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Completeness theorem for predicate calculus
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13.1 Theorem (G¨odel) Let Γ ⊆ Form(L), φ ∈ Form(L).
If Γ |= φ then Γ ⊢ φ.
Two additional assumptions:
First Step Call ∆ ⊆ Sent(L) consistent if for no sentence ψ, both ∆ ⊢ ψ and ∆ ⊢ ¬ψ.
13.2 enough to show (⋆) Every consistent set of sentences has a model.
i.e. ∆ consistent ⇒ there is an L-structure A such that A |= δ for every δ ∈ ∆.
Proof: Assume Γ |= φ and assume (⋆) ⇒ Γ ∪ {¬φ} has no model ⇒(⋆) Γ ∪ {¬φ} is not consistent ⇒ Γ ∪ {¬φ} ⊢ ψ and Γ ∪ {¬φ} ⊢ ¬ψ for some ψ ⇒DT Γ ⊢ (¬φ → ψ) and Γ ⊢ (¬φ → ¬ψ) for some ψ But Γ ⊢ ((¬φ → ψ) → ((¬φ → ¬ψ) → φ)) [taut.] ⇒ Γ ⊢ φ [2xMP] (^2) 13.
Third Step
to prove CT:
13.4 enough to show: Every maximal consistent witnessing set of sen- tences has a model.
For the proof of 13.4 we need 2 Lemmas:
13.5 Lemma If ∆ ⊆ Sent(L) is consistent, then for any sen- tence ψ, either ∆ ∪ {ψ} or ∆ ∪ {¬ψ} is consis- tent.
Proof: Exercise – as for Propositional Calcu- lus. 2.
13.6 Lemma Assume ∆ ⊆ Sent(L) is consistent, ∃xiψ ∈ Sent(L), ∆ ⊢ ∃xiψ, and cj is not occurring in ψ nor in any δ ∈ ∆.
Then ∆ ∪ {ψ[cj/xi]} is consistent.
By ∀, ∆ ⊢ ∀xi(ψ → χ) and ∆ ⊢ ∀xi(ψ → ¬χ) (note that xi 6 ∈ Free(δ) for any δ ∈ ∆ ⊆ Sent(L)).
Now: ⊢ (∀xi(A → B) → (∃xiA → B)) for any A, B ∈ Form(L) with xi 6 ∈ Free(B) (Exercise Sheet ♯ 4, (2)(i))
MP ⇒ ∆ ⊢ (∃xiψ → χ) and ∆ ⊢ (∃xiψ → ¬χ) (χ, ¬χ ∈ Sent(L), so xi 6 ∈ Free(χ))
By hypothesis, ∆ ⊢ ∃xiψ ⇒ by MP, ∆ ⊢ χ and ∆ ⊢ ¬χ contradicting consistency of ∆.
Proof of 13.4: Let ∆ be any consistent set of sentences.
to show: ∆ has a model assuming that any maximal consistent, witnessing set of sentences has a model.
By 13.3(a), ∆′^ is consistent and does not con- tain any c 2 m+1.
Let φ 1 , φ 2 , φ 3 ,... be an enumeration of Sent(L′^ ∪ {c 1 , c 3 , c 5 ,.. .}).
Construct finite sets ⊆ Sent(L′^ ∪{c 1 , c 3 , c 5 ,.. .})
Γ 0 ⊆ Γ 1 ⊆ Γ 2 ⊆...
such that ∆′^ ∪ Γn is consistent for each n ≥ 0 as follows:
Let Γn+1 := Γn+1/ 2 ∪ {ψ[c 2 m+1/xi]} ⇒ by Lemma 13.6, Γn+1 is consistent.
Let Γ := ∆′^ ∪ ⋃ n≥ 0 Γn.
⇒ Γ is maximal consistent (as in Propositional Calculus) and Γ is witnessing (by construction).
By assumption, Γ has a model, say A.
⇒ in particular, Γ |= δ for any δ ∈ ∆′
⇒ by Lemma 13.3(b), ∆ has a model
