Download Proof of Godel's Completeness Theorem: Model Existence for Maximal Consistent Sets and more Study notes Mathematics in PDF only on Docsity!
13.7 Theorem (CT after reduction 13.4)
Let Γ be a maximal consistent witnessing set
of sentences not containing a =. -symbol.
Then Γ has a model.
Proof:
Let A := {t ∈ Term(L) | t is closed}
(recall: t closed means no variables in t).
A will be the domain of our model A of Γ
(A is called term model).
For P = P (^) n( k) ∈ Pred(L) resp. f = f (^) n(k ) ∈
Fct(L) resp. c = cn ∈ Const(L) define the
interpretations PA resp. fA resp. cA by
PA(t 1 ,... , tk) holds :⇔ Γ ⊢ P (t 1 ,... , tk) fA(t 1 ,... , tk) := f (t 1 ,... , tk) cA := c
to show: A |= Γ (i.e. A |= Γ[v] for some/all assignments v in A: note that Γ contains only sentences).
Let v be an assignment in A, say v(xi) =: si ∈ A for i = 0, 1 , 2 ,.. ..
Claim 1: For any u ∈ Term(L): v˜(u) = u[~s/~x] (:= the closed term obtained by replacing each xi in u by si)
Proof: by induction on u
- u = xi ⇒ v^ ˜(u) = v(xi) = si = xi[si/xi] = u[~s/~x]
- u = c ∈ Const(L) ⇒ v^ ˜(u[~s/~x]) = v˜(u) = v(c) = cA
- u = f (t 1 ,... , tk) ⇒
v^ ˜(u) := fA(v˜(t 1 ),... , v˜(tk)) = fA(t 1 [~s/~x],... , tk[~s/~x]) by IH = f (t 1 [~s/~x],... , tk[~s/~x]) by def. offA = f (t 1 ,... , tk)[~s/~x] by def. of subst. = u[~s/~x] (^2) Claim 1
Induction Step
A |= ¬φ[v] iff not A |= φ[v] [def. of ‘|=’] iff not Γ ⊢ φ[~s/~x] [IH] iff Γ ⊢ ¬φ[~s/~x] [Γ max. cons.]
A |= (φ → ψ)[v] iff not A |= φ[v] or A |= ψ[v] [def. ‘|=’] iff not Γ ⊢ φ[~s/~x] or Γ ⊢ ψ[~s/~x] [IH] iff Γ ⊢ ¬φ[~s/~x] or Γ ⊢ ψ[~s/~x] [Γ max.] iff Γ ⊢ (¬φ[~s/~x] ∨ ψ[~s/~x]) [def. ‘⊢’] iff Γ ⊢ (φ[~s/~x] → ψ[~s/~x]) [taut.] iff Γ ⊢ (φ → ψ)[~s/~x] [def. subst.]
∀-step ‘⇒’
Suppose A |= ∀xiφ[v] (⋆)
but not Γ ⊢ (∀xiφ)[~s/~x]
⇒ Γ ⊢ (¬∀xiφ)[~s/~x] (Γ max.)
⇒ Γ ⊢ (∃xi¬φ)[~s/~x] (Exercise)
Now let φ′^ be the result of substituting each
free occurrence of xj in φ by sj for all j 6 = i.
⇒ (∃xi¬φ)[~s/~x] = ∃xi¬φ′
⇒ Γ ⊢ ∃xi¬φ′
Γ witnessing ⇒
Γ ⊢ ¬φ′[c/xi] for some c ∈ Const(L)
Define
v⋆(xj) :=
{ v(xj) if j 6 = i c if j = i and^ s
⋆ j :=
{ sj if j 6 = i c if j = i
⇒ ¬φ′[c/xi] = ¬φ[ s~⋆/~x]
⇒ Γ ⊢ ¬φ[ s~⋆/~x]
⇒ Γ |= ¬φ[v⋆] [IH]
But, by (⋆), A |= φ[v⋆]: contradiction.
So
Γ ⊢ ¬∀xi¬¬φ′
⇒ Γ ⊢ ¬∀xiφ′
⇒ Γ ⊢ (¬∀xiφ)[~s/~x]
⇒ Γ 6 ⊢ (∀xiφ)[~s/~x]
(^2) Claim 2
Now choose any φ ∈ Γ ⊆ Sent(L)
⇒ φ[~s/~x] = φ
⇒ A |= φ[v], i.e. A |= φ [Claim 2]
⇒ A |= Γ
13.8 Modification required for =. –symbol
Define an equivalence relation E on A by
t 1 Et 2 iff Γ ⊢ t 1 =. t 2
(easy to check: this is an equivalence relation, e.g. transitivity = (1)(ii) of sheet ♯ 4).
Let A/E be the set of equivalence classes t/E (with t ∈ A).
Define L-structure A/E with domain A/E by
PA/E(t 1 /E,... , tk/E) :⇔ Γ ⊢ P (t 1 ,... , tk) fA/E(t 1 /E,... , tk/E) := fA(t 1 ,... , tk) cA/E := cA/E
check: independence of representatitves of t/E (this is the purpose of Axiom A7).
Rest of the proof is much the same as before.
(^2) 13.
