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Compactness
First order logic
Alexander Clark
Autumn 2014
Compactness theorem
Compactness theorem
if every finite subset of a set Φ of wffs is satisfiable, then Φ is
satisfiable.
Alternative: contrapositive
If Φ α then there is a finite subset Φ0 ⊆ Φ such that Φ0 α.
Compactness
Definition
A set Φ of wffs is finitely satisfiable if every finite subset Σ of Φ
is satisfiable.
Compactness theorem
Φ is satisfiable if and only if it is finitely satisfiable.
Proof:
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⇒ is immediate.
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⇐ is immediate if Φ is finite
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So we just need to show that if Φ is infinite and finitely
satisfiable then it is satisfiable. We do this by constructing
a v that satisfies Φ.
Extending Φ
We enumerate all wff: α1 , α2 , . . .
Now we define the following enumeration of sets of wffs:
Φ0 = Φ
Φn+1
(
Φn ; αn+1
=
Φn ; (¬αn+1 )
if this is finitely satisfiable
otherwise
[
Φ̄ =
Φn
n∈N
Note: we are making things harder . . .
Claim 1
Φ̄ is finitely satisfiable.
We show that for all n, Φn is finitely satisfiable, by induction on
n.
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Base case: Φ0 is fs.
Inductive step: suppose Φn is fs. We want to show that
Φn+1 is.
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Suppose Φn is fs, but neither Φn ; αn+1 nor Φn ; ¬αn+1 is.
Then we have finite subsets of Φn , Γ1 , Γ2 such that Γ1 ; αn+1
is unsatisfiable and Γ2 ; ¬αn+1 is unsatisfiable.
This means that Γ1 ∪ Γ2 is unsatisfiable. (if v satisfied it,
what could v̄ (αn+1 ) be?) This contradicts the fs of Φn .
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This establishes that for all n, Φn is fs.
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Suppose Φ̄ was not fs; then there would be some finite
subset Γ that was not fs.
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But if it is finite, then it must be included in some Φn for
sufficiently large n, and this would mean that that Φn was
not fs. Contradiction.
Therefore Φ̄ is fs.
Claim 2
For every wff α, (¬α) ∈ Φ̄ iff α 6∈ Φ̄.
Proof:
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We can’t have {α, (¬α)} both in Φ̄ as this would be a finite
unsatisfiable set.
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Conversely, suppose α 6∈ Φ̄. α occurs somewhere in the
enumeration that we used to construct Φ̄. Say α = αn . If
αn is not in Φn , then (¬αn ) must be in Φn and thus in Φ̄.
Now we need to show that Φ̄ is satisfiable, which will imply that
Φ is, because Φ ⊆ Φ̄. We do this by explicitly constructing a
truth assignment v that satisfies Φ.
For every sentence symbol Ai we define
(
T if Ai ∈ Φ̄
v (Ai ) =
F if Ai 6∈ Φ̄
Note that this is the only possibility! No other truth assignment
could satisfy Φ̄, though there might be many ones that satisfy Φ.
Claim
For all wff α, v̄ (α) = T iff α ∈ Φ̄.
This implies that v satisfies Φ̄.
Prove that v̄ (α) = T iff α ∈ Φ̄
Proof by induction on complexity of α.
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Base case: α = Ai . True by construction of v .
Prove that v̄ (α) = T iff α ∈ Φ̄
Induction: negation
Suppose α = (¬β).
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v̄ (α) = T
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iff v̄ (β) = F (by definition of v̄ )
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iff β 6∈ Φ̄ (by induction)
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iff (¬β) ∈ Φ̄ (by maximality)
Induction: conjunction
Suppose α = (β ∧ γ).
We have to show:
v̄ (α) = T ⇔ (β ∧ γ) ∈ Φ̄
⇒
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v̄ (α) = T
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then v̄ (β) = T and v̄ (γ) = T
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then β ∈ Φ̄ and γ ∈ Φ̄
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Suppose (β ∧ γ) 6∈ Φ̄.
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Then (¬(β ∧ γ)) ∈ Φ̄.
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But
{β, γ, (¬(β ∧ γ))}
is unsatisfiable; which is a contradiction.
Induction: conjunction
Suppose α = (β ∧ γ).
We have to show:
v̄ (α) = T ⇔ (β ∧ γ) ∈ Φ̄
⇐
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If (β ∧ γ) ∈ Φ̄
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then β ∈ Φ̄ since we cannot have (¬β) in the set or it would
be unsatisfiable.
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And similarly γ ∈ Φ̄
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So v̄ (β) = T and v̄ (γ) = T (by induction)
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therefore v̄ (β ∧ γ) = T (by definition of v̄ ).
Analogously for the other connectives, where we have to show:
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β ∈ Φ̄ or γ ∈ Φ̄ iff (β ∨ γ) ∈ Φ̄
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β 6∈ Φ̄ or γ ∈ Φ̄ iff (β → γ) ∈ Φ̄
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β, γ ∈ Φ̄ or β, γ 6∈ Φ̄ iff (β ↔ γ) ∈ Φ̄
Therefore v satisfies Φ̄. Since Φ ⊆ Φ̄, v also satisfies Φ.
This finishes the proof of the compactness theorem.
Corollaries of the compactness theorem
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If Φ α then there is a finite subset Φ0 ⊆ Φ such that
Φ0 α.
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So if φ α then there is a tautology of the form
(α1 ∧ · · · ∧ αn ) → α such that {α1 , . . . , αn } ⊆ Φ.
Maximally satisfiable set
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Φ̄ is a maximally satisfiable set. Adding only one new wff to
Φ̄ would result in an unsatisfiable set (why?), hence the
maximality.
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It is satisfiable, but there is only one, unique, truth
assignment that satisfies this. (Because for any sentence
symbol Ai either Ai or (¬Ai ) is in Φ̄.
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So there is a one-one correspondence between truth
assignments and maximally satisfiable sets. In other
words: A set Γ is maximally satisfiable if and only it is of the
form
Γ = {α ∈ WFF | v̄ (α) = T}
for some truth assignment v .
We are now going to consider the definitions of the language
and its semantics from an abstract perspective. What are the
properties of the language that make the semantics possible?
The essential properties will turn out to be
1. the inductive definition of the language on the basis of a
set of sentence symbols,
2. The unique readability of the well-formed formulas.
We define abstractions of (1) in the notion of inductive sets that
are freely generated from the basic element (an abstraction of
(2)).
In that case, the recursion theorem guarantees that any
mapping v from the set of basic elements to a set V can be
homomorphically extended to a mapping v̄ mapping the whole
inductive set to V .
What is special about the language of Sentential Logic and its
truth assignments is that the latter are mappings to a set V with
two elements.
Abstract Induction
We have an initial set B ⊆ U and a set F of functions.
As an example F = {f , g}, f : U × U 7→ U, g : U 7→ U.
(Think of U as the set of all L-strings, B as the set of sentential
letters, f as ∧ and g as ¬.)
A subset S of U is closed under F if
x, y ∈ S ⇒ f (x, y ) ∈ S, g(x) ∈ S
A set C ⊆ U is inductive if
1. B ⊆ C
2. C is closed under f , g
Let C ∗ be the intersection of all inductive sets.
Fact: C ∗ is inductive and it is the smallest inductive set.
We call C ∗ the set generated from B by the functions in F.
C ∗ is freely generated from B by F if
1. The functions in F restricted to C ∗ are one-to-one
2. The ranges of the restrictions of the functions in F to C ∗
and B are mutually disjoint
Abstract Induction and the Definition of wffs
We now formulate the definition of the language of Sentential
Logic in the abstract framework.
Let B be the set of all sentence symbols, let U be the set of all
L-strings, and let F consist of the 5 formula building operations
1. E¬ (α) = (¬α)
2. E∧ (α, β) = (α ∧ β)
3. E∨ , E→ , E↔ similarly defined
Example
(A ∧ ((¬B) → C)) = E∧ (A, E→ (E¬ (A), C))
Note that if α is not a sentential symbol then either it is E¬ (β)
for some β or E (β, γ) for some binary connective . That is,
every wff α can be expressed in terms of formula building
operations and sentence symbols.
The language of sentential logic, that is, the set of all wffs,
corresponds to C ∗ , the intersection of all inductive sets w.r.t. B
and F. By the unique readability theorem C ∗ is freely generated
from the set of sentence symbols by the functions in F.
This guarantees the uniqueness of the extension of truth
assignment v to v̄ by the recursion theorem below.
Abstract Recursion
Let set C ∗ be generated from set B by the functions in F. (We
let F = {f , g} as under abstract induction)
To define a function h on C ∗ recursively we need
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Rules for computing h(x) for x ∈ B
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Rules for computing h(f (x, y )) using h(x) and h(y )
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Rules for computing h(g(x)) using h(x).
If we have these rules, then there is at most one function h
satisfying the requirements. If C ∗ is freely generated, then
there is exactly one such function.
Recursion theorem
Assume that a subset C ∗ of U is freely generated from B by f
and g. Assume that V is a set and F , G and h functions such
that
h : B 7→ V and F : V × V 7→ V and G : V 7→ V then there is a
unique function h̄ : C ∗ 7→ V such that:
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h̄(x) = h(x) for x ∈ B
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h̄(f (x, y )) = F (h̄(x), h̄(y )), for x, y ∈ C ∗
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h̄(g(x)) = G(h̄(x)), for c ∈ C ∗
Abstract Recursion and the Definition of the Semantics
In the recursive definition of the truth assignment v̄ we have:
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B=L
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C ∗ is the set of all wff
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V = {T, F}
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h = v , a function from B to V
Abstract Recursion and the Definition of the Semantics
With each formula building operation in F we associate an
operation on the set of truth values:
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With E¬ we have F¬ defined as F¬ (T) = F and
F¬ (F) = true
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With E∧ we have a binary function F∧ defined by the truth
x y F∧ (x, y )
T T T
table: T F F
F T F
F F F
The other functions F∨ , F→ , F↔ are defined analogously.
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Then v̄ is the unique function (h̄) from wffs to V guaranteed by
the recursion theorem.