Metatheory - University of Cambridge

6. Completeness 55
It will make our lives easier if we introduce a bit more terminology. Where
Ω are some sentences, say that Ω are jointly polarising iff both: 6
(p1) every sentence among Ω is either an atomic sentence or a negation of an
atomic sentence; and
(p2) if a sentence is among Ω, then its negation is not among Ω
Where Ω are jointly polarising, say that A is in the field of Ω iff A is an
atomic sentence such that either A or ¬A is among Ω.
It is worth pausing for a moment to think about how this terminology,
and this result, connects with SimpleSearch. When ∆ ⊬ ⊥, SimpleSearch
generates a proof-skeleton with at least one open line. From each open line,
we can read off some jointly polarising sentences, Ω. Moreover, when D is any
sentence among ∆, every atomic subsentence of D is in the field of Ω. At this
point, we claimed (in the proof of Lemma 6.6) that there is bound to be some
valuation that makes all of Ω true. That was right, and the more general result
is simply this very easy observation.
Lemma 6.8. Suppose Ω are jointly polarising. Then there is exactly one
valuation of the sentences in the field of Ω that makes every sentence among
Ω true.
Proof. Define a valuation, v, as follows: for each sentence A in the field of Ω,
• v makes A true if A is among Ω
• v makes A false if ¬A is among Ω
Given (p2), v is well-defined. Now, if B is a sentence among Ω, then B is either
A or ¬A, for some sentence A in the field of Ω. Either way, v makes B true.
Moreover, any valuation that assigned a different value to A would make B
false. Hence v is unique.
■
Continuing, though, to think through how SimpleSearch relates to this terminology:
the next stage in the proof of Lemma 6.6 was for us to show that the
(unique) valuation that makes all of Ω true also makes all of ∆ true too. This
required a bit of work (some of which I left as an exercise). But let me now
part company with the proof of Lemma 6.6, and instead make a much more
simple observation.
We are considering a situation where all of the atomic subsentences of some
sentence (in this case, D) are in the field of some jointly polarising sentences
(in this case, Ω). We can obtain a very easy and general result about such
situations.
Lemma 6.9. 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. Using Lemma 6.8, let v be the unique valuation of the sentences in the
field of Ω that makes all of Ω true . Either v makes A true, or v makes A false.
If v makes A true, then since v is the only valuation that makes all of Ω true,
Ω ⊨ A. If v makes A false, then it makes ¬A true, and similar reasoning shows
that Ω ⊨ ¬A.
■
6 NB: this terminology is not standard among logicians.