Metatheory - University of Cambridge

6. Completeness 59
if Γ are infinite, SimpleSearch will not terminate. Indeed, we will not even
get past Step 1 of SimpleSearch, since we will never finish writing all of Γ and
¬C as assumptions!
A similar problem arises for the more abstract proof of Lemma 6.2 offered
in §6.3. In the course of that proof, I assumed that only finitely many
atomic sentences occur as subsentences in the sentences among ∆, namely,
A 1 , A 2 , . . . , A m , and I used this assumption to select my jointly polarising sentences,
Ω m . But if ∆ are infinite, it might be that infinitely many atomic
sentences occur as subsentences among ∆, and my proof will break down.
Fortunately, this last problem can be repaired. In making our repair, the
crucial insight is that every TFL-proof is only ever finitely long. This insight
will allow us to prove Lemma 6.2 in full generality.
Proof of Lemma 6.2, in the fully general case. Suppose ∆ ⊬ ⊥; let A 1 , A 2 , . . .,
be all the atomic sentences that occur as subsentences in any of the sentences
among ∆. We define:
Base clause. If ∆, A 1 ⊬ ⊥, then let Ω 1 be A 1 . Otherwise, let Ω 1 be ¬A 1
Recursion clause. For each number n > 1:
• If ∆, Ω n , A n+1 ⊬ ⊥, then let Ω n+1 be all of Ω n and A n+1
• Otherwise, let Ω n+1 be all of Ω n and ¬A n+1
Just as before, we can prove by induction that:
Claim 1. ∆, Ω n ⊬ ⊥, for all n.
Now let Ω be all and only the sentences that are to be found among some (any)
Ω n . I claim:
Claim 2. ∆, Ω ⊬ ⊥
To establish this claim, suppose that ∆, Ω ⊢ ⊥. Since every TFL-proof is
only finitely long, there must be finitely many sentences, ±A i1 , ±A i2 , . . . , ±A ik ,
among Ω such that:
∆, ±A i1 , ±A i2 , . . . , ±A ik ⊢ ⊥
But then all of these sentences, ±A i1 , ±A i2 , . . . , ±A ik are among some Ω n .
(Specifically, they are in Ω n , when n is the largest number among i 1 , . . . , i k .)
So ∆, Ω n ⊢ ⊥, which contradicts Claim 1.
It follows from Claim 2 that Ω are jointly polarising. The remainder of the
proof now proceeds exactly as before.
■
In short, pursuing our more abstract proof-strategy allows us to offer a ‘completely
general’ proof of the Completeness Theorem. What is more, we can
obtain an important result as an immediate consequence.
Theorem 6.12. Compactness Theorem. Let ∆ be any sentences. Suppose
that, whenever you take only finitely many sentences of ∆, those finitely many
sentences are jointly consistent. Then ∆ themselves are jointly consistent.
Proof. We prove the contrapositive. Suppose that ∆ are jointly inconsistent.
By the (completely general) Completeness Theorem, ∆ ⊢ ⊥. Since every TFLproof
is only finitely long, there are finitely many sentences, ∆ 0 , among ∆ such
that ∆ 0 ⊢ ⊥. By the Soundness Theorem, ∆ 0 are jointly inconsistent. ■