Metatheory - University of Cambridge

6. Completeness 53
that we introduce as an assumption is shorter than the newly-expanded disjunction.
So, even though expanding a disjunction may increase the number of
unexpanded disjunctions, the unexpanded disjunctions keep getting shorter.
Eventually, then, these will dwindle into atomic sentences or negations of
atomic sentences, so that every disjunction has been expanded. 4
Step 5 is guaranteed to terminate, since as Step 4 terminated, there will
only be finitely many open lines, and inspecting whether or not ⊥I can be
applied on a given line is a bounded procedure (you only need to inspect all
the earlier lines). Step 6 is similarly guaranteed to terminate.
■
So we now know that, given any inputs, SimpleSearch will yield an output.
We next need to confirm that we can trust that output.
Lemma 6.5. If SimpleSearch outputs a proof or proof-skeleton whose last
line is ‘⊥’ on no assumptions other than ∆, then ∆ ⊢ ⊥
Proof sketch. By inspection, it is clear that SimpleSearch simply applies the
rules of the TFL-proof system (in a very specific order) to a proof that starts
with ∆. So if the proof-skeleton’s last line is ‘⊥’ on no assumptions other than
∆, then SimpleSearch has simply outputted a TFL-proof that ∆ ⊢ ⊥. ■
Lemma 6.6. If SimpleSearch outputs a proof or proof-skeleton whose last
line is not ‘⊥’ on no assumptions other than ∆, then:
1. there is at least one open line on the proof-skeleton; and
2. every open line yields (in the manner described) a valuation which
makes all of ∆ true; and
3. ∆ ⊬ ⊥.
Proof. I shall prove each point separately, on the assumption that
SimpleSearch terminates by outputting a proof-skeleton whose last line is
not ‘⊥’ on no assumptions other than ∆.
Proof of (1). There are two cases to consider:
Case 1: None of the CNF-sentences that we generated in Step 2 of
SimpleSearch contained any occurrences of ‘∨’. Then, since the last
line of our proof is not ‘⊥’, the last line of our proof-skeleton is open.
Case 2: At least one of the CNF-sentences that we generated in Step 2 of
SimpleSearch contained an occurrence of ‘∨’. Then, if every assumption
that we made when expanding disjunctions led to ‘⊥’, we would have
applied ∨E repeatedly until we obtained ‘⊥’ on no assumptions. So at
least one assumption did not lead to ‘⊥’, and so there is an open line.
Proof of (2). Let Ω be the sentences that we read off (in the manner
described earlier) from an open line on our proof (whose existence is guaranteed
by point (1)). They are all atomic sentences and their negations. Now, since
the line in question is open, this means that we cannot find both a sentence
and its negation among Ω, since otherwise we could have applied ⊥I. Hence
there is a valuation, v, which makes all of them true. 5
4 One way to make this rigorous would be to prove, by induction, that if ∆ contains
exactly n instances of ‘∨’, Step 3 will require you to introduce at most 2 n+1 − 2 assumptions.
5 For details of this last step, see Lemma 6.8, below.