Feb 15 Lecture MAT309, Winter 2018 Announcements: Problem Set #3 - - PowerPoint PPT Presentation

Feb 15 Lecture MAT309, Winter 2018 Announcements: Problem Set #3 posted on website. Due date: Tuesday, March 6 Midterm #2 will take place in class on Tuesday, March 13

Feb 15 Lecture MAT309, Winter 2018 Announcements: Problem Set #3 posted on website. Due date: Tuesday, March 6 Midterm #2 will take place in class on Tuesday, March 13 Completeness Theorem. If Σ | = ϕ, then Σ ⊢ ϕ. Follows from: Model Existence Theorem. Every consistent set of sentences has a model. (If Σ ⊢ ⊥, then there exists a model A | = Σ.)

Proof of the Model Existence Theorem

Let L be a countable language, and let Σ be a consistent set of L-sentences.

Let L be a countable language, and let Σ be a consistent set of L-sentences. We construct a language L′ = L ∪ {countably many new constant symbols} and a set of L′-sentences Σ such that (1) Σ ⊆ Σ, (2)

• Σ is consistent,

(3)

• Σ contains a Henkin axiom (∃xθ) → θx

c for each L′-sentence ∃xθ.

Let L be a countable language, and let Σ be a consistent set of L-sentences. We construct a language L′ = L ∪ {countably many new constant symbols} and a set of L′-sentences Σ such that (1) Σ ⊆ Σ, (2)

• Σ is consistent,

(3)

• Σ contains a Henkin axiom (∃xθ) → θx

c for each L′-sentence ∃xθ.

We then complete Σ to a maximal consistent set of L′-sentences Σ′ (by adding either ϕ or ¬ϕ one-at-a-time for each L′-sentence ϕ). In addition to properties (1)–(3), Σ′ has the property: (4) For every L′-sentence ϕ, ϕ ∈ Σ′ ⇐ ⇒ Σ′ ⊢ ϕ ⇐ ⇒ ¬ϕ / ∈ Σ′ ⇐ ⇒ Σ′ ⊢ ¬ϕ.

The L′-Structure A: We define an equivalence relation ∼ on {variable-free L′-terms} by t ∼ u

def

⇐ ⇒ (t = u) ∈ Σ′ ⇐ ⇒ Σ′ ⊢ (t = u).

The L′-Structure A: We define an equivalence relation ∼ on {variable-free L′-terms} by t ∼ u

def

⇐ ⇒ (t = u) ∈ Σ′ ⇐ ⇒ Σ′ ⊢ (t = u). CLAIM: ∼ is symmetric. PROOF: Assume t ∼ u. Then Σ′ ⊢ (t = u). We know ⊢ (∀x)(∀y)[(x = y) → (y = x)] (from Chapter 2). Therefore, ⊢ (t = u) → (u = t) by (Q1) axiom and (PC) rule. Therefore, Σ′ ⊢ (u = t) by (PC) rule; hence u ∼ t.

The L′-Structure A: We define an equivalence relation ∼ on {variable-free L′-terms} by t ∼ u

def

⇐ ⇒ (t = u) ∈ Σ′ ⇐ ⇒ Σ′ ⊢ (t = u). Let A be the following structure:

• A := {[t]∼ : t is a variable-free L′-term}

(universe of A)

• cA := [c]∼

for each constant symbol c

• f A([t1]∼, . . . , [tn]∼) := [ft1 . . . tn]∼

for each n-ary f

• ([t1]∼, . . . , [tn]∼) ∈ RA

def

⇐ ⇒ Rt1 . . . tn ∈ Σ′ for each n-ary R

The L′-Structure A: We define an equivalence relation ∼ on {variable-free L′-terms} by t ∼ u

def

⇐ ⇒ (t = u) ∈ Σ′ ⇐ ⇒ Σ′ ⊢ (t = u). Let A be the following structure:

• A := {[t]∼ : t is a variable-free L′-term}

(universe of A)

• cA := [c]∼

for each constant symbol c

• f A([t1]∼, . . . , [tn]∼) := [ft1 . . . tn]∼

for each n-ary f

• ([t1]∼, . . . , [tn]∼) ∈ RA

def

⇐ ⇒ Rt1 . . . tn ∈ Σ′ for each n-ary R NOTE: To show that f A and RA are well-defined, we must show that if ti ∼ ui for i = 1, . . . , n, then

• [ft1 . . . tn]∼ = [fu1 . . . un]∼ (that is, (ft1 . . . tn = fu1 . . . un) ∈ Σ′) and
• Rt1 . . . tn ∈ Σ′ ⇐

⇒ Ru1 . . . un ∈ Σ′.

Proposition 3.2.6. For every L′-sentence ϕ, we have ϕ ∈ Σ′ ⇐ ⇒ A | = ϕ.

Proposition 3.2.6. For every L′-sentence ϕ, we have ϕ ∈ Σ′ ⇐ ⇒ A | = ϕ. NOTE: This finishes the proof of the Model Existence Theorem (and hence also the Completeness Theorem):

• As a consequence of Prop 3.2.6, A |

= Σ′.

• Since Σ ⊂

Σ ⊂ Σ′, it follows that A | = Σ.

• Therefore, Σ has an L′-structure model.

(Recall that Σ is a set of L-sentences.)

• To show that Σ has an L-structure model, simply ignore the extra constant

symbols in A.

Proposition 3.2.6. For every L′-sentence ϕ, we have ϕ ∈ Σ′ ⇐ ⇒ A | = ϕ.

• PROOF. We argue by induction on the complexity1 of ϕ.

As usual, we consider five cases:

• 1. ϕ :≡ Rt1 . . . tn

(base case)

• 2. ϕ :≡ t1 = t2

(base case)

• 3. ϕ :≡ ¬α

(easy induction step)

• 4. ϕ :≡ α ∨ β

(easy induction step)

• 5. ϕ :≡ ∀xα

(non-trivial induction step)

1As defined by 2·#(∀ symbols in ϕ) + #(¬ symbols in ϕ) + #(∨ symbols in ϕ).

Proposition 3.2.6. For every L′-sentence ϕ, we have ϕ ∈ Σ′ ⇐ ⇒ A | = ϕ. CASE 1. ϕ :≡ Rt1 . . . tn We have ϕ ∈ Σ′ ⇐ ⇒ ([t1]∼, . . . , [tn]∼) ∈ RA ⇐ ⇒ A | = ϕ.

Proposition 3.2.6. For every L′-sentence ϕ, we have ϕ ∈ Σ′ ⇐ ⇒ A | = ϕ. CASE 2. ϕ :≡ t1 = t2 We use the fact (shown by induction) that s(t) = [t]∼ for every variable-free term t and s : Vars → A. It follows: A | = ϕ ⇐ ⇒ A | = ϕ[s] for every s : Vars → A ⇐ ⇒ s(t1) = s(t2) for every s : Vars → A ⇐ ⇒ [t1]∼ = [t2]∼ (since s(t) = [t]∼) ⇐ ⇒ (t1 = t2) ∈ Σ′ ⇐ ⇒ ϕ ∈ Σ′.

Proposition 3.2.6. For every L′-sentence ϕ, we have ϕ ∈ Σ′ ⇐ ⇒ A | = ϕ. CASE 3. ϕ :≡ ¬α A | = ϕ ⇐ ⇒ A | = α ⇐ ⇒ α / ∈ Σ′ (induction hypothesis applied to α) ⇐ ⇒ ¬α ∈ Σ′ (since α ∈ Σ′ ⇐ ⇒ ¬α ∈ Σ′) ⇐ ⇒ ϕ ∈ Σ′.

Proposition 3.2.6. For every L′-sentence ϕ, we have ϕ ∈ Σ′ ⇐ ⇒ A | = ϕ. CASE 4. ϕ :≡ α ∨ β Another trivial induction step.

Proposition 3.2.6. For every L′-sentence ϕ, we have ϕ ∈ Σ′ ⇐ ⇒ A | = ϕ. CASE 5. ϕ :≡ ∀xα ϕ ∈ Σ′ = ⇒ A | = ϕ

Proposition 3.2.6. For every L′-sentence ϕ, we have ϕ ∈ Σ′ ⇐ ⇒ A | = ϕ. CASE 5. ϕ :≡ ∀xα ϕ ∈ Σ′ = ⇒ A | = ϕ Assume ϕ ∈ Σ′. To show A | = ϕ, we must show A | = ϕ[s] for every s : Vars → A, that is, A | = α[s[x|a]] for every s : Vars → A and a ∈ A.

Proposition 3.2.6. For every L′-sentence ϕ, we have ϕ ∈ Σ′ ⇐ ⇒ A | = ϕ. CASE 5. ϕ :≡ ∀xα ϕ ∈ Σ′ = ⇒ A | = ϕ Assume ϕ ∈ Σ′. To show A | = ϕ, we must show A | = ϕ[s] for every s : Vars → A, that is, A | = α[s[x|a]] for every s : Vars → A and a ∈ A. Fix any s : Vars → A and a = [t]∼ ∈ A. We have A | = α[s[x|a]] ⇐ ⇒ A | = α[s[x|s(t)]] (since s(t) = [t]∼ = a) ⇐ ⇒ A | = αx

t [s]

(by Theorem 2.6.2) ⇐ ⇒ A | = αx

t

(since αx

t is a sentence).

Proposition 3.2.6. For every L′-sentence ϕ, we have ϕ ∈ Σ′ ⇐ ⇒ A | = ϕ. CASE 5. ϕ :≡ ∀xα ϕ ∈ Σ′ = ⇒ A | = ϕ Assume ϕ ∈ Σ′. To show A | = ϕ, we must show A | = ϕ[s] for every s : Vars → A, that is, A | = α[s[x|a]] for every s : Vars → A and a ∈ A. Fix any s : Vars → A and a = [t]∼ ∈ A. We have A | = α[s[x|a]] ⇐ ⇒ A | = α[s[x|s(t)]] (since s(t) = [t]∼ = a) ⇐ ⇒ A | = αx

t [s]

(by Theorem 2.6.2) ⇐ ⇒ A | = αx

t

(since αx

t is a sentence).

Since ϕ ∈ Σ′, we have Σ′ ⊢ ∀xα, hence Σ ⊢ αx

t by (Q1) axiom and (PC) rule.

(OBS: Since t is variable-free, it is substitutable for x.)

Proposition 3.2.6. For every L′-sentence ϕ, we have ϕ ∈ Σ′ ⇐ ⇒ A | = ϕ. CASE 5. ϕ :≡ ∀xα ϕ ∈ Σ′ = ⇒ A | = ϕ Assume ϕ ∈ Σ′. To show A | = ϕ, we must show A | = ϕ[s] for every s : Vars → A, that is, A | = α[s[x|a]] for every s : Vars → A and a ∈ A. Fix any s : Vars → A and a = [t]∼ ∈ A. We have A | = α[s[x|a]] ⇐ ⇒ A | = α[s[x|s(t)]] (since s(t) = [t]∼ = a) ⇐ ⇒ A | = αx

t [s]

(by Theorem 2.6.2) ⇐ ⇒ A | = αx

t

(since αx

t is a sentence).

Since ϕ ∈ Σ′, we have Σ′ ⊢ ∀xα, hence Σ ⊢ αx

t by (Q1) axiom and (PC) rule.

(OBS: Since t is variable-free, it is substitutable for x.) Therefore, αx

t ∈ Σ′. By induction hypothesis applied to αx t , it follows that

A | = αx

t . Hence, A |

= ϕ.

Proposition 3.2.6. For every L′-sentence ϕ, we have ϕ ∈ Σ′ ⇐ ⇒ A | = ϕ. CASE 5. ϕ :≡ ∀xα ϕ / ∈ Σ′ = ⇒ A | = ϕ

Proposition 3.2.6. For every L′-sentence ϕ, we have ϕ ∈ Σ′ ⇐ ⇒ A | = ϕ. CASE 5. ϕ :≡ ∀xα ϕ / ∈ Σ′ = ⇒ A | = ϕ Assume ϕ / ∈ Σ′. Then ¬ϕ ∈ Σ′, hence Σ′ ⊢ ¬ϕ :≡ ¬∀xα, hence Σ′ ⊢ ∃x¬α.

Proposition 3.2.6. For every L′-sentence ϕ, we have ϕ ∈ Σ′ ⇐ ⇒ A | = ϕ. CASE 5. ϕ :≡ ∀xα ϕ / ∈ Σ′ = ⇒ A | = ϕ Assume ϕ / ∈ Σ′. Then ¬ϕ ∈ Σ′, hence Σ′ ⊢ ¬ϕ :≡ ¬∀xα, hence Σ′ ⊢ ∃x¬α. The set Σ (and hence also Σ′) contains a Henkin axiom (∃x¬α) → (¬α)x

c.

Proposition 3.2.6. For every L′-sentence ϕ, we have ϕ ∈ Σ′ ⇐ ⇒ A | = ϕ. CASE 5. ϕ :≡ ∀xα ϕ / ∈ Σ′ = ⇒ A | = ϕ Assume ϕ / ∈ Σ′. Then ¬ϕ ∈ Σ′, hence Σ′ ⊢ ¬ϕ :≡ ¬∀xα, hence Σ′ ⊢ ∃x¬α. The set Σ (and hence also Σ′) contains a Henkin axiom (∃x¬α) → (¬α)x

c.

Hence Σ′ ⊢ (∃x¬α) → (¬α)x

• c. Therefore, by (PC) rule (modus ponens),

Σ′ ⊢ (¬α)x

c.

By the induction hypothesis applied to the sentence (¬α)x

c, we have A |

= (¬α)x

c.

Proposition 3.2.6. For every L′-sentence ϕ, we have ϕ ∈ Σ′ ⇐ ⇒ A | = ϕ. CASE 5. ϕ :≡ ∀xα ϕ / ∈ Σ′ = ⇒ A | = ϕ Assume ϕ / ∈ Σ′. Then ¬ϕ ∈ Σ′, hence Σ′ ⊢ ¬ϕ :≡ ¬∀xα, hence Σ′ ⊢ ∃x¬α. The set Σ (and hence also Σ′) contains a Henkin axiom (∃x¬α) → (¬α)x

c.

Hence Σ′ ⊢ (∃x¬α) → (¬α)x

• c. Therefore, by (PC) rule (modus ponens),

Σ′ ⊢ (¬α)x

c.

By the induction hypothesis applied to the sentence (¬α)x

c, we have A |

= (¬α)x

c.

It follows that A | = ∃x¬α, hence A | = ¬∃x¬α, hence A | = ϕ as required. Q.E.D.

Model Existence Theorem. Every consistent set of sentences has a model. Completeness Theorem. If Σ | = ϕ, then Σ ⊢ ϕ. Compactness Theorem. Σ | = ϕ iff Σ0 | = ϕ for some finite Σ0 ⊆ Σ.

The Compactness Theorem

COMPACTNESS THEOREM (Theorem 3.3.1) Let Σ be any set of

• formulas. Then Σ has a model (a.k.a. it is “satisfiable”) if, and only if, every

finite subset of Σ has a model.

COMPACTNESS THEOREM (Theorem 3.3.1) Let Σ be any set of

• formulas. Then Σ has a model (a.k.a. it is “satisfiable”) if, and only if, every

finite subset of Σ has a model.

• Proof. (⇒) trivial: If A |

= Σ, then A | = Σ0 for every finite Σ0 ⊆ Σ.

COMPACTNESS THEOREM (Theorem 3.3.1) Let Σ be any set of

• formulas. Then Σ has a model (a.k.a. it is “satisfiable”) if, and only if, every

finite subset of Σ has a model.

• Proof. (⇒) trivial: If A |

= Σ, then A | = Σ0 for every finite Σ0 ⊆ Σ. (⇐): Assume that Σ has no model. We will show that some finite Σ0 ⊆ Σ has no model.

COMPACTNESS THEOREM (Theorem 3.3.1) Let Σ be any set of

• formulas. Then Σ has a model (a.k.a. it is “satisfiable”) if, and only if, every

finite subset of Σ has a model.

• Proof. (⇒) trivial: If A |

= Σ, then A | = Σ0 for every finite Σ0 ⊆ Σ. (⇐): Assume that Σ has no model. We will show that some finite Σ0 ⊆ Σ has no model. Since Σ has no model, this means Σ | = ⊥. By the Completeness Theorem (or the contrapositive of the Model Existence Theorem), it follows that Σ ⊢ ⊥. That is, there exists a deduction (δ1, . . . , δn) of ⊥ from Σ.

COMPACTNESS THEOREM (Theorem 3.3.1) Let Σ be any set of

• formulas. Then Σ has a model (a.k.a. it is “satisfiable”) if, and only if, every

finite subset of Σ has a model.

• Proof. (⇒) trivial: If A |

= Σ, then A | = Σ0 for every finite Σ0 ⊆ Σ. (⇐): Assume that Σ has no model. We will show that some finite Σ0 ⊆ Σ has no model. Since Σ has no model, this means Σ | = ⊥. By the Completeness Theorem (or the contrapositive of the Model Existence Theorem), it follows that Σ ⊢ ⊥. That is, there exists a deduction (δ1, . . . , δn) of ⊥ from Σ. Let Σ0 := Σ ∩ {δ1, . . . , δn}. Observe that Σ0 is a finite set and Σ0 ⊢ ⊥ (since the same sequence (δ1, . . . , δn) is a deduction of ⊥ from Σ0). Therefore, Σ0 | = ⊥ by the Soundness Theorem. This means that Σ0 has no model, as required.

COMPACTNESS THEOREM (Theorem 3.3.1) Let Σ be any set of

• formulas. Then Σ has a model (a.k.a. it is “satisfiable”) if, and only if, every

finite subset of Σ has a model. Corollary 3.2.2. Σ | = ϕ iff Σ0 | = ϕ for some finite subset Σ0 of Σ.

COMPACTNESS THEOREM (Theorem 3.3.1) Let Σ be any set of

• formulas. Then Σ has a model (a.k.a. it is “satisfiable”) if, and only if, every

finite subset of Σ has a model. Corollary 3.2.2. Σ | = ϕ iff Σ0 | = ϕ for some finite subset Σ0 of Σ. Why is this called “Compactness Theorem”? Topological definition of a compact space: every open cover has a finite sub-

• cover. The Compactness Theorem implies compactness of a certain topological

space (whose points are the complete theories of a language L).

Applications of Compactness

• Definition. The theory of a structure A is the set Th(A) of all formulas true

in A: Th(A) := {ϕ : A | = ϕ}.

• Definition. The theory of a structure A is the set Th(A) of all formulas true

in A: Th(A) := {ϕ : A | = ϕ}.

• Definition. Two L-structures A and B are elementarily equivalent (denoted

A ≡ B) if Th(A) = Th(B).

• Definition. The theory of a structure A is the set Th(A) of all formulas true

in A: Th(A) := {ϕ : A | = ϕ}.

• Definition. Two L-structures A and B are elementarily equivalent (denoted

A ≡ B) if Th(A) = Th(B).

• Definition. If A is an LNT-structure such that A ≡ N, then we call A a

“model of arithmetic”. (This means that A is indistinguishable from N from the perspective of first-order logic.)

• Definition. The theory of a structure A is the set Th(A) of all formulas true

in A: Th(A) := {ϕ : A | = ϕ}.

• Definition. Two L-structures A and B are elementarily equivalent (denoted

A ≡ B) if Th(A) = Th(B).

• Definition. If A is an LNT-structure such that A ≡ N, then we call A a

“model of arithmetic”. (This means that A is indistinguishable from N from the perspective of first-order logic.) Theorem (Example 3.3.3). There exists a non-standard model of arith- metic (that is, a model of arithmetic which is non-isomorphic to the standard model N).

• Theorem. There exists a non-standard model of arithmetic.
• Proof. Let L := LNT ∪ {c} and let

Θ := Th(N) ∪ {0 < c, S0 < c, SS0 < c, . . . }. OBS: Every finite subset Θ0 ⊂ Θ has a model. Therefore, by the Compactness Theorem, Θ has a model A. The reduct A|L (forgetting the constant c) is elementarily equivalent to N. Note that A is non-isomorphic to N, since it contains “infinite” elements. This theorem shows that no set of axioms completely characterizes N up to isomorphism.