Metatheory - University of Cambridge

6. Completeness 44
6.2 An algorithmic approach
I start with the more concrete approach to Lemma 6.2. This works by specifying
an algorithm, SimpleSearch, which determines, for any arbitrary sentences,
whether or not those sentences are jointly contrary. (Iam assuming that we are
given only finitely many sentences. I shall revisit this point in §6.4, but shall
not mention it much until then.) Moreover, SimpleSearch two very desirable
properties. For any sentences, ∆, that we feed in:
(a1) if ∆ are jointly contrary, then the algorithm outputs a TFL-proof which
starts with ∆ as assumptions and terminates in ‘⊥’ on just those assumptions;
and
(a2) if ∆ are not jointly contrary, then the algorithm yields a valuation which
makes all of ∆ true together.
Clearly, any algorithm with both of these properties is independently interesting.
But the important thing, for our purposes, is that if we can prove that
there is an algorithm with property (a2), then we will have proved Lemma 6.2
(pause for a moment to check this!) and hence the Completeness Theorem. So,
my aim is to specify an algorithm, and then prove that it has these properties.
First steps towards an algorithm
To say that ∆ ⊢ ⊥ is to say that there is a TFL-proof, whose assumptions are
all among ∆, and which terminates in ‘⊥’ without any additional assumptions.
It obvious that, if ‘⊥’ appears anywhere in a TFL-proof, then the TFLproof
involves an application of the rule ⊥I. So our algorithm will have to look
for situations in which we might be able to apply the rule ⊥I.
Of course, to show that some sentences are jointly contrary, it is not enough
simply to find a situation in which we can apply ⊥I. For if we only obtain ‘⊥’
within the scope of additional assumptions, we will not have shown that our
original sentences are jointly contrary. Fortunately, there are rules that allow us
to take instances of ‘⊥’ that appear in the scope of additional assumptions, and
derive an instance of ‘⊥’ that appears within the scope of fewer assumptions.
As an example, consider the last line of this TFL-proof:
1 A ∨ B
2 (¬A ∧ ¬B)
3 ¬A ∧E 2
4 ¬B ∧E 2
5 A
6 ⊥ ⊥I 5, 3
7 B
8 ⊥ ⊥I 7, 4
9 ⊥ ∨E 1, 5–6, 7–8