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