Metatheory - University of Cambridge

6. Completeness 52
Step 1. Start a TFL-proof, with ∆ as your assumptions.
Step 2. For each sentence D among ∆, apply the algorithm which finds a
CNF-sentence D ∗ such that D ∗ ⊢ D. This will leave you with a single
TFL-proof where each D ∗ appears on some line.
Step 3: Now apply ∧E as often as you can to lines of your TFL-proof that
are in CNF, until any further application of ∧E would yield a sentence
already on your proof.
Step 4: Expand the first disjunction in CNF which you have not yet expanded.
More precisely: let (A ∨B) be the first disjunction in CNF that you have
not yet expanded and let l be any line in your proof-skeleton that is both
beneath the line on which the disjunction occurs; and which depends
upon (at least) the assumptions on which the disjunction occurs. Then,
beneath line l, introduce two new assumptions – A and B – separately,
and with an empty line beneath both. At this point, we say that the
disjunction ‘(A ∨ B)’ has been expanded.
Repeat this step until every disjunction in CNF has been expanded.
Step 5: Go through every open line and check to see whether ⊥I can be applied.
If so, apply it (filling in line numbers).
Step 6: Go through every open line and check to see whether ∨E can be applied
(allowing you to generate a line with ‘⊥’ on it). If so, apply it (filling in
line numbers). Repeat this step until ∨E cannot be applied any further.
Step 7: Output the proof or proof-skeleton. This terminates SimpleSearch.
This is the algorithm SimpleSearch.
Proving that SimpleSearch behaves as required
⊢
I now want to prove that SimpleSearch has the properties (a1) and (a2),
which I advertised for it.
The very first thing I need to do is to make sure that, when we run
SimpleSearch, there is no chance of our getting trapped in an infinite loop.
That is, we need a guarantee that SimpleSearch will output something in the
end, regardless of the inputs we feed it.
Lemma 6.4. SimpleSearch is guaranteed to terminate, on any inputs.
Proof sketch. Step 1 is guaranteed to terminate eventually, since it just involves
writing down finitely many sentences.
Step 2 is similarly guaranteed to terminate, since it only ever takes finitely
many steps to generate a CNF equivalent from each of the finitely many assumptions.
Step 3 is guaranteed to terminate, since are only finitely many conjunction
signs among the CNF sentences you have written down, and so you only need
to apply ∧E finitely many times.
Step 4 is the tricky step. The worry is that, in expanding a disjunction (in
CNF), you may have to introduce many new assumptions; and if one of the disjuncts
is itself a disjunction, this can increase the total number of unexpanded
disjunctions. 3 However, when we expand a disjunction, each of the disjuncts
3 To see this in action, see what happens when your initial assumptions are ‘(A ∨ B)’ and
‘((C ∨ D) ∨ (D ∨ E))’.