ Θ by ④, and thus A ∈ Ψ
by ⑤. ┤
Theorem 2.2.14 Let S be a normal system, and let Φ be a S-maximal
consistent set.
(1) □A ∈ Φ ⇔ for all S-maximal consistent set Ψ,
－
□ Φ ⊆ Ψ ⇒ A ∈ Ψ.
Namely,
－
□A ∈ Φ ⇔ ∀Ψ ∈ MCS(S) (□ Φ ⊆ Ψ ⇒ A ∈ Ψ).
(2) ◇A ∈ Φ ⇔ there is a S-maximal consistent set Ψ such that
◇+ Ψ ⊆ Φ and A ∈ Ψ.
Namely,
◇A ∈ Φ ⇔ ∃Ψ ∈ MCS(S) (◇+ Ψ ⊆ Φ and A ∈ Ψ).
Proof. (1) ‘⇒’: Assume that □ A ∈ Φ. Given any S-maximal
－
－
consistent set Ψ such that □ Φ ⊆ Ψ. Since □A ∈ Φ and □ Φ ⊆ Ψ,
it follows that A ∈ Ψ.
－
‘⇐’: Assume that for all S-maximal consistent set Ψ, if □ Φ ⊆ Ψ,
133
then A ∈ Ψ. This means that A belongs to each S-maximal consistent
－
set containing □ Φ as a subset. By the following exercise 2.2.31, we
－
－
have □ Φ ⊢ S A. So there are some B1, …, Bn ∈ □ Φ such that
⊢ S B1 ∧…∧ Bn → A.
By RK,
(☆) ⊢ S □B1 ∧…∧ □Bn → □A.
－
Since B1, …, Bn ∈ □ Φ, it follows that □B1, …, □Bn ∈ Φ. By (☆)
and Lemma 2.2.7(2)① and (5), □A ∈ Φ.
[The proof another of ‘⇐’: Assume that □A ∉ Φ. We will show
－
that there is a S-maximal consistent set Ψ such that □ Φ ⊆ Ψ and A ∉
Ψ. By Lindenbaum-Lemma, it suffices to show that
－
(%) □ Φ ∪ {¬A} is S-consistent.
Hypothesize that (%) does not hold. Then, by Consistency Lemma
－
2.2.2(11), there are some B1, …, Bn ∈ □ Φ such that
⊢ S B1 ∧…∧ Bn → A.
By RK,
(☆) ⊢ S □B1 ∧…∧ □Bn → □A.
Since □B1, …, □Bn ∈ Φ, so, by (☆) and Lemma 2.2.7(2)① and (5),
□A ∈ Φ, and thus we have a contradiction.]
(2) By (1). We leave the details of the proof as an exercise. ┤
Remark. The above (1) is the key to prove the completeness
theorem for a normal system (refer to 4.2.2 in Chapter 4). Note that
RK is only used in the proof of (1).
By 1.1.16 and 2.2.3, we have
Theorem 2.2.15 (Consistency Theorem) Let S be a system defined
134
in 1.1.9. Then S is consistent. ┤
By Maximal Consistent Set Lemma 2.2.7(3) and 2.2.16(2) below (a
version of Compactness Theorem), we can also prove 2.2.15, but
2.2.7(3) presupposes that there is a S-maximal consistent set Φ. So we
can prove that S is consistent: given any finite subset Ψ of Th(S). By
2.2.7(3), Ψ is a finite subset of Φ. Since Φ is S-consistent, by
Compactness Theorem, Ψ is S-consistent. By the arbitrariness of Ψ
and Compactness Theorem, Th(S) is S-consistent, so S is consistent.
But this proof is not general. Since if S is inconsistent but we do
not know this case, then by Consistency Lemma 2.2.2(5), any formula
set is not S-maximal consistent, so we can not use the above method
to prove 2.2.15.
Exercises 2.2
2.2.16 Prove that
(1) If Φ is S-consistent and Ψ ⊆ Φ, then Ψ is S-consistent.
(2) Φ is S-consistent ⇔ all subsets of Φ are S-consistent. ┤
2.2.17 Prove that
(1) Φ is S-consistent ⇔ CnS(Φ) is S-consistent.
(2) Φ is S-inconsistent ⇔ Φ ⊢ S ⊥.
(3) Φ is S-inconsistent ⇔ for all A ∈ ML, Φ ⊢ S A.
(4) Φ is S-consistent ⇔ Φ ⊬ S ⊥
⇔ there is some A such that Φ ⊬ S A.
(5) Assume that Φ is S-inconsistent. Then for all A ∈ ML, there is a
finite set Ψ ⊆ Φ such that ⊢ S ∧Ψ → A. ┤
135
2.2.18 Let Φ be S-consistent. Prove that Φ ∪ {A} or Φ ∪ {¬A} is Sconsistent. ┤
2.2.19 Prove that Φ is S-maximal consistent ⇔ the following
conditions are satisfied:
(1) Φ is S-consistent, and
(2) A ∉ Φ ⇒ Φ ∪ {A} is S-inconsistent for all formula A. ┤
2.2.20 Let S and S1 be two systems such that S ⊆ S1. Prove that if A is
S1-consistent, then A is also S-consistent. ┤
2.2.21 Prove that
(1) Φ is S-consistent ⇔ there is some A such that Φ ⊬ S A.
(2) Φ is S-consistent
⇔ there is not some A such that Φ ⊢ S A and Φ ⊢ S ¬A.
(3) Assume that Φ is S-consistent and Φ ⊢ S A. Then Φ ∪ {A} is Sconsistent. ┤
2.2.22 Let S be a monotonic system, and let Φ be a S-maximal
consistent set. Prove that
□A ∈ Φ ⇔ there is □B ∈ Φ such that ‖B‖⊆‖A‖. ┤
Definition 2.2.23 Let the English abbreviation of the axiom of the
form A → B mentioned before and below be X. We use Xc as the
English abbreviation of its converse formula B → A. For example, the
English abbreviation of the converse formula ◇p → □p of D is Dc.
┤
136
2.2.24 Prove that Dc, Tc, 4c, 5c, Bc, M and G are S5-consistent. ┤
2.2.25 Prove that
(1) Th(Tr) is a consistent set but is not a maximal set.
(2) Th(Ver) is a consistent set but is not a maximal set. ┤
2.2.26 Prove that if A is KW-consistent, then □¬A ∧ A is KWconsistent. ┤
2.2.27 (Problem) Add the inference rules in 1.1.7(2) into a system S
respectively. Is a S-maximal consistent set closed under which rules
and unclosed under which rules? ┤
2.2.28 Let S be a normal system and Φ a S-maximal consistent set.
Prove that for all n > 0,
－, n
(1) □nA ∈ Φ ⇔ ∀Ψ ∈ MCS(S) (□ Φ ⊆ Ψ ⇒ A ∈ Ψ ).
+, n
(2) ◇nA ∈ Φ ⇔ ∃Ψ ∈ MCS(S) (◇ Ψ ⊆ Φ and A ∈ Ψ ). ┤
2.2.29 Let S ⊇ T. If Φ is S-maximal consistent, then □ Φ is Sconsistent. ┤
－
2.2.30* Let a modal system S be consistent.
(1) Let |At| = ℵ0. Prove that there are exactly 2ℵ0 different Smaximal consistent sets.
(2) (Problem) Let |At| = 1. How many are there different Smaximal consistent sets?
(3) (Problem) Let |At| = n. How many are there different Smaximal consistent sets? ┤
137
2.2.31 Prove that the assumption of 2.2.9(2) is not essential, namely,
for all formula set Φ and formula A,
Φ ⊢ S A ⇔ ∀Ψ ∈ MCS(S) (Φ ⊆ Ψ ⇒ A ∈Ψ ). ┤
2.2.32 Assume that S is a normal system and Φ a S-maximal
consistent set such that ¬□⊥ ∈ Φ.
－
(1) □ Φ is S-consistent.
－
(2) There is a S-maximal consistent set Ψ such that □ Φ ⊆ Ψ. ┤
2.2.33 (Problem) Let S be consistent.
(1) Is {p} S-consistent?
(2) Is {¬p} S-consistent?
(3) Is {p ∧ q} S-consistent?
(4) Is {p ∨ q} S-consistent? ┤
2.2.34 Let S be a normal system and Φ a S-maximal consistent set
－
such that □ Φ is S-inconsistent. Prove that □A ∈ Φ for any
formula A. ┤
2.2.35 Construct a minimal system S such that the following
proposition (*) holds:
+
(*) If Φ is S-maximal consistent, then ◇ Φ is S-consistent. ┤
[By a minimal system S such that (*) holds we mean that S is a
subsystem of any system satisfying (*).]
2.2.36 Assume that S is a normal system and Φ a S-maximal
consistent set.
□⊥ ∈ Φ
⇔ there is not a S-maximal consistent set Ψ such that Φ ⊆ Ψ. ┤
138
Chapter 3 Relation Semantics and
Soundness Theorem
The relation semantics is the standard semantics of modal logic, since
it can characterizes most of modal systems. So, in this course, we also
use it as the basic semantics of studying modal systems.
In § 1 we will introduce the relation semantics, give out the
definitions of frame, model, frame class and model class as well as the
interrelated definitions, and show some basic properties of them.
In§2 we will discuss the correspondence relations between some
familiar modal formulas and first-order formulas.
In§3 we will introduce the soundness definition, and prove the
soundness theorems of some familiar systems.
In§4 we will use the countermodel method to answer the some
problems left in the above chapters, mostly show the proper inclusion
relations between the basic systems.
From this chapter to Chapter 5, if not especially point out, the
system S we mentioned always is a normal system in the following
sense: S = K + Φ and S is consistent.
139
§1 Relation Semantics
In this section we will introduce the relation semantics, give out the
definitions of frame, model, frame class and model class as well as the
interrelated definitions, and show some basic properties of them.
Definition 3.1.1
(1) (Frame Definition) F = <W, R> is a relation frame (Kripkeframe), denoted as F ∈ Fra, ⇔ the following conditions are satisfied:
① W is a nonempty set: W ≠ ∅.
Each element of W is also called a possible world or a point.
② R is a binary relation on W: R ⊆ W × W.
R is also called the accessibility relation.
(2) (Model Definition) M = <W, R, V> is a relation model (Kripkemodel), denoted as M ∈ Mod, ⇔ the following conditions are
satisfied:
① <W, R> is a relation frame, and
② V is a valuation mapping from At into ℘(W). In other words,
V assigns a subset V(pn) of W to each pn ∈ At, namely, V(pn) ⊆ W for
all pn ∈ At. ①
V is also called a valuation (assignment) of M.
(3) (Truth Set Definition) The truth sets of all compound formulas
in M are defined inductively as follows:
① V(¬A) = W − V(A),
② V(A ∧ B) = V(A) ∩ V(B),
①
Intuitively, V(pn) is the set of all possible worlds at which pn is true.
140
③ V(□A) = {w ∈ W | R(w) ⊆ V(A)},
where R(w) := {u ∈ W | wRu}. ①
(4) Assume that M = <W, R, V> and F = <W, R>. M is called a
model based on F or a model on F, V a valuation on F, F the frame
underlying M or the underlying frame of M or the frame of M. ┤
Remark. (Ⅰ) What is a possible world? This is a problem discussed
in the logic philosophy. We here just see it as an element in some
nonempty set.
V(A) is called the proposition that A denotes w.r.t. the current
model M. This a kind of point of view of relativizing propositions.
The above definition (3) is to extend the valuation mappings
defined on At to the set of all formulas ML, so (3), especially ③
where, is particularly important.
(Ⅱ) Fra is the class of all relation frames, Mod is the class of all
relation models.
Note that the frame classes and model classes we usually meet may
be proper classes (such as Fra and Mod), although the possible world
set of each frame or model is a set.
(Ⅲ) Since ∅ is a subset of any set, so, when R(w) = ∅, we always
have w ∈ V(□A).
By (3) and the simple knowledge of the set theory, we have
① w ∉ V(□A) ⇔ R(w) ∩ V(¬A) ≠ ∅,
② w ∈ V(◇A) ⇔ R(w) ∩ V(A) ≠ ∅,
③ w ∉ V(◇A) ⇔ R(w) ⊆ V(¬A).
We exemplify ②:
w ∈ V(◇A) ⇔ w ∈ V(¬□¬A)
⇔ w ∉ V(□¬A)
by 3.1.1(3)①
①
Intuitively, R(w) is the set of all worlds accessible from w.
141
⇔ R(w) ⊈ V(¬A)
by 3.1.1(3)③
⇔ R(w) ∩ V(A) ≠ ∅.
by the simple knowledge of the set theory
(Ⅳ) For ∨, →, ↔, we have
① V(A ∨ B) = V(A) ∪ V(B),
② V(A → B) = V(¬A) ∪ V(B),
③ V(A ↔ B) = (V(¬A) ∪ V(B)) ∩ (V(¬B) ∪ V(A)).
(Ⅴ) For simplicity, below, we often use w ⊨ A as w ∈ V(A) and w
⊭ A as w ∉ V(A) if no confusion will arise. So we have
① w ⊨ □A ⇔ for all u ∈ W (wRu ⇒ u ⊨ A),
② w ⊭ □A ⇔ there is some u ∈ W (wRu and u ⊭ A),
③ w ⊨ ◇A ⇔ there is some u ∈ W (wRu and u ⊨ A),
④ w ⊭ ◇A ⇔ for all u ∈ W (wRu ⇒ u ⊭ A).
The more simple denotation of (the denotation of the set theory) is:
①* w ⊨ □A ⇔ ∀u ∈ R(w) (u ⊨ A),
②* w ⊭ □A ⇔ ∃u ∈ R(w) (u ⊭ A),
③* w ⊨ ◇A ⇔ ∃u ∈ R(w) (u ⊨ A),
④* w ⊭ ◇A ⇔ ∀u ∈ R(w) (u ⊭ A).
(Ⅵ) Together a model definition and a truth set definition is called
a semantics, since herefrom we can interprete each formula in ML,
namely, reveal the meaning (proposition) of each formula w.r.t. a
model.
The semantics given by the above definitions is called the relation
semantics or Kripke-semantics.
(Ⅶ) For convenience, below we use M, N as any models; F, G as
any frames; M, N as any model classes; F, and G as any frame classes,
with or without subscripts, respectively.
142
Where no ambiguity is likely to arise, we often
use ‘w ∈ F’ as ‘w ∈ W such that F = <W, …>’,
use ‘w ∈ M’ as ‘w ∈ W such that M = <W, …>’.
(Ⅷ) □ is not seen as a truth-value function generally by the
relation semantics (refer to §4 of Chapter 1). This is why the
relation semantics becomes an important semantic means describing a
lot of systems with operators that are not seen as truth connectives,
such as temporal logic (tense logic, time logic), deonic logic,
epistemic logic (doxastic logic), dynamic logic, dynamic epistemic
logic (dynamic doxastic logic), conditional logic, intuitionist logic,
nonmonotonic modal logic, update logic, logic of provability, logic of
language, hybrid logic, arrow logic, logic of game, and so on.
Lemma 3.1.2 (Equivalence Lemma) Given any w ∈ M.
(1) w ⊭ □A ⇔ w ⊨ ◇¬A. (E(3) is expressed in the semantics)
(2) w ⊭ ◇A ⇔ w ⊨ □¬A. (E(4) is expressed in the semantics)
Proof. (1) Let M = <W, R, V>. It is easy to see that
w ⊭ □A ⇔ w ∉ V(□A)
Abbreviation
⇔ R(w) ∩ V(¬A) ≠ ∅　① in (Ⅲ) above
⇔ w ∈ V(◇¬A)
② in (Ⅲ) above
Abbreviation
⇔ w ⊨ ◇¬A.
The proof of (2) resembles (1), and prove it please. ┤
Convention and Definition 3.1.3 Given any w ∈ M.
(1) w ⊨ Φ denotes: Φ ⊨ A for all A ∈ Φ.
(2) A is true at w ⇔ w ⊨ A.
A is false at w ⇔ A is not true at w.
(3) Φ is true at w ⇔ w ⊨ Φ. ┤
143
In the following we also use w → u as wRu, use w ← u as uRw, use
w ↔ u as wRu and uRw, and use wο as w → w. This denotation can
denote some simple frames quite intuitively. On the side, we use wΦ
as the set Φ of all formulas true at w. This denotation and the above
denotations can denote a simple model quite intuitively as well. For
example, we use
wp, ¬ q ← u p, q → v¬ p, q
as the following model <W, R, V>:
W = {w, u, v},
R = {<u, w>, <u, v>},
V(p) = {w, u}, V(q) = {u, v}. ①
The model we constructed here can be seen as simplified: {p, q} ⊆
At, and V assigns any subset of W to any sentence symbol except p
and q in At.
In the following we often use the above simplified frames and
simplified models.
The above model can be denoted even by the following denotation:
{p, ¬q} ← {p, q} → {¬p, q},
namely, see a possible world as a maximal set w.r.t. At: Φ is a
maximal set w.r.t. At ⇔ Φ ⊆ At ∪ {¬pn | pn ∈ At} such that Φ
contains exactly one from pn, ¬pn for all pn ∈ At.
Definition 3.1.4 Given any M = <W, R, V>.
(1) A is satisfiable in M, denoted as A ∈ Sat(M), ⇔ V(A) ≠ ∅.
A is satisfiable ⇔ there is a N ∈ Mod such that A ∈ Sat(N).
(2) Φ is satisfiable in M ⇔ there is some w ∈ M such that w ⊨ Φ.
(3) A is valid in M, denoted as A ∈ Val(M) or M ⊨ A, ⇔ V(A) = W.
(4) Φ is valid in M, denoted as M ⊨ Φ, ⇔ Φ ⊆ Val(M).
①
Here we assume impliedly At = {p, q}.
144
We also say that M is a model of Φ or a Φ-model.
(5) M is a countermodel of A, denoted as M ⊭ A, ⇔ V(¬A) ≠ ∅.
We say that A has a countermodel ⇔ there is a model M such that
M is a countermodel of A.
(6) If A (or Φ) is valid in M, then A (or Φ) is also called true in M.
(7) Φ entails pointwise A, denoted as Φ ⊨ P A (for short Φ ⊨ A),
⇔ for all M ∈ Mod and w ∈ M (w ⊨ Φ ⇒ w ⊨ A).
(8) Φ entails modelwise A, denoted as Φ ⊨ M A,
⇔ for all M ∈ Mod (M ⊨ Φ ⇒ M ⊨ A). ┤
Remark. (Ⅰ) In (3) we defined the concept of model validity.
( Ⅱ ) The entailment-pointwise relation is stronger than the
It is easy to prove that
Corollary 3.1.5 M is a countermodel of A ⇔ ¬A is satisfiable in M.
┤
Definition 3.1.6 Let M be a model class.
(1) A is satisfiable in M, denoted as A ∈ Sat(M),
⇔ there is some M ∈ M such that A ∈ Sat(M).
(2) Φ is satisfiable in M
⇔ there is some M ∈ M such that Φ is satisfiable in M.
(3) A is valid in M, denoted as A ∈ Val(M) or M ⊨ A,
⇔ for all M ∈ M, M ⊨ A.
Here A is also called M-valid.
(4) Φ is valid in M, denoted as M ⊨ Φ, ⇔ Φ ⊆ Val(M).
145
(5) Φ entails A w.r.t. M, denoted as Φ ⊨ M A,
⇔ (M ⊨ Φ ⇒ M ⊨ A).
(6) Φ entails modelwise A w.r.t. M, denoted as Φ ⊨ M A,
⇔ for all model M ∈ M (M ⊨ Φ ⇒ M ⊨ A). ┤
Remark. (Ⅰ) by (1), we have
Φ is satisfiable in M
⇔ there is some M ∈ M such that Φ is satisfiable in M
⇔ there is some M ∈ M and w ∈ M such that w ⊨ Φ.
(Ⅱ) In (3) we defined the concept of model class validity.
Definition 3.1.7 Let F be a frame.
(1) We use Mod(F) as the class of all models on F.
(2) A is satisfiable in F, denoted as A ∈ Sat(F),
⇔ there is some M ∈ Mod(F) such that A ∈ Sat(M).
(3) Φ is satisfiable in F
⇔ there is some M ∈ Mod(F) such that Φ is satisfiable in M.
(4) A is valid in F or A is F-valid, denoted as A ∈ Val(F) or F ⊨ A,
⇔ for all M ∈ Mod(F), M ⊨ A.
(5) Φ is valid in F, denoted as F ⊨ Φ, ⇔ Φ ⊆ Val(F).
We also say that F is a frame of Φ or a Φ-frame.
(6) F is a counterframe of A, denoted as F ⊭ A,
⇔ there is some M ∈ Mod(F) such that M ⊭ A.
(7) Φ entails framewise A, denoted as Φ ⊨ F A,
⇔ for all frame F ∈ Fra (F ⊨ Φ ⇒ F ⊨ A). ┤
Remark. (Ⅰ) By (3), we have
146
Φ is satisfiable in F
⇔ there is some M ∈ Mod(F) such that Φ is satisfiable in M
⇔ there is some M ∈ Mod(F) and w ∈ M such that w ⊨ Φ.
(Ⅱ) In (3) we defined the concept of frame validity.
( Ⅲ ) The entailment-modelwise relation is stronger than the
Definition 3.1.8 Let F be a frame class.
(1) We use Mod(F) as the class of all models on all the frames in F:
Mod(F) = ∪ {Mod(F) | F ∈ F}.
(2) A is satisfiable in F, denoted as A ∈ Sat(F),
⇔ there is some F ∈ F such that A ∈ Sat(F).
(3) Φ is satisfiable in F
⇔ there is some F ∈ F such that Φ is satisfiable in F.
(4) A is valid in F or A is F-valid, denoted as A ∈ Val(F) or F ⊨ A,
⇔ for all F ∈ F, F ⊨ A.
(5) Φ is valid in F, denoted as F ⊨ Φ, ⇔ Φ ⊆ Val(F).
(6) Φ entails A w.r.t. F, denoted as Φ ⊨ F A,
⇔ (F ⊨ Φ ⇒ F ⊨ A).
(7) Φ entails framewise A w.r.t. F, denoted as Φ ⊨ F A,
⇔ for all frame F ∈ F (F ⊨ Φ ⇒ F ⊨ A). ┤
Remark. (Ⅰ) by (2), we have
Φ is satisfiable in F
⇔ there is some F ∈ F such that Φ is satisfiable in F
⇔ there is a F ∈ F and M ∈ Mod(F) such that Φ is satisfiable in M
⇔ there is some F ∈ F, M ∈ Mod(F) and w ∈ M such that w ⊨ Φ.
(Ⅱ) In (3) we defined the concept of frame class validity.
147
(Ⅲ) By the above several definitions, we have the following four
concepts of validities:
model validity,
model class validity,
frame validity,
frame class validity.
Model class validity is the most general one of the concepts, and
other concepts of validities are its special case:
model validity is validity of some single element model class,
frame validity is validity of the class of all model on some single
element frame class,
frame class validity is validity of the class of all classes of all
model on each frame in current frame class.
On the side, frame validity is also a special case of frame class
validity, namely,
frame validity is validity of some single element frame class.
We have defined a lot of entailment relations before, in other words,
semantics consequence relations or logical consequence relations.
Similarly, we can still define other entailment relations.
Note that
Φ ⊨ A, Φ ⊨ M A, Φ ⊨ F A, …, and
w ⊨ A, M ⊨ A, F ⊨ A, …
are two groups of concepts.
Definition 3.1.9 Let A1, …, An∕B be a inference rule and C a model
class or frame class.
(1) A1, …, An∕B is pointwise valid or validity-preserving pointwise
⇔ {A1, …, An} entails pointwise B (refer to 3.1.4(7)).
(2) A1, …, An ∕ B is modelwise valid or validity-preserving
modelwise ⇔ {A1, …, An} entails modelwise B (refer to 3.1.4(8)).
(3) A1, …, An ∕ B is framewise valid or validity-preserving
framewise ⇔ {A1, …, An} entails framewise B (refer to 3.1.7(7)).
148
(4) A1, …, An∕B is validity-preserving w.r.t. C ⇔ {A1, …, An}
entails B w.r.t. C (refer to 3.1.6(5) and 3.1.8(6)). ┤
Theorem 3.1.10 Let C be a model class or a frame class.
(1) A ∉ Val(C) ⇔ ¬A ∈ Sat(C).
(2) ¬A ∈ Val(C) ⇔ A ∉ Sat(C).
Proof. (1) ‘⇒’: Assume that A ∉ Val(C), then there is some x ∈ C
such that x ⊭ A.
If C is a model class, then x is a model, so there is some w ∈ x such
that w ⊭ A, and thus w ⊨ ¬A, and hence ¬A ∈ Sat(x), so ¬A ∈ Sat(C).
If C is a frame class, then x is a frame, so there is a model M on x
such that M ⊭ A. As what has been proved above, ¬A ∈ Sat(M), and
thus ¬A ∈ Sat(x), and hence ¬A ∈ Sat(C).
‘⇐’: Assume that ¬A ∈ Sat(C), then there is some x ∈ C such that
¬A ∈ Sat(x).
If C is a model class, then x is a model, so there is some w ∈ x such
that w ⊨ ¬A, and thus w ⊭ A, and hence A ∉ Val(x), so A ∉ Val(C).
If C is a frame class, then x is a frame, so there is a model M on x
such that ¬A ∈ Sat(M). As what has been proved above, A ∉ Val(M),
so A ∉ Val(x), and thus A ∉ Val(C).
The proof of (2) resembles the proof of (1). ┤
Definition 3.1.11 Let S be a modal system.
(1) A model M is a model for S, or simply, a S-model, denoted as M
∈ Mod(S), ⇔ M ⊨ Th(S).
(2) A frame F is a frame for S, or simply, a S-frame, denoted as F
∈ Fra(S), ⇔ F ⊨ Th(S). ┤
149
Remark. Mod(S) is the class of all S-models, and Fra(S) is the class
of all S-frames.
By the above several definitions, we have
Corollary 3.1.12
(1) Mod = Mod(Fra).
(2) Mod(F) ⊨ A ⇔ F ⊨ A for all frame class F.
(3) Mod(Fra) ⊨ A ⇔ Fra ⊨ A.
(4) Mod ⊨ A ⇔ Fra ⊨ A.
(5) Mod(Fra(S)) ⊆ Mod(S). ┤
Remark. Generally speaking,
Mod(S) ⊈ Mod(Fra(S)).
In the following we will prove it.
On the side, Mod(Fra(S))-validity (namely, Fra(S)-validity) and
Mod(S)-validity are not also equivalent, below we shall show it.
Exercises 3.1
3.1.13 Given any w ∈ M. Prove that
(1) w ⊨ □⊥ ⇔ R(w) = ∅.
(2) w ⊭ ◇T ⇔ R(w) = ∅.
(3) R(w) = ∅ ⇒ w ⊨ □A.
(4) R(w) = ∅ ⇒ w ⊭ ◇A.
(5) ∅ ⊨ ML ① where ∅ be the empty model class or the empty
①
By (5), we have Val(∅) = ML.
150
frame class.
(6) {⊥} ⊨ A.
(7) Let X, Y be two model classes or X, Y be two frame classes.
Prove that if X ⊆ Y, then Val(Y) ⊆ Val(X). ┤
3.1.14 Prove 3.1.2(2). ┤
3.1.15 Prove that
(1) M is a countermodel of A ⇔ ¬A is satisfiable in M (refer to
3.1.5).
(2) F is a counterframe of A ⇔ ¬A is satisfiable in F. ┤
3.1.16 Prove 3.1.10(2).
3.1.17 Prove that there is a frame F such that F is not a symmetric
frame (refer to 3.2.2 below), but there is a valuation V such that
<W, R, V> ⊨ B.
3.1.18 Prove that
(1) 5 is valid in any transitive and symmetric frame (refer to 3.2.2
below).
(2) ◇A∕A is validity-preserving w.r.t. Mod(D).
(3) □A → □B∕A → B is validity-preserving w.r.t. Fra(K).
(4) □A → □B∕A → B is validity-preserving w.r.t. Fra(D). ┤
3.1.19 Let A be a constant formula. Prove that A is true at all dead
points or A is false at all dead points (refer to 3.2.7 below). ┤
3.1.20 Prove that
151
(1) Φ
(2) Φ
(3) Φ
(4) Φ
⊨ A ⇒ Φ ⊨ M A.
⊨ M A ⇒ Φ ⊨ Mod A.
⊨ F A ⇒ Φ ⊨ Fra A.
⊨ M A ⇒ Φ ⊨ F A. ┤
3.1.21* (Popkorn,  (p.95)) Define □ωΦ as 2.1.9. Does the
following proposition hold?
(☆) □ωΦ ⊨ A ⇔ □ωΦ ⊨ M A ⇔ Φ ⊨ M A. ┤
3.1.22 (Problem) Define other entailment relations, and compare
them and the entailment relations defined before. ┤
3.1.23 Does the following proposition hold?
Given any model M = <W, R, V> and X ⊆ W, there is some A ∈
ML such that X = V(A). ┤
152
§2 Correspondence Theorem
In this section we will discuss the correspondence relations between
some familiar modal formulas and first-order formulas.
Definition and Convention 3.2.1
(1) X is a set of frame conditions or semantics conditions on a
frame F = <W, R> ⇔ each condition in X specifies some property of
R. ①
We here also say that F satisfies X.
(2) A model M satisfies a set of frame conditions X ⇔ its frame
satisfies X.
(3) If X can be denoted by a property ϕ (ϕ is usually a conjunction
of finite first-order sentences), then the model satisfying ϕ and the
frame satisfying ϕ are called ϕ-model and ϕ-frame, respectively.
A possible world w ∈ W is called a ϕ-point ⇔ w satisfies ϕ (strictly
speaking, w satisfies the formula obtained from ϕ by deleting the
quantifier(s) quantifying w). [So ϕ-model and ϕ-frame mean that their
each possible world satisfies ϕ or is a ϕ-point.]
(4) The class of all ϕ-models is called ϕ-model class, denoted as
Mod(ϕ). The class of all ϕ-frames is called ϕ-frame class, denoted as
Fra(ϕ). [For example, Fra(reflexive) denotes the class of all reflexive
frames.] ┤
Remark. (Ⅰ) Since Mod(ϕ) and Fra(ϕ) are the largest model class
①
Refer to 3.2.2 below.
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and the largest frame class satisfying ϕ, respectively, it follows that ϕmodel class validity defined in 3.1.6(3) is just ϕ-frame class validity
defined in 3.1.8(4). So, we may equate one of them with the another.
Of course, in the literature about modal logic, ϕ-validity usually
denotes ϕ-frame class validity. In this course, we also follow this
convention.
Fra(ϕ) is the semantics object we are interested in furthest, since it
allow all possible valuations, whereas the essence of a logic consists
in revealing the properties still preserving invariability after
evaluating any possible value to any non-logical symbol. So, a frame
class and a model class based on a frame class, compared with a
model class not based on a frame class, is a more important concept
characterizing a modal system from the point of view of philosophy.
(Ⅱ) ϕ-frame (class) validity and S-frame (class) validity (refer to
3.1.11) are two different groups of concepts. Look at their definitions,
it is easy to see that the former group of concepts are just defined by a
frame and a group of frame condition on it, so thay are pure semantic
concepts; the latter group of concepts are defined by general
properties of a frame, as well as all the theorem of current formal
system, so, in essence, this two groups of concepts are quite different.
In some cases, they are equivalent, namely, the valid formula sets
characterized respectively by them coincide, but, in other cases, not so.
In the following, we often use the latter group of concepts to show
the properties of a system w.r. t. some frame class (model class), for
example, soundness, completeness, and etc.; use the former group of
concepts to study the correspondence relation between a modal
formula and a first-order sentence (a closed formula).
Definition 3.2.2 Define the familiar frame conditions on F = <W, R>
as follows:
(seriality) ∀x∃y . xRy,
154
(reflexivity) ∀x . xRx,
(symmetry) ∀xy . xRy → yRx,
(transitivity) ∀xyz . xRy ∧ yRz → xRz,
(euclideanness) ∀xyz . xRy ∧ xRz → yRz. ┤
Remark. (Ⅰ) ‘.’ in the above formulas means that the scope of the
quantifiers before it is whole of the formula after it, so point ‘.’ can be
understood as a rightward quantification sign.
We usually omit the outermost universal quantifiers when
mentioning the above frame conditions. (Ⅱ) The above properties can be iterated. For example, R can be
both symmetric and transitive.
(Ⅲ) If R is reflexive, transitive and symmetric, then R is called
equivalence relation.
(Ⅳ) By 3.2.1, 3.2.2 and Remark above 3.2.2, given any x ∈ W, we
say that x is a serial point, if and only if, x satisfies ∃y .xRy, and thus
∃y ∈ W such that xRy; say that x is a reflexivel point, if and only if, x
satisfies xRx; ….
Note that in this course, we do not strictly differentiate between xRy
as a first-order formula and xRy as ‘x and y have relation R ’.
Corollary 3.2.3 In the first-order predicate calculus, we have:
(1) Reflexivity ⇒ seriality.
(2) Symmetry + transitivity ⇒ euclideanness.
(3) Euclideanness implies:
① ∀xy . xRy → yRy, (shift reflexivity ①)
② ∀xyz . xRy ∧ xRz → yRz ∧ zRy, (shift symmetry).
(4) Reflexivity + euclideanness ⇒ symmetry.
①
Shift reflexivity is also called secondary reflexivity.
155
(5) Symmetry + euclideanness ⇒ transitivity.
(6) Symmetry + transitivity ⇔ symmetry + euclideanness.
(7) Symmetry + transitivity + seriality
⇔ reflexivity + euclideanness.
(8) Reflexivity + symmetry + transitivity
⇔ reflexivity + euclideanness. ┤
Remark. Every frame condition in 3.2.2 can be expressed by a firstorder formula (sentence). This urges us to see a modal relation frame
<W, R> as a first-order relation model, and thus study the
correspondence relation between a modal formula and a first-order
formula (sentence). More detailedly, there is a primitive symbol set
{P} of a first-order language ① such that <W, R> is the abbreviation
of a first-order model <W, H> where W is a domain (a field of
individuals) and H (P) = R ⊆ W2. In this way, the frame conditions
given in 3.2.2 can be seen as the first-order sentences that is true in a
first-order model <W, R> .
Definition 3.2.4 Let A ∈ ML and ϕ be a (first-order) closed formula.
(1) A and ϕ (first-order) correspond ⇔ for any tuple <W, R>,
(☆) <W, R> ⊨ A ⇔ <W, R> ⊨ ϕ.
Here A is called the modally definable formula of ϕ, and ϕ is called
the first-order definable formula of A.
(2) A is first-order definable ⇔ there is a first-order formula
corresponding to A.
(3) ϕ is modally definable ⇔ there is a modal formula
corresponding to ϕ. ┤
①
The language can be with identity if need.
156
Remark. (Ⅰ) ‘<W, R> ⊨ A’ in the previous (☆) means that A is
valid in the modal relation frame <W, R>, whereas ‘<W, R> ⊨ ϕ’
means that ϕ is true in the first-order relation model <W, R>. Don’t
confuse them at any rate, please.
(Ⅱ) A frame condition usually can be denoted by a first-order
sentence, but there is a frame condition such that it cannot be denoted
by a first-order sentence (refer to 4.4.7).
Theorem 3.2.5 (Correspondence Theorem)
(1) Axiom D corresponds to seriality,
(2) Axiom T corresponds to reflexivity,
(3) Axiom B corresponds to symmetry,
(4) Axiom 4 corresponds to transitivity,
(5) Axiom 5 corresponds to euclideanness.
Proof. Given any tuple <W, R>.
(1) Assume that <W, R> ⊭ ∀x∃y . xRy. Then there is some w ∈ W
such that for any u ∈ W, ~ wRu. So R(w) = ∅, So, for any valuation V
on the modal frame <W, R>, w ⊨ □p and w ⊭ ◇p, and hence w ⊭ □p
→ ◇p, so <W, R> ⊭ D.
Assume that <W, R> ⊭ D. Then there is a model <W, R, V> on the
modal frame <W, R> (and thus, in essence, there is a valuation V on
the modal frame <W, R>) and w ∈ W such that w ⊭ □p → ◇p, so, by
Equivalence Lemma 3.1.2, it is easy to see that
w ⊨ □p and w ⊨ □¬p,
so, by Truth Set Definition,
R(w) ⊆ V(p) and R(w) ⊆ V(¬p).
It follows that
R(w) ⊆ V(p) ∩ V(¬p) = V(p ∧ ¬p).
157
Since V(p ∧ ¬p) = ∅, it follows that R(w) = ∅, hence for any u ∈ W,
~ wRu, and thus
<W, R> ⊭ ∀x∃y . xRy.
(2) Assume that <W, R> ⊭ ∀x . xRx. Then there is some w ∈ W such
that ~ wRw. Let V be a valuation on the modal frame <W, R> such that
V(p) = W − {w}. ① Since w ∉ R(w), so R(w) ⊆ V(p), and thus w ⊨ □
p, we have w ⊭ □p → p by w ⊭ p, so <W, R> ⊭ T.
Assume that <W, R> ⊭ T. Then there is a valuation V on the modal
frame <W, R> and w ∈ W such that w ⊨ □p and w ⊭ p. Since w ⊨ □p,
it follows that R(w) ⊆ V(p). Hypothesize that w ∈ R(w), then w ⊨ p,
and thus we have a contradiction. So <W, R> ⊭ ∀x . xRx.
(3) Assume that <W, R> ⊭ ∀xy . xRy → yRx. Then there are some
w, u ∈ W such that wRu and ~ uRw. Let V be a valuation on the modal
frame <W, R> such that
(☆) V(p) = {w}. ②
By (☆) and w ∉ R(u), it is easy to see that R(u) ⊆ V(¬p), so u ⊨ □¬p,
and thus u ⊭ ¬□¬p, namely, u ⊭ ◇p. Since wRu, it follows that w ⊭
□◇p. By (☆), w ⊭ p → □◇p, so <W, R> ⊭ B.
Assume that <W, R> ⊭ B. Then there is a valuation V on the modal
frame <W, R> and w ∈ W such that w ⊨ p and w ⊨ ◇□¬p. By the
latter, there is some u ∈ R(w) such that u ⊨ □¬p. Hypothesize that
uRw, then w ⊨ ¬p, contradicting w ⊨ p, so ~ uRw, and thus
<W, R> ⊭ ∀xy . xRy → yRx.
①
②
Such a valuation is also called p-maximal valuation.
Such a valuation is also called p-minimum valuation.
158
(4) Assume that <W, R> ⊭ ∀xyz . xRy ∧ yRz → xRz. Then there are
some w, u, v ∈ W such that wRu, uRv and ~ wRv. Let V be a valuation
on the modal frame <W, R> such that
① V(p) = W − {v}.
Since ~ wRv, it follows by ① that
② w ⊨ □p.
Since uRv, it follows by ① that
③ u ⊭ □p.
By wRu and ③, w ⊭ □□p, so <W, R> ⊭ 4 by ②.
Assume that <W, R> ⊭ 4. Then there is a valuation V on the modal
frame <W, R> and w ∈ W such that
④ w ⊨ □p, and
⑤ w ⊨ ◇◇¬p.
By ⑤, there is some u ∈ R(w) such that u ⊨ ◇¬p. So there is some v
∈ R(u) such that v ⊨ ¬p. Hypothesize that v ∈ R(w), then w ⊭ □p,
contradicting ④, so ~ wRv, and thus
<W, R> ⊭ ∀xyz . xRy ∧ yRz → xRz.
(5) Assume that <W, R> ⊭ ∀xyz . xRy ∧ xRz → yRz. Then there are
some w, u, v ∈ W such that wRu, wRv and ~ uRv. Let V be a valuation
on the modal frame <W, R> such that
① V(p) = {v}.
Since wRv, it follows by ① that
② w ⊨ ◇p.
Since v ∉ R(u), it follows by ① that R(u) ⊆ V(¬p), so u ⊨ □¬p, and
thus u ⊭ ¬□¬p, and hence u ⊭ ◇p, by wRu, w ⊭ □◇p, so <W, R>
⊭ 5 by ②.
159
Assume that <W, R> ⊭ 5. Then there is a valuation V on the modal
frame <W, R> and w ∈ W such that
③ w ⊨ ◇p, and
④ w ⊨ ◇¬◇p.
By ④, there is some u ∈ R(w) such that
⑤ u ⊨ ¬◇p.
By ③, there is some v ∈ R(w) such that
⑥ v ⊨ p.
Hypothesize that v ∈ R(u). Then u ⊨ ◇p by ⑥, contradicting ⑤. So
~ uRv, and thus
<W, R> ⊭ ∀xyz . xRy ∧ xRz → yRz. ┤
By Correspondence theorem, it is clear that:
Corollary 3.2.6
(1) The character axioms (except Axiom K) of the basic systems
are all first-order definable.
(2) The first-order formulas given in 3.2.2 are all modally definable.
┤
Remark. (Ⅰ) By 3.3.10 below, Axiom K is valid in any frame, so
we can use a first-order formula ∀x .x = x or ∀xy .xRy ∨ ¬xRy to
correspond to K.
( Ⅱ ) There are many modal formulas that are not first-order
definable, ① Contrarily, there are many first-order formulas that are
①
Such as Axiom W. Refer to Example 3.9 (p.133) of Blackburn, de
Rijke and Venema
.
160
not modally definable as well. Modal logicians specially establish the
correspondence theory to study this aspect of contents.
The following definition shall be used below:
Definition 3.2.7 Given any F = <W, R>.
(1) Given any w, u ∈ W, wR n u is inductively defined as follows:
① wR 0u ⇔ u = w,
② wR n + 1 u ⇔ there is some v ∈ W (wRv and vRnu).
If wR n u, then we say that there is a n-step-R-chain from w to u.
(2) Let w ∈ W.
① w is an dead point ⇔ R(w) = ∅,
② w is an end point ⇔ for all u ∈ W (wRu ⇒ u = w),
③ w is an exactly-reflexive point ⇔ R(w) = {w},
④ w is an universal point ⇔ R(w) = W.
(3) R is the universal relation ⇔ R = W2. ┤
Remark. Note that
w is an end point ⇔ w is a dead point or exactly-reflexive point.
Exercises 3.2
3.2.8 Prove 3.2.3. ┤
3.2.9 Consider whether the following frames satisfy reflexivity,
seriality, symmetry, transitivity and euclideanness, respectively:
(1) w.
(2) wο.
(3) w → u.
161
(4) w ↔ u.
(5) w → uο.
(6) wο → u. ┤
3.2.10 Find out the first-order formulas corresponding to the
following modal formulas, respectively.
(1) Dc, Tc, Bc, 4c and 5c.
(2) Tr and Ver. ┤
3.2.11 Find out the first-order formulas corresponding to the
following modal formulas, respectively.
(G)
◇□p → □◇p.
(MV) □p ∨ ◇□p.
(Lem) □(□p → q) ∨ □(□q → p).
(4n)
□p → □n p.
mn
(4 ) □m p → □n p.
(Vern) □n p.
(RL) ◇p ∧ ◇q → ◇(p ∧ q) ∨ ◇(◇p ∧ q) ∨ ◇(p ∧ ◇q).
(w5) p → ◇p → □◇p.
(w5#) p → ◇¬p → □◇¬p.
(f)
p ∧ ◇□q → □(◇p ∨ q).
(Mk) □(□□p → □q) → □p → q. ┤
3.2.12 Find out the first-order formulas corresponding to the
following modal formulas, respectively.
(1) □(□p → p),
(2) □p → □◇□p,
(3) □(◇□p → p),
(4) □(p ∧ □p → q) ∨ □(q ∧ □q → p),
162
(5) ◇(p ∧ □q) → □(p ∨ ◇q),
(6) □p → □(p → □p),
(7)* □(□(p → □p) → p) → p,
(8) ◇p ∧ ◇q → ◇(p ∧ q),
(9) ◇□p → p → □p,
(10) ◇T → (□p → p),
(11) p → □◇◇p. ┤
(Grz)
3.2.13 Prove that
(1) G corresponds to the first-order formula found out in 3.2.11.
(2) Lem corresponds to the first-order formula found out in 3.2.11.
┤
3.2.14 (Problem)*
(1) Does McK correspond to finality ①:
∀x∃y . xRy ∧ ∀z . yRz → z = y?
(2) Does McK correspond to other first-order formulas? ┤
①
Each world accesses to an end-point.
163
§3 Soundness Theorem
In this section we will introduce the soundness definition, and prove
the soundness theorems of some familiar systems.
Definition 3.3.1 Let C be a model class or a frame class. A modal
system S is (weak) sound w.r.t. C ⇔ Th(S) ⊆ Val(C). ┤
Remark. It is easy to see that
S is sound w.r.t. C ⇔ C ⊨ Th(S).
Lemma 3.3.2 ((Weak) Soundness Equivalence Lemma) Let C be a
model class or a frame class, and let S be a modal system. Then the
following two propositions are equivalent:
(1) Th(S) ⊆ Val(C).
(2) Given any A ∈ ML, if A ∈ Sat(C), then A is S-consistent.
Proof. (1) ⇒ (2): Assume that (1) holds. Given any A ∈ Sat(C), by
3.1.10(2), ¬A ∉ Val(C). By (1), ¬A ∉ Th(S), so, by Consistency
Lemma 2.2.2(7), A is S-consistent.
(2) ⇒ (1): Assume that (2) holds. Given any A ∈ Th(S), then by
2.2.2(9), ¬A is S-inconsistent, so, by the assumption, ¬A ∉ Sat(C), So
A ∈ Val(C) by 3.1.10(1). ┤
Remark. (Ⅰ) Since ‘A is S-consistent’ is the abbreviation of ‘{A} is
S-consistent’, it follows that the above (2) cannot be simply written as
‘Sat(C) ⊆ CS(S) ’.
164
(Ⅱ) By the previous lemma, to prove that S is sound w.r.t. C, it
suffices to show that for all A ∈ ML
A is satisfiable w.r.t. C ⇒ A is S-consistent.
In the following we will define a kind of ‘absolute’ soundness:
Definition 3.3.3 A system S is sound ⇔ there is a nonempty class of
S-frames. ┤
Lemma 3.3.4
(1) If S is sound, then S is consistent.
(2) Let F be a nonempty frame class. If S is sound w.r.t. F, then S
is consistent.
Proof. (1) Assume that S is sound, then, by 3.3.3, there is a
nonempty class F of S-frames. Hypothesize that S is inconsistent,
then by Consistency Lemma 2.2.2(3), ⊥ ∈ Th(S). Since F ⊨ Th(S), it
follows that F ⊨ ⊥, whereas this is impossible.
(2) By (1). ┤
Remark. By the above results, if we show that S is sound then show
that S is consistent. This is a kind of semantics proof method differing
from the method given in 2.2.3(2). Of course, the semantics method is
not more operable than the latter (the proof theory method) when we
actually prove the consistency of S.
In the following we will consider the soundness theorem of the
basic systems. For this, we first consider the validity of inference rules:
Theorem 3.3.5
(1) MP is pointwise valid (refer to 3.1.9).
165
(2) MP is modelwise valid, framewise valid and is validitypreserving for any model class or frame class.
(3) RN is modelwise valid.
(4) RN is framewise valid, and thus is validity-preserving for any
model class or frame class.
Proof. (1) Given any M = <W, R, V> ∈ Mod and w ∈ W such that
w ∈ V(A → B) ∩ V(A).
Since V(A → B) = V(¬A) ∪ V(B) and w ∉ V(¬A), it follows that w ∈
V(B).
(2) By (1) and 3.1.20.
(3) Given any M = <W, R, V> ∈ Mod and A such that M ⊨ A, then
V(A) = W. Given any w ∈ W, it is easy to see that R(w) ⊆ V(A), so w
⊨ □A. By the arbitrariness of w, M ⊨ □A.
(4) By (3) and 3.1.20. ┤
Remark. (Ⅰ) Please show that RN is not pointwise valid.
(Ⅱ) From the point of view of the model theory, MP is validitypreserving at each point, namely, the validity it preserves is just
related to the current world, and not related to other worlds. It is clear
that RN does not have this property. So the two rules are quite
different .
Definition 3.3.6 A is substitution valid in M ⇔ each substitution
instance of A is valid in M. ┤
Theorem 3.3.7
(1) Rule US is framewise valid, and thus is validity-preserving for
any frame class.
(2) Let M be any model. If each formula of Φ is substitution valid
in M, then K + Φ is also valid in M.
166
Proof. (1) We argue via contraposition. Hypothesize that there is a
frame F = <W, R> such that US is not validity-preserving w.r.t. F,
then there is a formula A and a substitution mapping σ such that
① F ⊨ A, and
② F ⊭ Aσ.
By ②, there is a model M = <W, R, V> and w ∈ W such that
③ w ∉ V(Aσ).
We construct a new model N = <W, R, V1> by M such that V1 is just as
V, except
④ V(σ(pn)) = V1(p n) for every sentence symbol pn in A.
We will show that
⑤ V(Aσ) = V1(A).
By induction on A. If A is some pn, then the result we will prove
holds clearly by ④ and σ(pn) = pnσ.
If A = ¬B or A = B ∧ C, then by the induction hypothesis
(hypotheses), it is easy to obtain the result we will prove.
Assume that A = □B, then by the induction hypothesis, we have
⑥ V(Bσ) = V1(B).
Given any u ∈ W, by ⑥, we have
R(u) ⊆ V(Bσ) ⇔ R(u) ⊆ V1(B).
So
⑦ u ∈ V(□(Bσ)) ⇔ u ∈ V1(□B) for all u ∈ W.
Since □(Bσ) = (□B)σ by 1.1.5, so, by ⑦,
V((□B)σ) = V1(□B).
Thus we get ⑤.
We have w ∉ V1(A) By ③ and ⑤, so N ⊭ A, and thus <W, R>
(2) Assume that each formula of Φ is substitution valid in M. By
167
3.3.25 (1), every theorem of K is valid in any model. So, it suffices to
show that
the formulas that are obtained by MP, RN or US from the formulas
that are substitution valid in M are still substitution valid in M.
Namely,
if A and B are substitution valid in M and C is obtained by MP from
A and B, then C is still substitution valid in M; and
if A is substitution valid in M and A* is obtained by RN or US from
A, then A* is still substitution valid in M.
We leave the details of the proof as an exercise. ┤
Remark. By the proof of the above result (2), if we rather use axiom
schemata than axioms (so we do not need US) when constructing a
normal system S in Chapter 1, then the inference rules of S are all
modelwise valid.
Corollary 3.3.8
(1) MP, RN and US are framewise valid.
(2) Let Axiom(S) be the character axiom set of S, and F a frame.
F ⊨ Axiom(S) ⇒ F ⊨ Th(S).
[This means that a Axiom(S)-frame is always a S-frame.]
Proof. (1) By 3.3.5(2) and (4) as well as 3.3.7(1).
It is easy to show by (1) that (2) holds. ┤
Remark. (Ⅰ) By (2), without loss of generality, we can regard a
Axiom(S)-frame as a S-frame.
(Ⅱ) The above theorem do not means that Rule US can always be
validity-preserving for any model. This means that a Axiom(S)-model
is not always a S-model. In fact, we have the following
counterexample:
168
Counterexample 3.3.9
(1) It is not necessary that US is validity-preserving for models, and
thus there are some S and M such that
M ⊨ Axiom(S) and M ⊭ Th(S).
(2) There are S and A such that A is a theorem of S and A has a
Axiom(S)-countermodel (namely, there is some Axiom(S)-model such
that it is a countermodel of A).
Proof. (1) Construct M as follows:
wp, q → up, ¬q.
So
w ⊨ □p → p, u ⊨ □p → p and u ⊭ □q → q.
So □p → p is valid in M, but its substitution instance □q → q is not
valid in M. So there is a system <T> = PC + T such that
M ⊨ Axiom(<T>) and M ⊭ Th(<T>).
(2) It is easy to see that □q → q is a theorem of the system <T>,
but by the proof of (1), □q → q has a Axiom(<T>)-countermodel.
┤
Remark. The above counterexample show that, when a theorem set
(let alone a formula set) Φ is valid in a model, it is not necessary that
a formula derived by US from Φ is also valid in this model. So, from
the point of view of the model theory, US is a rule that is quite
different from MP and RN. By the previous result, we can say that,
US does not preserve model validity, and thus does not preserve
model class validity.
Sum up the above results, for MP, RN and US, we have the
following table: where + means that a current rule has a current
property, and − means that a current rule does not have a current
169
property:
inference
rule
MP
RN
US
pointwise
validity
+
−
−
modelwise
validity
+
+
−
framewise
validity
+
+
+
Theorem 3.3.10
(1) Every axiom of K is valid in any frame.
(2) Every primitive rule of K is validity-preserving in any frame.
(3) Every theorem of K is valid in any frame.
Proof. (1) Let <W, R> be any frame. It is clear that every axiom of PC
is valid in <W, R>, so it suffices to show that the character axiom K is
valid in <W, R>.
Hypothesize that the result we will show does not hold, then there
are some <W, R, V> and w ∈ W such that
① w ⊨ □(p → q), and
② w ⊭ □p → □q.
By ② and Equivalence Lemma (3.1.2(1)), it is easy to see that
③ w ⊨ □p, and
④ w ⊨ ◇¬q.
By ④, there is some u ∈ R(w) such that
⑤ u ⊨ ¬q.
On the other hand, by ①, ③ and u ∈ R(w), we have u ⊨ p → q
and u ⊨ p, so, by 3.3.5(1), u ⊨ q, contradicting ⑤.
(2) By 3.3.8(1).
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(3) By (1) and (2). ┤
Remark. By the previous result, we no longer consider Axiom K
when below consider the validity of the character axioms of a normal
system.
As far as we know, each correspondence theorem is established
w.r.t. a frame, below we will show that each T-frame is a reflexive
frame, each reflexive frame is a T-frame, …
With respect to a model, can we say it in this way? The answer is
negative. For example, even if each reflexive model is a T-model, it is
not also necessary that each T-model is a reflexive model. In the
following is an example:
Example 3.3.11 There is a T-model which is not reflexive.
Proof. Consider a frame <W, R> = {w ↔ u}. Define M = <W, R, V>
as follows:
w ⊨ pn ⇔ u ⊨ pn for all pn ∈ At.
By induction on the formulas. It is easy to prove that
(1) w ⊨ A ⇔ u ⊨ A for all A ∈ ML.
We will show that
(2) w ⊨ □A → A for all A ∈ ML.
Assume that w ⊨ □A. Since wRu, it follows that u ⊨ A, hence w ⊨ A
by (1), so (2) holds.
By the symmetry of w and u, we can prove that
(3) u ⊨ □A → A for all A ∈ ML.
So far we have proved that each substitution instance of T is valid
in M, by 3.3.7(2), it is easy to see that M is a T-model. On the other
hand, it is clear that it is not a reflexive model. ┤
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Theorem 3.3.12 Let S = K + A1 + … + An, and let ϕ1, …, ϕn be the
first-order closed formulas corresponding to A1, …, An, respectively.
Let ϕ = ϕ1 ∧…∧ ϕn. Then
F ⊨ Th(S) ⇔ F ⊨ ϕ.
[As we remarked above, F on the left is a modal frame, F on the right
is a first-order model. ]
Proof. ‘⇒’: Assume that F ⊨ Th(S), then F ⊨ A1, …, F ⊨ An, by the
correspondence relations, F ⊨ ϕ1, …, F ⊨ ϕn, so F ⊨ ϕ.
‘⇐’: Assume that F ⊨ ϕ, then F ⊨ ϕ1, …, F ⊨ ϕn, by the
correspondence relations, we have F ⊨ A1, …, F ⊨ An, and thus F ⊨
A1 ∧…∧ An, by (3.3.10(1) and) 3.3.8(2), we get what we will show.
┤
Corollary 3.3.13
(1) Fra(K) = Fra.
(2) Fra(D) = Fra(seriality).
(3) Fra(T) = Fra(reflexivity).
(4) Fra(B) = Fra(reflexivity, symmetry).
(5) Fra(S4) = Fra(reflexivity, transitivity).
(6) Fra(S5) = Fra(reflexivity, euclideanness).
(7) Fra(S5) = Fra(reflexivity, symmetry, transitivity).
Proof. (1) - (6) By the above results and Correspondence Theorem
3.2.5, Definition 3.1.11 and the denotation in (Ⅱ) below 3.2.1.
(7) By (6) and 3.2.3(8). ┤
Remark. As far as we know, a reflexive, symmetric and transitive
relation is an equivalence relation. By the equivalence relation on a
current possible world set, we can partition the set into as a class of
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equivalent classes, so, by (7), each S5-frame <W, R> is a frame
constituted by some equivalent classes as subsets of W.
Theorem 3.3.14
(1) The basic systems and Tr are sound.
(2) The basic systems and Tr are consistent.
Proof. (1) By the previous corollary (1), all frame are K-frame,
whereas by the previous corollary (2) - (6), when S is a basic system
except K, it is easy to show that there always exists a nonempty frame
class Fra(S), so S is always sound.
Please prove that Tr is sound.
(2) By (1) and 3.3.4(1). ┤
Remark. In 2.2.15, we proved that the basic systems are consistent.
But as we remarked before, the method used in 2.2.15 is a proof
theory method, whereas the method used in the previous theorem is a
model theory method.
Theorem 3.3.15 (Frame Soundness Theorem)
(1) Th(K) ⊆ Val(Fra).
(2) Th(D) ⊆ Val(Fra(seriality)).
(3) Th(T) ⊆ Val(Fra(reflexivity)).
(4) Th(B) ⊆ Val(Fra(reflexivity, symmetry)).
(5) Th(S4) ⊆Val(Fra(reflexivity, transitivity)).
(6) Th(S5) ⊆ Val(Fra(reflexivity, euclideanness)).
(7) Th(S5) ⊆ Val(Fra(equivalence)).
Proof. We will prove (5) by giving an example. Given any A ∈
Th(S4), then Fra(S4) ⊨ A. By 3.3.13(5),
Fra(reflexivity, transitivity) ⊨ A,
so A ∈ Val(Fra(reflexivity, transitivity)). ┤
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Remark. Generally, if we want to prove independently that a
system S is sound w.r.t. a frame class Fra(ϕ), ① then it suffices to
show that all the axioms of S are valid in any frame F in Fra(ϕ) and
all the inference rules of S are validity-preserving w.r.t. F.
Exercises 3.3
3.3.16 Prove that
Th(S4.2) ⊆ Val(Fra(reflexivity, transitivity, convergence)). ┤
3.3.17
F is a single reflexive point frame ⇔ F = {wο}.
F is a single dead point frame ⇔ F = {w}.
Prove that
(1) Tr is sound w.r.t. the class of all single reflexive point frames.
(2) Ver is sound w.r.t. the class of all single dead point frames. ┤
3.3.18 Prove that
(1) Ver is sound.
(2) Ver is consistent. ┤
3.3.19
(1) Prove that RE and RER are modelwise valid.
(2) Are RE and RER pointwise valid? ┤
3.3.20
①
Here, ‘independently’ means that we do not use a result like 3.3.13.
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(Ⅰ) Do the converse propositions of (1) - (4) of 3.1.20 hold?
(Ⅱ) Do the following entailment have the deduction property,
namely, do the following propositions hold?
(1) Φ ∪ {A} ⊨ B ⇒ Φ ⊨ A → B.
(2) Φ ∪ {A} ⊨ M B ⇒ Φ ⊨ M A → B.
(3) Φ ∪ {A} ⊨ F B ⇒ Φ ⊨ F A → B. ┤
Definition 3.3.21 A modal system S is Halldén-complete ⇔ Given
any A and B such that they do not contain the same sentence symbol,
then
⊢ S A ∨ B ⇔ ( ⊢ S A or ⊢ S B). ┤
3.3.22(Chagrov, Zakharyaschev,  (p.90)) Prove that K is not
Halldén-complete.
[Note that PC is Halldén-complete]. ┤
3.3.23(Chagrov, Zakharyaschev,  (p.88)) Prove that K is not
Post-complete. ┤
3.3.24 (Problem) Do the following Strong Soundness Theorems
hold?
(1) Φ ⊢ K A ⇒ Φ ⊨ A.
(2) Φ ⊢ K A ⇒ Φ ⊨ M A.
(3) Φ ⊢ K A ⇒ Φ ⊨ F A.
(4) Φ ⊢ K A ⇒ Φ ⊨ A. ┤
3.3.25
(1) Prove that every theorem of K is valid in any model.
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(2) (Problem) Does the following proposition hold?
(☆) Is each theorem of K is true at any possible world of any
model? ┤
3.3.26 (Problem)
(1) Is there a K4-model which is not transitive?
(2) Is there a K5-model which is not euclidean? ┤
3.3.27 (Problem)
(1) Does the following proposition hold?
－
Φ is S-consistent ⇒ □ Φ is also S-consistent.
(2) If (1) does not hold, then it holds w.r.t. which condition? ┤
3.3.28 (Problem about Strong Soundness Equivalence Lemma) Let
C be a model class or a frame class. For all Φ ⊆ ML and A ∈ ML, we
say that Φ entails pointwise A w.r.t. C, denoted by Φ ⊨ C, P A: Given
any x ∈ C and w ∈ x (and any model on x, if x is a frame),
w ⊨ Φ ⇒ w ⊨ A.
(1) For the following two propositions, does one implies another?
Is one equivalent to another?
① For all Φ ⊆ ML and A ∈ ML, Φ ⊢ S A ⇒ Φ ⊨ C, P A.
② For all Φ ⊆ ML, if Φ is satisfiable in C, ① then Φ is Sconsistent.
(2) For the following two propositions, does one implies another?
Is one equivalent to another?
① For all Φ ⊆ ML and A ∈ ML, Φ ⊢ S A ⇒ Φ ⊨ C, P A.
② For all Φ ⊆ ML, if Φ is satisfiable in C, then Φ is S①
Refer to 3.1.6 (3) and 3.1.8 (2).
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consistent. ┤
3.3.29 (Problem) Let ∅ be the empty model class or the empty frame
class. Is S sound w.r.t. ∅? ┤
3.3.30 Prove that
Φ is satisfiable in M and M is a S-model ⇒ Φ is S-consistent. ┤
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§4 Countermodel Method
In this section we will use the countermodel method to answer the
some problems left in the above chapters, mostly show the proper
inclusion relations between the basic systems.
Definition 3.4.1 Given any system S and formula A.
(1) M is a S-countermodel of A
⇔ M ⊨ Th(S) and M ⊭ A
⇔ M is both a S-model and is a countermodel of A.
(2) Let ϕ be a (semantic) property. M is a ϕ-countermodel of A
⇔ M satisfies ϕ and M ⊭ A. ┤
Lemma 3.4.2
(1) (Countermodel Lemma) If there is a S-countermodel of A,
then A ∉ Th(S).
(2) (Counterframe Lemma) If there is some Axiom(S)counterframe of A, then A ∉ Th(S).
Proof. (1) Assume that there is a S-countermodel M of A, then
① M ⊨ Th(S), and
② M ⊭ A.
Hypothesize that A ∈ Th(S), then M ⊨ A by ①, contradicting ②.
(2) Assume that there is some Axiom(S)-counterframe F of A, then
① F ⊨ Axiom(S), and
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② F ⊭ A.
By ① and 3.3.8(2), F ⊨ Th(S). Hypothesize that A ∈ Th(S), then F
Remark. (Ⅰ) By the proof of 3.3.9(2), even if A has a Axiom(S)countermodel, A can still be a theorem of S.
(Ⅱ) We usually use Counterframe Lemma (3.4.2(2)) to show the
some A concerned is not in Th(S). Of course, this need the related
results of Correspondence Theorem 3.2.5.
Theorem 3.4.3 (Countermodel Theorem)
(1) There is a K-countermodel of D.
(2) There is a D-countermodel of T.
(3) There is a T-countermodel of B.
(4) There is a T-countermodel of 4.
(5) There is a S4-countermodel of 5.
(6) There is a B-countermodel of 5.
(7) There is a S5-countermodel of p → □p, and thus there is a S5countermodel of Axiom Tr.
Proof. (1) By 3.3.10(3), any model is a K-model. So it suffices to
construct an appropriate non-serial model. Given any single dead
point frame w and any model M on the frame, it is easy to see that M
is not serial, and hence M ⊨ □p and M ⊭ ◇p, so
M ⊭ □p → ◇p.
(2) Construct an appropriate serial and non-reflexive model as
follow:
w¬ p → uοp.
It is easy to see that the frame of the model is a serial frame, so, by
3.3.13(2), the frame is a D-frame, hence the model is a D-model.
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Since w ⊨ □p and w ⊭ p, it follows that
w ⊭ □p → p.
(3) Construct an appropriate reflexive and non-symmetric model as
follow:
wοp → uο¬ p.
Since R is reflexive, it is easy to see by 3.3.13(3) that the model is a
T-model.
Since u ⊭ p and R(u) = {u}, it follows that u ⊭ ◇p, so w ⊭ □◇p
by wRu. Since w ⊨ p, it follows that
w ⊭ p → □◇p.
(4) Construct an appropriate reflexive and non-transitive model as
follow:
wο p → uοp → vο¬ p.
It is easy to see that the frame of the model is a reflexive frame, so, by
3.3.13(3), it is easy to see that the model is a T-model.
Since v ⊭ p, it follows that u ⊭ □p by uRv, so w ⊭ □□p by wRu.
Since w ⊨ □p, it follows that
w ⊭ □p → □□p.
(5) Construct an appropriate reflexive and transitive but noneuclideanness model as follow:
wο p → uο p → vο¬ p
Since wRv and wRu but ~ vRu, it follows that w is not an
euclideanness point, so R does not satisfy euclideanness. But it is easy
to see that R is reflexive and transitive, so, by 3.3.13(5), it is easy to
see that the model is a S4-model.
Since v ⊭ ◇p by R(v) = {v}, so w ⊭ □◇p by wRv. Since w ⊨ ◇p,
it follows that
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w ⊭ ◇p → □◇p.
(6) Construct an appropriate reflexive and symmetric but noneuclidean model as follow:
vο¬ p ↔wο¬ p ↔ uο p.
Since wRv and wRu but ~ vRu, it follows that w is not an
euclideanness point, so R does not satisfy euclideanness. But it is easy
to see that the frame of the model is reflexive and symmetric frame, so,
by 3.3.13(4), it is easy to see that the model is a B-model.
Since v ⊭ ◇p, it follows that w ⊭ □◇p by wRv. Since w ⊨ ◇p, it
follows that
w ⊭ ◇p → □◇p.
(7) Construct an appropriate reflexive, transitive and symmetric
model as follow:
wο p ↔ uο¬ p.
It is easy to see that the frame of the model is reflexive, transitive and
symmetric frame ①, so, by 3.3.13(7), it is easy to see that the model is
a S5-model.
Since w ⊨ p and w ⊭ □p by wRu, it follows that
w ⊭ p → □p. ┤
Remark. Axiom T is a more intuitive axiom:
A necessary true proposition is always true (A thing that is
necessary is always a thing that is material).
But we can still find a countermodel of T. This show that the
relation semantics is a very generalized semantics, it can also describe
things beyond our intuitions.
By the related results of Chapter 1, there is an inclusion relation
①
In fact, this is an universal frame, so R is an equivalence relation.
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between each pair of the systems mentioned below, therefore, by
Countermodel Theorem 3.4.3 and Countermodel Lemma 3.4.2(1) (as
well as Definition 1.1.13), there is a proper inclusion relation
between each pair of the systems:
Theorem 3.4.4 (Proper Extension Theorem)
(1) K ⊂ D.
(2) D ⊂ T.
(3) T ⊂ B.
(4) T ⊂ S4.
(5) S4 ⊂ S5.
(6) B ⊂ S5.
(7) S5 ⊂ Tr. ┤
Definition 3.4.5 Let A be an axiom of S.
A is independent in S ⇔ Th(S − A) ⊂ Th(S). ┤
Remark. By 3.4.4, we have the related independence theorems. For
example,
D is independent in D.
The countermodel method can still be applied to prove other results
except applied to prove the proper inclusion relations between
systems. In the following we will use this method to prove that
System T has countable infinite modal symbol strings that are not
equivalent to each other:
Theorem 3.4.6 System T has countable infinite modal symbol strings
that are not equivalent to each other.
182
Proof. First we will prove
(☆) □n p → □n + m p ∉ Th(T) for all n < ω and 1 ≤ m < ω.
Given any n < ω and 1 ≤ m < ω, construct a countermodel M as
follows:
<W, R> = {wοi → wοi + 1 | i < n + m},
V(p) = {w0, …, wn + m − 1}.
That is,
wο0, p → wο1, p →…→ wοn + m − 1, p → wοn + m, ¬ p.
Since R is reflexive, it is easy to see by 3.3.13(3) that M is a T-model.
Since wi ⊨ p for all i ≤ n, it follows that w0 ⊨ □n p. Since wn + m ⊭ p,
it follows that w0 ⊭ □n + m p, so w0 ⊭ □n p → □n + m p.
Hence there is a T-countermodel of □n p → □n + m p. According to
Countermodel Lemma 3.4.2(1), we have (☆).
By (☆), for any 1 ≤ m < ω, □n + m is not reducible to □n in T (by
Definition 1.3.16(1)), so T has countable infinite modal symbol
strings that are not equivalent to each other. ┤
In Chapter 1 we proved that S4 and S5 have at most 14 and 6
modal symbol strings that are not equivalent to each other,
respectively. Now we will show that S4 and S5 have at least 14 and 6
modal symbol strings that are not equivalent to each other,
respectively:
Theorem 3.4.7
(1) Modal(S4) ≥ 14, and thus Modal(S4) = 14 (by 1.3.18).
(2) Modal(S5) ≥ 6, and thus Modal(S5) = 6 (by 1.3.20 (1)).
Proof. (1) It suffices to show that 14 the modal symbol strings listed
in the proof of 1.3.18 are not equivalent to each other.
First, as the proof of 1.3.18 (b), we have proved that if s is a modal
183
symbol string, then sA ↔ ¬sA is not a theorem of any consistent
system. So, from two lines of modal symbol strings listed in the
beginning of the proof of 1.3.18, each pair of symbol strings that
constitute upright are always not equivalent.
Given any other pair of the modal symbol strings, say, □◇ and
◇□, we will construct a S4-countermodel of □◇p ↔ ◇□p, and
thus, by Countermodel Lemma 3.4.2(1), show that □◇p ↔ ◇□p is
not a theorem of S4:
For this, it suffices to construct an appropriate reflexive and
transitive model. Given a universal model:
wοp ↔ uο¬ p.
By 3.3.13(5), it is easy to see that the model is a S4-model. Since R
is universal relation, it follows that
w ⊨ ◇p and u ⊨ ◇p,
so w ⊨ □◇p. But u ⊭ p, so w ⊭ □p and u ⊭ □p, and thus w ⊭ ◇□
p. Hence
w ⊭ □◇p → ◇□p.
The proofs of other any pair of the modal symbol strings are also
similar.
The proof of (2) is similar to (1). ┤
Exercises 3.4
3.4.8 Prove that A ∨ B∕◇A ∨ ◇B is not a rule of K.
3.4.9 Prove that
(1) Mod(Fra(S4)) ≠ Mod(S4),
(2) Mod(Fra(S5)) ≠ Mod(S5). ┤
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3.4.10
(1)
① Prove that for any A, ◇A ∉ Th(K).
② Prove that ◇A∕A is a virtual rule of K.
③ Is ◇A∕A a strong rule of K?
④ Is □A∕A a strong rule of K?
⑤ Is □A → □B∕A → B a strong rule of K?
⑥ Is ◇A → ◇B∕A → B a strong rule of K?
Prove that
(2) □(p ∨ q) → □p ∨ □q ∉ Th(K).
(3) 4 ∉ Th(B).
(4) B ∉ Th(S4).
(5) Grz ∉ Th(S4).
(6)* T ∉ Th(KW).
(7) ◇A∕A is not a weak rule of T.
(8) □A∕A is not a weak rule of KB. ┤
3.4.11 Prove that D and B have countable infinite modal symbol
strings that are not equivalent to each other. ┤
3.4.12 Prove that KG has countable infinite modal symbol strings that
are not equivalent to each other. ┤
3.4.13 Prove that
(1) Modal(K5) ≥ 14, and thus Modal(K5) = 14 (by 1.3.41 (2)).
(2) Modal(K45) ≥ 10, and thus Modal(K5) = 10 (by 1.3.41 (3)).
(3) Modal(KD5) ≥ 10, and thus Modal(K5) = 10 (by 1.3.41 (4)).
(4) Modal(K4B) ≥ 10, and thus Modal(K5) = 10 (by 1.3.41 (5)).
185
(5) Modal(KD45) ≥ 6, and thus Modal(K5) = 6 (by 1.3.41 (6)). ┤
3.4.14 Prove that □A → □B∕A → B is not validity-preserving
w.r.t. Fra(T). ┤
Definition 3.4.15 A modal system S is local finite ⇔ Given any
sentence symbols q1, …, qn, finite formulas can be generated only
from q1, …, qn such that the formulas are not equivalent to each other
in S. ┤
3.4.16 Prove that Systems K, D, T and B are not local finite.
[Note that PC is local finite.] ┤
3.4.17 Prove that S4.3.2 ⊂ S4.4 (refer to 1.5.7 (2)). ┤
186
Chapter 4 Canonical Model and
Completeness Theorem
The canonical model method (is also called Henkin-method) is a
important method proving completeness theorems in modal logics.
This thought in the back of the method is to offer the components of
constructing a model by ML itself. For example, constructing the
model, all maximal consistent sets are used as possible worlds, a
relation defined between associated formula sets is used as an
accessibility relation.
In this chapter, the major result we will prove is: for all (consistent
normal) system S, there is the canonical model MS for S such that
(Ⅰ) Val(MS) = Th(S), ①
(Ⅱ) The canonical frame FS of MS belongs to Fra(ϕ),
where ϕ is a conjunction of all the frame conditions the character
axioms ② of S correspond to.
By ( Ⅰ ) and ( Ⅱ ), we will show that the following frame
completeness theorem:
(☆) Val(Fra(ϕ)) ⊆ Th(S).
Proof. Given any A ∈ Val(Fra(ϕ)), by (Ⅱ), we have A ∈ Val(FS), so
①
②
By (Ⅰ), it is easy to see that MS is a very special model.
As mentioned before, don’t consider Axiom K.
187
A ∈ Val(MS), and thus A ∈ Th(S) by (Ⅰ). ①
So, now, in order to prove (☆), it suffices to show that (Ⅰ) and (Ⅱ)
hold. We can uniformly and one-off prove (Ⅰ) (refer to below 4.2.3),
so what is left just prove (Ⅱ) respectively w.r.t. different systems.
In§1 we will present the completeness definition, and then prove
Completeness Lemma.
In§2 we will introduce the canonical model for a system, prove
Canonical Model Elementary Theorem and (Ⅰ) above, and herefrom
we get easily that all (consistent normal) systems are model complete.
In§3 we will show that the basic systems have frame completeness
theorems.
In§4 we will give out or prove the characterization theorems for
other important systems by the results in the previous section.
In§5 we will show that the characterization theorem for the
generalization system KG
①
mn
jk
.
Note that we just need Val(MS) ⊆ Th(S).
188
§1 Completeness Definition
In this section we will present the completeness definition, and then
prove Completeness Lemma.
Definition 4.1.1 Let C be a model class or a frame class. A modal
system S is (weak) complete w.r.t. C ⇔ Val(C) ⊆ Th(S). ┤
Lemma 4.1.2 ((Weak) Completeness Equivalence Lemma) Let C
be a model class or a frame class, and let S be a modal system. Then
the following two propositions are equivalent:
(1) Val(C) ⊆ Th(S).
(2) Given any A ∈ ML, A is S-consistent ⇒ A ∈ Sat(C).
Proof. (1) ⇒ (2): Assume that (1) holds. Given any A ∈ ML such that
A is S-consistent. By Consistency Lemma 2.2.2(7), ¬A ∉ Th(S), so,
by (1), ¬A ∉ Val(C). By 3.1.10(2), A ∈ Sat(C).
(2) ⇒ (1): Assume that (2) holds. Given any A ∈ Val(C). By
3.1.10(1), ¬A ∉ Sat(C). By (2), ¬A is S-inconsistent, so, by 2.2.2(9),
A ∈ Th(S). ┤
Remark. (Ⅰ) Since ‘A is S-consistent’ is the abbreviation of ‘{A} is
S-consistent’, it follows that (2) cannot be simply written as CS(S) ⊆
Sat(C).
(Ⅱ) By the previous result, in order to prove that S is complete
w.r.t. C, it suffices to show that for all A ∈ ML,
A is S-consistent ⇒ A satisfiable in C.
In the following we shall often do so.
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Definition 4.1.3
(1) A model class M characterizes S, denoted as M ► S or S ◄ M,
⇔ Th(S) = Val(M).
(2) A frame class F characterizes S, denoted as F ► S or S ◄ F,
⇔ Th(S) = Val(F).
(3) A system S is model class complete
⇔ there is a nonempty model class M such that M ► S.
Such a M is also called a characterization model class for S.
If M = {M}, then S is called model complete.
Such a M is called a characterization model for S.
(4) A system S is frame class complete ⇔ there is a nonempty frame class F such that F ► S.
Such a F is called a characterization frame class for S.
If F = {F}, then S is called frame complete.
Such a F is called a characterization frame for S. ┤
Remark. By the concept of frame class complete, we can define a
kind of ‘absolute’ completeness and incompleteness:
Definition 4.1.4
(1) S is complete ⇔ S is frame class complete.
(2) S is incomplete ⇔ S is not frame class complete. ┤
Exercises 4.1
4.1.5 Prove that S is complete ⇔ S ◄ Fra(S). ┤
4.1.6 (Problem about Strong Completeness Lemma) Let C be a
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model class or a frame class.
(1) Consider the following two propositions, does one of
imply another? does one of them be equivalent to another?
① For all Φ ⊆ ML and A ∈ ML, Φ ⊨ C, P A ⇒ Φ ⊢ S A. ①
② For all Φ ⊆ ML, if Φ is satisfiable in C, ② then Φ
consistent.
(2) Consider the following two propositions, does one of
imply another? does one of them be equivalent to another?
① For all Φ ⊆ ML and A ∈ ML, Φ ⊨ C, P A ⇒ Φ ⊢ S A.
② For all Φ ⊆ ML, if Φ is satisfiable in C, then Φ
consistent. ┤
them
is Sthem
is S-
4.1.7 (Problem) Let ∅ be the empty model class or the empty frame
class.
(1) Is a consistent system complete w.r.t. ∅?
(2) Is an inconsistent system complete w.r.t. ∅? ┤
4.1.8 (Problem) Can the empty model class or the empty frame class
characterize a consistent system? Can they characterize an
inconsistent system? ┤
①
②
Φ ⊨ C, P A is defined as 3.3.28.
Refer to 3.1.6 (3) and 3.1.8 (2).
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§2 Canonical Model and Elementary Theorem
In this section we will introduce the canonical model for a system,
prove Canonical Model Elementary Theorem and (Ⅰ) before, and
herefrom we get easily that all (consistent normal) systems are model
complete.
By 2.2.9(3), we have
⊢ S A ⇔ ∀w ∈ MCS(S) (A ∈ w).
To prove the proposition (Ⅰ) in front of this chapter, we will
construct the canonical model <WS, RS, VS> for S such that it has
correspondingly the the following property:
⊢ S A ⇔ ∀w ∈ WS (w ∈ VS(A)) for all A ∈ ML
The idea occurs to us by the previous expression: WS should be
defined as MCS(S), simultaneously, VS should have the following
property:
w ∈ VS(A) ⇔ A ∈ w for all w ∈ WS and A ∈ ML.
But by Canonical Model Elementary Theorem we will prove below, it
suffices to define VS as:
w ∈ VS(pn) ⇔ pn ∈ w for all w ∈ WS and pn ∈ At.
Now, what we will do is to define RS. RS is required to have the
following property:
□A ∈ w ⇔ ∀u ∈ WS (wRS u ⇒ A ∈ u) for all w ∈ WS.
By 2.2.14(1), it suffices to define RS as follows:
－
wRS u ⇔ □ w ⊆ u.
By the above idea, we introduce the following definition:
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Definition 4.2.1
(1) FS is the canonical frame for S ⇔ FS = <WS, RS> such that
① WS = MCS(S), ① and
－
② wRS u ⇔ □ w ⊆ u for all w, u ∈ WS.
(2) MS is the canonical model for S ⇔ MS = <FS, VS> such that
③ VS is a mapping from At into ℘(WS) such that
w ∈ VS(pn) ⇔ pn ∈ w for all w ∈ WS and pn ∈ At. ┤
Remark. (Ⅰ) By the above ③,
VS(pn) =‖pn‖S for all pn ∈ At. ②
(Ⅱ) We often omit the subscript ‘S’ if no confusion will arise.
(Ⅲ) By 2.2.30, if |At| = ℵ0, then the cardinality of WS is 2ℵ0.
(Ⅳ) It is easy to see that the canonical frame F = <W, R> for S and
the canonical model M = <W, R, V> for S are both unique. But there
can be other models on F.
(Ⅴ) Note that in the canonical model for S, maximal consistent
sets provide the bridge between syntax and semantics that facilitates
the completeness theorems we shall prove after.
Theorem 4.2.2 (Canonical Model Elementary Theorem) Let M =
<W, R, V> is the canonical model for S. Then for all A ∈ ML,
(1) A ∈ w ⇔ w ∈ V(A) for all w ∈ W.
(2) V(A) =‖A‖.
Proof. (1) By induction on A.
Case 1 A = pn ∈ At: by the definition of V.
①
Since the definition of model requires that each set of possible worlds is
nonempty, so, in the front of Chapter 3, we presuppose that S is consistent
(refer to 2.2.9(1)).
②
Refer to 2.2.8.
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Case 2 A = ¬B or A = B ∧ C: By Maximal Consistent Set Lemma
2.2.7 (1) - (2) and the induction hypotheses.
Case 3 A = □B: we have
□B ∈ w
－
⇔ ∀u ∈ W (□ w ⊆ u ⇒ B ∈ u)
by 2.2.14(1)
⇔ ∀u ∈ W (wRu ⇒ B ∈ u)
by Definition 4.2.1
by the induction hypothesis,
⇔ ∀u ∈ W (wRu ⇒ u ∈ V(B))
by the Truth Set Definition
⇔ w ∈ V(□B).
(2) is gotten immediately by (1). ┤
Remark. (Ⅰ) 2.2.14(1) is the key to the proof of Canonical Model
Elementary Theorem (refer to Remark below 2.2.14).
(Ⅱ) Canonical Model Elementary Theorem means: ‘to belong to’
w.r.t. a modal system is exactly ‘to be satisfiable’ w.r.t. the semantics.
More loosely, Canonical Model Elementary Theorem means:
To belong to is exactly to be point-true.
Corollary 4.2.3 Let M be the canonical model for S. Then
Th(S) = Val(M),
and thus
M is the characterization model for S.
Proof. Let M = <W, R, V>.
‘⊆’: Assume that A ∈ Th(S), then by Maximal Consistent Set
Lemma 2.2.7(3), A is in each w ∈ W, namely, A ∈ ∩W. So, by
Canonical Model Elementary Theorem, w ∈ V(A) for all w ∈ W,
namely, V(A) = W, and thus A ∈ Val(M).
‘⊇’: Assume that A ∉ Th(S), then, by Consistency Lemma 2.2.2(8),
{¬A} is S-consistent, so, by Lindenbaum-Lemma, there is some w ∈
W such that ¬A ∈ w, by 2.2.7(1), A ∉ w, by Canonical Model
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Elementary Theorem, w ∉ V(A), so A ∉ Val(M). ┤
Remark. The above corollary show that the set of all formula valid
in the canonical model for S is exactly Th(S). Thus we get (Ⅰ) in
front of this chapter. This result also shows that
each non-theorem of S is not valid in the canonical model for S.
So we have
Corollary 4.2.4
(1) The canonical model for S is a S-countermodel of A ∉ Th(S).
(2) (Model Completeness Theorem) S is model complete. ┤
Remark. By the previous result, we proved that
all (consistent normal) system is model complete.
In next section, we will show that some familiar systems are still
frame (class) complete. ┤
Exercises 4.2
4.2.5
(1) (Blackburn, de Rijke and Venema, , p199) Prove that
each system S is strong complete w.r.t. its canonical model M in the
following sense: each S-consistent set is satisfiable in M.
(2) The result in (1) shows that there is a implication relation
between which provability consequence relation and which semantics
consequence relation (entailment relation)? ┤
4.2.6 Let <W, R> be the canonical frame for S. Then for all w ∈ W,
(1) □A ∈ w ⇔ ∀u ∈ W (wRu ⇒ A ∈ u).
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(2) □A ∈ w and wRu ⇒ A ∈ u.
(3) ◇A ∈ w ⇔ ∃u ∈ W (wRu and A ∈ u).
(4) ◇A ∈ w ⇒ ∃u ∈ W (wRu).
(5) A ∉ u and wRu ⇒ □A ∉ w.
(6) A ∈ u and wRu ⇒ ◇A ∈ w.
(7) □A ∉ w ⇔ ∃u ∈ W (wRu and A ∉ u).
(8) ◇A ∉ w ⇔ ∀u ∈ W (wRu ⇒ A ∉ u). ┤
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§3 Completeness Theorem
In this section we will show that the basic systems have frame
completeness theorems.
Lemma 4.3.1 ((Weak) Completeness Lemma) Let MS and FS be the
canonical model for S and the canonical frame for S, respectively.
(1) Let M is a model class and MS ∈ M. Then Val(M) ⊆ Th(S).
(2) Let F is a frame class and FS ∈ F. Then Val(F) ⊆ Th(S).
Proof. (1) Given any A ∈ Val(M). Since MS ∈ M, it follows that A ∈
Val(MS), so A ∈ Th(S) by corollary 4.2.3.
(2) Given any A ∈ Val(F). Since FS ∈ F, it follows that A ∈
Val(FS), and thus A ∈ Val(MS), so A ∈ Th(S) by 4.2.3. ┤
Remark. By the previous lemma, in order to show that S is
complete w.r.t. a frame class Fra(ϕ), it suffices to show that the
canonical frame FS = <WS, RS> for S belongs to Fra(ϕ), namely,
prove (Ⅱ) in front of this chapter. For this, it suffices to verify
RS has the corresponding property ϕ.
Lemma 4.3.2
(1) <WK, RK> ∈ Fra.
(2) <WD, RD> ∈ Fra(seriality),
(3) <WT, RT> ∈ Fra(reflexivity),
(4) <WB, RB> ∈ Fra(reflexivity, symmetry),
(5) <WS4, RS4> ∈ Fra(reflexivity, transitivity),
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(6) <WS5, RS5> ∈ Fra(reflexivity, euclideanness).
Proof. (1) By the definition of the canonical frame, it holds clearly.
(2) Verify that RD is serial. Given any w ∈ WD, by D(1) of 1.3.7 and
Maximal Consistent Set Lemma 2.2.7(3), ◇(p → p) ∈ w, by 4.2.6(4),
we have wRDu.
[We can also prove it directly: Given any w ∈ WD, by Lindenbaum－
Lemma, it suffices to show that □ w is D-consistent. Hypothesize
－
－
that □ w is not D-consistent, then there are some A1, …, An ∈ □ w
such that, in D, ⊢ ¬(A1 ∧…∧ An), and thus ⊢ A1 ∧…∧ An → ⊥, by
RK,
⊢ □A1 ∧…∧ □An → □⊥.
Since □A1, …, □An ∈ w, it follows that □⊥∈ w. On the other
hand, it is easy to see by 1.3.30 (Ⅱ) that ¬□⊥ ∈ w, contradicting the
D-maximal consistency of w.]
(3) Verify that RT is reflexive. Given any w ∈ WT, it is easy to see
that for all A ∈ ML, □A → A ∈ w, so, by 2.2.7(6)①,
□A ∈ w ⇒ A ∈ w for all A ∈ ML.
－
So it is easy to see that □ w ⊆ w, and thus wRT w.
(4) Verify that RB is reflexive and symmetric. By (3), it suffices to
show that RB is symmetric: Given any w, u ∈ WB, assume that wRBu,
we will show that uRBw. For this, it suffices to show that
① □A ∈ u ⇒ A ∈ w for all A ∈ ML.
We argue via contraposition: Assume that A ∉ w, then ¬A ∈ w. Since
¬A → □¬□A ∈ Th(B), it follows that □¬□A ∈ w. By wRBu and
4.2.6(2), ¬□A ∈ u, and thus □A ∉ u.
(5) Verify that RS4 is reflexive and transitive. By (3), it suffices to
show that RS4 is transitive: Given any w, u, v ∈ WS4, assume that
wRS4u and uRS4v. We will show that wRS4v. For this, it suffices to
show that
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② □A ∈ w ⇒ A ∈ v for all A ∈ ML.
Given any A ∈ ML such that □A ∈ w. Since □A → □□A ∈
Th(S4), it follows that □□A ∈ w. Since wRS4u, it follows that □A
∈ u, hence, by uRS4v, A ∈ v.
(6) Verify that RS5 is reflexive and euclidean. By (3), it suffices to
show that RS5 is euclidean: Given any w, u, v ∈ WS5, assume that
wRS5u and wRS5v. We will show that uRS5v. For this, it suffices to
show that
③ □A ∈ u ⇒ A ∈ v for all A ∈ ML.
Given any A ∈ ML such that □A ∈ u. Then ◇□A ∈ ◇+ u. Since
wRS5u, by 2.2.11(1), it is easy to see that ◇+ u ⊆ w, and thus ◇□A
∈ w. Since
◇□A → □A ∈ Th(S5),
it follows that □A ∈ w. Since wRS5v, it follows that A ∈ v. ┤
By 4.3.2 and 4.3.1(2) as well as the Remark below 4.3.1, it is easy
to see that:
Theorem 4.3.3 (Frame Completeness Theorem)
(1) Val(Fra) ⊆ Th(K).
(2) Val(Fra(seriality)) ⊆ Th(D).
(3) Val(Fra(reflexivity)) ⊆ Th(T).
(4) Val(Fra(reflexivity, symmetry)) ⊆ Th(B).
(5) Val(Fra(reflexivity, transitivity)) ⊆ Th(S4).
(6) Val(Fra(reflexivity, euclideanness)) ⊆ Th(S5). ┤
Using the completeness theorems, we can still prove other results.
Definition 4.3.4 (Chagrov, Zakharyaschev,  (p.16)) A modal
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system S is a 0-reducible system ⇔ given any A ∉ Th(S), there is a
substitution instance A* of A such that A* is a constant formula (refer
to 1.3.8) and A* ∉ Th(S). ┤
Theorem 4.3.5 (Chagrov, Zakharyaschev,  (p. 88)) K is not 0reducible.
Proof. Given the model
u¬p ← w¬p → vp,
and A = □(□⊥ → p) ∨ □(□⊥ → ¬ p). Then, by 3.1.13(1),
u ⊨ □⊥ ∧ ¬ p and v ⊨ □⊥ ∧ p,
and thus
w ⊨ ◇(□⊥ ∧ ¬ p) ∧ ◇(□⊥ ∧ p),
so w ⊭ A, and hence A ∉ Th(K) by Countermodel Lemma 3.4.2(1).
On the other hand, given any substitution instance of A
A* = □(□⊥ → CF) ∨ □(□⊥ → ¬CF)
such that CF is a constant formula. It is easy to see that A* is a
constant formula, and thus it suffices to show that A* ∈ Th(K).
Hypothesize that A* ∉ Th(K). By the completeness theorem
4.3.3(1), A* is not valid in some frame F. So there is some w0 ∈ F and
the valuation V on F such that w0 ⊭ A*, and thus
w0 ⊨ ◇(□⊥ ∧ ¬CF) ∧ ◇(□⊥ ∧ ¬¬CF),
and thus there are some w1, w2 ∈ R(w0) such that
w1 ⊨ □⊥ ∧ ¬CF and w2 ⊨ □⊥ ∧ ¬¬CF.
This shows that F has two dead points such that the constant formula
CF is true at one dead point and false at another dead point,
Exercises 4.3
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4.3.6 Prove that the following frame completeness theorems:
(☆) Val(Fra(ϕ)) ⊆ Th(K + A), where
(1) A = MV: □p ∨ ◇□p and
ϕ = ‘each world is a dead point or accesses to a dead point’.
(2) A = w5: p → ◇p → □◇p and
ϕ = ∀xyz . xRy ∧ xRz → zRy ∨ zRx.
(3) A = w5#: p → ◇¬p → □◇¬p and
ϕ = ∀xy . xRy ∧ x ≠ y → ∀z . xRz → zRy.
[Hint: first prove that ◇□p → q → □(p ∨ q) ∈ Th(K + w5). ]
(4) A = p → □◇◇p and ϕ = ∀xy . xRy → yR2x.
(5) A = ◇□p → ◇p and ϕ = ∀xy . xRy → ∃z . xRz ∧ yRz.
(6) A = ◇□⊥ ∨ (p → □◇p) and
ϕ = ‘x accesses to a dead point or (xRy ⇒ yRx) ’. ┤
4.3.7 Prove that the following frame completeness theorems:
(☆) Val(Fra(ϕ)) ⊆ Th(T + A), where
(1) A = Mk: □(□□p → □q) → □p → q and
ϕ = ∀x∃y . xRy ∧ yRx ∧ ∀z . yR2z → xRz.
(2) A = □p → □◇□p and
ϕ = ∀xy . xRy → ∃z . yRz ∧ ∀z1 . zRz1 → xRz1.
(3) A = p ∧ q ∧ ◇(p ∧ ¬q) → □p and
ϕ = ‘each world at most accesses to another world except it’. ┤
4.3.8 Prove that the following frame completeness theorem:
(☆) Val(Fra(ϕ)) ⊆ Th(K4 + A), where
(1)* A = Lem0: □(p ∧ □p → q) ∨ □(q ∧ □q → p) and
ϕ = ∀xyz . xRy ∧ xRz ∧ y ≠ z → yRz ∨ zRy.
(2)* A = p → □(◇p → p) and
201
ϕ = ∀xyz . xRy ∧ yRz → x = y ∨ y = z.
(3) A = ◇(p ∧□q) → □(p ∨ ◇q) and
ϕ = ∀xyz . xRy ∧ xRz ∧ y ≠ z → ∃x0 . yRx0 ∧ zRx0
(weak convergence). ┤
4.3.9 Prove that <WS5, RS5> ∈ Fra(equivalent). ┤
4.3.10 Let S be a (consistent normal) system, and ⊨ A mean ∅ ⊨ A.
Do the following (strong) completeness hold?
(1) Φ ⊨ A ⇒ Φ ⊢ K A.
(2) Φ ⊨ A ⇒ Φ ⊢ S A.
(3) Φ ⊨ A ⇒ Φ ⊢ K A.
(4) Φ ⊨ A ⇒ Φ ⊢ S A.
(5) ⊨ A ⇒ ⊢ K A.
(6) ⊨ A ⇒ ⊢ S A.
(7) Φ ⊨ F A ⇒ Φ ⊢ K A.
(8) Φ ⊨ F A ⇒ Φ ⊢ S A.
(9) Φ ⊨ F A ⇒ Φ ⊢ K A.
(10) Φ ⊨ F A ⇒ Φ ⊢ S A. ┤
4.3.11(Problem) When does the weak completeness imply the strong
completeness, namely, when does (1) imply (2)?
(1) ⊨ A ⇒ ⊢ S A.
(2) Φ ⊨ A ⇒ Φ ⊢ S A where Φ ≠ ∅. ┤
4.3.12 Is there a strong soundness and a strong completeness w.r.t. the
following systems and frame classes, respectively? If there is, which?
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How many?
(1) K and Fra.
(2) D and Fra(seriality).
(3) T and Fra(reflexivity).
(4) B and Fra(reflexivity, symmetry).
(5) S4 and Fra(reflexivity, transitivity).
(6) S5 and Fra(reflexivity, euclideanness). ┤
4.3.13* (Problem) Are the other basic systems 0-reducible? ┤
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§4 Characterization Theorem
In this section we will give out or prove the characterization theorems
for other important systems by the results in the previous section.
Theorem 4.4.1 (Characterization Theorem) Let S be a basic system,
Axiom(S) the set of all character axioms ① of S, and ϕ a conjunction
of all first-order formulas of corresponding to the axioms respectively
in Axiom(S). Then S ◄ Fra(ϕ).
Proof. By Frame Soundness Theorem (3.3.15) and Frame
Completeness Theorem (4.3.3). ┤
Remark. (Ⅰ) Canonical Model Elementary Theorem shows that S
is complete w.r.t. the canonical frame FS = <W S, R S>. But, Canonical
Model Elementary Theorem can not show that S is sound w.r.t. FS,
since such a soundness requires:
(☆) Th(S) is valid in all model on FS, namely, FS is a S-frame.
Now, as far as I can see, no each system has the property (☆), so we
say that Characterization Theorem cannot be obtained directly from
Canonical Model Elementary Theorem.
(Ⅱ) We can see from the previous result that there is an important
difference between the canonical model and the canonical frame: let S
be a system, the canonical model for S is a characterization model for
S, but this cannot infer that the canonical frame F for S is a
characterization frame for S, since it is not necessary that the
①
By 3.3.10(1), we need not consider Axiom K.
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canonical frame for S is a S-frame. Of course, if it is (namely, it is a
so-called canonical system), then F must be a characterization frame
for S (refer to 4.4.22).
(Ⅲ) A system S can be characterized by different model classes,
also by different frame classes (Sometimes, there are a lot of such
classes). For example, T can be characterized by the class of all
reflexive frames, as what shall be showed in the next chapter, T can
still be characterized by the class of all finite reflexive frames.
Corollary 4.4.2
(1) The largest characterization model class for S is Mod(S).
(2) Let S be a canonical system, namely, the canonical frame for S
is a S-frame. Then the largest characterization frame class for S is
Fra(S).
Proof. (1) First we will show that
① Mod(S) ► S.
Since Mod(S) ⊨ Th(S), it follows that
② Th(S) ⊆ Val(Mod(S)).
It is clear that MS ∈ Mod(S), so, by 4.3.1,
③ Val(Mod(S)) ⊆ Th(S).
By ② and ③, we have ①.
Finally, we will prove the largestness. Given any characterization
model class M for S. It suffices to show that M ⊆ Mod(S).
Given any M ∈ M, since M ► S, it follows that M ⊨ Th(S), and
thus M ∈ Mod(S).
(2) This proof resembles (1). We leave the details of the proof as an
exercise. ┤
In the following we further realize an important property of the
canonical frame for a modal system.
205
Let R ⊆ W2, by 3.2.7(3),
R is not the universal relation
⇔ there are some w, u ∈ W such that ~ uRw.
Lemma 4.4.3 Let S be a modal system. The canonical frame for S is
not universal frame.
Proof. Let <W, R> be the canonical frame for S. We will show that
there are some w, u ∈ W such that ~ uRw, For this, we will prove a
stronger proposition:
① Given any w ∈ W, there is some u ∈ W such that ~ uRw.
Given any w ∈ W. Let
p,
p ∈ w;
A=
¬p, p ∉ w.
It is easy to see that
② ¬A ∉ w.
First we will show that
③ ◇A ∉ Th(S).
Hypothesize that ③ doesn’t hold.
Case 1 ◇p ∈ Th(S): By US, ◇¬T ∈ Th(S), by LMC, we have
¬ □ T ∈ Th(S). By Normality, □ T ∈ Th(S), contradicting the
consistency of S.
Case 2 ◇¬p ∈ Th(S): By US, ◇¬T ∈ Th(S). As Case 1, we
Thus we get ③.
By ③ and Consistency Lemma 2.2.2(8), {¬◇A} is S-consistent.
By Lindenbaum-Lemma, there is some u ∈ W such that ¬◇A ∈ u.
Since ¬◇A → □¬A ∈ Th(S), it follows that by Lemma 2.2.7(5),
－
－
□¬A ∈ u, so ¬A ∈ □ u. By ②, it is easy to see that □ u ⊈ w, and
206
hence ~ uRw. ┤
By the previous lemma, it is easy to prove:
Corollary 4.4.4
(1) There is some formula A such that ◇A ∉ Th(S).
(2) S5 is characterized by the class of all non-universal equivalent
frames. ① ┤
In the following we will prove Characterization Theorems for the
degenerate system Tr and Ver.
Definition 4.4.5 Given any frame <W, R> and w ∈ W.
(1) w is an exactly-reflexive world ⇔ ∀u ∈ W (wRu ⇔ u = w).
(2) <W, R> is an exactly-reflexive frame ⇔ each w ∈ W is an
exactly-reflexive point.
(3) <W, R> is a dead point frame ⇔ each w ∈ W is a dead point,
namely,
∀u ∈ W . ~ wRu. ┤
Remark. (Ⅰ) It is easy to see that exactly-reflexive points and dead
points are first-order definable, by the denotation of the set theory, we
more simply have:
w is an exactly-reflexive point ⇔ R(w) = {w},
w is a dead point ⇔ R(w) = ∅,
<W, R> is a dead point frame ⇔ R = ∅.
(Ⅱ) exactly-reflexive points and dead points are two types of very
Note that the part of soundness of the corollary is not a corollary of
Lemma 4.4.3, another proof must be given out, but it is easy.
①
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special points: each of them does not access to any other point.
Conversely, the points not accessing to any other point are just them.
Lemma 4.4.6
(1) The canonical frame for Tr is an exactly-reflexive frame.
(2) The canonical frame for Ver is a dead point frame.
Proof. (1) It suffices to show that the canonical frame <W, R> for Tr
satisfies
∀xy . xRy ↔ x = y. ①
For this, first we will show that
－
① □ w ⊆ u ⇒ w = u for all w, u ∈ W.
－
Assume that □ w ⊆ u. By the anti-chain property of maximal
consistent sets 2.2.7(7), it suffices to show that w ⊆ u. Given any A ∈
w. Since A ↔ □A is a theorem of Tr, it follows that □A ∈ w. Since
－
□ w ⊆ u, it follows that A ∈ u.
Next we will show that
－
② w = u ⇒ □ w ⊆ u for all w, u ∈ W.
－
Assume that w = u. Given any A ∈ □ w. Then □A ∈ w. Since w =
u, it follows that □A ∈ u. Since A ↔ □A is a theorem of Tr, it is
easy to see that A ∈ u.
(2) Let F = <W, R> be the canonical frame for Ver. Hypothesize
that there are some w, u ∈ W such that wRu. Since a theorem □⊥ of
Ver is in w, it follows that ⊥∈ u, on the other hand, by Maximal
Consistent Set Lemma 2.2.7(3), ¬⊥∈ u, contradicting the S-maximal
consistency of u. ┤
Definition 4.4.7 Given any frame F = <W, R>.
①
Note that ∀xy . xRy ↔ x = y is equivalent to ∀x . xRx ∧∀xy . xRy → x = y.
208
(1) R is cohesive ⇔ R satisfies the following condition:
(ϕ) ∀x∃y . x ≠ y ∧ ∃n < ω . xRny ∨ yRnx. ①
Otherwise, R is called non-cohesive.
(2) F is cohesive ⇔ R is cohesive.
Otherwise, F is called non-cohesive. ┤
Remark. R is non-cohesive
⇔ ∃w ∈ W∀u ∈ W (w ≠ u ⇒ ∀n < ω (~ wRnu and ~ uRnw)).
By the previous lemma, we have
Corollary 4.4.8 The canonical frame for Tr and the canonical frame
for Ver are not cohesive frame. ┤
In the following we will prove Characterization Theorem for Tr
and Ver.
Theorem 4.4.9 (Characterization Theorem)
(1) Tr ◄ Fra (∀xy . xRy ↔ x = y).
(2) Ver ◄ Fra(∀xy . ¬xRy).
Proof. Soundness Theorems for Tr and Ver hold clearly. By 4.4.6, it
is easy to get Completeness Theorems for Tr and Ver hold. ┤
It is easy to prove that
Theorem 4.4.10 (Frame Soundness Theorem)
(1) Tr is sound w.r.t. the class of all single reflexive point frames.
Refer to the definition of Rn in 3.2.7. Note that ϕ is not a first-order
formula. Such a formula is usually called a weak second-order formula.
①
209
(2) Ver is sound w.r.t. the class of all single dead point frames. ┤
As far as we know, there is not a system for which the canonical
frame is a single element frame (a single point frame), since the
cardinality of the possible world set of the canonical frame is 2ℵ0 (as
usual, let |At| = ℵ0, refer to 2.2.30). But we have proved that each
world of the canonical frame for Tr is exactly-reflexive, and thus does
not access any other point. So if A is valid in all model on a single
reflexive point frame, then A is valid in the frame consisting of all
reflexive points, and thus A is valid in the canonical frame for Tr, so,
by the corollary 4.2.3 of Canonical Model Elementary Theorem, A is
a theorem of Tr.
Similarly, we have proved that each world of the canonical frame
for Ver is a dead point. So if A is valid in the all model on a single
dead point frame, then A is valid in the frame consisting of all dead
point, and thus A is valid in the canonical frame for Ver, so, by the
corollary of Canonical Model Elementary Theorem, A is a theorem of
Ver.
So, without loss of generality, we can take the canonical frame for
Tr and the canonical frame for Ver as the collection of the
corresponding single point frames, and thus we have a
characterization theorem in the sense of the following ‘weakenss’:
Theorem 4.4.11 (Weak Characterization Theorem)
(1) Tr is characterized by a single reflexive point frame
<{w}, <w, w>>.
(2) Ver is characterized by a single dead point frame <{w}, ∅>. ┤
Remark. (Ⅰ) looking on Tr and Ver from the above point of view,
we understand why they are degenerate systems. Since, w.r.t. the
210
above two single point frames, we can not describe exactly the logical
meaning of modal symbols.
It is easy to see that in the sense of isomorphism, there are exactly
two different single point frames, one of them is a single reflexive
point frame, and another is a single dead point frame. In other words,
there are exactly two types of frames such that in such a frame, each
world does not access to another world, and thus such a frame can be
seen as a single point frame whereas not influence the truth-value of
any formula.
Now, from the point of view of the model theory, we understand
why there are exactly two consistently degenerate ways for a system,
and thus understand more profoundly the associated results of the
proof theory given in§4 of Chapter 2.
(Ⅱ) Now, as far as I can see, Ver is not only characterized by the
canonical frame F for Ver, but is also characterized by a single dead
point frame G. But there is an important difference between F and G:
there is a model on F (namely, the canonical model for Ver)
characterizes Ver, whereas there is not a model on G characterizing
Ver, since, let <{w}, ∅, V> be any model on a single dead point
frame <{w}, ∅>, for all p ∈ At, we have
V(p) = {w} or V(¬p) = {w},
and thus hypothesize that <{w}, ∅, V> characterizes Ver, then p or
¬p is a theorem of Ver. By US, it is easy to prove that ⊥ is a
theorem of Ver, contradicting Consistency of Ver.
The case of Tr is also similar. Tr is characterized by the canonical
model for Tr, but is not characterized by any model on a single
reflexive point frame.
Now we can say that we understand more comprehensively and
deeply the degenerate systems Tr and Ver.
211
In the following we consider Characterization Theorem for S4family. For this, first we will prove a lemma:
Lemma 4.4.12 Let <W, R> be the canonical frame for S. Then for all w
∈ W,
w is an end point ① ⇔ {A → □A | A ∈ ML} ⊆ w.
Proof. ‘⇐’: Assume that A → □A ∈ w for all A ∈ ML, and wRu. We
will show that w = u.
Hypothesize that w ≠ u, then w ⊈ u by the anti-chain property of
maximal consistent sets, and thus there is some B ∈ w such that B ∉ u.
Since B → □B ∈ w, it follows that □B ∈ w. Since wRu, it follows
that B ∈ u, and thus we have a contradiction.
‘⇒’: Assume that there is some A ∈ ML such that A → □A ∉ w.
Then A ∈ w but □A ∉ w. So, by the latter and 4.2.6 (7), there is
some u ∈ W such that wRu and A ∉ u, and thus w ≠ u by A ∈ w. In
this way, we show that there is some u ∈ W such that wRu and w ≠ u,
and thus w is not an end point. ┤
Theorem 4.4.13
(1) The relation of the canonical frame for S4.1 satisfies finality:
∀x∃y . xRy ∧ ∀z . yRz → z = y.
(2) The relation of the canonical frame for S4.2 satisfies
convergence:
∀xyz . xRy ∧ xRz → ∃x0 . yRx0 ∧ zRx0.
(3) The relation of the canonical frame for S4.3 satisfies forward
connectedness:
∀xyz . xRy ∧ xRz → yRz ∨ zRy.
Proof. (1) Let <W, R> be the canonical frame for S4.1. We will show
①
By 3.2.7(2), w ∈ W is an end-point ⇔ for all u ∈ W (wRu ⇒ u = w).
212
that R satisfies finality: given any w ∈ W, there is some u ∈ W such
that
① wRu, and
② ∀v ∈ W (uRv ⇒ v = u).
Given any w ∈ W. First we will show that
－
③ Φ = □ w ∪ {A → □A | A ∈ ML} is S4.1-consistent.
Hypothesize that ③ does not hold. Then there are some A1, …, An ∈
－
□ w and B1, …, Bm ∈ ML such that, in S4.1,
⊢ ¬(A1 ∧…∧ An ∧ (B1 → □B1) ∧…∧ (Bm → □Bm)),
by RPC,
⊢ A1 ∧…∧ An → ¬((B1 → □B1) ∧…∧ (Bm → □Bm)),
by RK and LMC of S4.1,
⊢ □A1 ∧…∧ □An → ¬◇((B1 → □B1) ∧…∧ (Bm → □Bm)),
Since □A1, …, □An ∈ w, it follows that
¬◇((B1 → □B1) ∧…∧ (Bm → □Bm)) ∈ w,
by 1.5.1(4), contradicting w is S4.1-consistent. So ③ holds.
By ③ and Lindenbaum-Lemma, there is some u ∈ W such that u
－
⊇ Φ. So we have □ w ⊆ u, and thus wRu. Since for all A ∈ ML, A
→ □A ∈ u, by the previous lemma, it is easy to see that u is an end
point.
(2) Let <W, R> be the canonical frame for S4.2. Given any w, u, v
∈ W such that wRu and wRv. It suffices to show that
① there is some w0 ∈ W such that uRw0 and vRw0.
－
－
Let Φ = □ u ∪ □ v. In order to prove ①, by Lindenbaum-Lemma,
it suffices to show that
② Φ is S4.2-consistent.
Hypothesize that ② does not hold. Then there are some A1, …, An ∈
－
－
□ u and B1, …, Bm ∈ □ v such that, in S4.2,
213
⊢ ¬(A1 ∧…∧ An ∧ B1 ∧…∧ Bm),
Let A = A1 ∧…∧ An and B = B1 ∧…∧ Bm. Then by Theorem R of S4.2,
it is easy to prove that there are some □A ∈ u and □B ∈ v such that
⊢ ¬(A ∧ B), and thus ⊢ A → ¬B. By Rules RM◇ and LMC of S4.2,
③ ⊢ ◇A → ¬□B.
Since wRu, by 2.2.11(1), ◇+ u ⊆ w, so we have ◇□A ∈ w by □A
∈ u. By (a substitution instance of) Axiom G, □◇A ∈ w. Since wRv,
it follows that ◇A ∈ v, so ¬□B ∈ v by ③, contradicting □B ∈ v
and the S4.2-consistency of v, and thus ② holds.
(3) Let <W, R> be the canonical frame for S4.3. Given any w, u, v
∈ W such that wRu and wRv. We will show that
(☆) uRv or vRu.
－
－
Hypothesize that the result doesn’t hold. Then □ u ⊈ v and □ v ⊈
u, and thus there are some A, B ∈ ML such that
① □A ∈ u,
② A ∉ v,
③ □B ∈ v, and
④ B ∉ u.
By ① and ④ as well as Maximal Consistent Set Lemma 2.2.7(6)①,
⑤ □A → B ∉ u.
By ② and ③ as well as 2.2.7(6)①,
⑥ □B → A ∉ v.
By wRu and ⑤ as well as 4.2.6 (5),
⑦ □(□A → B) ∉ w.
By wRv and ⑥ as well as 4.2.6 (5),
⑧ □(□B → A) ∉ w.
By ⑦ and ⑧ as well as 2.2.7(2)②,
214
⑨ □(□A → B) ∨ □(□B → A) ∉ w,
contradicting any substitution instance of Axiom Lem is in w. ┤
Remark. ( Ⅰ ) Please, by the previous theorem, prove the
corresponding characterization theorem.
(Ⅱ) In the proofs of (2) and (3) above, when we verify R is
convergent and forward connected, except the theorems and derived
rules of K, just use (substitution instances of) G and Lem, and do not
use Axiom T or 4. This shows that System KG is complete w.r.t. the
class of all convergent frames, and System KLem is complete w.r.t.
the class of all forward connected frames.
(Ⅲ) The solution of 3.2.14 is that McK does not correspond to any
first-order formula, and thus does not correspond to finality especially.
How do you understand the unification of the results and (1) above?
As for Strong Characterization Theorem for a system, we do not
detailedly discuss it. The problem is more complex, as there are a lot
of provability consequences and semantics consequences. Some of
them have the strong completeness, some have the strong soundness,
and other not.
The implication relations we have showed or let the reader consider
between provability consequences and semantics consequences can be
seen as the discussions of these aspects (strong completeness and
strong soundness). For example, if 3.3.24(4) and 4.3.10(1) all holds,
then we have the following strong characterization theorem:
Φ ⊨ A ⇔ Φ ⊢ K A.
In the following we more generally prove Strong Characterization
Theorem for a system.
215
Definition 4.4.14
(1) Φ entails pointwise A w.r.t. S, denoted as Φ ⊨ S, P A or Φ ⊨ S A,
⇔ for all S-model M and w ∈ M (w ⊨ Φ ⇒ w ⊨ A).
(2) Φ entails modelwise A w.r.t. S, denoted as Φ ⊨ S, M A,
⇔ for all S-model M (M ⊨ Φ ⇒ M ⊨ A).
(3) Φ entails framewise A w.r.t. S, denoted as Φ ⊨ S, F A,
⇔ for all S-frame F (F ⊨ Φ ⇒ F ⊨ A). ┤
Remark. Let X ∈ {P, M, F}. It is easy to see that
Φ ⊨ X A ⇔ Φ ⊨ K, X A.
This shows that ⊨ S, X is a generalization of ⊨ X.
Lemma 4.4.15
(1)
① Φ ⊨ S, P A ⇔ Φ ∪ Th(S) ⊨ P A.
② Φ ⊨ S, M A ⇔ Φ ∪ CA(S) ⊨ M A where CA(S) is the set of all
substitution instances of all character axioms of S.
③ Φ ⊨ S, F A ⇔ Φ ∪ Axiom(S) ⊨ F A.
(2) Φ ⊨ S, P A ⇒ Φ ⊨ S, M A ⇒ Φ ⊨ S, F A.
Proof. (1) ① holds clearly. By 3.3.7(2), we get easily ②. By
3.3.8(2), we get easily ③.
(2) By (1) and 3.1.20. ┤
Definition 4.4.16 Let □ωΦ be defined as 2.1.9 such that Φ is Sconsistent. MS,Φ = <WS,Φ, RS,Φ VS,Φ> is the canonical model
containing Φ for S ⇔ the following conditions are satisfied:
(1) WS,Φ = {w ∈ MCS(S) | □ωΦ ⊆ w}.
,
216
(2) wRS,Φ u ⇔ □ w ⊆ u for all w, u ∈ WS,Φ.
(3) V S,Φ is a mapping from At into ℘(WS,Φ) such that:
w ∈ VS,Φ(pn) ⇔ pn ∈ w for all w ∈ WS,Φ and pn ∈ At. ┤
－
We often omit the subscript ‘S,Φ’ if no confusion will arise.
Lemma 4.4.17 Let W = WS,Φ and R = RS,Φ be defined as above. Then
for all w ∈ W and formula A,
(☆) □A ∈ w ⇔ ∀u ∈ W (wRu ⇒ A ∈ u).
Proof. ‘⇒’: It holds clearly.
‘⇐’: Assume that
① wRu ⇒ A ∈ u for all u ∈ W.
Let □ωΦ be defined as Definition 2.1.9. First we will show that
－
② □ωΦ ∪ □ w ⊆ u ⇒ A ∈ u for all u ∈ W.
－
－
Given any u ∈ W such that □ωΦ ∪ □ w ⊆ u. So □ w ⊆ u. By the
definition of R and ①, we have A ∈ u, and thus ② holds.
Next we will show that :
－
③ □ωΦ ∪ □ w ⊢ S A.
－
If □ωΦ ∪ □ w is S-inconsistent, then we have ③ by 2.2.17(3).
－
－
Assume that □ωΦ ∪ □ w is S-consistent but □ωΦ ∪ □ w ⊬ S A.
－
ω
By 2.2.4(1), □ Φ ∪ □ w ∪ {¬A} is S-consistent. By Lindenbaum－
Lemma, there is a S-maximal consistent set u such that □ωΦ ∪ □ w
∪ {¬A} ⊆ u. So u ∈ W and A ∉ u, contradicting ②, and hence we
have ③.
－
By ③, there are some A1, …, An ∈ □ωΦ and B1, …, Bm ∈ □ w
such that, in S,
⊢ A1 ∧…∧ An ∧ B1 ∧…∧ Bm → A.
By RK,
217
④ ⊢ □A1 ∧…∧ □An ∧ □B1 ∧…∧ □Bm → □A.
For each 1 ≤ i ≤ n, □Ai ∈ □ωΦ ⊆ w by the constitutions of □ωΦ
and W, so
⑤ □A1 ∧…∧ □An ∈ w.
－
On the other hand, since B1, …, Bm ∈ □ w, it follows that
⑥ □B1 ∧…∧ □Bm ∈ w.
By ④ - ⑥, it is easy to see that □A ∈ w. ┤
The above theorem resembles 2.2.14(1) or 4.2.6 (1).
Theorem 4.4.18 (Elementary Theorem of the canonical model
containing Φ for S) Let M = <W, R, V> be the canonical model
containing Φ for S. Then
(☆) A ∈ w ⇔ w ∈ V(A) for all A ∈ ML and w ∈ W.
Proof. By induction on A.
Case 1 A = p ∈ At: by the definition of V.
Case 2 A = ¬B or A = B ∧ C: by 2.2.7(1) - (2) and the induction
hypotheses.
Case 3 A = □B: we have that
□B ∈ w
⇔ ∀u ∈ W (wRu ⇒ B ∈ u)
by the previous Lemma
by the induction hypothesis
⇔ ∀u ∈ W (wRu ⇒ u ∈ V(B))
⇔ w ∈ V(□B).
by Truth Set Definition ┤
Lemma 4.4.19 Let M = <W, R, V> be the canonical model containing
Φ for S. Then
(☆) M ⊨ Th(S) ∪ □ωΦ.
Proof. Since for all w ∈ W, it follows that Th(S) ∪ □ωΦ ⊆ w, by the
previous theorem, the result we will prove holds clearly. ┤
218
Remark. The above lemma shows that the canonical model
containing Φ for S is a S-model.
Theorem 4.4.20 (Popkorn,  (p.127))
The following propositions are equivalent:
(1) □ωΦ ⊢ S A.
(2) Φ ⊢ S A.
(3) □ωΦ ⊢ S A.
(4) Φ ⊨ S, M A.
(5) □ωΦ ⊨ S, M A.
(6) M ⊨ A where M is the canonical model containing Φ for S.
Proof. By 2.1.9(4), it is easy to see that (1) ⇔ (2) ⇔ (3).
As usual proof of soundness, it is easy to prove that (2) ⇒ (4), (3)
⇒ (5).
Since Φ ⊆ □ωΦ, it is easy to prove that (4) ⇒ (5).
By the previous Lemma, it is easy to prove that (5) ⇒ (6).
Finally, we will show that (6) ⇒ (1). Assume that (6). There are
two cases to consider:
Case 1 □ωΦ is S-inconsistent: by 2.2.17(3), □ωΦ ⊢ S A.
Case 2 □ωΦ is S-consistent: by (6) and Elementary Theorem of
the canonical model containing Φ for S, A belongs to each S-maximal
consistent set containing □ωΦ as a subset. By 2.2.9(2),
□ωΦ ⊢ S A. ┤
Theorem 4.4.21 (Strong Characterization Theorem) Let S be a
system. Then
Φ ⊢ S A ⇔ Φ ⊨ S, M A. ┤
219
Exercises 4.4
4.4.22 Prove that if S is a canonical system, namely, the canonical
frame F for S is a S-frame, then F is a characterization frame for S.
┤
4.4.23 Let A be the converse formula of a character axiom of a basic
system. Prove that the relation of the canonical frame for a system
containing A satisfies the frame condition A corresponds to. ┤
4.4.24 Prove that K + □p ∧ p → □□p is characterized by the class
of all frames satisfying the following condition:
(weak transitivity) ∀xyz . xRy ∧ yRz ∧ x ≠ z → xRz. ┤
4.4.25 Using the canonical model method, show that
(1) G ∈ Th(KB).
(2) G ∈ Th(T + Lem). ┤
[It suffices to show that in a canonical model,
(1) A symmetric relation is convergent.
(2) A reflexive and forward connected relation is convergent.
Refer to 4.4.13(3).]
4.4.26 Prove that Th(S + Φ) ⊆ CnS(Φ) do not hold generally. ┤
4.4.27
(1) Prove that Mod(S) ⊨ A ⇔ A ∈ Th(S).
(2) Prove that the following propositions are equivalent:
① S has a weak rule A1, …, An∕A.
220
② Mod(S) ⊨ A1 ∧…∧ An ⇒ Mod(S) ⊨ A.
(3) Do the following propositions hold?
① A∕B is a weak rule of S ⇔ (Mod(S) ⊨ A ⇒ Mod(S) ⊨ B).
② A∕B is a weak rule of S ⇔ (Fra(S) ⊨ A ⇒ Fra(S) ⊨ B). ┤
Definition 4.4.28 □ ωΦ is defined as 2.1.9. Do the following
propositions hold?
(1) Let S be a system containing Axiom 4. Then
□+ Φ ∪ Φ ⊢ S A ⇔ Φ ⊢ S A.
(2) □ωΦ ⊨ S, P A ⇔ □ωΦ ⊨ S, M A ⇔ Φ ⊨ S, M A. ┤
Definition 4.4.29 A has a derivation for US from Φ in S, denoted as
Φ ⊢ S, US A, ⇔ there are some A1, …, An = A such that for all 1 ≤ i ≤ n,
one of the following condition holds:
(1) Ai is an axiom of S;
(2) Ai ∈ Φ;
(3) There are some j, k < i such that Ai is obtained from Aj and Ak by
MP;
(4) There is some j < i such that Ai is obtained from Aj by RN;
(5) There is some j < i such that Ai is obtained from Aj by US. ┤
4.4.30 Do the following propositions hold?
(1) Φ ⊢ S, US A ⇒ Φ ⊨ S, F A.
(2)* Φ ⊨ S, F A ⇒ Φ ⊢ S, US A. ┤
4.4.31* Prove that S5 ◄ Fra(universal). ┤
221
4.4.32* (Problem) Can you find a method such that it can match each
entailment relation (a semantics consequence) to a syntax
consequence (a consequence of the form ⊢ X or ⊢ X ) such that the
corresponding characterization theorem and strong characterization
theorem hold? ┤
4.4.33 Do the following propositions hold?
(1) If Φ ⊨ A, then there is a finite set Ψ ⊆ Φ such that Ψ ⊨ A.
(2) If Φ ⊨ M A, then there is a finite set Ψ ⊆ Φ such that
Ψ ⊨ M A.
(3) If Φ ⊨ F A, then there is a finite set Ψ ⊆ Φ such that
Ψ ⊨ F A. ┤
4.4.34 Do the following propositions hold?
(1) If Φ is S-consistent, then Φ is satisfiable in Mod(S).
(2) If Φ is satisfiable in Mod(S), then Φ is S-consistent.
(3) If Φ is S-consistent, then Φ is valid in some S-model.
(4) If Φ is valid in some S-model, then Φ is S-consistent. ┤
4.4.35 Let S be a (consistent normal) system. Do the following
propositions hold?
(1) Φ ⊨ A ⇔ Φ ⊢ K A.
(2) Φ ⊨ A ⇒ Φ ⊢ K A ⇒ Φ ⊢ K A ⇒ Φ ⊨ M A ⇒ Φ ⊨ Mod A.
(3) Φ ⊨ A ⇒ Φ ⊢ K A ⇒ Φ ⊢ K A
⇒ Φ ⊨ M A ⇒ Φ ⊨ F A ⇒ Φ ⊨ Fra A.
(4) Φ ⊨ A ⇒ Φ ⊢ S A ⇒ Φ ⊢ S A
where S is a normal (may be inconsistent) system. ┤
222
§5 Characterization Theorem of KG mnjk
In this section we will show that Characterization Theorem for the
mn
generalization system KG jk .
Definition 4.5.1
mn
(1) The generalization axiom G jk is the following formula:
mn
(G jk ) ◇m□n p → □j◇k p where n, m, j, k < ω.
mn
(2) KG jk := K + G
mn
jk
. ┤
Remark. (Ⅰ) A lot of modal formulas we are familiar with are
mn
instances of G jk :
(G 01
) □p → ◇p
01
(= D),
(G ) ◇p → □p
(= Dc),
10
10
01
00
00
10
(G ) □p → p
(= T),
(G ) p → □p
(= Tc),
00
(G 11 ) p → □◇p (= B),
(G 01
) □p → □□p (= 4),
20
02
(G 10
) □□p → □p
(= 4c),
(G ) ◇p → □◇p
(= 5),
10
11
223
(G 11
) ◇□p → □◇p (= G),
11
…
mn
It is clear that the generalization power of G jk is very strong, but
there are also a lot of familiar modal formulas that are not the
instances of it, such as M, C, N, K, 5c, Bc and McK.
mn
(Ⅱ) Loosely speaking, KG jk is a generalization of basic systems
(refer to 4.5.9 below).
In this section we will show that Characterization Theorem for
mn
KG jk . Lemmon and Scott proved foremost this theorem in their
.
Consider the following first-order formula:
mn
(rg jk ) ∀xyz . xRmy ∧ xR jz → ∃x0 . yRnx0 ∧ zRkx0.
This condition is called generalization convergence or m-n-j-kconvergence: if w accesses to u by m steps and to v by j steps, then
there is some w0 such that u accesses to w0 by n steps, and v accesses
to w0 by k steps.
Lemma 4.5.2 Let M = <W, R, V> and w ∈ W. Given any n < ω.
(1) w ∈ V(□nA) ⇔ ∀u ∈ W (wRn u ⇒ u ∈ V(A)),
(2) w ∈ V(◇nA) ⇔ ∃u ∈ W (wRn u and u ∈ V(A)).
Proof. (1) By induction on n < ω.
Let n = 0. Then
w ∈ V(□0A) ⇔ w ∈ V(A)
⇔ ∀u ∈ W . w = u ⇒ u ∈ V(A)
⇔ ∀u ∈ W . wR0u ⇒ u ∈ V(A).
Let n > 0. Inductively hypothesize that the result we will show
holds for any n < ω. Then
224
w ∈ V(□n + 1A) ⇔ w ∈ V(□□nA)
⇔ ∀v ∈ W . wRv ⇒ v ∈ V(□nA)
⇔ ∀v ∈ W . wRv ⇒ ∀u ∈ W . vRn u ⇒ u ∈ V(A)
⇔ ∀v ∈ W ∀u ∈ W . wRv ⇒ vRn u ⇒ u ∈ V(A)
⇔ ∀u ∈ W ∀v ∈ W . wRv and vRn u ⇒ u ∈ V(A)
⇔ ∀u ∈ W . ∃v ∈ W (wRv and vRn u) ⇒ u ∈ V(A)
⇔ ∀u ∈ W . wRn + 1u ⇒ u ∈ V(A),
where third ‘⇔’ is by the induction hypothesis, fourth ‘⇔’ and sixth
‘⇔’ are by the following first-order valid formula, respectively:
① (α → ∀uβ) ↔ ∀u (α → β) where u is not free in α,
② ∀v (α → β) ↔ (∃vα → β) where v is not free in β .
(2) This proof resembles (1). ┤
Definition 4.5.3 (Abbreviation Definition) Given <W, R> and n < ω.
Rn(w) := {u ∈ W | wRn u}. ┤
Remark. Note that R0(w) = {w}.
Using Rn(w), Lemma 4.5.2 can be restated as:
Lemma 4.5.2# Let M = <W, R, V> and w ∈ W. Given any n < ω.
(1) w ∈ V(□nA) ⇔ Rn(w) ⊆ V(A),
(2) w ∈ V(◇nA) ⇔ Rn(w) ∩ V(A) ≠ ∅. ┤
In the following we first prove that
Theorem 4.5.4 (Correspondence Theorem) G
rg
mn
jk
.
Proof. Given any <W, R>. We will show that
225
mn
jk
corresponds to
mn
(☆) <W, R> ⊨ G jk ⇔ <W, R> ⊨ rg
‘⇒’: Assume that <W, R> ⊭
mn
rg jk
mn
jk
.
. Then there are some w, u, v ∈
W such that
(1) wRmu, wR jv, and
(2) Rn(u) ∩ Rk(v) = ∅.
Let V be a valuation on <W, R> such that
V(p) = Rn(u).
It suffices to show that
(3) w ∈ V(◇m□np), and
(4) w ∈ V(◇j□k¬p) (namely, w ∉ V(□j◇kp)).
By (1) and 4.5.2(2), it suffices to show that
(5) u ∈ V(□np), and
(6) v ∈ V(□k¬p).
By (2) and the constitution of V(p), we have
Rn(u) ⊆ V(p) and Rk(v) ⊆ V(¬p),
and thus we have (5) and (6) by 4.5.2#.
mn
‘⇐’: Assume that <W, R> ⊭ G jk . Then there is a valuation V on
<W, R> and w ∈ W such that (3) and (4) hold. We will show that
(7) there are u, v ∈ W such that (1) and (2) holds.
By (3), (4) and 4.5.2(2), we have that
(8) there are u, v ∈ W such that (1), (5) and (6) hold.
We will show that (2) holds. Hypothesize that (2) does not hold. Then
there is some w0 ∈ W such that
w0 ∈ Rn(u) ∩ Rk(v),
By (5) and (6), w0 ∈ V(p) and w0 ∈ V(¬p), and thus we have a
Remark. By the previous correspondence theorem, by an effective
226
method, we can give out the first-order formula corresponding to an
mn
instance of G jk .
Example 4.5.5 The first-order formula corresponding to p → □□p
is:
∀xyz . xRy ∧ yRz → x = z.
mn
Proof. By rg jk , we have
so
so
so
so
so
∀xyz . xR0y ∧ xR2z → ∃u . yR0u ∧ zR0u,
∀xyz . x = y ∧ xR2z → ∃u . y = u ∧ z = u,
∀xyz . x = y ∧ xR2z → y = z,
∀xz . xR2z → x = z,
∀xz . ∃y(xRy ∧ yRz) → x = z, (refer to 3.2.7(1))
∀xyz . xRy ∧ yRz → x = z. (refer to ② in the proof of 4.5.2) ┤
Remark. Note that an instance of the previous formula is ∀xy.xRy
→ x = y, and thus it is different from symmetry.
Using the above method, please write out the first-order formulas to
which the modal formulas in Remark (Ⅰ) below 4.5.1 correspond.
Theorem 4.5.6 (Frame Soundness Theorem)
mn
mn
Th(KG jk ) ⊆ Val(F(rg jk )).
Proof. By the part of ‘⇐’ of (☆) in the proof of 4.5.4. ┤
Lemma 4.5.7 Let <W, R, V> be the canonical model for a system.
Given any w, u ∈ W and n < ω.
－, n
(1) wRn u ⇔ □ w ⊆ u. ①
①
Refer to 2.2.10.
227
(2) □nA ∈ w ⇔ ∀u ∈ W . wRn u ⇒ A ∈ u.
(compare it with 2.2.14(1) or 4.2.6 (1))
+, n
n
(3) wR u ⇔ ◇ u ⊆ w.
(4) ◇nA ∈ w ⇔ ∃u ∈ W . wRn u and A ∈ u.
(compare it with 2.2.14(2) or 4.2.6 (3))
Proof. (1) Given any w, u ∈ W. By induction on n < ω. Assume that n
= 0, then wRn u is w = u, and
－
□ , 0 w = {A | □0A ∈ Φ} = {A | A ∈ w} = w.
So (1) we will show becomes
w = u ⇔ w ⊆ u.
The part of ‘⇒’ holds clearly. By the anti-chain property of maximal
consistent sets (2.2.7(7)), the part of ‘⇐’ holds clearly as well.
In the following we consider the case of n > 0.
‘⇒’: By 4.2.6 (1), by induction on, it is easy to prove the part of
‘⇒’. We leave the details of the proof as an exercise.
‘⇐’: We will show that
－, n
① □ w ⊆ u ⇒ wRnu.
Inductively hypothesize that ① holds for any w, u ∈ W and n < ω.
And assume that
－, n + 1
w ⊆ u.
② □
We will show that wRn + 1u. For this, it suffices to show that
③ there is some v ∈ W such that wRv and vRnu. (refer to 3.2.7(1))
By ② and 2.2.13 as well as the induction hypothesis (① holds for
n), it is easy to see that ③ holds.
By 2.2.28(1) and (1), we get (2) easily.
By (1) and 2.2.11(2), we have (3).
By (3) and 2.2.28(2), we get (4) easily. ┤
Theorem 4.5.8 (Frame Completeness Theorem)
228
Val(Fra(rg
mn
jk
mn
)) ⊆ Th(KG jk ).
Proof. Let <W, R> be the canonical frame for KG
show that R satisfies
mn
rg jk
mn
jk
, it suffices to
: Given any w, u, v ∈ W such that wRmu and
wR jv. We will show that
① there is some w0 ∈ W such that uRn w0 and vRk w0.
－, n
－, k
Let Φ = □ u ∪ □ v. First we will show that
mn
② Φ is KG jk -consistent.
Hypothesize that ② does not hold, then there are some A1, …, Ai ∈
□
－,
n
u and B1, …, Bh ∈ □
－,
k
v such that, in KG
mn
jk
,
⊢ A1 ∧…∧ Ai → ¬(B1∧…∧Bh).
Let A = A1 ∧…∧ Ai and B = B1 ∧…∧ Bh. Then ⊢ A → ¬B. Using k
times RM◇ and LMC,
③ ⊢ ◇kA → ¬□kB.
Since □nA1, …, □nAi ∈ u and □kB1, …, □kBh ∈ v, by Theorem R,
it follows that
④ □nA ∈ u, and
⑤ □kB ∈ v.
So, by ④ and wRmu as well as the previous Lemma (4), ◇m□nA ∈
w. Since ⊢ G
mn
jk
, it follows that □j◇kA ∈ w. Since wR jv, by the
previous lemma (2), it follows that ◇kA ∈ v. By ③ and Lemma
2.2.7(5), ¬ □ kB ∈ v, contradicting ⑤ and the KG
mn
jk
-maximal
consistency of v.
Thus we get ②, and then, by Lindenbaum-Lemma, there is some
w0 ⊇ Φ. By the construction of Φ and the previous lemma (1), it is
229
easy to see that ① holds. ┤
mn
Corollary 4.5.9 Let Γ be a finite set of formulas of the form G jk ,
then K + Γ is complete system. ┤
By 6.4.7 in Section 6, the minimal system the relation semantics
(namely, Kripke-semantics) can characterize is K. In the following we
introduce simply the indexed relation semantics which can
characterize the proper subsystems of K.
Definition 4.5.10 (Marek and Truszczynski, , p. 200 – 201)
(1) (Frame Definition) F = <W, {RA | A ∈ ML}> is an indexed
relation frame, denoted as F ∈ IFra, ⇔ the following conditions are
satisfied:
① W is a nonempty set,
② RA is a binary relation on W for all A ∈ ML.
(2) (Model Definition) M = <W, {RA | A ∈ ML}, V> is an indexed
relation model, denoted as M ∈ IMod, ⇔ the following conditions
are satisfied:
① <W, {RA | A ∈ ML}> is an indexed relation frame, and
② V is a valuation mapping from At into ℘(W).
(3) (Truth Set Definition) The truth sets of all compound formulas
in M ∈ IMod are defined inductively as follows:
① V(¬A) = W − V(A),
② V(A ∧ B) = V(A) ∩ V(B),
③ V(□A) = {w ∈ W | RA(w) ⊆ V(A)},
where RA(w) := {u ∈ W | wRAu}. ┤
Theorem 4.5.11 (Model Class Characterization Theorem) (Marek
230
and Truszczynski, , p. 209) N = PC + RN is characterized by
Imod. ┤
Adding corresponding frame conditions, we can prove
corresponding model class characterization theorems for
corresponding systems. For example, for
t = N + T,
b=N+B
tr = N + 4, and
e = N + 5,
we have the corresponding frame conditions ①as follows:
(reflexivity) ∀x . xRA x for all A ∈ ML,
(symmetry) ∀xy . xR¬□A y → yRA x for all A ∈ ML,
(transitivity) ∀xyz . xR□A y ∧ yRA z → xRA z for all A ∈ ML,
(euclideanness) ∀xyz . xR¬□A y ∧ xRA z → yRA z for all A ∈ ML.
The reader can consider corresponding frame class characterization
theorem for corresponding systems.
Exercises 4.5
4.5.12 Prove that the following rule is an inference rule of KG
(RG
mn
jk
mn
jk
:
) A → B∕◇m□nA → □j◇kB where n, m, j, k < ω. ┤
4.5.13 As the proof of 4.5.5, find out the first-order formulas to
correspond to 1-0-1-0-convergence, 0-0-1-0-convergence, and 0-2-1Strictly speaking, such conditions should be called semi-frame conditions as
they deal with formulas, and thus deal with a modal language.
①
231
0-convergence, respectively. ┤
4.5.14
(1) K + ◇□p → ◇p is characterized by the class of all frames
satisfying which frame condition?
(2) K + □p → ◇◇p is characterized by the class of all frames
satisfying which frame condition?
(3) K + ◇(□p → p) is characterized by the class of all frames
satisfying which frame condition? ┤
4.5.15 Prove that
(1) K + □l (◇m□n p → □j◇k p) is characterized by the class of
all frames satisfying the following frame condition:
∀xyzw . xR l y ∧ yR mz ∧ yR j w → ∃x0 . zR n x0 ∧ wR k x0.
(2) K + ◇l T → (◇m□n p → □j◇k p ) is characterized by the
class of all frames satisfying the following frame condition:
∀xyzw . xR l y ∧ xRm z ∧ xR jw → ∃x0 . zR n x0 ∧ wR k x0. ┤
232
Chapter 5 Finite Model Property
and Decidablilty
In this chapter, we will show that a lot of systems (including KW)
have the finite model property, and they are characterized by a finite
frame class. If these system are still finite axiomatizable, then we can
still show that they are decidable.
To prove that a system S has the finite model property, there are
usually 4 methods: the normal form method (refer to  of Zhou
Beihai (周北海)), the filtration method, the minimal canonical model
method and the algebra method. But the method logicians use most
broadly is still the filtration method, and thus, in this chapter, we
introduce mostly the filtration method, and deal with the minimal
canonical model method, since the latter is also a kind of very
convenient method, and, more importantly, the reader can compare
the differences between the canonical model and the minimal
canonical model as well as their functions.
In§1 we give out the definitions of the finite model property and
the finite frame property.
In§2 we give out Filtration Definition, prove Filtration Elementary
Theorem, and show that the finite model property and the finite frame
property are two equivalent concepts, that the basic systems have the
finite model property. Finally, we still introduce the definition of
Lemmon-filtration, and by this definition show that K4 has the finite
model property.
233
In§3 we give out the minimal canonical model method and show
by this method that the basic systems and KW have the finite model
property.
In § 4 we will discuss the relation between the finite model
property and the decidablilty of a system.
In§5 we give out a system that does not have the finite model
property.
234
§1 Finite Model Property
and Finite Frame Property
In this section we give out the definitions of the finite model property
and the finite frame property.
Definition and Abbreviation 5.1.1
(1) M is a finite model ⇔ the possible world set of M is finite.
(2) F is a finite frame ⇔ the possible world set of F is finite.
(3) In the following we use Mod f and Fra f as the class of all finite
models and the class of all finite frames, respectively. We use Mod f
(ϕ) and Fra f (ϕ) as the class of all finite models satisfying property ϕ
and the class of all finite frames satisfying property ϕ, respectively.┤
Definition 5.1.2
(1) A system S has the finite model property ⇔ there is a class M ⊆
Mod f such that M ► S.
We here also say that M characterizes finitely S.
(2) A system S has the finite frame property ⇔ there is a class F ⊆
Fra f such that F ► S.
We here also say that F characterizes finitely S. ┤
Definition 5.1.3 Given any A ∉ Th(S).
235
(1) A has a finite S-countermodel ⇔ there is a finite model M such
that M is a S-model and is a countermodel of A, namely,
① M ⊨ Th(S), and
② M ⊭ A.
(2) A has a finite S-counterframe ⇔ there is a finite frame F such
that F is a S-frame and is a counterframe of A, namely,
① F ⊨ Th(S), and
② F ⊭ A. ┤
Lemma 5.1.4
(1) S has the finite model property ⇔ each A ∉ Th(S) has a finite
S-countermodel.
(2) S has the finite frame property ⇔ each A ∉ Th(S) has a finite Scounterframe.
Proof. (1) ‘⇒’: Assume that M ⊆ Mod f and M characterizes S. Given
any A ∉ Th(S), then A ∉ Val(M), and thus there is a finite model M ∈
M such that M ⊨ Th(S) and M ⊭ A.
‘⇐’: Assume that each A ∉ Th(S) has a finite S-countermodel M.
Let M be the class of all finite S-countermodels of all non-theorems of
S:
M = {M | M is a finite S-countermodel of A for some A ∉ Th(S)}.
It is clear that M ⊆ Mod f , so it suffices to show that
(☆) S ◄ M.
Given any M ∈ M, we have M ⊨ Th(S), and thus Th(S) ⊆ Val(M).
On the other hand, given any A ∉ Th(S), by the assumption, A has a
finite S-countermodel M, and thus M ∈ M. Since M is a countermodel
of A, A ∉ Val(M), and hence Val(M) ⊆ Th(S).
(2) This proof resembles (1). We leave the details of the proof as an
236
exercise. ┤
Exercises 5.1
5.1.5 (Problem) Let S1 and S2 be two modal systems such that S1 ⊂
S2. Do the following propositions hold?
(1) If S2 has the finite model property, then S1 has the finite model
property.
(2) If S1 has the finite model property, then S2 has the finite model
property. ┤
5.1.6 Prove that 5.1.4(2). ┤
5.1.7 (Problem) Does the inconsistent system have the finite model
property? ┤
237
§2 Filtration and Finite Model Property In this section we give out Filtration Definition, prove Filtration
Elementary Theorem, and show that the finite model property and the
finite frame property are two equivalent concepts, that the basic
systems have the finite model property. Finally, we still introduce the
definition of Lemmon-filtration, and by this definition show that K4
has the finite model property.
Before introducing filtration method, we first describe intuitively
the filtration method.
As far as we know, although it is not the case that each system is
characterized by a frame class, but each such system is characterized
by some model class. This means that each A ∉ Th(S) is not valid in
some S-model <W, R, V>. (Refer to 4.2.4 (1)) Of course, this model
can be infinite model. But since the subformula set Sub(A) of A is
finite, we can partition W into finite equivalent classes w.r.t. Sub(A),
and take each equivalent class as a possible world, and thus construct
a finite model M* = <W*, R*, V*> such that A is not valid in M* as
well, and M* is still S-model. It is clear that, if we can prove it for any
A ∈ ML, then can also prove that S has the finite model property.
In the following we introduce formally the filtration method.
Definition 5.2.1
(1) Given any Φ and M = <W, R, V>. Define relation ≈Φ as follows:
238
for all w, u ∈ W,
w ≈Φ u ⇔ given any A ∈ Φ (w ∈ V(A) ⇔ u ∈ V(A)).
(2) Let |w|Φ = {u ∈ W | w ≈Φ u}.
(3) for any X ⊆ W, Let |X|Φ = {|w|Φ | w ∈ X}.
(4) |w|Φ is called the equivalence class w.r.t. Φ, ① and w is called a
representative of |w|Φ. ┤
Remark. (Ⅰ) w ≈Φ u means that w and u cannot differentiate the
truth-value of formulas in Φ.
(Ⅱ) We often omit the subscript ‘Φ’ if no confusion will arise.
It is easy to prove that
Corollary 5.2.2
(1) w ∈ |w|.
(2) |w| = |u| ⇔ w ≈ u.
(3) |w| ≠ |u| ⇒ |w| ∩ |u| = ∅.
(4) ≈ is an equivalence relation. ┤
Remark. By the previous result, it is easy to see that |W|Φ partition
W into disjoint equivalent classes w.r.t. Φ. If Φ is finite, then |W|Φ is
finite, since there are just finite different valuations for a set of finite
formulas. In the following is a more concrete result:
Lemma 5.2.3 Let Card(Φ) = n. ② Then Card(|W|Φ) ≤ 2n.
Proof. By induction on n.
Let n = 0. Then Φ = ∅. By definition, for any w, u ∈ W, we always
①
②
By 5.2.2(4) below.
Card(Φ) represents the cardinality of Φ.
239
have w ≈ u. So |W|Φ = {W} (a single element set), and hence
Card(|W|Φ) = Card({W}) = 1 ≤ 20.
Let n = 1. Then Φ = {A}. It is easy to see that W can be at most
partitioned into two equivalent classes:
|w|Φ = V(A) and |u|Φ = V(¬A).
Since one of them can be the empty set, it follows that Card(|W|Φ) ≤ 21.
Assume that Card(Φ) = Card(Ψ ∪ {A}) = k + 1 such that Card(Ψ) =
k and A ∉ Ψ. By the induction hypothesis,
Card(|W|Ψ) ≤ 2k.
Given any |w|Ψ ∈ |W|Ψ. It is easy to see that w.r.t. A, |w|Ψ can be at
most partitioned into two equivalent classes |w|Ψ,1 and |w|Ψ,2 such that
A is true at all elements in one equivalent class of them, and false at
all elements in another equivalent class of them, namely,
|w|Ψ,1 = |w|Ψ ∩ V(A) and |w|Ψ,2 = |w|Ψ ∩ V(¬A).
By the arbitrariness of |w|Ψ and Card(|W|Ψ) ≤ 2k, it is easy to see that
Card(|W|Φ) ≤ 2·2k = 2k + 1. ┤
Remark. (Ⅰ) According to the previous proof, by adding a formula
to Φ at a time, each original equivalent class can be at most
partitioned into two equivalent classes (one of them can be empty).
(Ⅱ) If Φ is infinite set, then the cardinality of |W|Φ can be infinite
or finite. Show it by giving an example, please.
By 1.1.4, we have
Sub(Φ) := ∪{Sub(A) | A ∈ Φ}.
Definition 5.2.4 Let Φ = Sub(Φ). M* = <W*, R*, V*> is a Φ-filtration
(model) of M = <W, R, V> ⇔ M* is a relation model such that the
following conditions are satisfied:
(1) W* = |W|Φ;
240
(2) R* is a binary relation on W* such that the following conditions
are satisfied (We here also say that R* is suitable): for all w, u ∈ W,
①(Minimal Condition) If wRu, then |w|R*|u|,
②(Maximal Condition) If |w|R*|u|, then
for all □A ∈ Φ (w ∈ V(□A) ⇒ u ∈ V(A));
(3) V * is a valuation mapping from At into ℘(W*) such that
V*(pn) = {|w| ∈ W* | w ∈ V(pn)} for all pn ∈ At ∩ Φ. ┤
Remark. (Ⅰ) By 5.2.1(3), the previous (3) can be simply denoted as:
(3′) V * is a valuation mapping from At into ℘(W*) such that
V*(pn) = |V(pn)| for all pn ∈ Φ. (Ⅱ) Since the above filtration is still a Kripke-relation model, it
follows that the semantic concepts defined as before (for example, the
concepts of truth set and various validities) can be directly applied to
the corresponding concepts w.r.t. filtration.
(Ⅲ) The above Φ is just like a filter, the possible world set of the
original model is filtered into some equivalent classes by Φ.
(Ⅳ) Note that R* is not unique: although M and Φ are fixed, but
there can be a lot of suitable relations R*.
(Ⅴ) In his  (p.45), Chellas said that the idea of a Φ-filtration
does not depend upon Φ being closed under subsentences. … In
general, when speaking of a Φ-filtration we presuppose that Φ is
closed under subsentences.
By 5.2.3, if Φ is finite, then every Φ-filtration of M is always finite.
In the following we will prove an important theorem:
Theorem 5.2.5 (Filtration Elementary Theorem) Let Φ = Sub(Φ).
Let <W*, R*, V*> be a Φ-filtration of a model <W, R, V>, and A ∈ Φ.
Then
241
(☆) |w| ∈ V*(A) ⇔ w ∈ V(A) for all |w| ∈ W*.
Proof. It is easy to see that A ∈ Sub(Φ). By induction on A, we will
show that (☆).
Case 1 A = pn ∈ At: then by the definition of filtration, we get
easily (☆).
Case 2 A = ¬B: then B ∈ Sub(Φ). By the induction hypothesis,
|w| ∈ V*(B) ⇔ w ∈ V(B) for all |w| ∈ W*,
and thus
|w | ∉ V*(B) ⇔ w ∉ V(B) for all |w| ∈ W*.
so
|w| ∈ V*(¬B) ⇔ w ∈ V(¬B) for all |w| ∈ W*.
Case 3 A = B ∧ C: then B, C ∈ Sub(Φ). By the induction
hypotheses,
|w| ∈ V*(B) ⇔ w ∈ V(B) for all |w| ∈ W*, and
|w| ∈ V*(C) ⇔ w ∈ V(C) for all |w| ∈ W*,
and thus for all |w| ∈ W*,
|w| ∈ V*(B) and |w| ∈ V*(C) ⇔ w ∈ V(B) and w ∈ V(C).
So
|w| ∈ V*(B ∧ C) ⇔ w ∈ V(B∧C) for all |w| ∈ W*.
Case 4 A = □B: then B ∈ Sub(Φ). We will show that
① |w| ∈ V*(□B) ⇔ w ∈ V(□B) for all |w| ∈ W*.
‘⇒’: Given any |w| ∈ W* such that w ∉ V(□B). Then by Truth Set
Definition of M, there is some u ∈ W such that wRu and u ∉ V(B). By
the induction hypothesis, |u| ∉ V*(B). Since wRu, by Minimal
Condition of R*, it follows that |w|R*|u|, and thus by the Truth Set
Definition of M*, |w| ∉ V*(□B).
‘⇐’: Given any |w| ∈ W* such that w ∈ V(□B). We will show that
② |w| ∈ V*(□B).
To prove ②, by the Truth Set Definition of M*, given any |u| ∈
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W* such that |w|R*|u|, it suffices to show that |u| ∈ V*(B). Since w ∈
V(□B), by |w|R*|u| and Maximal Condition of R*, it follows that u ∈
V(B), and thus by the induction hypothesis, |u| ∈ V*(B). ┤
Remark. In Filtration Elementary Theorem, we use Sub(Φ) as a
filter according to the following fact: whether A ∈ Φ is valid in any
model depends just on the truth-value of subformula of A, and does
not depend on the truth-value of other formulas. For example, whether
□A is true at w depends just on whether A is true at the worlds
accessible from w.
It is clear that
Corollary 5.2.6 Let Φ = Sub(Φ).
(1) Let M* be a Φ-filtration of M. Then for all A ∈ Φ,
M* ⊨ A ⇔ M ⊨ A.
(2) Let M be a model class and Fil(M, Φ) the class of all Φfiltrations of all models in M, namely,
Fil(M, Φ) = {M* | M ∈ M and M* is a Φ-filtration of M}.
Then
(☆) Fil(M, Φ) ⊨ A ⇔ M ⊨ A for all A ∈ Φ. ┤
Corollary 5.2.7
(1) Assume that M is a countermodel of A. Then each Sub(A)filtration of M is a finite countermodel of A.
(2) Each formula having a countermodel has a finite countermodel.
Proof. (1) Let M = <W, R, V>. Then by the assumption, there is some
w ∈ W such that w ∉ V(A). Assume that M* = <W*, R*, V*> is a
Sub(A)-filtration of M. Since A ∈ Sub(A), by Filtration Elementary
Theorem, it follows that there is some |w| ∈ W* such that |w| ∉ V*(A),
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and thus M* is a countermodel of A. Since Sub(A) is finite, it follows
that M* is finite.
(2) By (1). ┤
Lemma 5.2.8 Let Φ = Sub(Φ) and M* = <W*, R*, V*> be a Φfiltration of M = <W, R, V>. Then
(1) The converse of Minimal Condition implies Maximal Condition,
(2) The converse of Maximal Condition implies Minimal Condition.
Proof. (1) Assume that the converse of Minimal Condition holds.
Given any |w|R*|u| and □A ∈ Φ such that w ∈ V(□A). By |w|R*|u|
and the converse of Minimal Condition, we have wRu. By w ∈ V(□
A), we have u ∈ V(A).
(2) Assume that the converse of Maximal Condition holds, namely,
(☆) ∀□A ∈ Φ (w ∈ V(□A) ⇒ u ∈ V(A)) ⇒ |w|R*|u|.
Assume that wRu. We will show that |w|R*|u|. Since wRu, by Truth
Set Definition, it follows that
∀□A ∈ ML (w ∈ V(□A) ⇒ u ∈ V(A)).
So
∀□A ∈ Φ (w ∈ V(□A) ⇒ u ∈ V(A)).
Hence we have |w|R*|u| by (☆). ┤
Remark. By the previous result, a method of proving that R* is
suitable is to prove that Minimal Condition and its converse both hold,
another method of proving that R* is suitable is to prove that Maximal
Condition and its converse both hold.
Lemma 5.2.9 Let Φ = Sub(Φ) and M = <W, R, V> be any model.
Define W* and V* as before, and define R* on W* such that R*
satisfies one of the following conditions:
(1) |w|R*|u| ⇔ there is some w0 ∈ |w| and u0 ∈ |u| such that w0Ru0.
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(2) |w|R*|u| ⇔ ∀□A ∈ Φ (w ∈ V(□A) ⇒ u ∈ V(A)).
Then M* = <W*, R*, V*> is a Φ-filtration of M.
Proof. It suffices to show that R* is suitable.
Assume that R* satisfies the condition (1). First we will show that
R* satisfies Minimal Condition of Filtration Definition. Assume that
wRu. By 5.2.2(1), w ∈ |w| and u ∈ |u|, so |w|R*|u| by (1).
Next we will show that R* satisfies Maximal Condition of
Filtration Definition: Assume that |w|R*|u|, then by (1), there are some
w0 ∈ |w| and u0 ∈ |u| such that w0Ru0. Given any □A ∈ Φ such that w
∈ V(□A), we will show that u ∈ V(A). Since w ∈ V(□A) and w0 ∈
|w|, it follows that w0 ∈ V(□A). By w0Ru0, we have u0 ∈ V(A). Since
u0 ∈ |u| and A ∈ Φ, it follows that u ∈ V(A).
Assume that R* satisfies the condition (2). By a half of (2) (the part
of ‘⇒’), it is easy to see that R* satisfies Maximal Condition of
Filtration Definition. By another half of (2) (the part of ‘⇐’) and
5.2.8(2), it is easy to see that R* satisfies Minimal Condition of
Filtration Definition. ┤
Remark. The filtration satisfying the previous condition (1) is called
the finest filtration, in essence, it requires that Minimal Condition and
its converse hold simultaneously. The filtration satisfying the previous
condition (2) is called the coarsest filtration, it actually requires that
Maximal Condition and its converse hold simultaneously.
Theorem 5.2.10 (Canonical Model Filtration Theorem) (Segerberg,
 (p.67)) Let Φ = Sub(Φ) and M = <W, R, V> be the canonical
model for S. Assume that M* = <W*, R*, V*> is a Φ-filtration of M.
Then M* is the finest filtration of M.
Proof. It suffices to show that 5.2.9(1) holds.
The proof of ‘⇐’ of 5.2.9 (1): Since M* is a Φ-filtration of M, by
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Minimal Condition, the result we will prove holds clearly.
The proof of ‘⇒’ of 5.2.9 (1): According to Theorem 7.5 in 
(p.67) of Segerberg, we have that M* is a S-model. By Filtration
Elementary Theorem and Canonical Model Elementary Theorem, we
have
① |w| ∈ V*(A) ⇔ A ∈ w for all w ∈ W and A ∈ Φ.
Given any |w|, |u| ∈ W* such that |w|R*|u|. Let
w0 = {A ∈ ML | |w| ∈ V*(A)} and
u0 = {A ∈ ML | |u| ∈ V*(A)}.
First we will show that
② w0 and u0 are S-maximal consistent.
By symmetry, it suffices to show that w0 is S-maximal consistent.
Since w0 is satisfiable in M* and M* is a S-model, so w0 is Sconsistent by 3.3.30.
Next we will show that w0 is maximal. Given any A ∈ Φ such that
A ∉ w0. By the constitution of w0, we have |w| ∉ V*(A). By Truth Set
Definition, |w| ∈ V*(¬A), by the constitution of w0, ¬A ∈ w0.
By ②, it is easy to see that w0, u0 ∈ W. Now we will show that
③ w0 ∈ |w| and u0 ∈ |u|.
We first will show that w0 ∈ |w|, namely,
④ A ∈ w0 ⇔ A ∈ w for all A ∈ Φ.
Given any A ∈ Φ. Then
by the constitution of w0
A ∈ w0 ⇔ |w| ∈ V*(A)
by ①
⇔ A ∈ w.
By symmetry, it is easy to see that u0 ∈ |u|.
Finally, it suffices to show that
⑤ w0Ru0.
Given any □A ∈ w0. Then |w| ∈ V*(□A) by the constitution of w0.
Since |w|R*|u|, it follows that |u| ∈ V*(A), by the constitution of u0, A
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∈ u0. Thus we get that □ w0 ⊆ u0, so ⑤ holds. ┤
－
Remark. The above theorem can be expressed loosely as:
an appropriate filtration of a canonical model is the finest filtration.
The proof of the following Lemma needs other technique, so we
here do not give out its proof:
Lemma 5.2.11 (Segerberg, ) Let S be a system. If A has a finite
S-countermodel, then A has a finite S-counterframe. ┤
In the following we will show that the finite model property and the
finite frame property are two equivalent concepts.
Theorem 5.2.12 (Segerberg, , p.33, or Blackburn, de Rijke and
Venema, , p147) Let S be a system. Then
S has the finite model property ⇔ S has the finite frame property.
Proof. ‘⇒’: Assume that S has the finite model property, then for any
A ∉ Th(S), A has a finite S-countermodel, by the previous lemma, A
has a finite S-counterframe, and thus S has the finite frame property.
‘⇐’: Assume that S has the finite frame property, then for any A ∉
Th(S), A has a finite S-counterframe F. It is easy to see that there is a
model M on F such that M is a finite S-countermodel of A. ┤
Theorem 5.2.13 If a system S is characterized by a class of finite
models, then S is characterized by a class of finite frames.
Proof. By the previous lemma and Definition 5.1.2. ┤
In the following we will show that the basic systems has the finite
model property.
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First we show the clue of the proof. By the corollary 4.2.4(1) of
Canonical Model Elementary Theorem, if A ∉ Th(S), then A has Scountermodel M. So, by 5.2.7(1), each Sub(A)-filtration of M is a
finite countermodel of A. Hence
each A ∉ Th(S) has a finite countermodel.
Whether does this mean that S have the finite model property? The
answer is negative, since, for a system S having the finite model
property, by 5.1.4 (1), each A ∉ Th(S) not only has a finite
countermodel, but also, more importantly, has a finite S-countermodel.
We do not still prove this. By far, we proved only that: if M is a Scountermodel of A, then we can always construct a or many finite
model(s) M* from M (namely, Sub(A)-filtration(s) of M) such that M*
is a countermodel of A, but we do not still prove that M* is a S-model.
For a certain system S, it is not trivial to show that there is a
filtration M* for any model M of a model class characterizing S such
that M* is a S-model. Since, just like what we shall see, there is some
system S that does not have the finite model property. This means that
there is some A ∉ Th(S) such that there is not a S-model such that it is
a finite countermodel of A, although A has a infinite countermodel.
Of course, for most of systems we have met, they do have the finite
model property. Using the following method, we shall show that if M
is a model class characterizing S, then, for all model M ∈ M, there is
some filtration of M such that its frame is a S-frame. For this, it
suffices to show that the relation R* of this filtration satisfies the
frame condition(s) we require such that the frame satisfying this
condition is a S-frame. Although we cannot prove that the relation R*
of each filtration of a given model satisfies the requirement, but, just
like what we have showed, it suffices to show that there is at least a
filtration having such a property. Of course, for some system S, we
can still prove a more strong result: S is characterized by a model
class M such that the relation R* of each filtration of each model in M
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satisfies the associated the requirement. For example, Systems K, D
and T have such a property.
Lemma 5.2.14 Let Φ = Sub(Φ) and M* = <W*, R*, V*> be a Φfiltration of M = <W, R, V>. Then
(1) If M is serial, then M* is also serial,
(2) If M is reflexive, then M* is also reflexive.
Proof. (1) Assume that M is serial. Given any |w| ∈ W*, it is clear that
w ∈ W. Since R is serial, it follows that there is some u ∈ W such that
wRu, and thus by the constitution of W*, there is some |u| ∈ W*. So
we have |w|R*|u| by Minimal Condition of R*.
(2) Assume that M is reflexive. Given any |w| ∈ W*, it is clear that
w ∈ W. Since R is reflexive, it follows that wRw, so |w|R*|w| by
Minimal Condition of R*. ┤
Remark. The previous results are quite stronger, since for transitive
model, symmetric model and euclideanness model, we do not have
such results. Please show it by giving an example.
Theorem 5.2.15 (Finite Model Property Theorem) The basic
systems have the finite model property.
Proof. (1) Consider K: since each frame is a K-frame, it follows that
each filtration of any model w.r.t. Φ = Sub(Φ) is a model on some Kframe.
(2) Consider D: since D ◄ Mod(serial), it suffices to show that
each filtration M* of a serial model M w.r.t. Φ = Sub(Φ) is also serial.
But the previous lemma (1) has showed it.
(3) Consider T: since T ◄ Mod(reflexive), it suffices to show that
each filtration M* of a reflexive model M w.r.t. Φ = Sub(Φ) is also
reflexive. But the previous lemma (2) has showed it.
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(4) Consider S4: as far as I can see, S4 ◄ Mod(reflexive,
transitive), but we cannot prove that each filtration of each reflexive
and transitive model is also transitive. But is, just like what we said in
the above Remark, we do not need do so. In order to prove that S4 has
the finite model property, it suffices to show that given any S4-model
M = <W, R, V>, w.r.t. Φ = Sub(Φ), we can construct some filtration
M* = <W*, R*, V*> of M such that M* is reflexive and transitive:
① W* and V* are defined as in Filtration Definition,
② R* is defined as: for all |w|, |u| ∈ W*,
(☆) |w|R*|u| ⇔ ∀□A ∈ Φ (w ∈ V(□A) ⇒ u ∈ V(□A)).
It is easy to prove that R* is reflexive and transitive.
We will show that M* = <W*, R*, V*> is a Φ-filtration of M.
Show that R* satisfies Minimal Condition: Given any w, u ∈ W
such that wRu. By (☆), it suffices to show that the right side of (☆)
holds. Given any □A ∈ Φ such that w ∈ V(□A). By the validity of
substitution instances of Axiom 4, w ∈ V(□□A). Since wRu, it
follows that u ∈ V(□A).
Show that R* satisfies Maximal Condition: Given any w, u ∈ W
and □A ∈ Φ such that |w|R*|u| and w ∈ V(□A). By (☆), u ∈ V(□A).
By the validity of substitution instances of Axiom T, u ∈ V(A).
From the proof of S4 having the finite model property, we can see,
the key to the proof is to construct a proper R* such that:
(Ⅰ) R* is suitble in the sense of Filtration Definition, and thus
make M* such obtained be a filtration,
(Ⅱ) R* satisfies the frame conditions we require.
Please prove by the above method that B and S5 have the finite
model property. ┤
In the following we will consider a kind of more special filtrations:
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Definition 5.2.16 Let Φ = Sub(Φ). M* = <W*, R*, V*> is a LemmonΦ-filtration of M = <W, R, V> ⇔ the following conditions are
satisfied:
(1) W* and V* are defined as in 5.2.4, and
(2) for all |w|, |u| ∈ W*,
|w|R*|u| ⇔ ∀□A ∈ Φ (w ∈ V(□A) ⇒ u ∈ V(□A ∧ A)). ┤
Lemma 5.2.17 Let M be a transitive model. Then a Lemmon-Φfiltration of M is a Φ-filtration of M and it is transitive.
Proof. Let M* = <W*, R*, V*> be a Lemmon-Φ-filtration of M = <W,
R, V>. By the previous definition, it is easy to show that M* is
transitive. So below it suffices to show that M* is a Φ-filtration of M.
It is clear that R* satisfies Maximal Condition. So below it suffices
to show that R* satisfies Minimal Condition. Assume that wRu. We
will show that |w|R*|u|. Given any □A ∈ Φ such that w ∈ V(□A), it
suffices to show that u ∈ V(□A ∧ A).
Since w ∈ V(□A), by the transitivity of R, all substitution instances
of Axiom 4 are valid in M, and thus w ∈ V(□□A), so
w ∈ V(□□A ∧ □A).
Since wRu, it follows that u ∈ V(□A ∧ A). ┤
Corollary 5.2.18 K4 has the finite model property.
Proof. Given any A ∉ Th(K4). Let M = <W, R, V> be the canonical
model for K4, then M is a countermodel of A.
Let M* = <W*, R*, V*> be a Lemmon-Sub(A)-filtration of M.
Since M is transitive, it follows by the previous lemma that M* is a
Sub(A)-filtration of M such that R* is transitive, and thus
① M* ⊨ Th(K4).
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Since M ⊭ A, by the corollary 5.2.6(1) of Filtration Elementary
Theorem, it follows that
② M* ⊭ A. ┤
Now we understand more deeply that for a system, there can be a
lot of different characterization frame classes. We use System T as an
example. In Chapter 4, we showed that T ◄ Fra(reflexive) and T ◄
FT (where FT is the canonical frame for T). In this section we have
showed that T ◄ Fra f (reflexive). But we have
Fra f (reflexive) ⊂ Fra (reflexive), and
{FT} ∩ Fra f (reflexive) = ∅!
Exercises 5.2
5.2.19 Prove 5.2.2. ┤
5.2.20 Prove that Systems B and S5 have the finite model property.
┤
5.2.21 Prove that
(1) S4.1, S4.2 and S4.3 have the finite model property.
(2) MV = K + □p ∨ ◇□p has the finite model property.
[Hint: Given a countermodel of A, by it construct a Sub(A ∧ □⊥)filtration.] ┤
5.2.22 Let Φ = Sub(Φ). Prove that any Φ-filtration of a universal
model is also universal. ┤
5.2.23 Let Mod(universal) be the class of all universal models, and
Mod f (universal) the class of all finite universal models. Prove that
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Val(Mod(universal)) = Val(Mod f (universal)). ┤
5.2.24 Give out the following examples:
(1) A non-symmetric filtration of a symmetric model.
(2) A non-transitive filtration of a transitive model.
(3) A non-euclideanness filtration of a euclideanness model. ┤
5.2.25 Using Lemmon-filtration, prove that
(1) K4G has the finite model property.
(2) S4.2 has the finite model property. ┤
5.2.26 (Problem)
(1) Let Φ = Sub(Φ). Is a Φ-filtration of the canonical model for a
system a S-model?
(2) Is a Sub(Th(S))-filtration of the canonical model for a system a
S-model? ┤
5.2.27 (Small Model Theorem) Let A be satisfiable. Then A is
satisfiable in a model with no more than 2Card(Sub(A)) worlds. ┤
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§3 Minimal Canonical Model
and Finite Model Property
In this section we give out the minimal canonical model method and
show by this method that the basic systems and KW have the finite
model property.
As far as we know, the canonical model for a modal system is not a
finite model, so we cannot directly use the canonical model method to
show that the system has the finite model property. But logicians
introduced the concept of minimal canonical model, resolved this
problem.
Definition and Convention 5.3.1
(1) Sub + (A) := Sub(A) ∪ {¬B | B ∈ Sub(A)}.
(2) Let S be consistent. Φ is an A-S-maximal consistent set ⇔ Φ ⊆
Sub + (A) such that the following conditions are satisfied:
①(A-maximality) for all B ∈ Sub(A), B ∈ Φ or ¬B ∈ Φ,
②(S-consistency) for all finite set Ψ ⊆ Φ, ⊬ S ¬∧Ψ. ┤
Remark. (Ⅰ) In (2), ② is equivalent to the following condition:
②′ ⊬ S ¬∧Φ.
(Ⅱ) It is clear that Sub + (A) is finite. So, for any A-S-maximal
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consistent set Φ, Φ is finite and thus Th(S) ⊈ Φ. Since the power set
of Sub + (A) is also finite, it follows that the set of all A-S-maximal
consistent sets is also finite.
Lemma 5.3.2 Let Φ be an A-S-maximal consistent set.
(1) Assume that B ∈ Sub(A). Then exactly one of B and ¬B is in Φ.
(2) Assume that B ∧ C ∈ Sub(A). Then
B ∧ C ∈ Φ ⇔ (B ∈ Φ and C ∈ Φ).
Proof. (1) By the previous definition ①, there is at least one of B and
¬B belonging to Φ. We will show that
(☆) there is at most one of B and ¬B belonging to Φ.
Hypothesize that (☆) doesn’t hold, namely, {B, ¬B} ⊆ Φ. It is
easy to see that ⊢ S ¬(B ∧ ¬B), contradicting the S-consistency
defined in 5.3.1(2)②.
(2) ‘⇒’: Assume that B ∧ C ∈ Φ. Hypothesize that B ∉ Φ. Since B
∧ C ∈ Sub(A), it follows that B ∈ Sub(A). By B ∉ Φ and (1), ¬B ∈ Φ.
So
{B ∧ C, ¬B} ⊆ Φ and ⊢ S ¬(B ∧ C ∧ ¬B),
contradicting the S-consistency of Φ, and hence B ∈ Φ, and thus we
have a contradiction. Similarly, we can show that C ∈ Φ.
‘⇐’: Assume that B, C ∈ Φ but B ∧ C ∉ Φ. Since B ∧ C ∈ Sub(A),
it follows that ¬(B∧C) ∈ Φ by B ∧ C ∉ Φ and (1). So
{B, C, ¬(B ∧ C)} ⊆ Φ.
But
⊢ S ¬(B ∧ C ∧ ¬(B ∧ C)),
contradicting the S-consistency of Φ, and thus B ∧ C ∈ Φ. ┤
Lemma 5.3.3 (A-Lindenbaum-Lemma) Let Φ ⊆ Sub + (A) be Sconsistent, then there is some A-S-maximal consistent set Ψ ⊇ Φ.
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Proof. Construct Ψ as follows: first list the all formulas of Sub + (A):
B1, …, Bn.
Let Ψ0 = Φ. For 0 ≤ i < n, let
Ψi ∪ {Bi + 1} is S-consistent;
Ψi ∪ {Bi + 1},
Ψi + 1 =
Ψi ∪ {¬Bi + 1}, Ψi ∪ {Bi + 1} is S-inconsistent.
We will show that each Ψi is S-consistent. It is easy to see that Ψ0 is
S-consistent. Inductively hypothesize that Ψk is S-consistent. We are
going to show that Ψk + 1 is S-consistent. For this, consider the Sconsistency of Ψk ∪ {Bk + 1}.
Case 1 Ψk ∪ {Bk + 1} is S-consistent: so Ψk + 1 = Ψk ∪ {Bk + 1}, and
thus Ψk + 1 is S-consistent.
Case 2 Ψk ∪ {Bk + 1} is S-inconsistent: so Ψk + 1 = Ψk ∪ {¬Bk + 1}.
Hypothesize that Ψk + 1 is S-inconsistent, then by the assumption of
this case and the hypothesis, we have
Ψk ∪ {Bk + 1} is S-inconsistent, and
Ψk ∪ {¬Bk + 1} is S-inconsistent,
so, by Consistency Lemma 2.2.2(10) − (11), there are finite sets Θ1,
Θ2 ⊆ Ψk such that
⊢ S ∧Θ1 → ¬Bk + 1 and
⊢ S ∧Θ2 → Bk + 1.
So ⊢ S ¬∧(Θ1 ∪ Θ2), and thus Ψk is S-inconsistent, contradicting
the induction hypothesis. Thus we show that Ψk + 1 is S-consistent.
Let Ψ = Ψn. Then Ψ is S-consistent.
It is easy to show that Ψ is an A-S-maximal consistent set. ┤
In the following we will define the A-the canonical model for S.
Note that the definition of such a model does not only depend on A,
but also depend on the characteristics of S. That is, we do not have a
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general definition of a minimal canonical model:
Definition 5.3.4 Let S be a basic system such that A ∉ Th(S). MA, S =
<WA, S, RA, S, VA, S> is the A-canonical model for S (for short the
minimal canonical model for S) ⇔ the following conditions are
satisfied:
(1) WA, S is the set of all A-S-maximal consistent sets. ①
(2) Given any w, u ∈ WA, S.
If S is K, T or D, then define RA, S as follows:
－
wRA, S u ⇔ □ w ⊆ u.
If S is B, then define RA, S as follows:
－
－
wRA, S u ⇔ (□ w ⊆ u and □ u ⊆ w).
If S is S4, then define RA, S as follows:
wRA, S u ⇔ □w ⊆ u. ②
If S is S5, then define RA, S as follows:
wRA, S u ⇔ □w = □u.
(3) Given any w ∈ WA, S. For pn ∈ Sub(A), let
w ∈ VA, S (pn) ⇔ pn ∈ w;
for pn ∈ At − Sub(A), define VA, S(pn) as any subset of WA, S. ┤
Remark. (Ⅰ) We often omit the subscript ‘A, S’ if no confusion
will arise.
(Ⅱ) It is easy to see that the A-canonical model for S is a relation
model.
Theorem 5.3.5 (Minimal Canonical Model Elementary Theorem
for Basic Systems) Let S be a basic system such that A ∉ Th(S), and
①
②
As Remark (Ⅱ) below 5.3.1, the cardinality of WA, S is finite.
By 2.2.10, □w := {□A | □A ∈ w}.
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let M = <W, R, V> be the A-canonical model for S. Then for all w ∈ W,
(☆) w ∈ V(B) ⇔ B ∈ w for all B ∈ Sub(A).
Proof. Given any w ∈ W.
Case 1 B = pn ∈ Sub(A): by definition, we get immediately (☆).
Case 2 B = ¬C ∈ Sub(A): then C ∈ Sub(A). According to the
induction hypothesis and Lemma 5.3.2(1), it is easy to obtain (☆).
Case 3 B = C ∧ D ∈ Sub(A): then C, D ∈ Sub(A). According to
the induction hypotheses and Lemma 5.3.2(2), it is easy to obtain (☆).
Case 4 B = □C ∈ Sub(A): then C ∈ Sub(A).
‘⇐’: Assume that □C ∈ w: We will show that w ∈ V(□C). Given
any u ∈ W such that wRu, it suffices to show that u ∈ V(C).
If S is K, T or D, then by the property of R defined in 5.3.4, C ∈ u.
By the induction hypothesis, u ∈ V(C).
If S is S4, then by the property of R defined in 5.3.4, □C ∈ u by
□C ∈ w. First we will show that C ∈ u. Hypothesize that the result
doesn’t hold, then by the A-maximality, ¬C ∈ u, so {□C, ¬C} ⊆ u.
By the S4-consistency of u, we have ⊬ S4 ¬(□C ∧ ¬C), and thus ⊬ S4
□C → C, contradicting that all substitution instances of Axiom T are
the theorems of S4. So we have that C ∈ u. By the induction
hypothesis, u ∈ V(C).
Please show the cases of S = B and S = S5.
‘⇒’: Assume that □C ∉ w: then ¬□C ∈ w.
Assume that S ∈ {K, T, D}. First we will show that
－
① □ w ∪ {¬C} is S-consistent.
Hypothesize that the result doesn’t hold, then there are some D1, …,
－
Dn ∈ □ w such that, in S,
⊢ D1 ∧…∧ Dn → C,
by RK,
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⊢ □D1 ∧…∧ □Dn → □C,
so
⊢ ¬(□D1 ∧…∧ □Dn ∧ ¬□C).
It is easy to see that □D1, …, □Dn, ¬□C ∈ w, contradicting the Sconsistency of w defined in 5.3.1(2)②. Thus we obtained ①.
－
Since □ w ∪ {¬C} ⊆ Sub + (A), by ① and the A-LindenbaumLemma, there is some A-S-maximal consistent set u ∈ W such that
－
□ w ∪ {¬C} ⊆ u.
Since S ∈ {K, T, D}, it follows that wRu and ¬C ∈ u. By the latter, C
∉ u. Since □C ∈ Sub(A), it follows that C ∈ Sub(A), and thus by the
induction hypothesis, u ∉ V(C). By Truth Set Definition about □, w
∉ V(□C).
Assume that S is S4. First we will show that
② □w ∪ {¬C} is S4-consistent.
Hypothesize that the result doesn’t hold, then there are some □D1,
…, □Dn ∈ □w such that, in S4,
⊢ □D1 ∧…∧ □Dn → C,
by RK,
⊢ □□D1 ∧…∧ □□Dn → □C,
Since ⊢ 4, it follows by US and RPC that
⊢ □D1 ∧…∧ □Dn → □C,
and thus
⊢ ¬(□D1 ∧…∧ □Dn ∧ ¬□C).
It is easy to see that □D1, …, □Dn, ¬□C ∈ w by the constitution
of □w and the assumption, contradicting the S4-consistency of w
defined in 5.3.1(2)②. Thus we obtained ②.
Since □w ∪ {¬C} ⊆ Sub + (A), by ② and the A-Lindenbaum-
259
Lemma, there is some A-S4-maximal consistent set u ∈ W such that
□w ∪ {¬C} ⊆ u.
So wRu and ¬C ∈ u, and thus C ∉ u. Since C ∈ Sub(A), by the
induction hypothesis, it follows that u ∉ V(C), and thus w ∉ V(□C).
Please show the cases of S = B and S = S5. ┤
Now, by the minimal canonical model method, we prove newly that
the basic systems have the finite model property.
Theorem 5.3.6 (Finite Model Property Theorem) The basic
systems have the finite model property.
Proof. Let S be a basic system. Given any A ∉ Th(S), then by
Consistency Lemma 2.2.2(8), {¬A} is S-consistent. Let M = <W, R,
V> be the A-canonical model for S. Since {¬A} ⊆ Sub + (A), by the ALindenbaum-Lemma, it follows that there is some w ∈ W such that
¬A ∈ w, by Minimal Canonical Model Elementary Theorem 5.3.5, w
∉ V(A), and thus A has a finite countermodel M. Presuppose that
(☆) R satisfies the frame conditions to which the character axioms
of S correspond.
By Correspondence Theorem, it is easy to see that M ⊨ Th(S). So A
has a finite S-countermodel M. Thus S has the finite model property.
So below it suffices to verify (☆).
Consider S = K. It is easy to see that the A-canonical frame of K is
a relation frame.
Consider S = T. We will show that R is reflexive. For this, it
suffices to show that for all w ∈ W,
－
① □ w ⊆ w.
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Hypothesize that □ w ⊈ w. ① Hence □ w ≠ ∅, so there is some B
∈ Sub(A) such that □B ∈ w but B ∉ w. Since B ∈ Sub(A), it follows
that ¬B ∈ w by A-maximality, and thus {□B, ¬B} ⊆ w, by the Sconsistency defined in 5.3.1(2)②, ⊬ T ¬(□B ∧ ¬B), contradicting
⊢ T □B → B.
Consider S = D. It suffices to show that R is serial. Hypothesize
that R is not serial, then there is some w ∈ W such that there is not
－
some u ∈ W such that □ w ⊆ u, but by the A-Lindenbaum-Lemma,
－
this means that □ w is D-inconsistent. So there are some □B1, …,
□Bn ∈ w such that, in D, ⊢ ¬(B1 ∧…∧ Bn), hence, by RN, we have
⊢ □¬(B1 ∧…∧ Bn).
So, by Axiom D, we have
⊢ ◇¬(B1 ∧…∧ Bn),
and thus by LMC and Theorem R,
⊢ ¬(□B1 ∧…∧ □Bn),
contradicting □B1, …, □Bn ∈ w and the D-consistency of w.
Consider S = S4. We defined R w.r.t. S4 as follows:
wRu ⇔ □w ⊆ u.
It is clear that R is reflexive and transitive.
Please prove the cases of S = B and S = S5. ┤
－
－
The finite model property and the completeness are two properties
that are correlated nearly. The completeness can be obtained from the
finite model property:
Theorem 5.3.7 Assume that S has the finite model property. Then S is
①
Note that if □－w = ∅, then it is easy to see that ① holds.
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complete (in the sense of 4.1.4(1)).
Proof. By the assumption, given any A ∉ Th(S), there is a finite Scountermodel of A, and thus there is a finite S-counterframe of A. Let
F be the class of all S-finite frame, it is easy to see that F ► S. ┤
Finally, we will show that KW has the finite model property.
As far as we know, we cannot use the canonical model method to
prove that KW is complete. But this does not mean that it is not
complete. In the following we will use the finite model property to
prove that KW is complete.
Definition 5.3.8 Let M = <W, R, V>.
(1) R is irreflexive ⇔ ∀x ∈ W . ~ xRx. ①
(2) R is a strict partial ordering ⇔ R is transitive and irreflexive.
┤
Theorem 5.3.9
(1) A strict partial ordering model is a KW-model.
(2) KW has the finite model property.
(3) KW ◄ M f (strict partial ordering).
Proof. (1) We leave the proof as an exercise.
(2) Given any A ∉ Th(KW). It suffices to show that
(☆) A has a finite KW-countermodel.
Construct a KW-finite model <W, R, V> as follows:
① W is the set of all A-KW-maximal consistent sets;
② For w, u ∈ W, define
－
wRu ⇔ □w ∪ □ w ⊆ u and ∃□C ∈ u (□C ∉ w);
③ For pn ∈ Sub(A), define
①
Note that irreflexivity corresponds to no modal formula.
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w ∈ V(pn) ⇔ pn ∈ w,
for pn ∈ At − Sub(A), define V(pn) as any subset of W.
Please verify that M is transitive and irreflexive finite model, and
thus by (1), M is KW-model.
We will show that Minimal Canonical Model Elementary Theorem
for KW: given any w ∈ W, it suffices to show that
④ w ∈ V(B) ⇔ B ∈ w for all B ∈ Sub(A).
By the proof of 5.3.5, it suffices to consider the case of B = □C.
‘⇐’: Assume that □C ∈ w. We will show that w ∈ V(□C). Given
any u ∈ R(w). By the definition of R, C ∈ u. Since C ∈ Sub(A), it
follows that u ∈ V(C) by the induction hypothesis.
‘⇒’: Assume that □C ∉ w: then ¬□C ∈ w. Let
－
Φ = □w ∪ □ w ∪ {□C, ¬C}.
We will show that
⑤ Φ is KW-consistent.
Without loss of generality, we assume that □w = {□C1, …, □Cn}.
Hypothesize that ⑤ does not hold, then, in KW,
⊢ □C1 ∧…∧ □Cn ∧ C1 ∧…∧ Cn → □C → C,
by RK,
⊢ □□C1 ∧…∧ □□Cn ∧ □C1 ∧…∧ □Cn → □(□C → C),
by 1.5.2(Ⅰ) (1), 4 is a theorem of KW, and thus
⊢ □C1 ∧…∧ □Cn → □(□C → C),
by Axiom W,
⑥ ⊢ □C1 ∧…∧ □Cn → □C.
On the other hand, since {□C1, …, □Cn, ¬□C} ⊆ w, by the AKW-maximal consistency of w, we have
⑦ ⊬ ¬(□C1 ∧…∧ □Cn ∧ ¬□C),
contradicting ⑥. Thus we get ⑤.
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Since Φ ⊆ Sub + (A), by ⑤ and A-Lindenbaum-Lemma, there is
some A-KW-maximal consistent set u ∈ W such that Φ ⊆ u. Herefrom
it is easy to see that
－
□w ⊆ u, □ w ⊆ u, □C ∈ u.
So wRu by the above results and the assumption that □C ∉ w. On
the other hand, ¬C ∈ u by Φ ⊆ u, so C ∉ u. But C ∈ Sub(A), so u ∉
V(C) by the induction hypothesis, and thus w ∉ V(□C). Hence we get
④.
Since A ∉ Th(KW), then by Consistency Lemma 2.2.2(8), {¬A} is
KW-consistent. Since A ∈ Sub(A), by A-Lindenbaum-Lemma, there is
some w ∈ W such that ¬A ∈ w, by Minimal Canonical Model
Elementary Theorem for KW, it is easy to see that w ∉ V(A).
Thus we have proved that A has a finite KW-countermodel.
(3) We get easily the soundness by (1). The completeness is obtaind
by (2) and 5.3.7. ┤
Remark. (Ⅰ) By Minimal Canonical Model Elementary Theorem
and the previous proof, to prove by the minimal canonical model
method that S has the finite model property, the key to define a proper
relation R on the set of all A-S-maximal consistent sets such that:
(1) The frame so obtained is a S-frame,
(2) Minimal Canonical Model Elementary Theorem for S holds.
(Ⅱ) In Chapter 4, we proved Characterization Theorem for the
familiar systems by the canonical model method. In above section, we
proved by the filtration method that the systems have the finite model
property, and thus, by 5.3.7, Finite Characterization Theorem for the
systems. In this section we proved by the minimal canonical model
method that the systems have the finite model property, and thus, by
5.3.7 again, Finite Characterization Theorem for the systems. This
shows that we can obtain Characterization Theorems by different
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methods.
(Ⅲ) For the minimal canonical model method, there is a shortage:
there is not an uniform method to define the minimal canonical model
for the familiar systems, and thus there is not an uniform method
prove Minimal Canonical Model Elementary Theorem for them. But
is, for the method, there is an virtue: it is more intuitive and simpler.
Exercises 5.3
5.3.10 Complete the unproved part of 5.3.5 - 5.3.6, 5.3.9. ┤
5.3.11 Prove by the minimal canonical model method that K4 has the
finite model property and K4 is characterized by the class of all finite
transitive models. ┤
5.3.12(Rybov, ) Prove by the finite model property that
(1) □◇A → □◇B ∈ Th(K4) ⇔ ◇A → ◇B ∈ Th(S5).
(2) □◇A ∨ ◇□⊥∈ Th(K4) ⇔ ◇A ∈ Th(S5). ┤
5.3.13*(Rybov, ) Let S ⊇ K4 has the finite model property, and
A be the formula obtained by substituting the formulas of the form
□◇B for all sentence symbols in some formula C. Prove that S + A
has the finite model property. ┤
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§4 Finite Model Property and Decidablilty
In this section we will discuss the relation between the finite model
property and the decidablilty of a system.
Definition 5.4.1 A system S is decidable ⇔ there is an effective
program such that, given any A ∈ ML, it can determine in finite steps
whether A is a theorem for S. ┤
Remark. The concept of decidability above can be interpreted as:
given any A ∈ ML, for the question ‘whether A belongs to Th(S)’,
there is an arithmetic that can answer ‘yes’ or ‘no’ in finite steps. This
arithmetic is similar to an ideal computer. If the question ‘whether A
belongs to Th(S)’ is input in it, then, in finite steps, the operation
about the question would stop finally, the computer would output the
By studying, it is found that there is a certain connection between
the finite model property and decidablilty. Concretely say, together
finite model property and finite axiomatizability ① constitutes a
sufficient condition of decidablilty: if a system S has the finite model
property and is finite axiomatizable, then S is decidable.
Definition 5.4.2 A modal system S is finite axiomatizable ⇔ there is
a finite set of axioms Φ ⊆ ML and a finite set of inference rules R
such that S = <Φ; R>. ┤
①
Refer to the following definition.
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Lemma 5.4.3 Let R be a set of finite inference rules, and Φ = {A1, …,
An, …} such that for all 1 ≤ i < ω, the following independence
condition holds:
Φn = {A1, …, An} ⇒ An + 1 ∉ Th(<Φn; R>).
Then S = <Φ; R> is not finite axiomatizable. ┤
In the following we consider only the finite axiomatizable problem
of normal systems, so the previous definition can restated as:
Definition 5.4.4 A (normal) system S is finite axiomatizable ⇔ S = K
+ Γ such that Γ is finite. ┤
Theorem 5.4.5 (Decidablilty Theorem) Let S be a system such that
(1) S is finite axiomatizable, and
(2) S has the finite model property.
Then S is decidable.
Proof. As far as we know, a formal proof of each theorem of S is a
finite long formula sequence, whereas ML is countable infinite, so the
set of all finite subsets of ML is also countable infinite, and thus the
set of all formal proofs of all theorems of S is also countable infinite,
and hence all the formal proofs can be listed effectively as the
following sequence:
① Pr0, …, Prn, ….
Since S is finite axiomatizable, by 5.4.4, there is a finite set Γ such
that S = K + Γ. So, by 3.3.8(2), for any frame F,
(#) F ⊨ Th(S) ⇔ F ⊨ Γ.
If F is still a finite frame, then there is a finite (naturally, is also
267
effective) program to check whether the finite formulas in Γ are valid
in F, and thus, by (#), to determine whether F is S-frame. It is clear
that, if we not consider isomorphic frames, then there is an effective
program generates all finite frames with some fixed order, and hence
we can enumerate all finite frames as follows:
② F0, …, Fn, ….
In the following we give out an effective program such that for all
A ∈ ML, in finite steps, for the question whether A is a theorem of S,
give out an answer ‘yes’ or ‘no’:
Start: to 0th step.
nth step where n < ω:
If n is an even number, ① then check whether Prn∕2 is a formal
proof of A in S. If is, then enter into the following Stop 1, otherwise,
enter into n + 1th step.
If n is an odd number, then check whether F(n − 1) ∕ 2 is a Scounterframe of A. If is, then enter into the following Stop 2,
otherwise, enter into n + 1th step.
Stop 1: stop further execution and answer ‘yes’.
Stop 2: stop further execution and answer ‘no’.
Hence the execution track of the above program is:
→Pr0, Pr1, Pr2, …
↓↗↓↗↓↗↓
F0, F1, F2, ….
By the above program, if A ∈ Th(S), then a formal proof of A
would finally occur in ①, namely, the program can finally find it in
①. If A ∉ Th(S), then since S has the finite model property, then A
has a finite S-counterframe, and thus this frame would finally occur in
②, namely, the program can finally find it in ②.
①
0 is an even number.
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For each A ∈ ML, whether A ∈ Th(S) or A ∉ Th(S). So, whether a
formal proof of A would occur in ① in finite steps, or a finite Scounterframe of A would occur in ② in finite steps (Of course, this
two cases would not occur simultaneously). So the above program
would stop finally by finite step executions, and thus output the
answer ‘yes’ or ‘no’, and then determine whether each formula in ML
is a theorem of S. ┤
Remark. (Ⅰ) that a system S has been proved to be decidable does
not mean that given actually any formula, it can be decided whether it
belongs to Th(S). Usually, we should still, by special properties of S,
devise some operable methods to decide whether a formula belongs to
Th(S). For example, we can decide this by the semantics map method.
The reader can refer to  of Zhou Beihai (周北海).
(Ⅱ) There are some systems that have the finite model property but
are not decidable. Refer to  of Iskrd.
( Ⅲ ) ‘finite model property + finite axiomatizable’ is just a
sufficient condition of decidablilty, but is not a necessary condition.
In fact, there are some finite axiomatizable systems that are decidable
but have no the finite model property.
(Ⅳ) There is also some decidable system that have the finite model
property but is not finite axiomatizable. Refer to  of Cresswell.
Exercises 5.4
5.4.6 Prove 5.4.3. ┤
5.4.7*(Rybov, ) Let S ⊇ K4 be decidable, and A be the formula
obtained by substituting the formulas of the form □◇B for all
269
sentence symbols in some formula C. Prove that S + A is also
decidable. ┤
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§5 A System without
the Finite Model Property
In this section we give out a system that does not have the finite
model property.
In his , Minson proved that there is a system that does not
have the finite model property. This system, now, is denoted as Mk =
T + Mk where
(Mk) □(□□p → □q) → □p → q.
Theorem 5.5.1 (Correspondence Theorem) Mk corresponds to the
following first-order formula:
(mk) ∀x∃y . xRy ∧ yRx ∧ ∀z . yR2z → xRz.
Proof. Given any tuple <W, R>.
Assume that <W, R> ⊭ mk. Then
(1) There is some w ∈ W for any u ∈ W, if wRu and uRw, then there
is some v ∈ W such that uR2v and ~ wRv.
Given any u ∈ W such that wRu and uRw. Then
(2) There is some v ∈ W such that uR2v and ~ wRv.
Let V be a valuation on a modal frame <W, R> such that
(3) V(p) = R(w), and
(4) V(q) = W − {w}.
By (3),
(5) w ⊨ □p.
By (3) and ~ wRv in (2), v ⊨ ¬p, so, by uR2v in (2), u ⊭ □□p, and
271
thus u ⊨ □□p → □q. By wRu and the arbitrariness of u,
(6) w ⊨ □(□□p → □q).
But by (4), w ⊭ q, and thus <W, R> ⊭ Mk by (5) and (6).
Assume that <W, R> ⊭ Mk. Then there is a valuation V on a modal
frame <W, R> and w ∈ W such that
(7) w ⊨ □(□□p → □q),
(8) w ⊨ □p, and
(9) w ⊭ q.
We will show that (1) holds. Given any u ∈ W such that wRu and uRw,
then by wRu and (7),
(10) u ⊨ □□p → □q.
By (9) and uRw, u ⊭ □q, and thus by (10), u ⊭ □□p, so there is
some v ∈ R2(u) such that v ⊭ p. By (8), ~ wRv, so (1) holds, and thus
<W, R> ⊭ mk. ┤
Theorem 5.5.2 (Characterization Theorem for Mk)
Mk ◄ Fra(mk, reflexive).
Proof. Soundness Proof. By the previous correspondence theorem.
Completeness Proof. (By 4.3.7 (1). In the following we will give
out the outline of the proof.) Assume that <W, R> is the canonical
frame for Mk. As usual, it suffices to show that R satisfies mk. Given
－
any w ∈ W, assume that Φ = □ w ∪ ◇+ w ∪ □□w is not Mkconsistent, then by Theorem R, there are some □A, B1, …, Bn, □D
∈ w such that, in Mk,
⊢ ¬(A ∧ ◇B1 ∧…∧ ◇Bn ∧ □□D),
By RPC,
⊢ A → □□D → □¬B1 ∨…∨ □¬Bn,
272
By 1.2.22(Ⅰ) (2), it is clear that
⊢ A → □□D → □(¬B1 ∨…∨ ¬Bn),
By RM,
⊢ □A → □(□□D → □(¬B1 ∨…∨ ¬Bn)),
Since □A ∈ w, it follows that
□(□□D → □(¬B1 ∨…∨ ¬Bn)) ∈ w,
and thus by Mk,
□D → ¬B1 ∨…∨ ¬Bn ∈ w.
Since □D ∈ w, ¬B1 ∨…∨ ¬Bn ∈ w, so ¬(B1 ∧…∧ Bn) ∈ w,
contradicting B1, …, Bn ∈ w and the Mk-maximal consistency of w.
Thus we showed that Φ is Mk-consistent. So, by Lindenbaum－
Lemma, there is some u ∈ W such that Φ ⊆ u. Since □ w ⊆ u, it
follows that
⑧ wRu.
Since ◇+ w ⊆ u, it follows that
⑨ uRw.
Finally, we will show that
⑩ uR2v ⇒ wRv for all v ∈ W.
Given any v ∈ W such that uR2v. We will show that wRv. Given any
□A ∈ w. Since □□w ⊆ u, it follows that □□A ∈ u, so A ∈ v.
－
Thus we have that □ w ⊆ v, and thus wRv. Hence we showed ⑩.
By ⑧ - ⑩, it is easy to see that R satisfies mk.
In the following we will show that Mk does not have the finite
model property:
Definition 5.5.3 Let <W, R> be a frame, and 3 ≤ n < ω.
w1, …, wn ∈ W is a length n non-transitive chain ⇔ the following
conditions are satisfied:
273
(1) wi ≠ wj for 1 ≤ i ≠ j ≤ n,
(2) wi Rwi + 1 for 1 ≤ i < n,
(3) ~ w1Rwi for 2 < i ≤ n. ┤
Remark. A length n ≥ 3 non-transitive chain can be denoted
intuitively as
w1 → w2 → w3 →…→ wn,
where w1 except accesses to w2 (and possibly accesses to itself), does
not accessing any other world in the chain.
Definition 5.5.4 (Hughes-Cresswell, )
<W, R> is a recession frame ⇔ the following conditions are
satisfied:
(1) W = ω is the set of all natural numbers,
(2) R is a relation on W such that for all w, u ∈ W,
wRu ⇔ w ≤ u + 1. ┤
Remark. The above (2) means: each natural number n accesses to
itself, to the direct predecessor of n, and to each natural number larger
than n.
Lemma 5.5.5 A recession frame is not transitive.
Proof. Let <W, R> be a recession frame. Since 2R1 and 1R0 but ~ 2R0
(as 2 ' 1), it follows that R is not transitive. ┤
Theorem 5.5.6 Mk does not have the finite model property.
Proof. First we will show that :
(1) each finite reflexive Mk-frame is transitive.
Let <W, R> be a finite reflexive Mk-frame.
We will show that R is transitive. Hypothesize that it is not, then
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there is a length n ≥ 3 non-transitive chain in the finite set W: w1, …,
wn. Since W is finite, without loss of generality, we can assume that
there is not a non-transitive chain longer than the above chain in W.
Let <W, R, V> be a model such that p ∈ At is false at w3, …, wn,
and true at any other worlds; q ∈ At is false at w1, and true at any
other worlds:
ο
ο
ο
ο
w1 p, ¬q →w2 p, q →w3 ¬p, q →…→ wn ¬p, q.
It is easy to see that
(2) w1 ∉ V(□p → q).
Given any w ∈ W such that w1Rw and w1 ≠ w. Then we have
ο
ο
ο
ο
ο
w p, q ← w1 p, ¬q →w2 p, q →w3 ¬p, q →…→ wn ¬p, q.
Consider the truth of □(□□p → □q) at w1.
Case 1 ~ wRw1: Since q is false at w1 only, it follows that w ∈
V(□q), thus w ∈ V(□□p → □q).
Case 2 there is some 1 < i ≤ n such that wRwi: then w ∉ V(□□p)
(when i = n, we need reflexivity), so w ∈ V(□□p → □q).
Case 3 (not (Case 1 or Case 2)) wRw1 and for all 1 < i ≤ n, ~
wRwi: then w, w1, …, wn constitute a length > n non-transitive chain,
and thus we have a contradiction, so it is impossible that this case
occurs.
Hence, for all w ∈ R(w1), w ∈ V(□□p → □q), and thus
w1 ∈ V(□(□□p → □q)).
So, by (2), it is easy to see that w1 ∉ V(Mk), and thus we have a
contradiction. So <W, R> is transitive, and thus we have (1).
Next we will show that
(3) 4 ∉ Th(Mk).
For this, we will construct a reflexive and not transitive infinite Mkframe <W, R> such that 4 is not valid in it.
Let <W, R> be a recession frame. Hypothesize that Mk is false at
275
some n ∈ W, then
(4) n ∈ V(□(□□p → □q)),
(5) n ∈ V(□p), and
(6) n ∉ V(q).
For each k ≥ n − 1, we have n ≤ k + 1, and thus nRk, by (5), k ∈ V(p).
So it is easy to show that for all j ≥ n, R(j) ⊆ V(p), thus j ∈ V(□p),
and hence
(7) n + 1 ∈ V(□□p).
So, by (4) and nRn + 1, n + 1 ∈ V(□□p → □q), hence, by (7),
(8) n + 1 ∈ V(□q),
Since n + 1Rn, by (8), it follows that n ∈ V(q), contradicting (6). Thus
we get that <W, R> ⊨ Mk, and thus <W, R> is Mk-frame by 3.3.8(2).
By 5.5.5, a recession frame is not transitive, so 4 is not valid in it.
(In fact, it suffices to let p be false at 0 and true at other worlds, and
thus can see that 4 is not true at 2.) So (3) holds.
Finally, we will show that
(9) Mk does not have the finite model property.
Hypothesize that (9) doesn’t hold. By 5.2.12, Mk has the finite frame
property. So, by 5.1.2(2),
(10) there is a class F ⊆ Fra f such that F ► Mk.
By the constitution of Mk = T + Mk and (1), it is easy to see that all
the frames in F are transitive, so
F ⊨ □p → □□p.
By (10), it is easy to show that 4 ∈ Th(Mk), contradicting (3). So (10)
does not hold, and thus (9) holds. ┤
Exercises 5.5
276
5.5.7* Prove that T + □(□□p → □□□p) → □p → □□p does
not have the finite model property. ┤
277
Chapter 6 Neighborhood Semantics
From the above chapters, we can see, the relation semantics cannot
describe proper subsystems of K logically, because, from the point of
view of logical systems, we actually have E ⊂ M ⊂ R ⊂ K. But by
the relation semantics we cannot prove it and the corresponding
Characterization Theorem. Do Systems E, M and R have the
soundness theorem and completeness theorem w.r.t. which semantics?
In this chapter we introduce the neighborhood semantics to solve
the above problem. Of course, we can also introduce other semantics,
such as the algebra semantics, the probability semantics and the game
semantics.
In § 1 we will introduce the neighborhood semantics, prove
Correspondence Theorem for the neighborhood semantics, Frame
Soundness Theorem and Proper Inclusion Theorem for proper
subsystems of K w.r.t. the neighborhood semantics. Finally, we shall
give out three types of important neighborhood frame transformations:
supplementation, intersection closure and quasi-filtering.
In§2 we will show that Canonical Model Elementary Theorem and
Characterization Theorem for proper subsystems of K w.r.t. the
neighborhood semantics.
In § 3 we define filtrations of neighborhood frames and
neighborhood models, prove Filtration Elementary Theorem and
Finite Characterization Theorem for proper subsystems of K w.r.t. the
neighborhood semantics, and thus obtain the finite model property of
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proper subsystems of K.
In § 4 we will show the relation between the neighborhood
semantics and the relation semantics. We will show that the relation
semantics is a subsemantics of the neighborhood semantics, the class
of all normal neighborhood frames is equivalent to the class of all
relation frame, and E and K are the widths of the neighborhood
semantics and the relation semantics, respectively. Finally, we will
show that compared with the relation semantics, the neighborhood
semantics has larger characterization width and larger normal
characterization depth.
279
§1 Neighborhood Semantics
In this section we will introduce the neighborhood semantics, prove
Correspondence Theorem for the neighborhood semantics, Frame
Soundness Theorem and Proper Inclusion Theorem for proper
subsystems of K w.r.t. the neighborhood semantics. Finally, we shall
give out three types of important neighborhood frame transformations:
supplementation, intersection closure and quasi-filtering.
Definition 6.1.1
(1) (Frame Definition) F = <W, N> is a neighborhood frame,
denoted as F ∈ FN, ⇔ the following conditions are satisfied:
① W is a nonempty set: W ≠ ∅, and
② N is a mapping from W into ℘(℘(W)), namely, N is a
mapping from W into the set of subsets of W:
N(w) ⊆ ℘(W) for all w ∈ W.
N is also called a neighborhood mapping.
(2) (Model Definition) M = <W, N, V> is a neighborhood model,
denoted as M ∈ MN, ⇔ <W, N> ∈ FN and V is a mapping from At
into ℘(W).
(3) (Truth Set Definition) The truth sets of all compound formulas
in M are defined inductively as follows:
① V(¬A) = W − V(A),
② V(A ∧ B) = V(A) ∩ V(B),
③ V(□A) = {w ∈ W | V(A) ∈ N(w)}.
(4) Assume that M = <W, N, V> and F = <W, N>. M is called a
280
model on F or a model based on F, V a valuation on F, F a frame of
M. ┤
Remark. (Ⅰ) If we see W as a point set, then for all w ∈ W, N(w)
resembles the neighborhood set of the point w in topology where each
X ∈ N(w) is called a neighborhood of w. This may be why N is called
a neighborhood mapping. The intuitive meaning of N(w) can be
understood as follows:
every proposition that is necessarily true at w is in N(w).
(Ⅱ) The semantics given out by the previous definition is called
the neighborhood semantics, it foremost came from  of
Montague and  of Scott, so the semantics is also called
Montague-Scott-Semantics. Afterward, Segerberg amend this
semantics further in his , Hansson and Gärdenfors obtained
some new results in their , so did Gerson in his , ,
[1975A] and . Chellas summarized the results more detailedly
in his .
(Ⅲ) By the previous semantics, please prove
V(◇A) = {w ∈ W | V(¬A) ∉ N(w)}.
(Ⅳ) Since ∅ ⊆ ℘(W), it follows that ∅ ∈ ℘(℘(W)), and thus it is
possible that N(w) = ∅.
(Ⅴ) As before, FN and MN denote the class of all neighborhood
frames and the class of all neighborhood models, respectively.
(Ⅵ) As in Chapter 3, we can define a series of semantic concepts,
such as the concepts of model validity, model class validity, …. So if
not especially mention, below semantics concept are always defined
as in Chapter 3.
( Ⅶ ) A neighborhood mapping N can also be represented
equivalently as a binary relation RN ⊆ W × ℘(W) as follows:
X ∈ N(w) ⇔ w RN X.
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Namely, N(w) = {X ⊆ W | w RN X}. So the above ③ can also be
represented equivalently as
V(□A) = {w ∈ W | w RN V(A)}.
From this, we can see a connection between the neighborhood
semantics and the relation semantics.
Lemma 6.1.2
(1) MP is pointwise valid.
(2) RE is modelwise valid, but is not pointwise valid.
(3) US is framewise valid, but is not modelwise valid.
(4) M, C, N ∉ Val(FN).
Proof. (1) holds clearly.
(2) Given any model M = <W, N, V> on any <W, N> ∈ FN such
that V(A ↔ B) = W. Then V(A) = V(B), and hence
V(A) ∈ N(w) ⇔ V(B) ∈ N(w) for all w ∈ W.
So
w ∈ V(□A) ⇔ w ∈ V(□B) for all w ∈ W.
So V(□A ↔ □B) = W, and thus RE is modelwise valid.
We will show that RE is not pointwise valid. Define M = <{w, u},
N, V> as a neighborhood model such that
V(p) = {w}, V(q) = {w, u}, N(w) = {{w}}.
It is easy to see that V(p) ∈ N(w) and V(q) ∉ N(w), so
w ∈ V(p ↔ q), w ∉ V(□p ↔ □q).
(4) Consider M: construct <W, N, V> as follows:
W = {w, u}, N(w) = {∅},
V(p) = {w}, V(q) = {u}.
Then V(p ∧ q) = ∅ ∈ N(w), so w ∈ V(□(p ∧ q)). But
V(p), V(q) ∉ N(w),
so w ∉ V(□p) and w ∉ V(□q), and thus w ∉ V(□p ∧ □q).
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Please show that C, N ∉ Val(FN). ┤
Definition 6.1.3 (Chellas,  (p.224)) Let F = <W, N> be a
neighborhood frame. Define frame conditions on F as follows: given
any w ∈ W and X, Y ⊆ W,
(nm) X ∩ Y ∈ N(w) ⇒ X, Y ∈ N(w).
(nc) X, Y ∈ N(w) ⇒ X ∩ Y ∈ N(w).
(nn) W ∈ N(w).
(nd) X ∈ N(w) ⇒ X ∉ N(w)
where X is the abbreviation of W − X. ①
(nt) X ∈ N(w) ⇒ w ∈ X.
(nb) w ∈ X ⇒ {u ∈ W | X ∉ N(u)} ∈ N(w).
(n4) X ∈ N(w) ⇒ {u ∈ W | X ∈ N(u)} ∈ N(w).
(n5) X ∉ N(w) ⇒ {u ∈ W | X ∉ N(u)} ∈ N(w). ┤
Remark. (Ⅰ) The small letter strings on the left of the previous
frame conditions represent the names of the frame conditions where
latter letters represent the small letter of the names of corresponding
axioms M, C, N, D, T, B, 4 and 5, respectively. For example, we shall
show that (nm) corresponds to M.
Please compare the frame conditions (nd) - (n5) with the frame
conditions in 3.2.2. It is easy to see that the frame conditions
expressed by the neighborhood semantics, when characterizing some
axioms, are not as intuitive as the frame conditions expressed by the
relation semantics.
(Ⅱ) we can also combine (nm) with (nc) as follows:
(nr) X ∩ Y ∈ N(w) ⇔ X, Y ∈ N(w).
①
Such a X is called a relative complement.
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Definition 6.1.4 Let A is a above modal formula and x a above frame
condition. A corresponds to x ⇔ for any neighborhood frame <W, N>,
<W, N> ⊨ A ⇔ N satisfies x. ┤
Remark. Note that the concept of correspondence here and the
concept of correspondence in Chapter 3 are a little different (refer to
3.2.4(1)). Since the frame conditions where are usually represented by
first-order formulas, whereas the frame conditions here are
represented by formulas of the set theory.
Theorem 6.1.5 (Correspondence Theorem)
(1) M corresponds to (nm).
(2) C corresponds to (nc).
(3) N corresponds to (nn).
Proof. (1) Given any F = <W, N>. Assume that
F ⊭ □(p ∧ q) → □p ∧ □q.
Then there is a valuation V on F and w ∈ W such that
① w ⊨ □(p ∧ q) and w ⊭ □p ∧ □q.
By the left expression of ①, we have
② V(p ∧ q) = V(p) ∩ V(q) ∈ N(w).
By the right expression of ①, w ⊭ □p or w ⊭ □q, so
③ V(p) ∉ N(w) or V(q) ∉ N(w).
By ② and ③, it is easy to see that N does not satisfy (nm).
Contrarily, assume that N does not satisfy (nm). Then there are
some X, Y ⊆ W such that
④ X ∩ Y ∈ N(w), but X ∉ N(w) or Y ∉ N(w).
Let V be a valuation on F such that X = V(p) and Y = V(q). By ④, it
is easy to see that
w ⊨ □(p ∧ q), and w ⊭ □p or w ⊭ □q.
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So F ⊭ □(p ∧ q) → □p ∧ □q.
(2) This proof resembles (1).
(3) holds clearly. ┤
Lemma 6.1.6 (nm) and the following frame condition are equivalent:
(nrm) X ∈ N(w) and X ⊆ Y ⇒ Y ∈ N(w).
Proof. Assume that (nm) holds. Given any X, Y ⊆ W such that X ∈
N(w) and X ⊆ Y. Then X = X ∩ Y, by (nm), Y ∈ N(w).
Assume that (nrm) holds. Given any X, Y ⊆ W such that X ∩ Y ∈
N(w). Since X ∩ Y ⊆ X and X ∩ Y ⊆ Y, by (nrm), it follows that X, Y
∈ N(w). ┤
Remark. (Ⅰ) The above result is very natural, since it is just a
semantic expression of System M = <RM>.
(Ⅱ) Condition (nrm) means that N(w) is closed under supersets.
Since Monotonic Rule RM is validity-preserving w.r.t. (nrm), we also
call (nrm) Monotonicity Condition.
In this chapter, if not especially illuminate, we always presuppose
that S is a consistent system.
In the following FN(S) is defined as the class of all Sneighborhood frames.
By Correspondence Theorem, we have the following result.
Corollary 6.1.7
(1) FN(E) = FN.
(2) FN(M) = FN(nm).
(3) FN(M) = FN(nrm).
(4) FN(EC) = FN(nc).
(5) FN(EN) = FN(nn). ┤
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Let X be a set of frame conditions on a neighborhood frame, we
shall use FN(X ) as the class of all neighborhood frames satisfying X ,
and thus Val(FN(X )) as the set of all formulas valid in FN(X ).
By the previous result, it is clear that:
Theorem 6.1.8 (Frame Soundness Theorem)
(1) Th(E) ⊆ Val(FN).
(2) Th(M) ⊆ Val(FN(nm)).
(3) Th(M) ⊆ Val(FN(nrm)).
(4) Th(R) ⊆ Val(FN(nm, nc)).
(5) Th(R) ⊆ Val(FN(nr)).
(6) Th(K) ⊆ Val(FN(nr, nn)). ┤
In the following we will give out frame conditions by filters:
Definition 6.1.9 Given any neighborhood model <W, N, V> and w ∈
W.
(1) N(w) is a trival filter ⇔ N(w) = {W}.
(2) N(w) is a quasi-filter ⇔ N(w) satisfies (nr).
(3) N(w) is a filter ⇔ N(w) satisfies (nrm), (nc) and (nn).
(4) N(w) is a ultrafilter ⇔ N(w) is filter and satisfies the following
condition:
(☆) X ∈ N(w) or X ∈ N(w) for all X ⊆ W. ┤
Theorem 6.1.10 The following eight systems are not identical
pairwise:
E, M, EC, EN, R, MN, ECN, EMCN.
Proof. Consider the following three neighborhood models:
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(1) W = {w}, N(w) = ∅, V(p) = ∅.
(2) W = {w, u}, N(w) = N(u) = {{w}, {u}, W},
V(p) = {w}, V(q) = {u}.
(3) W = {w, u}, N(w) = N(u) = {∅, W},
V(p) = {w}, V(q) = {u}.
It is easy to see that N in first model satisfies (nm) and (nc), but not
satisfy (nn), so Axiom N is not valid in the model.
It is easy to see that N in second model satisfies (nm) and (nn). On
the other hand,
{w}, {u} ∈ N(w),
{w} ∩ {u} = ∅ ∉ N(w),
so N not satisfy (nc), and thus Axiom C is not valid in the model.
It is easy to see that N in third model satisfies (nc) and (nn). On the
other hand,
{w} ∩ {u} = ∅ ∈ N(w),
{w}, {u} ∉ N(w),
so N not satisfy (nm), and thus Axiom M is not valid in the model.
By the above countermodels, we get easily the result we will prove.
┤
By 1.2.13(2), 1.2.16(2), 1.3.1(3) and (5), and the above result, we
have
Theorem 6.1.11 (Proper Extension Theorem) E ⊂ M ⊂ R ⊂ K. ┤
Definition 6.1.12 Let F = <W, N> be a neighborhood frame.
F is a normal frame ⇔ the following conditions are satisfied:
(1) N satisfies (nm), and
(2) ∩N(w) ∈ N(w) for all w ∈ W. ┤
Lemma 6.1.13 Let <W, N> be a neighborhood frame. Then the
following (1) and (2) are equivalent:
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(1) <W, N> is a normal frame.
(2) X ∈ N(w) ⇔ ∩N(w) ⊆ X.
Proof. The proof of (1) ⇒ (2): Assume that <W, N> is a normal frame.
‘⇒’: Assume that X ∈ N(w). By a basic fact in the set theory,
∩N(w) ⊆ X.
‘⇐’: Assume that ∩N(w) ⊆ X. By (1) and 6.1.12(2), ∩N(w) ∈ N(w).
Since N satisfies (nm), so N also satisfies (nrm) by 6.1.6, and thus X
∈ N(w).
The proof of (2) ⇒ (1): Assume that (2) holds.
We will show that N satisfies 6.1.12(1): By 6.1.6, it suffices to
show that N satisfies (nrm): Given any X ∈ N(w) and X ⊆ Y. By the
former and (2), ∩N(w) ⊆ X, so ∩N(w) ⊆ Y, hence Y ∈ N(w) by (2)
again.
We will show that N satisfies 6.1.12(2): Since ∩N(w) ⊆ ∩N(w), by
(2), we have ∩N(w) ∈ N(w). ┤
By the previous lemma, we have the following definition:
Definition 6.1.14 (Chellas,  (p. 222)) Let <W, N> be a
neighborhood frame.
(1) A neighborhood frame <W, M> is the augmentation of <W, N>
⇔ M(w) = {X ⊆ W | ∩N(w) ⊆ X} for all w ∈ W.
(2) <W, N> is an augmentation frame ⇔ <W, N> is its own
augmentation, namely,
(#) N(w) = {X ⊆ W | ∩N(w) ⊆ X} for all w ∈ W.
[That is, 6.1.13(2) holds.] ┤
Remark. (Ⅰ) By 6.1.13 and 6.1.14,
F is a normal frame ⇔ F is an augmentation frame.
(Ⅱ) Accordingly, we can define the augmentation of a model and
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an augmentation model.
In the following we will consider the relation between normal
frames and filter frames:
Lemma 6.1.15 F is a normal frame ⇒ F is a filter frame.
Proof. Assume that F = <W, N> is a normal frame. By 6.1.6, N
satisfies (nrm), so it suffices to show that N satisfies (nc) and (nn).
The verification of (nc): Assume that X, Y ∈ N(w). By a basic fact
in the set theory, it is easy to see that,
∩N(w) ⊆ X ∩ Y,
so, by the previous lemma (2), X ∩ Y ∈ N(w).
The verification of (nn): Since ∩N(w) ⊆ W, by the previous lemma
(2), we have W ∈ N(w). ┤
Remark. If <W, N> is a normal frame, then N satisfies (nrm) and
6.1.12(2), but this means that N(w) is closed under any intersection
(refer to 6.1.23(2)), and thus we have the previous lemma. But, it is
not necessary that a filter frame is a normal frame, except all filters in
it are all finite sets, or other sets that have a certain closed property.
Counterexample 6.1.16 there is a non-normal filter frame.
Proof. Construct a frame <W, N> as follows:
(1) W is the set of all real numbers,
(2) N(w) = {X ⊆ W | ∃u ∈ W (w < u and (w, u) ⊆ X } for all w ∈ W.
Here each X ∈ N(w) is a superset of the open interval (w, u)
constituted by w and some real number u > w.
First we will show that N(w) is a filter for all w ∈ W.
It is clear that N(w) satisfies (nrm) and (nn). So it suffices to show
that N(w) satisfies (nc). Assume that X, Y ∈ N(w). Then there are two
289
open intervals (w, u) and (w, v) such that
① (w, u) ⊆ X, (w, v) ⊆ Y.
Without loss of generality, we assume that u ≤ v. So (w, u) ⊆ (w, v),
and thus by the right expression of ①, (w, u) ⊆ Y. By the left
expression of ①, it is easy to see that (w, u) ⊆ X ∩ Y, and hence, by
(2), X ∩ Y ∈ N(w).
We will show that N(w) is not normal. By Remark (Ⅰ) below
6.1.14, it suffices to show that 6.1.12(2) does not hold, namely, we
will show that
② ∩N(w) ∉ N(w) for some w ∈ W.
Given any w ∈ W, by the properties of real numbers, there is not the
smallest open interval (w, u) such that w < u, so
③ ∩N(w) = ∅.
On the other hand, there is not a real number u > w such that (w, u) =
∅, so every set in N(w) is not the empty set, and thus
④ ∅ ∉ N(w).
By ③ and ④, we get easily ②. ┤
Corollary 6.1.17 Let FN(normal) be the class of all normal frames
and FN(filter) the class of all filter frames. Then
FN(normal) ⊂ FN(filter). ┤
In the following we will give out three types of simple frame
transformations (operations on frames): supplementation, intersection
closure and quasi-filtering.
First we will give out the definitions of the above two types of
frame transformations:
Definition 6.1.18 (Chellas,  (p. 216)) Let F = <W, N> be a
neighborhood frame.
290
(1) F + = <W, N + > is the supplementation of F ⇔ for all w ∈ W,
N (w) is the superset closure of N(w), namely, for all X ⊆ W,
X ∈ N + (w) ⇔ ∃Y ∈ N(w) (Y ⊆ X).
(2) F is a supplementation frame ⇔ it is its own supplementation,
namely, for all w ∈ W,
N(w) = N + (w).
(3) F － = <W, N －> is the (finite) intersection closure of F ⇔ for
all w ∈ W, N －(w) is the (finite) intersection closure of N(w), namely,
for all X ⊆ W,
X ∈ N －(w) ⇔ ∃n > 0 ∃X1, …, Xn ∈ N(w) (X = X1 ∩…∩ Xn).
(4) F is a intersection closure frame ⇔ it is its own intersection
closure, namely, for all w ∈ W,
N(w) = N －(w). ┤
+
Remark. Accordingly, we can define the supplementation of a
model and a supplementation model, and the intersection closure of a
model and an intersection closure model, respectively
Lemma 6.1.19 Let <W, N> be a neighborhood frame. Then
(1) N(w) ⊆ N + (w) for all w ∈ W.
(2) N(w) ⊆ N －(w) for all w ∈ W.
(3) (F +) － = (F －) +.
Proof. (1) Given any X ∈ N(w), X ⊆ X, so X ∈ N + (w).
(2) Given any X ∈ N(w), X = X ∩ X, so X ∈ N －(w).
(3) For each w ∈ W, it suffices to show that
① N + －(w) = N －+ (w).
‘⊆’: Given any X ∈ N + －(w). It is easy to see that there is some n
> 0 and X1, …, Xn ∈ N + (w) such that X = X1 ∩…∩ Xn, so there are
291
some Y1, …, Yn ∈ N(w) such that
Y1 ⊆ X1, …, Yn ⊆ Xn.
So Y1 ∩…∩ Yn ∈ N －(w). Since
Y1 ∩…∩ Yn ⊆ X1 ∩…∩ Xn = X,
it follows that X ∈ N － + (w).
‘⊇’: Given any X ∈ N －+ (w). Then there is some Y ∈ N －(w) such
that Y ⊆ X, so there is some n > 0 and Y1, …, Yn ∈ N(w) such that Y =
Y1 ∩…∩ Yn, and thus
② Y1 ∩…∩ Yn ⊆ X.
Let X1 = Y1 ∪ X, …, Xn = Yn ∪ X. Then
③ Y1 ⊆ X1, …, Yn ⊆ Xn.
So X1, …, Xn ∈ N + (w). Since
X1 ∩…∩ Xn = (Y1 ∪ X) ∩ …∩ (Yn ∪ X)
= (Y1 ∩…∩ Yn) ∪ X,
it follows by ② that X = X1 ∩…∩ Xn, so X ∈ N +－(w). ┤
By the previous lemma, we have the following definition:
Definition 6.1.20 (Chellas,  (p. 217)) Let F = <W, N> be a
neighborhood frame.
±
±
(1) F = <W, N > is the quasi-filtering of F ±
⇔ N (w) = N + －(w) = N －+ (w) for all w ∈ W.
(2) F is a quasi-filtering frame ⇔ it is its own quasi-filtering. ┤
Remark. Accordingly, we can define the quasi-filtering of a model
and a quasi-filtering model.
Note that a quasi-filtering frame is a frame that is closed under
supplementation and intersection closure.
292
Theorem 6.1.21
(1) □(p ∧ q) → □p ∧ □q is valid in all supplementation frames.
(2) □p ∧ □q → □(p ∧ q) is valid in all intersection closure
frames.
(3) □(p ∧ q) ↔ □p ∧ □q is valid in all quasi-filtering frames.
┤
Exercises 6.1
6.1.22
(1) Let <W, N, V> be a neighborhood model. Show that
V(◇A) = {w ∈ W | V(¬A) ∉ N(w)}.
(2) Prove the unproved part of 6.1.2(3) and (4).
(3) Verify that (nd), (nt), (nb), (n4) and (n5) in 6.1.3 correspond to
Axiom D, T, B, 4 and 5, respectively. ┤
6.1.23 Prove the
(1) N(w) is a filter ⇔ N(w) is a nonempty quasi-filter.
(2) Assume that <W, N> is a normal frame. Then for all w ∈ W,
N(w) is closed under any intersection. ┤
6.1.24 Prove 6.1.21. ┤
6.1.25 Find out the neighborhood frame conditions of the following
formulas correspond to:
(1) ◇T (= P),
(2) □(p ∨ q) → □p ∨ □q (= F),
293
(3) □(□p → p)
(4) G. ┤
(= U),
6.1.26 Let FN(x filter) be the class of all neighborhood frames such
that each N(w) of each frame in FN(x filter) is a x filters. Prove that
(1) Th(EN) ⊆ Val(FN(trivial filter)).
(2) Th(R) ⊆ Val(FN(quasi-filter)).
(3) Th(K) ⊆ Val(FN(filter)).
(4) Th(K + Dc) ⊆ Val(FN(ultrafilter)). ┤
6.1.27 Prove that there not a formula of the form □A such that it is a
theorem of Systems E, M or R. ┤
6.1.28 Given any neighborhood frame <W, N>. Define N n as follows:
for all w ∈ W,
(1) X ∈ N 0(w) ⇔ w ∈ X,
(2) X ∈ N n + 1(w) ⇔ {u ∈ W | X ∈ N n (u)} ∈ N(w).
Let <W, N, V> be a neighborhood model. For each w ∈ W, prove that
(Chellas,  (p.211)):
(3) w ∈ V(□nA) ⇔ V(A) ∈ N n (w),
(4) w ∈ V(◇nA) ⇔ V(¬A) ∉ N n (w). ┤
6.1.29(Chellas,  (pp.225 - 226)) Prove G
mn
jk
is sound w.r.t. the
class of the neighborhood frame satisfies the following condition:
mn
(ng jk ) {u ∈ W | X ∈ N n (u)} ∉ N m(w)
⇒ {u ∈ W | X ∉ N k(u)} ∈ N j(w). ┤
6.1.30 Let F be a neighborhood frame.
294
(1) If F is a supplementation frame, then F + is also a
supplementation frame;
(2) If F is an intersection closure frame, then F － is also an
intersection closure frame. ┤
6.1.31 (Problem) Let F = <W, N> be a neighborhood frame. Whether
the following proposition are equivalent?
±
(1) F = F .
(2) F satisfies the following condition: for all w ∈ W and X, Y ⊆ W,
(nk) X ∪ Y ∈ N(w) and X ∈ N(w) ⇒ Y ∈ N(w). ┤
6.1.32 Let <W, N, V> be a supplementation model. Prove that for all w
∈ W,
w ∈ V(□A) ⇔ ∃X ∈ N(w) (X ⊆ V(A)). ┤
6.1.33 Prove that
(1) Modal(R5) = 10.
(2) Modal(MD5) = 10.
(3) Modal(ED45) = 6. ┤
6.1.34
(1) Prove that (p ↔ q) → (□p ↔ □q) is not a theorem of E.
(2) Does the following proposition hold?
Φ ⊢ E A ⇒ Φ ⊢ E A. ┤
6.1.35 Like in Chapter 3, w.r.t. the neighborhood semantics, define
the corresponding semantics consequence, and hence consider
whether the following propositions and its converse propositions hold?
(1) Φ ⊨ A ⇒ Φ ⊨ M A.
295
(2) Φ ⊨ M A ⇒ Φ ⊨ MN A.
(3) Φ ⊨ F A ⇒ Φ ⊨ FN A.
(4) Φ ⊨ M A ⇒ Φ ⊨ F A. ┤
296
§2 Characterization Theorem
In this section we will show that Canonical Model Elementary
Theorem and Characterization Theorem for proper subsystems of K
w.r.t. the neighborhood semantics.
The ideas and skills used below resemble the ones used in the
relation semantics.
Definition 6.2.1
(1) FS = <WS, NS> be a canonical (neighborhood) frame for S ⇔
the following conditions are satisfied:
① WS = MCS(S), and
② NS is a mapping from WS into ℘(℘(WS)) such that
□A ∈ w ⇔‖A‖S ∈ NS(w) for all w ∈ WS. ①
(2) MS = <WS, NS, VS> is a canonical (neighborhood) model for S
⇔　<WS, NS> is a canonical frame for S, and
③ V S(pn) =‖pn‖S for all pn ∈ At. ┤
Remark. (Ⅰ) As before, we often omit the subscript ‘S’ if no
confusion will arise. (Ⅱ) We will show that ② is a well-definition. Assume that
‖A‖ = ‖B‖,
then by 2.2.9(4), it is easy to show that ⊢ S A ↔ B, so, by RE,
①
By 2.2.8, ‖A‖S := {w ∈ WS | A ∈ w}.
297
⊢ S □A ↔ □B.
By Maximal Consistent Set Lemma 2.2.7(3),
□A ↔ □B ∈ w for all w ∈ W S,
so by 2.2.7(6)②,
□A ∈ w ⇔ □B ∈ w for all w ∈ WS.
And hence, by the above ②, we have
‖A‖∈ N(w) ⇔ ‖B‖∈ N(w).
(Ⅲ) The existence of a mapping N satisfying ② is given as
follows:
Definition 6.2.2 (Chellas,  (p.254)) Let F = <W, N> be a
canonical frame for S.
(1) F is the smallest canonical frame for S ⇔ for all w ∈ W,
N(w) = {‖A‖| □A∈w}.
(2) F is the largest canonical frame for S ⇔ for all w ∈ W,
N(w) = {‖A‖| □A ∈ w}
∪ {X ⊆ W | ∀A ∈ ML (X ≠‖A‖)}. ┤ ①
Remark. Note that the concept of canonical frames defined in 6.2.1
and the concept of the smallest canonical frame defined in 6.2.2(1) are
different, because N(w)’s expressed in the following ① and ② are
different:
① □A ∈ w ⇔‖A‖S ∈ NS(w) for all w ∈ WS.
② NS(w) = {‖A‖S | □A ∈ w} for all w ∈ WS.
This is because there are also uncountable subsets of WS such that
they cannot be expressed as subsets of the form ‖A‖S (Since WS is
not countable, whereas ML is countable), but, in ①, this specification
①
Refer to 3.1.23.
298
is not given out, so there are uncountable NS(w)’s satisfying ① (refer
to Exercise 6.2.12 below), and thus there are uncountable canonical
neighborhood frames, whereas NS(w) in ② is uniquely specified and
still the smallest. But for a consistent modal system S, w.r.t. the
relation semantics, there is just the unique canonical (relation) frame
for S.
It is easy to see that, ② is an instance of ①.
On the other hand, ① cannot be also weakened as the ‘⇒’ or ‘⇐’
of ①. Since ① means: for all A ∈ ML,
□A ∈ w ⇒‖A‖S ∈ NS(w);
□A ∉ w ⇒‖A‖S ∉ NS(w).
Lemma 6.2.3 Let <W, N> be the canonical frame for S. For each w ∈
W, the following propositions are equivalent:
(1) □A ∈ w ⇔ ‖A‖∈ N(w).
(2) ◇A ∈ w ⇔ ‖A‖∉ N(w).
Proof. The proof of (1) ⇒ (2):
◇A ∈ w
by the abbreviation definition
⇔ ¬□¬A ∈ w
⇔ □¬A ∉ w
by the S-maximality and consistency of w
⇔ ‖¬A‖∉ N(w)
by (1)
⇔ ‖A‖∉ N(w).
Final ‘⇔’ holds is because‖A‖and‖¬A‖are a partition of W, so
‖¬A‖= W −‖A‖.
The proof of (2) ⇒ (1):
□A ∈ w
⇔ ¬◇¬A ∈ w
by E(3), US and 2.2.7(6) ⇔ ◇¬A ∉ w
by the S-maximality and consistency of w
299
⇔ ‖¬A‖∈ N(w)
by (2)
⇔ ‖A‖∈ N(w). ┤
Theorem 6.2.4 (Canonical Model Elementary Theorem)
Let <W, N, V> be a canonical model for S. Then for all A ∈ ML,
(1) A ∈ w ⇔ w ∈ V(A) for all w ∈ W.
(2) V(A) =‖A‖.
Proof. (1) By induction on A. It suffices to consider the case of A =
□B (The rest of the cases is as what has been proved before). By the
induction hypothesis,
B ∈ w ⇔ w ∈ V(B) for all w ∈ W.
So V(B) =‖B‖, and thus
‖B‖∈ N(w) ⇔ V(B) ∈ N(w),
and hence, by 6.2.1 ② and Truth Set Definition of □B, we have
□B ∈ w ⇔ w ∈ V(□B).
(2) is gotten immediately by (1). ┤
As before, it is easy to see that:
Theorem 6.2.5 (Model Completeness Theorem) Let M be a
canonical model for S. Then Th(S) = Val(M). ┤
Remark. By this theorem, a canonical neighborhood model for S is
also a characterization model for S.
Definition 6.2.6 S is a canonical system (w.r.t. the neighborhood
semantics) ⇔ a canonical frame for S is a S-frame. ┤
Remark. Refer to (Ⅱ) below 4.4.1.
300
Theorem 6.2.7 (Characterization Theorem)
(1) E ◄ MN.
(2) E ◄ FN.
(3) E is a canonical system w.r.t. the neighborhood semantics.
Proof. (1) We get easily the soundness by 6.1.8(1). In the following
we will prove the completeness. Given any A ∉ Th(E), by Model
Completeness Theorem 6.2.5, A is not valid in a canonical model for
E, and thus we have the completeness.
(2) We leave the proof as an exercise.
(3) We leave the proof as an exercise. ┤
Remark. Let X be a nonempty subset of the frame conditions given
in 6.1.3, and Γ the axiom set corresponding to X . We will consider
the following Characterization Theorem
E + Γ ◄ FN(X ).
By the above results (include the related exercises), the soundness
and model completeness have been proven. So, to the end of this
section, we will prove frame completeness. As we remarked above, it
suffices to show that some canonical frame for S belongs to FN(X),
and thus to construct an appropriate canonical frame for S as it is
possible occurs such a case:
not all canonical frames for S belong to FN(X ).
For example, by the following exercise 6.2.15, we are told that the
smallest canonical frame for a monotonic system is not a
supplementation frame, namely, it not satisfy (nm). But the following
theorem shows that the supplementation of the smallest canonical
frame for a monotonic system S is a canonical frame for S.
Lemma 6.2.8 Let F = <W, N> be the smallest canonical frame for S.
301
(1) If S is a monotonic system, then the supplementation of F is a
canonical frame for S.
(2) If S has Theorem C, then N satisfies (nc).
(3) If S is a normal system, then the augmentation of F is a
canonical frame for S.
Proof. By the assumption and 6.2.2(1), we have that
(☆) N(w) = {‖A‖| □A ∈ w} for all w ∈ W.
(1) Assume that S is monotonic and F + = <W, N + > is the
supplementation of F. By 6.1.18(1) and (☆),
① X ∈ N + (w) ⇔ there is some □A ∈ w such that‖A‖⊆ X.
By 6.2.1(1), it suffices to show that for all A ∈ ML and w ∈ W,
② □A ∈ w ⇔‖A‖∈ N + (w).
‘⇒’: Assume that □A ∈ w. It follows that‖A‖∈ N(w) by (☆).
But N(w) ⊆ N + (w) by 6.1.19(1), so‖A‖∈ N + (w).
‘⇐’: Assume that ‖A‖∈ N + (w). By ①, there is some □B ∈ w
such that‖B‖⊆‖A‖, by 2.2.9(4), ⊢ S B → A. Since S is monotonic,
by RM,
⊢ S □B → □A.
Since □B ∈ w, by 2.2.7(5), it follows that □A ∈ w.
(2) Assume that S has Theorem C. Given any X, Y ∈ N(w), by (☆),
it follows that there are some □A, □B ∈ w such that
X =‖A‖and Y =‖B‖.
Since □A ∧ □B ∈ w, by Theorem C, we have □(A ∧ B) ∈ w, by
(☆), we have‖A∧B‖∈ N(w). Since
‖A∧B‖=‖A‖∩‖B‖= X ∩ Y,
we have X ∩ Y ∈ N(w).
(3) Assume that S is normal and G = <W, M> is the augmentation
of F. By 6.2.1(1), it suffices to show that
302
③ □A ∈ w ⇔‖A‖∈ M(w) for all A ∈ ML and w ∈ W.
By the augmentation definition 6.1.14(1), it suffices to show that
④ □A ∈ w ⇔ ∩N(w) ⊆‖A‖for all A ∈ ML and w ∈ W.
‘⇒’: Assume that □A ∈ w. Given any u ∈ ∩N(w). Since □A ∈ w,
by (☆),‖A‖∈ N(w), and thus u ∈‖A‖.
‘⇐’: Assume that ∩N(w) ⊆‖A‖. We will show that □A ∈ w.
First we will show that
－
⑤ u ∈ ∩{‖A‖| □A ∈ w} ⇔ □ w ⊆ u.
－
Assume that u ∈ ∩{‖A‖| □A ∈ w}. Given any B ∈ □ w, then
□B ∈ w. By the assumption, u ∈‖B‖, so B ∈ u.
－
Assume that □ w ⊆ u. Then
⑥ □A ∈ w ⇒ A ∈ u for all A ∈ ML.
So
⑦ □A ∈ w ⇒ u ∈‖A‖ for all A ∈ ML,
and thus u ∈ ∩{‖A‖| □A ∈ w}.
Thus we get ⑤.
By the assumption ∩N(w) ⊆‖A‖and (☆), we have
⑧ ∩{‖A‖| □A ∈ w} ⊆‖A‖.
On the other hand, by 2.2.14 (1),
－
⑨ □A ∈ w ⇔ ∀u ∈ W (□ w ⊆ u ⇒ A ∈ u).
In order to show that □A ∈ w, by ⑨, it suffices to show that
－
⑩ ∀u ∈ W (□ w ⊆ u ⇒ A ∈ u).
－
－
Given any u ∈ W such that □ w ⊆ u. By □ w ⊆ u and ⑤, we
have
u ∈ ∩{‖A‖| □A ∈ w},
By ⑧, u ∈‖A‖, so A ∈ u. and thus ⑩ holds. ┤
Lemma 6.2.9 Let F = <W, N> be any canonical frame for S. If S has
Theorem N, then N satisfies (nn).
303
Proof. Assume that S has Theorem N. Then for all w ∈ W, □T ∈ w,
so, by 6.2.1(1)②, ‖T‖∈ N(w), whereas ‖T‖ = W, and thus N
satisfies (nn). ┤
By the above two lemmas, please prove that
Theorem 6.2.10 (Characterization Theorem)
(1) M ◄ FN(supplementation).
(2) R ◄ FN (supplementation and intersection closure).
(3) K ◄ FN(augmentation ). ┤
Usually，a frame class is a proper class (such as the class of all
neighborhood frames), but we can also consider frame sets rather than
proper classes. For this, assume that S is characterized by a frame
class F, then for all A ∉ Th(S), there is a counterframe FA ∈ F of A.
By a basic fact of the set theory, there are at most countable infinite
non-theorems of S, so {FA | A ∉ Th(S)} is a set formed by at most
countable infinite S-frames.
We can still go further: for all A ∉ Th(S), let the above FA = <WA,
NA>. Define G = <U, M> as follows:
(1) U = {<A, w> | A ∉ Th(S) and w ∈ WA},
(2) M (<A, w>) = {{<A, u> | u ∈ X}: <A, w> ∈ U and X ∈ NA (w)}.
Accordingly, define the concepts of satiablility and etc. of the
above models, we have
Theorem 6.2.11 (Characterization Theorem) (Segerberg, 
(p.28))
(1) G ► S.
(2) S is characterized by a single frame. ┤
304
Exercises 6.2
6.2.12
(1) Show (2) – (3) of 6.2.7.
(2) Show (1) – (3) of 6.2.10. ┤
6.2.13 (Chellas,  (p.255)) Prove that the following two
propositions are equivalent:
(1) F = <W, N> is a canonical frame for S.
(2) W = MCS(S), and for all w ∈ W,
N(w) = {‖A‖| □A ∈ w} ∪ X,
where X ⊆ {X ⊆ W | ∀A ∈ ML (X ≠‖A‖)}. ┤
6.2.14 Let M = <W, N, V> be a canonical model for S. Prove that
(☆) ‖□nA‖= {w ∈ W |‖A‖∈ N n (w)} for all n < ω. ┤
6.2.15 Let F = <W, N> be the smallest canonical frame for a
monotonic system. Prove that F is not a supplementation frame. ┤
6.2.16 Prove that
(1) EN ◄ FN(trivial filter).
(2) R ◄ FN(quasi-filter).
(3) K ◄ FN(filter).
(4) K + Dc ◄ FN(ultrafilter). ┤
6.2.17
(1) Let S has Theorem T, and <W, N> be the smallest canonical
frame for S. Prove that N satisfies (nt).
305
(2) Let S has Theorem 5, and <W, N> be the largest canonical
frame for S. Prove that N satisfies (n5).
(3) Deleting the above ‘smallest’ and ‘largest’, whether the results
still hold. ┤
6.2.18 Prove that
(1) ET ◄ FN(nt).
(2) E5 ◄ FN(n5).
(3) E45 ◄ FN(n4, n5).
mn
mn
(4) EG jk ◄ FN(ng jk ).
(5) MD ◄ FN(nm, nd). ┤
6.2.19
(1) Given any <W, N>. Do the following conditions correspond to
which modal axioms, respectively?
∅ ∈ N(w) for all w ∈ W.
W ∉ N(w) for all w ∈ W.
N(w) = ℘(W) for all w ∈ W.
N(w) = ∅ for all w ∈ W.
(2) Prove Characterization Theorem w.r.t. the above conditions.
Note that in what kind of case, the system that is characterized is
inconsistent?
[Hint: in order to prove the completeness of E + □p, we will show
that the augmentation of a canonical frame for E + □p satisfies the
condition corresponding to □p. ] ┤
6.2.20 Given any <W, N>.
(1) For each w ∈ W, that N(w) is a proper filter, ideal, proper ideal,
trivial ideal or prime ideal corresponds to which modal axiom,
306
respectively?
(2) Prove Characterization Theorem w.r.t. the previous condition.
Note that in what kind of case, the system that is characterized is
inconsistent? ┤
6.2.21 Let S be consistent. Prove that
Φ ⊨ A ⇔ Φ ⊢ E A. ┤
307
§3 Filtration and Finite Model Property
In this section we define filtrations of neighborhood frames and
neighborhood models, prove Filtration Elementary Theorem and
Finite Characterization Theorem for proper subsystems of K w.r.t. the
neighborhood semantics, and thus obtain the finite model property of
the proper subsystems of K.
Definition 6.3.1 Let Φ = Sub(Φ).
(1) Let M = <W, N, V> be a neighborhood model. Define the
equivalent class |w|Φ and the equivalent class set |X|Φ w.r.t. Φ as in
5.2.1. In the following we shall omit subscript Φ.
M* is a Φ-filtration (neighborhood model) of M
⇔ M* = <W*, N*, V*> is a neighborhood model such that the
following conditions are satisfied:
① W* = |W|.
② N* is a mapping from W* into ℘(℘(W*)) such that
|V(A)| ∈ N*(|w|) ⇔ V(A) ∈ N(w) for all □A ∈ Φ.
③ V * is a (valuation) mapping from At into ℘(W*) such that
V*(pn) = {|w| ∈ W* | w ∈ V(pn)} for all pn ∈ At ∩ Φ.
(2) A filtration <W*, N*, V*> is the finest filtration ⇔ the
following conditions are satisfied:
N*(|w|) = {|V(A)| | □A ∈ Φ and V(A) ∈ N(w)}.
(3) Let F be a neighborhood frame. F* is a Φ-filtration frame of F
⇔ there are some model M on F and Φ-filtration M* of M such that
F* is a frame of M*. ┤
308
(Ⅰ) Please give out the definition of the coarsest filtration.
(Ⅱ) By the proof of 5.2.3, it is easy to show that if Φ is finite, then
Φ-filtration is also finite; if Card(Φ) = n, then the cardinality of the
possible world sets of the Φ-filtration is at most 2n.
Theorem 6.3.2 (Filtration Elementary Theorem) Let Φ = Sub(Φ)
and M* = <W*, N*, V*> be a Φ-filtration of M = <W, N, V>.
(1) |w| ∈ V*(A) ⇔ w ∈ V(A) for all |w| ∈ W* and A ∈ Φ.
(2) M* ⊨ A ⇔ M ⊨ A for all A ∈ Φ.
Proof. (1) By induction on A ∈ Φ. If A ∈ At or A is a formula of the
form ¬B, B ∧ C, then, as before, it is easy to show (1).
Consider A = □B. By the induction hypothesis,
① |w| ∈ V*(B) ⇔ w ∈ V(B) for all |w| ∈ W*.
So
② V*(B) = |V(B)|.
Hence, for all |w| ∈ W*,
|w| ∈ V*(□B)
⇔ V*(B) ∈ N*(|w|)
by Truth Set Definition
by ②
⇔ |V(B)| ∈ N*(|w|)
⇔ V(B) ∈ N(w)
by 6.3.1(1) ②
⇔ w ∈ V(□B).
by Truth Set Definition
(2) is gotten immediately by (1). ┤
Corollary 6.3.3 Let Φ = Sub(Φ), M be a neighborhood model class
and Fil(M, Φ) be the class of all Φ-filtrations of all models in M.
Then
(☆) Fil(M, Φ) ⊨ A ⇔ M ⊨ A for all A ∈ Φ. ┤
309
Theorem 6.3.4 Let Φ = Sub(Φ) and M* be the Φ-finest filtration of M.
Then
(1) If M is a supplementation model, then M* + is a Φ-filtration of
M.
(2) If M is an intersection closure model, then M* － is a Φfiltration of M.
±
(3) If M is a quasi-filter model, then M* is a Φ-filtration of M.
Proof. Please prove that (1) - (2).
We will show that (3). Let M* = <W*, N*, V*> be a Φ-finest
±
±
filtration of a quasi-filter model M = <W, N, V> and M* = <W*, N* ,
V*> be a quasi-filtering of M* (refer to 6.1.20).
Given any □A ∈ Φ and w ∈ W, it suffices to show that
±
(☆) V(A) ∈ N(w) ⇔ |V(A)| ∈ N* (|w|).
‘⇒’: Assume that V(A) ∈ N(w). Since M* is a Φ-filtration of a
quasi-filter model M, by 6.3.1(1)②, it follows that |V(A)| ∈ N*(|w|).
±
As M* is a quasi-filtering of M*, |V(A)| is its own supplementation,
and also an intersection of two itself, so, by 6.1.20 (1),
±
|V(A)| ∈ N* (|w|). ①
±
‘⇐’: Assume that |V(A)| ∈ N* (|w|). Since M* is the Φ-finest
filtration of M = <W, N, V> and □A ∈ Φ, so, by the assumption and
±
N* (|w|) = N *－+ (w), there are some □A1, …, □An ∈ Φ such that
① |V(A1)|, …, |V(An)| ∈ N*(|w|), and
② |V(A1)| ∩ …∩ |V(An)| ⊆ |V(A)|.
By ① and 6.3.1(1)②, we have
③ V(A1), …, V(An) ∈ N(w).
Since Φ = Sub(Φ), it follows that A1, …, An, A ∈ Φ, by ② and
Filtration Elementary Theorem, it is easy to show that
①
We can also get this easily by 6.1.19 (1) - (2).
310
④ V(A1) ∩ …∩ V(An) ⊆ V(A).
By 6.1.9(2), a quasi-filter satisfies (nr) and thus (nc), so, by ③, we
have
⑤ V(A1) ∩ …∩ V(An) ∈ N(w).
Since a quasi-filter also satisfies (nm), and hence, by 6.1.6, a quasifilter satisfies (nrm), and thus by ④ and ⑤, V(A) ∈ N(w). ┤
Theorem 6.3.5 Let Φ = Sub(Φ) such that □T ∈ Φ, and let M* be a
Φ-filtration of a neighborhood model M. If M satisfies (nn), then M*
satisfies (nn). ┤
Theorem 6.3.6 (Finite Characterization Theorem) (Chellas, 
(p.265)) Let MNf (x filter) be the class of all finite neighborhood
models such that each N(w) of each model in MNf (x filter) is a x
filters.
(1) E ◄ MNf ,
(2) M ◄ MNf (supplementation).
(3) R ◄ MNf (quasi-filter).
(4) K ◄ MNf (filter).
Proof. The soundness is obtained easily by 6.2.10 and 6.1.26. We will
prove the completeness:
(1) Given any A ∉ Th(E), then by 6.2.7(1), A is false in some
neighborhood model M. Let M* be Sub(A)-filtration of M. By
Filtration Elementary Theorem (6.3.2(2)), A is also false in M*.
(2) Given any A ∉ Th(M), then by 6.2.10(1), it is easy to see that A
is false in some supplementation model M. Let M* + be the
supplementation of the Sub(A)-finest filtration M* of M, then by the
definition of supplementation, it is easy to see that M* + is still a finite
311
model. By 6.3.4(1), M* + is also a Sub(A)-filtration of M, so, by
Filtration Elementary Theorem, A is also false in M* +.
(3) This proof resembles the proof of (2), but to use 6.2.16(2) and
6.3.4(3) rather than 6.2.10(1) and 6.3.4(1). We leave the details of the
proof as an exercise.
(4) Given any A ∉ Th(K), then by 6.2.16(3), A is false in some
±
filter model M. Let M* be the quasi-filtering of Sub({A, □T})finest filtration M* of M (refer to Remark below 6.1.20(1)), then it is
±
±
easy to show that M* is a finite quasi-filter model. By 6.3.4(3), M*
is also a Sub({A, □T})-filtration of M. Since, M is a filter model, so,
±
by Definition 6.1.9(3), M satisfies (nn), so, by 6.3.5, M* satisfies
±
(nn), and hence M* is also a filter model. On the other hand, since
±
M* is a Sub({A, □T})-filtration of M, by Filtration Elementary
±
Theorem, A is also false in M* . ┤
Remark. In this way, we show that the above systems have the
finite model property w.r.t. the neighborhood semantics, as they is
finite axiomatizable, so it is easy to show that the above systems are
decidable.
Exercises 6.3
6.3.7 Prove that 6.3.4(1) - (2), 6.3.6 (3) and 6.3.5. ┤
312
§4 Neighborhood Semantics
vs Relation Semantics
In this section we will show the the relation between the
neighborhood semantics and the relation semantics. We will show that
the relation semantics is a subsemantics of the neighborhood
semantics, the class of all normal neighborhood frames is equivalent
to the class of all relation frame, and E and K are the widths of the
neighborhood semantics and the relation semantics, respectively.
Finally, we will show that compared with the relation semantics, the
neighborhood semantics has larger characterization width and larger
normal characterization depth.
Definition 6.4.1
(1) A model class M pointwise implies a model class N, denoted as
M →P N, ⇔ Given any M = <W, *, VM> ∈ M such that * is R or N,
there is some N = <W, *1, VN> ∈ N such that *1 is R or N, and
(☆) w ∈ VM(A) ⇔ w ∈ VN(A) for all A ∈ ML and w ∈ W.
(2) Let C and D be any two model classes or frame classes.
① C (one by one) implies D, denoted as C → D,
⇔ ∀x ∈ C ∃y ∈ D . Val(x) = Val(y).
② C and D are equivalent, denoted as C ↔ D,
⇔ C → D and D → C. ┤
Lemma 6.4.2
(1) Mod →P MN.
313
(2) Mod → MN.
(3) Fra → FN.
Proof. (1) Given any relation model M = <W, R, VM>, define a
neighborhood model N = <W, N, VN> as follows:
① N(w) = {X ⊆ W | R(w) ⊆ X} for all w ∈ W, and
② VM = VN.
Given any A ∈ ML, it suffices to show that
③ w ∈ VM(A) ⇔ w ∈ VN(A) for all w ∈ W.
If A = pn ∈ At, then by ②, we have
④ w ∈ VM(pn) ⇔ w ∈ VN(pn) for all w ∈ W.
If A has the form ¬B or B ∧ C, then by the induction hypotheses we
get easily the result we will prove.
Let A = □B, then
w ∈ VM(□B)
⇔ R(w) ⊆ VM(B)
by Truth Set Definition of the relation semantics
by Induction Hypothesis
⇔ R(w) ⊆ VN(B)
⇔ VN(B) ∈ N(w) by ①
⇔ w ∈ VN(□B).
by Truth Set Definition of the neighborhood semantics
(2) and (3) hold clearly by (1). We leave the details of the proof as
exercises. ┤
Theorem 6.4.3 (Subsemantics Theorem) (Gerson, , (p.17))
In the following sense, the relation semantics is a subsemantics of the
neighborhood semantics:
(☆) There is a mapping h from Fra to FN such that:
(1) h is one-to-one, ① and
①
h is one-to-one ⇔ (h(x) = h(y) ⇒ x = y for all x, y ∈ Fra).
314
(2) for all F ∈ Fra, there is some h(F) ∈ FN such that
Val(F) = Val(h(F)).
Proof. Given any relation frame F = <W, R>, define a neighborhood
mapping N from W into ℘(℘(W)) as follows
① NR(w) = {X ⊆ W | R(w) ⊆ X} for all w ∈ W.
Let h (F) = <W, NR>, then h(F) is a neighborhood frame. According to
the the proof of the previous lemma, it is easy to prove (2), so, below,
it suffices to prove (1).
Given any F1 = <W1, R1> and F2 = <W2, R2> such that h(F1) = h(F2).
Then W1 = W2 and NR1 = N R2. So for all w ∈ W1 = W2,
{X ⊆ W1 | R1(w) ⊆ X} = {X ⊆ W2 | R2(w) ⊆ X},
so R1(w) = R2(w), and thus for all u ∈ W,
wR1u ⇔ wR2u.
So R1 = R2 and W1 = W2, and thus F1 = F2. ┤
Theorem 6.4.4 (Gerson, ) FN(T) does not imply Fra: there is a
T-neighborhood frame F such that there is not a relation frame G such
that Val(F) = Val(G).
Proof. This proof is too long, omitted here, please refer to Gerson’s
. ┤
Theorem 6.4.5 (Equivalence Theorem for Normal Frame Class)
Let NFN be the class of all normal neighborhood frames. Then
NFN ↔ Fra.
Proof. First we will show that NFN → Fra. Given any F = <W, N> ∈
NFN. Define
① R = {<w, u> ∈ W2 | u ∈ ∩N(w)}.
So
② R(w) = ∩N(w) for all w ∈ W.
It is easy to see that G = <W, R> ∈ Fra. It suffices to show that
315
③ Val(F) = Val(G).
For this, we will show that a stronger result (pointwise result): given
any A ∈ ML, M = <W, N, V> on F and N = <W, R, V′′ > on G,
④ w ∈ V(A) ⇔ w ∈ V′′(A) for all w ∈ W.
By induction on A. We here consider the case of A = □B only:
w ∈ V(□B)
⇔ R(w) ⊆ V(B)
by Truth Set Definition of the relation semantics
⇔ ∩N(w) ⊆ V(B)
by ②
⇔ ∩N(w) ⊆ V′′(B) by Induction Hypothesis
by 6.1.13 as <W, N> is a normal frame
⇔ V′′(B) ∈ N(w)
⇔ w∈V′′(□B).
by Truth Set Definition of the neighborhood semantics
Next we will show that Fra → NFN. Let G = <W, R> ∈ Fra.
Define
⑤ N(w) = {X | R(w) ⊆ X} for all w ∈ W.
Then it is easy to see that
⑥ ∩N(w) = R(w).
By ⑤ and ⑥, it is easy to show that N(w) satisfies 6.1.12(1) - (2),
so F = <W, N> ∈ NFN. As we remarked above, it suffices to show
that given any A ∈ ML, N = <W, R, V> on G and M = <W, N, V′′ > on
F,
⑦ w ∈ V(A) ⇔ w ∈ V′′(A) for all w ∈ W.
By induction on A. We here consider the case of A = □B only:
w ∈ V(□B)
⇔ R(w) ⊆ V(B)
by Truth Set Definition of the relation semantics
⇔ V(B) ∈ N(w)
by ⑤
⇔ V′′(B) ∈ N(w)
by Induction Hypothesis
⇔ w ∈ V′′(□B).
by Truth Set Definition of the neighborhood semantics ┤
316
Definition 6.4.6
(1) (Gerson,  (p.32)) A system S is the width of a semantics Y
⇔ S is characterized by the class of all frames given out by Y (namely,
the ⊆-largest frame class of Y ).
(2) Assume that the widths of semantics Y1 and Y 2 are S1 and S2,
respectively. Y1 has larger characterization width than Y2 ⇔ S1 ⊂ S2.
(3) Let ϕ is a property. Y1 has larger ϕ-characterization depth than
Y2 ⇔ there is a system S such that
there is the class of all frames satisfying ϕ of Y1 such that the class
characterizes S, but
there is not the class of all frames satisfying ϕ of Y2 such that the
class characterizes S. ┤
By the previous definition, we get immediately:
Theorem 6.4.7
(1) E is the width of the neighborhood semantics.
(2) K is the width of the relation semantics.
(3) The neighborhood semantics has larger characterization width
than the relation semantics. ┤
Theorem 6.4.8 (Gerson,  (p.80)) The neighborhood semantics
has larger normal-characterization depth than the relation semantics:
there is a system S such that NFN characterizes S but Fra does not
characterize S.
Proof. This proof is too long, omitted here, please refer to Gerson’s
. ┤
Remark.
There
are
still
other
317
differences
between
the
neighbourhood semantics and the relational semantics. For example,
according to Gerson’s [1975a], he gave out a supplementation system
of S4 such that it is complete for the neighbourhood semantics but
incomplete for the relational semantics.
Exercises 6.4
6.4.9 Do the following propositions hold? Where the superscript R
denotes the consequence relations of the relation semantics, and the
superscript N denotes the consequence relations of the neighborhood
semantics.
(1) Φ ⊨ R A ⇒ Φ ⊨ N A.
(2) Φ ⊨ N A ⇒ Φ ⊨ R A.
(3) Φ ⊨ R M A ⇒ Φ ⊨ N M A.
(4) Φ ⊨ N M A ⇒ Φ ⊨ R M A.
(5) Φ ⊨ R F A ⇒ Φ ⊨ N F A.
(6) Φ ⊨ N F A ⇒ Φ ⊨ R F A.
(7) Φ ⊨ Fra A ⇒ Φ ⊨ FN A.
(8) Φ ⊨ FN A ⇒ Φ ⊨ Fra A. ┤
6.4.10 Do the following propositions hold?
(1) MN →P Mod.
(2) MN → Mod.
(3) FN → Fra. ┤
6.4.11 Prove that Fra ∩ FN ≠ ∅. ┤
318
6.4.12* Let SUB be the set of all substitution mappings. If ③ in
Definition 3.1.1 (3) or ③ in Definition 6.1.1 (3) is replaced by the
following truth set definitions, respectively, then what can we obtain?
V(□A) = {w ∈ W | R(w) ⊆ V(Aσ) for all σ ∈ SUB},
V(□A) = {w ∈ W | R(w) ⊆ V(Aσ) for some σ ∈ SUB},
V(□A) = {w ∈ W | V(Aσ) ∈ N(w) for all σ ∈ SUB},
V(□A) = {w ∈ W | V(Aσ) ∈ N(w) for some σ ∈ SUB}. ┤
319
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[2004a] Comparing Consequence Relations of Modal Logic 模态逻

(Logic and Cognition) (online journal), 2004 (1).
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ZHOU Beihai (周北海)
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Blackburn, P., de Rijke, M. and Venema, Y.
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Boolos, G.
 The Logic of Provability. Cambridge University Press, 1993.
Bull, R.A. and Segerberg, K.
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Chellas, B.F.
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Dissertation, Simon Fraser University, 1974.
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logic. The Journal of Symbolic Logic, vol. 40, 141-148, 1975.
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semantics but incomplete for the relational semantics. Studia
Logica, vol. 34, 333-342, 1975.
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frame. Zeitschrift für Mathematische Logik und Grundlagen
der Mathematik, vol. 22, 29-34, 1976.
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323
Postscript
The following students has been taught by this course:
The graduate students of logic specialty of Philosophy
Department of Peking University in the year of 2001,
The students of Logic Class of Philosophy Pepartment of Sun
Yat-sen University and the graduate students of logic specialty of
Institute of Logic and Cognition of the University in the years of
2002-2007.
I am grateful to all students who seriously attended class and
completed exercises. The following students of Sun Yat-sen
University gave me profundity impressions:

I, depending on these students, in this course, more detailedly
interpreted things understood difficultly, or amended my clerical error
even things I could not have understood distinctly and my errors.
Dr. 郭 佳 宏 became a teacher of Philosophy Department of
Beijing Normal University after graduated from Sun Yat-sen
University in 2006. During teaching this course for the graduate
students of logic specialty of the department in first term of the year
of 2006, he found and told me that there are a certain clerical errors
and shortages in this course. I also want to express my gratitude to
him for his criticism and advice.
Finally, the point I need emphasis is that if the reader find that there
324
is still error and shortage in this course, please give me any criticism
and advice in any form, so that I can amend it at any moment. My email address is:
[email protected]
[email protected]
Modal logic introduced in this course is quite simple. If the reader
still want to study deeply modal logic, I recommend  of
Blackburn, de Rijke and Venema.
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