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Lecture 8
Modal Logic
1
Review of Greek Lower-Case
Letters
✤
Recall: we define formula as follows:
✤
✤
✤
We use lower-case greek letters
for arbitrary formulas:
✤
ϕ or φ, phi: pronounced “phee”
ψ, psi: pronounced “sai”
χ, chi; pronouced “kai”
α, alpha
β, beta
γ, gamma
δ, delta
ρ, rho, pronounced “row”
θ, theta
the basic propositional letters
p,q,r, . . . . are formulas
if ϕ is a formula, so is ¬ϕ, □ϕ, ⬦ϕ
✤
if ϕ, ψ, are formulas, then so are
(ϕ ∧ ψ), (ϕ ∨ ψ), (ϕ → ψ), and (ϕ ↔ ψ).
✤
Nothing else is a formula.
2
Outline of Today’s Lecture
✤
I. Hilbert-Style Deductive System for Propositional Logic
✤
II. Hilbert-Style Deductive System for Modal Propositional Logic
3
Outline of Today’s Lecture
✤
I. Hilbert-Style Deductive System for Propositional Logic
✤
II. Hilbert-Style Deductive System for Modal Propositional Logic
4
Two Kinds of Deductive Systems
for Propositional Logic
✤
One basic kind of deductive
system for propositional logic is
a natural deduction system. In
this kind of system, there is an
“introduction” rule for each
connective and an
“elimination” rule for each
connective. For instance, the
introduction rule for “and”
might say: if you can deduce ϕ
and if you can deduce ψ, then
you can deduce ϕ ∧ ψ.
5
✤
While natural deduction
systems are easy to learn, they
have lots of rules, and there is
no real analogue for modal
logic.
✤
Another kind of deductive
system is a Hilbert-style
deductive system. It has very few
rules and compensates by
having more axioms.
A Hilbert-Style Deductive System
for Propositional Logic
✤
The three parts of the system:
✤
I. A set of definitions and
tautologies, which serve as our
axioms.
✤
II. A set of rules, which say things
like “if such and such is a
theorem, then this related thing is
as well.”
✤
✤
III. A specification of what counts
as a proof in the system.
6
Let’s begin with the Part III.
Because it’s the simplest. A proof
of ϕ, or a demonstration that ⊢ϕ,
consists of a numbered list of
well-formed formulas such that
✤
the list ends with ϕ
✤
each formula on the list is an
axiom (from Part I) or an
earlier theorem,
or follows from some of the
earlier lines by the rules (from
Part II).
Description of Part I
✤
We assume, when we are setting up
the deductive system, that all the
connectives besides arrow and
negation are defined in terms of
arrow and negation as follows (cf.
section 2.1 of Sider):
✤
ϕ ∧ ψ is defined as ~(ϕ➝~ψ)
✤
ϕ ∨ ψ is defined as ~ϕ➝ψ
✤
✤
ϕ ↔ ψ is defined as (ϕ ➝ ψ) ∧ (ψ ➝ ϕ)
7
Practically, this means that we can
just state a small set of tautologies
concerning arrow and negation, and
not have to worry about stating
specific axioms concerning “or” and
“and” and “biconditional.”
Description of Part I, Continued
✤
✤
✤
✤
The tautologies of the deductive system are
the following list from section 2.6 of Sider.
✤
ϕ➝(ψ➝ϕ)
✤
ϕ ➝ ( ψ ➝ χ)) ➝ ((ϕ ➝ ψ ) ➝ (ϕ ➝ χ)) PL2
You can draw a truth-table to check that
each of these is a validity.
✤
(~ψ ➝ ~ϕ)➝((~ψ ➝ ϕ) ➝ ψ)
The “PL” stands for “propositional logic.”
✤
The way to think about PL1 is: if you assume ϕ, then whatever else ψ you
assume, you still have ϕ.
The way to view PL2 is to think of the
“ϕ➝” as distributing through the
conditional ψ➝χ.
✤
The way to think about PL3 is to think of
~ψ➝~ϕ as ψ∨~ϕ, and to think of ~ψ➝ϕ as
ψ∨ϕ. Hence, an equivalent formulation of
PL3 would be:
(ψ∨~ϕ)➝((ψ∨ϕ) ➝ ψ)
8
PL1
PL3
Description of Part II
Part II of the deductive system is a
set of rules, which say things like
“if such and such has a proof,
then this related thing has a proof
as well.”
✤
✤
This means: if you have ϕ➝ψ at
line ℓ𝓁, and you have ϕ at line ℓ𝓁’,
then you may write ψ at any
subsequent line ℓ𝓁’’.
✤
ℓ𝓁.
ϕ➝ψ
✤
ℓ𝓁’.
ϕ
✤
ℓ𝓁’’.
ψ
The first rule is:
✤
ϕ➝ψ
ϕ
ψ
MP
The MP stands for “modus
ponens”.
9
MP, ℓ𝓁, ℓ𝓁’.
Description of Part II
✤
The second rule is that you can
substitute defined notions.
✤
For instance, you can do:
✤
✤
ℓ𝓁. (ϕ➝ψ) ∧ (ψ➝ϕ)
✤
✤
ℓ𝓁’. (ϕ ↔ ψ)
✤
(Defn of ↔, ℓ𝓁).
✤
✤
You don’t need to substitute
everywhere either.
10
For instance, suppose you had a
proof that started out like:
ℓ𝓁. (~ϕ➝ψ) ∧ (~ϕ➝ψ)
ℓ𝓁’. (ϕ ∨ ψ)∧ (~ϕ➝ψ)
(Defn, ℓ𝓁)
In moving from ℓ𝓁 to ℓ𝓁’, we
substitute the defined notion ϕ∨ψ
in the first conjunct but not in the
second conjunct of line ℓ𝓁.
Theorem 1 + Discussion
✤
Let’s look at an example
derivation of ϕ➝(ψ ∨ ϕ).
✤
1. ϕ➝(~ψ➝ϕ)
2. ϕ➝(ψ ∨ ϕ)
✤
✤
✤
(PL1)
(Defn 1)
Note that line 1 is an instance of
PL1 since we substitute in ~ψ
for ψ.
Of course, in general, you
should not confuse ψ and ~ψ.
11
The idea of the axioms is that
they hold for all formulas, and
so PL1 ϕ➝(ψ➝ϕ) has all the
following instances:
✤
ϕ➝(~ψ➝ϕ)
✤
(~ϕ ∧ ψ) ➝(~ψ➝ (~ϕ ∧ ψ))
✤
ϕ➝(ϕ➝ϕ)
✤
~ϕ➝(ψ➝~ϕ)
Discussion, Continued
✤
But there’s a general warning
that should be given here:
✤
✤
✤
In doing this kind of
substitution, you have to
substitute uniformly.
For instance, PL1 ϕ➝(ψ➝ϕ)
does not have the following
substitution instance:
(~ϕ ∧ ψ) ➝(ψ➝ϕ)
For, you substituted in (~ϕ ∧ ψ)
for ϕ only in the first instance.
12
The way to remember this is as
follows:
✤
When doing substitution
instances of axioms (or
theorems), you must
substitute uniformly.
✤
When doing substitution
instances of definitions, you
may substitute selectively.
Example of Finding Substitution
Instance of an Axiom
✤
Consider PL3:
(~ψ ➝ ~ϕ)➝((~ψ ➝ ϕ) ➝ ψ)
✤
What do you get if you substitute in ϕ ∧ ψ for ψ?
✤
A. (~ψ ➝ ~(ϕ ∧ ψ))➝((~ψ ➝ (ϕ ∧ ψ)) ➝ ψ)
✤
B. (~(ϕ ∧ ψ) ➝ ~ϕ)➝((~(ϕ ∧ ψ) ➝ ϕ) ➝ (ϕ ∧ ψ))
✤
C. (~ ϕ ∧ ψ ➝ ~ϕ)➝((~ψ ➝ ϕ) ➝ ψ)
✤
D. (~ψ ➝ ~ϕ)➝((~ψ ➝ ϕ) ➝ ϕ ∧ ψ)
13
Theorem 2
✤
Prove: ϕ ➝ ϕ
✤
Proof:
✤
1. ϕ ➝ (ϕ ➝ ϕ)
PL1
✤
2. ϕ ➝ ((ϕ ➝ϕ) ➝ ϕ)
PL1
✤
3. [ϕ ➝ ((ϕ ➝ϕ) ➝ ϕ)] ➝ [(ϕ ➝ (ϕ ➝ϕ)) ➝ (ϕ ➝ ϕ) ]
PL2
✤
4. (ϕ ➝ (ϕ ➝ϕ)) ➝ (ϕ ➝ ϕ)
MP, 2, 3
✤
5. ϕ ➝ ϕ
MP, 1, 4
14
Theorem 3
✤
Let’s look at an example
derivation of ~~ϕ ∨ ~ϕ.
✤
1. ~ϕ➝~ϕ
2. ~~ϕ ∨ ~ϕ
✤
(prev.)
(Defn, 1)
15
In the first line, we appealed to
the previous result Theorem 2.
Of course, as we stated it there,
the conclusion of Theorem 2
read:
ϕ ➝ ϕ
But again, the idea is that if you
have a result, you can substitute
uniformly.
Completeness Theorem
✤
The important result about the
Hilbert-style deductive calculus
itself is the completeness theorem:
✤
A formula ϕ of propositional
logic is has a proof if and only if
ϕ is a tautology.
✤
Hence if you can use a truthtable to test that a formula is a
tautology, then it has a proof in
the Hilbert-style system.
16
✤
It’s beyond the scope of the
methods we’ve developed to
actually prove the completeness
theorem.
✤
But we can use this result to
justify that anything we know
to be a tautology of ordinary
propositional logic has a proof
in the Hilbert-style system.
✤
The following slides contain a
big long list of tautologies.
A Big List of Tautologies
✤
#1. ϕ➝(ψ➝ϕ)
PL1
✤
#2. ϕ ➝ ( ψ ➝ χ)) ➝ ((ϕ ➝ ψ ) ➝ (ϕ ➝ χ))
PL2
✤
#3. (~ψ ➝ ~ϕ)➝((~ψ ➝ ϕ) ➝ ψ)
PL3
✤
#4. (ϕ ∧ ψ) ➝ ϕ
(analogue of “and” elim.)
✤
#5. (ϕ ∧ ψ) ➝ ψ
(analogue of “and” elim.)
✤
#6. (ϕ ➝ (ψ ➝ (ϕ ∧ ψ))
(analogue of “and” intro.)
✤
#7. (ϕ ➝ ψ) ➝ ((ϕ ➝ χ) ➝ (ϕ ➝ (ψ ∧ χ) ))
(analogue of “and” intro, with hypothesis ϕ)
✤
#8. (ϕ ➝ (ψ ➝ χ)) ➝ ( (ϕ ∧ ψ) ➝χ )
(analogue of “and” elim)
17
A Big List of Tautologies,
Continued
✤
#9. ϕ ➝ (ϕ ∨ ψ)
(analogue of “or” intro.)
✤
#10. ψ ➝ (ϕ ∨ ψ)
✤
#11. (ϕ ➝ χ) ➝ ( (ψ➝ χ) ➝ ((ϕ ∨ ψ) ➝ χ)) )
(analogue of “or” elimination)
✤
#12. (ϕ ∧ ψ) ↔ (ψ ∧ ϕ)
(commutativity of conjunction)
✤
#13. (ϕ ∨ ψ) ↔ (ψ ∨ ϕ)
(commutative of disjunction)
✤
#14. ( ϕ ∧ ( ψ ∧ χ )) ↔ ( (ϕ ∧ ψ) ∧ χ )
(associativity of conjunction)
✤
#15. ( ϕ ∨ ( ψ ∨ χ )) ↔ ( (ϕ ∨ ψ) ∨ χ )
(associativity of disjunction)
(analogue of “or” intro.)
18
A Big List of Tautologies,
Continued
✤
#16. ϕ ↔ ~~ ϕ
(Double negation Law 1)
✤
#17. ϕ ∨ ~ ϕ
(Double negation Law 2)
✤
#18. (ϕ ∨ ψ) ↔ ~(~ ϕ ∧ ~ ψ)
(DeMorgan Law 1)
✤
#19. (ϕ ∧ ψ) ↔ ~(~ ϕ ∨ ~ψ)
(DeMorgan Law 2)
✤
#20. (ϕ ∧ (ψ ∨ χ)) ↔ (ϕ ∧ ψ ) ∨ (ϕ ∧ χ))
(Distribution Law 1)
✤
#21. (ϕ ∨ (ψ ∧ χ)) ↔ (ϕ ∨ ψ ) ∧ (ϕ ∨ χ))
(Distribution Law 2)
19
A Helpful Derived Rule
✤
Here’s a helpful derived rule:
If ϕ is a theorem, and if ψ is a theorem, then ϕ ∧ ψ is a
theorem.
✤
Why should we trust this rule?
✤
Well, suppose that we had
(1) ϕ
assumption
(2) ψ
assumption
✤
From our list (#6) we have
(3) (ϕ ➝ (ψ ➝ (ϕ ∧ ψ)) (PL)
20
✤
Then from MP we obtain:
(4) ψ ➝ (ϕ ∧ ψ)
MP 1, 3
✤
And from MP we obtain
(5) ϕ ∧ ψ
MP 2, 4
Another Helpful Derived Rule
✤
Here’s a helpful derived rule:
If ϕ ➝ ψ is a theorem, and if ψ ➝ χ is a theorem, then ϕ ➝ χ
is a theorem.
✤
Why should we trust this rule?
✤
Well, suppose that we had:
(1) ϕ ➝ ψ
assumption
(2) ψ ➝ χ
assumption
✤
From our derived rule, we have
(3) (ϕ ➝ ψ) ∧ (ψ ➝ χ)
DR 1,2
21
✤
From truth-tables, we know
that the following is a
tautology:
((ϕ ➝ ψ) ∧ (ψ ➝ χ)) ➝ (ϕ ➝χ) Hence by the completeness
theorems, we can put it on a line
and justify it by ‘prop logic’:
(4) ((ϕ ➝ ψ) ∧ (ψ ➝ χ)) ➝ (ϕ ➝χ) PL
✤
Then we can argue by MP:
(5) ϕ ➝χ
MP 3, 4
Summary of Axioms, Rules, Derived Rules,
and Theorems of Hilbert-Style System
Axioms:
✤
Derived Rules:
ϕ
✤
ϕ➝(ψ➝ϕ)
PL1
ϕ ➝ ( ψ ➝ χ)) ➝ ((ϕ ➝ ψ ) ➝ (ϕ ➝ χ)) PL2
(~ψ ➝ ~ϕ)➝((~ψ ➝ ϕ) ➝ ψ)
PL3
ϕ∧ψ
ϕ ∧ ψ is defined as ~(ϕ➝~ψ)
ϕ ∨ ψ is defined as ~ϕ➝ψ
ϕ ↔ ψ is defined as (ϕ ➝ ψ) ∧ (ψ ➝ ϕ)
ψ➝χ
ϕ ➝χ
Abbreviated in proofs as ‘DR’,
simply because context allows us
to figure out which one is intended.
MP
ψ
Abbreviated in proofs as ‘MP’
ϕ➝ψ
Rules:
Modus Ponens:
ϕ➝ψ
ϕ
✤
ψ
22
Outline of Today’s Lecture
✤
I. Hilbert-Style Deductive System for Propositional Logic
✤
II. Hilbert-Style Deductive System for Modal Propositional Logic
23
The K-Deductive System
✤
Today we describe the so-called Kdeductive system for Modal
Propositional Logic.
✤
The three parts of the system:
✤
I. A set of definitions and tautologies,
which serve as our axioms.
✤
II. A set of rules, which say things like
“if such and such is a theorem, then
this related thing is as well.”
✤
III. A specification of what counts as a
proof in the system.
24
✤
Part I expands the list of
tautologies by the K-axiom;
and by a definition of
diamond in terms of box.
✤
Part II expands the list of
rules by a single rule, known
as the necessitation rule.
✤
Part III is exactly like the
deductive system for
ordinary propositional logic.
The K-axiom
✤
Our first axiom is the K-axiom,
where the ‘K’ stands for Kripke.
✤
It say:
□( ϕ → ψ ) → (□ϕ → □ψ)
✤
✤
Intuitively, it’s saying that
whatever necessarily follows
from a necessary truth is itself
necessary.
✤
Or, in a word, necessary truth is
closed under logical
consequence.
That is, it says: if it’s necessary
that ϕ implies ψ, then necessarily
ϕ implies necessarily ψ.
✤
25
Or, more colloquially, if ϕ→ψ had to be the case, and if
ϕ had to be the case, then also ψ
had to be the case.
Example of Finding Substitution
Instance of an Axiom
✤
Consider the K-axiom: □( ϕ → ψ ) → (□ϕ → □ψ)
✤
What do you get if you substitute in ϕ ∨ ψ for ψ?
✤
A. □( ϕ → ψ ) → (□ϕ → □(ϕ ∨ ψ))
✤
B. □( (ϕ ∨ ψ) → ψ ) → (□(ϕ ∨ ψ) → □ψ)
✤
C. □( ϕ → (ϕ ∨ ψ) ) → (□ϕ → □ψ)
✤
D. □( ϕ → (ϕ ∨ ψ) ) → (□ϕ → □(ϕ ∨ ψ))
26
The Necessitation Rule
✤
✤
The necessitation rule says that
if ϕ is a theorem, then also □ϕ is
a theorem.
✤
So what does this rule say?
✤
It says that if ϕ is proven, then
so is □ϕ.
✤
The necessitation rule is not
saying that this is a tautology:
Visually, it looks like this:
ϕ
□ϕ
NEC
(*)
ϕ ➝ □ϕ
Note: (*) is usually false: just
because Bill went to beach
doesn’t mean he had to.
27
A Helpful Derived Rule
✤
It’s helpful to note that we have
the following derived rule:
if ϕ ➝ ψ is a theorem, then so is
□ϕ ➝ □ψ.
✤
Then by the Necessity Rule NEC,
we know that the following is a
theorem:
(2)
□(ϕ ➝ ψ)
NEC, 1
✤
Why should we trust this rule?
✤
✤
Well, suppose that know that the
following is proven:
(1)
ϕ➝ψ
assumption
Now, one instance of the K-axiom
is the following:
(3) □(ϕ ➝ ψ) ➝ (□ϕ ➝ □ψ) K
✤
So by modus ponens rule from
(2) and (3), the following is a
tautology:
(4)
(□ϕ ➝ □ψ) MP 2,3
28
First Theorem
✤
Our first theorem is:
□(ϕ ∧ ψ) ➝ (□ϕ ∧ □ψ)
✤
Now recall this tautology of
propositional logic (#7 on list):
(ϕ ➝ ψ) ➝ ((ϕ ➝ χ) ➝ (ϕ ➝ (ψ ∧ χ) ))
✤
Here’s the proof.
✤
First, we have that the following
are theorems of prop. logic:
(1)
(ϕ ∧ ψ) ➝ ϕ
PL
(2)
(ϕ ∧ ψ) ➝ ψ
PL
✤
Hence, we have the following:
(5) (□(ϕ∧ψ) ➝ □ϕ) ➝
((□(ϕ∧ψ) ➝ □ψ) ➝ (□(ϕ∧ψ) ➝ (□ϕ ∧ □ψ ) ) PL
✤
Then by derived rule, we have the
following are theorems:
(3) □(ϕ∧ψ) ➝ □ϕ
DR, 1
(4) □(ϕ∧ψ) ➝ □ψ
DR, 2
29
Then by two applications of MP:
(6) ((□(ϕ∧ψ) ➝ □ψ) ➝ (□(ϕ∧ψ) ➝ (□ϕ ∧ □ψ ) )) MP 3,6
(7) (□(ϕ∧ψ) ➝ (□ϕ ∧ □ψ ) )) MP 4,7
First Theorem
✤
On the previous slide, we had the commentary mixed in with the proof. The proof itself would just look like
the following:
✤
Prove in the K-deductive system: □(ϕ ∧ ψ) ➝ (□ϕ ∧ □ψ).
✤
Proof:
✤
(1) (ϕ ∧ ψ) ➝ ϕ
PL
✤
(2) (ϕ ∧ ψ) ➝ ψ
PL
✤
(3) □(ϕ∧ψ) ➝ □ϕ
DR, 1
✤
(4) □(ϕ∧ψ) ➝ □ψ
DR, 2
✤
(5) (□(ϕ∧ψ) ➝ □ϕ) ➝ ((□(ϕ∧ψ) ➝ □q) ➝ (□(ϕ∧ψ) ➝ (□ϕ ∧ □ψ ) ))
PL
✤
(6) ((□(ϕ∧ψ) ➝ □ψ) ➝ (□(ϕ∧ψ) ➝ (□ϕ ∧ □ψ ) ))
MP 3,5
✤
(7) (□(ϕ∧ψ) ➝ (□ϕ ∧ □ψ ) ))
MP 4,6
30
Second Theorem
✤
Our second theorem:
(□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ)
✤
Recall the following analogue
of the introduction rule for
‘and’ (#6 on list):
(1) ϕ ➝ (ψ ➝ (ϕ ∧ ψ))
✤
✤
By applying derived rule to it,
(2) □ϕ ➝ □(ψ➝ (ϕ ∧ ψ)) DR 1
✤
Then by our derived rule on p.
19 of the slides today, we get:
(4) □ϕ ➝ (□ψ➝ □(ϕ ∧ ψ)) DR 2,3
✤
The following is an instance of
a tautology from the list (#8):
(5) (□ϕ ➝ (□ψ ➝ □(ϕ ∧ ψ))) ➝
( (□ϕ ∧ □ψ) ➝□(ϕ ∧ ψ) ) PL
✤
One instance of K is (3) □(ψ➝ (ϕ ∧ ψ)) ➝ (□ψ➝ □(ϕ ∧ ψ)) K
31
(6) Then by MP, we get:
(□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ) MP 4,5
Second Theorem
✤
Prove in the K-deductive system: (□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ)
✤
Proof:
✤
(1) ϕ ➝ (ψ ➝ (ϕ ∧ ψ))
PL
✤
(2) □ϕ ➝ □(ψ➝ (ϕ ∧ ψ))
DR 1
✤
(3) □(ψ➝ (ϕ ∧ ψ)) ➝ (□ψ➝ □(ϕ ∧ ψ))
K
✤
(4) □ϕ ➝ (□ψ➝ □(ϕ ∧ ψ))
DR 2,3 ✤
(5) (□ϕ ➝ (□ψ ➝ □(ϕ ∧ ψ))) ➝ ( (□ϕ ∧ □ψ) ➝□(ϕ ∧ ψ) )
PL
✤
(6) (□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ)
MP 4,5
32
Third theorem. Putting the first
two together.
✤
So we just proved:
□(ϕ ∧ ψ) ➝ (□ϕ ∧ □ψ)
(□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ)
✤
Now let’s prove:
□(ϕ ∧ ψ) ↔ (□ϕ ∧ □ψ)
✤
First, we list our earlier results,
and just justify them as previous
result:
(1) □(ϕ ∧ ψ) ➝ (□ϕ ∧ □ψ) prev.
(2) (□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ) prev.
33
✤
Then we use the derived rule
from p. 18 of the slides to obtain
the conjunction of the two
previous lines:
(3) ((□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ))
∧ ((□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ)) DR 1,2
✤
Then we recall from p. 6 of the
slides that p↔q is defined as (p➝q) ∧ (q➝p). So we can write:
(4) □(ϕ ∧ ψ) ↔ (□ϕ ∧ □ψ) Defn, 3
Third Theorem
✤
Prove in the K-deductive system: □(ϕ ∧ ψ) ↔ (□ϕ ∧ □ψ)
✤
Proof:
✤
(1) □(ϕ ∧ ψ) ➝ (□ϕ ∧ □ψ)
prev.
✤
(2) (□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ)
prev.
✤
(3) ((□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ)) ∧ ((□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ))
DR 1,2
✤
(4) □(ϕ ∧ ψ) ↔ (□ϕ ∧ □ψ)
Defn, 3
34
A fourth theorem
✤
Our final theorem is:
(□ϕ ∨ □ψ) ➝ □(ϕ ∨ ψ)
✤
We have the following tautologies
of propositional logic:
(1) ϕ ➝ (ϕ ∨ ψ)
PL
(2) ψ ➝ (ϕ ∨ ψ)
PL
✤
Applying our derived rule to
these, we get
(3) □ϕ ➝ □(ϕ ∨ ψ) DR 1 (4) □ψ ➝ □(ϕ ∨ ψ)
DR 2
✤
Then we look at our big list, and
we find (#11) the following:
(ϕ ➝ χ) ➝ ( (ψ➝ χ) ➝ ((ϕ ∨ ψ) ➝ χ)) ) A substitution instance of this is
the following:
(5) (□ϕ ➝ □(ϕ ∨ ψ)) ➝ [ (□ψ ➝ □(ϕ ∨ ψ)) ➝
((□ϕ ∨ □ψ) ➝ □(ϕ ∨ ψ))] PL
✤
35
Then by modus ponens, we get:
(6) (□ψ ➝ □(ϕ ∨ ψ) ) ➝
((□ϕ ∨ □ψ) ➝ □(ϕ ∨ ψ)) MP 3,5
(7) (□ϕ ∨ □ψ) ➝ □(ϕ ∨ ψ) MP 4,6
Fourth Theorem
✤
Prove in the K-deductive system: (□ϕ ∨ □ψ) ➝ □(ϕ ∨ ψ)
✤
Proof:
✤
(1) ϕ ➝ (ϕ ∨ ψ)
PL
✤
(2) ψ ➝ (ϕ ∨ ψ)
PL
✤
(3) □ϕ ➝ □(ϕ ∨ ψ)
DR 1
✤
(4) □ψ ➝ □(ϕ ∨ ψ)
DR 2
✤
(5) (□ϕ ➝ □(ϕ ∨ ψ)) ➝ [ (□ψ ➝ □(ϕ ∨ ψ)) ➝ ((□ϕ ∨ □ψ) ➝ □(ϕ ∨ ψ))]
✤
(6) (□ψ ➝ □(ϕ ∨ ψ) ) ➝ ((□ϕ ∨ □ψ) ➝ □(ϕ ∨ ψ))
MP 3,5
✤
(7) (□ϕ ∨ □ψ) ➝ □(ϕ ∨ ψ)
MP 4,6
36
PL
Summary of Axioms, Rules, Derived
Rules, and Theorems of System K
K-axiom:
□( ϕ → ψ ) → (□ϕ → □ψ)
Abbreviation in proofs: ‘K’
✤
□ϕ
□ϕ→□ψ
Abbreviation in proofs: DR
ϕ
ϕ→ψ
Necessitation Rule:
✤
Derived Rule:
✤
NEC
✤
Abbreviation in proofs: ‘NEC’
37
Theorems:
1. □(ϕ ∧ ψ) ➝ (□ϕ ∧ □ψ)
2. (□ϕ ∧ □ψ) ➝ □ (ϕ ∧ ψ)
3. □(ϕ ∧ ψ) ↔ (□ϕ ∧ □ψ)
4. (□ϕ ∨ □ψ) ➝ □(ϕ ∨ ψ)
Costs and Benefits of the
Deductive System
✤
One huge cost of the deductive
system is that you have to be
able to do clever substitutions
on really long formulas, and
you have to remember what
you proved earlier, so that you
don’t have to recreate proofs on
the fly.
38
✤
One benefit of the deductive
system is that you don’t have to
mention possible worlds at all.
Hence, we have a route to the
notions of necessity and
possibility which does not
depend on the notion of
possible world.
✤
Compare: proving things about
geometry using a deductive
system vs. using intuitions
about space.
What do I need to know about the deductive
system (for the purposes of this class)?
✤
You need to know the following
things:
✤
1. What the K-axiom and the
Necessitation rule say.
✤
2. The additional axioms that
we will go over next time,
which are specific to the
systems D, T, B, S4, S5.
✤
✤
3. How to read proofs in the
deductive system.
39
You don’t need to know the
following things:
✤
1. How to create/write
proofs in the deductive
system.
✤
2. How to tell automatically
whether any of the big long
formulas we went over today
are tautologies of
propositional logic.
Ω
40