Computability and Logic

290 ARITHMETICAL DEFINABILITY
23.4 Lemma. If p ⊩ S and q extends p, then q ⊩ S.
Proof: Suppose that p ⊩ S and q extends p. The proof that q ⊩ S is by induction
on complexity of S. The atomic case has two subcases. If S is an atomic sentence
of L, then since p ⊩ S, S is true in N , and since S is true in N , q ⊩ S. IfS is an
atomic sentence of form Gt, then since p ⊩ S, Gm is in p, where m is the denotation
of t in N , and since q extends p, Gm is also in q and q ⊩ Gt. IfS is (B ∨ C), then
since p ⊩ S, either p ⊩ B or p ⊩ C; so by the induction hypothesis, either q ⊩ B or
q ⊩ C, and so q ⊩ (B ∨ C). If S is ∃xB(x), then since p ⊩ S,wehavep ⊩ B(m) for
some m; so by the induction hypothesis, q ⊩ B(m) and q ⊩ ∃xB(x). Finally, if S is
∼B, then since p ⊩ S, no extension of p forces B; and then, since q is an extension
of p, every extension of q is an extension of p, so no extension of q forces B, and so
q ⊩ ∼B.
Two observations, not worthy of being called lemmas, follow directly from the
preceding lemma. First, if p ⊩ B, then p ⊩ ∼∼B; for any extension of p will force
B, hence no extension of p will force ∼B. Second, if p ⊩ ∼B and p ⊩ ∼C, then
p ⊩ ∼(B ∨ C); for every extension of p will force both ∼B and ∼C, and so will
force neither B nor C, and so will not force (B ∨ C).
A more complicated observation of the same kind may be recorded here for future
reference, concerning the sentence
(∗)
∼(∼(∼B ∨∼C) ∨∼(B ∨ C))
which is a logical equivalent of ∼(B ↔ C). Suppose p ⊩ B and p ⊩ ∼C. Then
p ⊩ (∼B ∨∼C), so by our first observation in the preceding paragraph, p ⊩
∼∼(∼B ∨∼C). Also p ⊩ (B ∨ C), so p ⊩ ∼∼(B ∨ C). Hence by our second observation,
p ⊩ (*). Similarly, if p ⊩ ∼B and p ⊩ C, then again p ⊩ (*).
23.5 Lemma. If S is a sentence of L, then for every p, p ⊩ S if and only if N |= S.
Proof: The proof again is by induction on the complexity of S.IfS is atomic, the
assertion of the lemma holds by the first clause in the definition of forcing. If S is
(B ∨ C), then p ⊩ S if and only if p ⊩ B or p ⊩ C, which by the induction hypothesis
is so if and only N |= B or N |= C, which is to say, if and only if N |= (B ∨ C). If
S is ∃xB(x), the proof is similar. If S is ∼B, then p ⊩ S if and only if no extension
of p forces B, which by the induction hypothesis is so if and only if it is not the case
that N |= B, which is to say, if and only if N |= ∼ B.
Forcing is a curious relation. Since ∅ does not contain any sentence Gn, for no n
does ∅ force Gn, and therefore ∅ does not force ∃xGx. But ∅ does force ∼∼∃xGx!
For suppose some p forces ∼∃xGx. Let n be the least number such that ∼Gn is not
in p. Let q be p ∪{Gn}. Then q is a condition, q extends p, and q forces Gn, soq
forces ∃xGx. Contradiction. Thus no p forces ∼∃xGx, which is to say, no extension
of ∅ forces ∼∃xGx,so∅ forces ∼∼∃xGx.
We are going to need some more definitions. Let A be a set of numbers. First,
we call a condition pA-correct if for any m, ifGm is in p, then m is in A, while
if ∼Gm is in p, then m is not in A. In other words, p is A-correct if and only if