Review of Model Existence Theorem MAT309, Fall 2019 Completeness - PowerPoint PPT Presentation

Review of Model Existence Theorem MAT309, Fall 2019 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 .) 1

Completeness Theorem. If Σ | = ϕ , then Σ ⊢ ϕ . Model Existence Theorem. If Σ �⊢ ⊥ , then ∃ a model A | = Σ . Proof of Completeness Thm from Model Existence Thm. 1. Suffices to prove Completeness Theorem in the case where Σ is a set of sentences and ϕ is a sentence . (Follows from the fact that { α } ⊢ ∀ α and {∀ xα } ⊢ α for all formulas α .) 2. Assume Σ | = ϕ . Then A | = Σ ⇒ A | = ϕ for all structures A . Therefore, Σ ∪ {¬ ϕ } has no models. By Model Existence Theorem, Σ ∪ {¬ ϕ } ⊢ ⊥ . By Deduction Theorem, Σ ⊢ ( ¬ ϕ → ⊥ ). By (PC) rule, Σ ⊢ ϕ (via the tautology ( ¬ P → False ) → P )). Q.E.D. 2

Proof of the Model Existence Theorem 3

Let L be a countable language. (The case of uncountable languages uses Zorn’s Lemma, as explained in tutorial.) Let Σ be a consistent set of L -sentences. We will construct a model A | = Σ. 4

Let L be a countable language. (The case of uncountable languages uses Zorn’s Lemma, as explained in tutorial.) Let Σ be a consistent set of L -sentences. We will construct a model A | = Σ. We first construct a language L ′ = L∪{ countably many new constant symbols } and a set of L ′ -sentences � Σ such that Σ ⊆ � (1) Σ, � (2) Σ is consistent, � Σ contains a Henkin axiom ( ∃ xθ ) → θ x c for each L ′ -sentence ∃ xθ . (3) 5

Let L be a countable language. (The case of uncountable languages uses Zorn’s Lemma, as explained in tutorial.) Let Σ be a consistent set of L -sentences. We will construct a model A | = Σ. We first construct a language L ′ = L∪{ countably many new constant symbols } and a set of L ′ -sentences � Σ such that Σ ⊆ � (1) Σ, � (2) Σ is consistent, � Σ contains a Henkin axiom ( ∃ xθ ) → θ x c for each L ′ -sentence ∃ xθ . (3) Σ to a maximal consistent set of L ′ -sentences Σ ′ (by adding We then complete � either ϕ or ¬ ϕ one-at-a-time for each L ′ -sentence ϕ ). In addition to properties (1)–(3), Σ ′ has the property: For every L ′ -sentence ϕ , (4) ϕ ∈ Σ ′ ⇐ ⇒ Σ ′ ⊢ ϕ ⇐ ∈ Σ ′ ⇐ ⇒ Σ ′ �⊢ ¬ ϕ. ⇒ ¬ ϕ / 6

The L ′ -Structure A : We define an equivalence relation ∼ on { variable-free L ′ -terms } by ⇒ ( t = u ) ∈ Σ ′ ⇐ ⇒ Σ ′ ⊢ ( t = u ) . def t ∼ u ⇐ 7

The L ′ -Structure A : We define an equivalence relation ∼ on { variable-free L ′ -terms } by ⇒ ( t = u ) ∈ Σ ′ ⇐ ⇒ Σ ′ ⊢ ( t = u ) . def t ∼ u ⇐ CLAIM: ∼ is an equivalence relation. PROOF: Here is the argument that ∼ is symmetric: 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 . 8

The L ′ -Structure A : We define an equivalence relation ∼ on { variable-free L ′ -terms } by ⇒ ( t = u ) ∈ Σ ′ ⇐ ⇒ Σ ′ ⊢ ( t = u ) . def t ∼ u ⇐ Let A be the following structure: • A := { [ t ] ∼ : t is a variable-free L ′ -term } (universe of A ) • c A := [ c ] ∼ for each constant symbol c • f A ([ t 1 ] ∼ , . . . , [ t n ] ∼ ) := [ ft 1 . . . t n ] ∼ for each n -ary f def ⇒ Rt 1 . . . t n ∈ Σ ′ • ([ t 1 ] ∼ , . . . , [ t n ] ∼ ) ∈ R A ⇐ for each n -ary R 9

The L ′ -Structure A : We define an equivalence relation ∼ on { variable-free L ′ -terms } by ⇒ ( t = u ) ∈ Σ ′ ⇐ ⇒ Σ ′ ⊢ ( t = u ) . def t ∼ u ⇐ Let A be the following structure: • A := { [ t ] ∼ : t is a variable-free L ′ -term } (universe of A ) • c A := [ c ] ∼ for each constant symbol c • f A ([ t 1 ] ∼ , . . . , [ t n ] ∼ ) := [ ft 1 . . . t n ] ∼ for each n -ary f def ⇒ Rt 1 . . . t n ∈ Σ ′ • ([ t 1 ] ∼ , . . . , [ t n ] ∼ ) ∈ R A ⇐ for each n -ary R NOTE: To show that f A and R A are well-defined, we must show that if t i ∼ u i for i = 1 , . . . , n , then • [ ft 1 . . . t n ] ∼ = [ fu 1 . . . u n ] ∼ (that is, ( ft 1 . . . t n = fu 1 . . . u n ) ∈ Σ ′ ) and • Rt 1 . . . t n ∈ Σ ′ ⇐ ⇒ Ru 1 . . . u n ∈ Σ ′ . 10

Proposition 3.2.6. For every L ′ -sentence ϕ , we have ϕ ∈ Σ ′ ⇐ ⇒ A | = ϕ. 11

Proposition 3.2.6. For every L ′ -sentence ϕ , we have ϕ ∈ Σ ′ ⇐ ⇒ A | = ϕ. The Model Existence Theorem follows from Prop 3.2.5 as follows: = Σ ′ . • As a consequence of Prop 3.2.6, A | • Since Σ ⊂ � Σ ⊂ Σ ′ , it follows that A | = Σ. • If we ignore that interpretation of constant symbols in L ′ \ L , then the “reduct” A| L is a model of Σ in the original language L . 12

Proposition 3.2.6. For every L ′ -sentence ϕ , we have ϕ ∈ Σ ′ ⇐ ⇒ A | = ϕ. PROOF. We argue by induction on the complexity of ϕ . As usual, we consider five cases: 1. ϕ : ≡ Rt 1 . . . t n (base case) 2. ϕ : ≡ t 1 = t 2 (base case) 3. ϕ : ≡ ¬ α (easy induction step) 4. ϕ : ≡ α ∨ β (easy induction step) 5. ϕ : ≡ ∀ xα (non-trivial induction step) 13

Proposition 3.2.6. For every L ′ -sentence ϕ , we have ϕ ∈ Σ ′ ⇐ ⇒ A | = ϕ. CASE 1. ϕ : ≡ Rt 1 . . . t n We have ϕ ∈ Σ ′ ⇐ ⇒ ([ t 1 ] ∼ , . . . , [ t n ] ∼ ) ∈ R A ⇐ ⇒ A | = ϕ. 14

Proposition 3.2.6. For every L ′ -sentence ϕ , we have ϕ ∈ Σ ′ ⇐ ⇒ A | = ϕ. CASE 2. ϕ : ≡ t 1 = t 2 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 ( t 1 ) = s ( t 2 ) for every s : Vars → A ⇐ ⇒ [ t 1 ] ∼ = [ t 2 ] ∼ (since s ( t ) = [ t ] ∼ ) ⇒ ( t 1 = t 2 ) ∈ Σ ′ ⇐ ⇒ ϕ ∈ Σ ′ . ⇐ 15

Proposition 3.2.6. For every L ′ -sentence ϕ , we have ϕ ∈ Σ ′ ⇐ ⇒ A | = ϕ. CASE 3. ϕ : ≡ ¬ α A | = ϕ ⇐ ⇒ A �| = α ∈ Σ ′ (induction hypothesis applied to α ) ⇐ ⇒ α / ⇒ ¬ α ∈ Σ ′ (since α ∈ Σ ′ ⇐ ⇒ ¬ α �∈ Σ ′ ) ⇐ ⇒ ϕ ∈ Σ ′ . ⇐ 16

Proposition 3.2.6. For every L ′ -sentence ϕ , we have ϕ ∈ Σ ′ ⇐ ⇒ A | = ϕ. CASE 4. ϕ : ≡ α ∨ β Another trivial induction step. 17

Proposition 3.2.6. For every L ′ -sentence ϕ , we have ϕ ∈ Σ ′ ⇐ ⇒ A | = ϕ. ϕ ∈ Σ ′ = CASE 5. ϕ : ≡ ∀ xα ⇒ A | = ϕ 18

Proposition 3.2.6. For every L ′ -sentence ϕ , we have ϕ ∈ Σ ′ ⇐ ⇒ A | = ϕ. ϕ ∈ Σ ′ = CASE 5. ϕ : ≡ ∀ xα ⇒ 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. 19

Proposition 3.2.6. For every L ′ -sentence ϕ , we have ϕ ∈ Σ ′ ⇐ ⇒ A | = ϕ. ϕ ∈ Σ ′ = CASE 5. ϕ : ≡ ∀ xα ⇒ 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 ) = α x ⇐ ⇒ A | t [ s ] (by Theorem 2.6.2) = α x (since α x ⇐ ⇒ A | t is a sentence). t 20

Proposition 3.2.6. For every L ′ -sentence ϕ , we have ϕ ∈ Σ ′ ⇐ ⇒ A | = ϕ. ϕ ∈ Σ ′ = CASE 5. ϕ : ≡ ∀ xα ⇒ 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 ) = α x ⇐ ⇒ A | t [ s ] (by Theorem 2.6.2) = α x (since α x ⇐ ⇒ A | t is a sentence). t Since ϕ ∈ Σ ′ , we have Σ ′ ⊢ ∀ xα , hence Σ ⊢ α x t by (Q1) axiom and (PC) rule. (OBS: Since t is variable-free, it is substitutable for x .) 21

Proposition 3.2.6. For every L ′ -sentence ϕ , we have ϕ ∈ Σ ′ ⇐ ⇒ A | = ϕ. ϕ ∈ Σ ′ = CASE 5. ϕ : ≡ ∀ xα ⇒ 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 ) = α x ⇐ ⇒ A | t [ s ] (by Theorem 2.6.2) = α x (since α x ⇐ ⇒ A | t is a sentence). t Since ϕ ∈ Σ ′ , we have Σ ′ ⊢ ∀ xα , hence Σ ⊢ α x t by (Q1) axiom and (PC) rule. (OBS: Since t is variable-free, it is substitutable for x .) Therefore, α x t ∈ Σ ′ . By induction hypothesis applied to α x t , it follows that = α x A | t . Hence, A | = ϕ . 22

Proposition 3.2.6. For every L ′ -sentence ϕ , we have ϕ ∈ Σ ′ ⇐ ⇒ A | = ϕ. ∈ Σ ′ = CASE 5. ϕ : ≡ ∀ xα ϕ / ⇒ A �| = ϕ 23

Proposition 3.2.6. For every L ′ -sentence ϕ , we have ϕ ∈ Σ ′ ⇐ ⇒ A | = ϕ. ∈ Σ ′ = CASE 5. ϕ : ≡ ∀ xα ϕ / ⇒ A �| = ϕ ∈ Σ ′ . Then ¬ ϕ ∈ Σ ′ , hence Σ ′ ⊢ ¬ ϕ : ≡ ¬∀ xα , hence Σ ′ ⊢ ∃ x ¬ α . Assume ϕ / 24