Metatheory - University of Cambridge

6. Completeness 56
Now, when we are in the business of proving the Completeness Theorem, our
aim is to tie ‘⊨’ to ‘⊢’. And so, in this regard, it is very striking that there is a
perfect proof-theoretic analogue of Lemma 6.9:
Lemma 6.10. Suppose Ω are jointly polarising and that every atomic subsentence
of A is in the field of Ω. Then either Ω ⊢ A or Ω ⊢ ¬A.
Proof. By (strong) induction on the length of A.
Fix any sentence A, and assume that all of A’s atomic subsentences are in
the field of Ω. For induction, suppose that, where B is any sentence shorter
than A and where all of B’s atomic subsentences are among the sentences that
jointly span Ω, either Ω ⊢ B or Ω ⊢ ¬B. There are now six cases to consider:
Case 1: A is atomic. Then either A is among Ω (in which case Ω ⊢ A, by
the rule R), or ¬A is among Ω (in which case Ω ⊢ ¬A, by the rule R).
So the atomic case is straightforward.
Case 2: A is ¬C . Suppose Ω ⊬ A, i.e. Ω ⊬ ¬C . Then Ω ⊢ C , by the
induction hypothesis. It follows that Ω ⊢ ¬¬C (just expand the proof of
C from Ω by supposing ¬C and then using ⊥I and ¬I). So Ω ⊢ ¬A. So
either Ω ⊢ A or Ω ⊢ ¬A.
Case 3: A is (C ∧ D). Suppose Ω ⊬ A, i.e. Ω ⊬ C ∧ D. Then either
Ω ⊬ C , or Ω ⊬ D. For if Ω ⊢ C and Ω ⊢ D, then Ω ⊢ C ∧D (just combine
the two proofs and then use ∧I). So, by the induction hypothesis, either
Ω ⊢ ¬C , or Ω ⊢ ¬D. Either way, Ω ⊢ ¬(C ∧ D). (Showing this is left as
an exercise.) That is, Ω ⊢ ¬A. So either Ω ⊢ A, or Ω ⊢ ¬A.
Case 4: A is (C ∨ D). Suppose Ω ⊬ A, i.e. Ω ⊬ C ∨ D. Then Ω ⊬ C and
Ω ⊬ D. So, by hypothesis, Ω ⊢ ¬C and Ω ⊢ ¬D. So Ω ⊢ ¬(C ∨ D).
(Showing this is left as an exercise). That is, Ω ⊢ ¬A. So either Ω ⊢ A,
or Ω ⊢ ¬A.
Case 5: A is (C → D). I leave this as an exercise.
Case 6: A is (C ↔ D). I leave this as an exercise.
This completes the (strong) induction on length.
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We here have a very tight connection between the behaviour of ‘⊨’ and ‘⊢’ in
the case of jointly polarising sentences. In fact, if we like, we can prove a kind
of mini-version of the Soundness and Completeness Theorems:
Lemma 6.11. Suppose Ω are jointly polarising and that every atomic subsentence
of A is in the field of Ω. Then Ω ⊢ A iff Ω ⊨ A.
Proof. Left-to-right. If Ω ⊢ A, then Ω ⊨ A by the Soundness Theorem.
Right-to-left. I prove the contrapositive, i.e. that if Ω ⊬ A then Ω ⊭ A.
So suppose Ω ⊬ A. Then Ω ⊢ ¬A, by Lemma 6.10. So Ω ⊨ ¬A, by the
Soundness Theorem. But in that case, Ω ⊭ A, since otherwise we would have
that Ω ⊨ (A ∧ ¬A), contradicting Lemma 6.8.
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What Lemma 6.11 tells us, in essence, is the following: if you pin down what is
happening with atomic sentences and their negations, then you determine all
the semantic and proof-theoretic consequences; moreover, these consequences
coincide. (It is no wonder, then, that SimpleSearch worked precisely by breaking
down sentences into pathways through TFL-proof-skeletons that consist of