Computability and Logic

292 ARITHMETICAL DEFINABILITY
Case 4. S is ∃xB(x). Then A FORCES S iff some A-correct p forces ∃xB(x), iff
for some A-correct p there is an m such that p forces B(m), iff for some m there is
an A-correct p such that p forces B(m), iff for some m, A forces B(m), iff (by the
induction hypothesis) for some m, NA G |= B(m), iff N A G |= ∃xB(x).
Case 5. S is ∼B.NosetFORCES both B and ∼B. Since A is generic, A FORCES at
least one of B or ∼B. Hence A FORCES ∼B iff it is not the case that A FORCES B,iff
(by the induction hypothesis) not NA G |= B,iffN A G |= ∼ B.
The last fact about generic sets that we have to prove is that none of them is
arithmetical.
23.8 Lemma. No generic set is arithmetical.
Proof: Suppose otherwise. Then there is a generic set A and a formula B(x)ofL
such that for every n, n is in A if and only if N |= B(n). So NA
G |= ∀x(Gx ↔ B(x)) or
|= ∼∃xF(x), where F(x) is the following logical equivalent of ∼(Gx ↔ B(x)):
N G A
∼(∼(∼Gx ∨∼B(x)) ∨∼(Gx ∨ B(x))).
By Lemma 23.7, A FORCES ∼∃xF(x), so some A-correct p forces ∼∃xF(x), so for
no q extending p and no n does q force F(n), which is to say
(∗)
∼(∼(∼Gn ∨∼B(n)) ∨∼(Gn ∨ B(n))).
Let k be the least number such that neither Gk nor ∼Gk is in p. Define a condition
q extending p by letting q = p ∪{Gk} if N |= ∼ B(k) and letting q = p ∪{∼Gk}
if N |= B(k). In the former case, q ⊩ Gk, while by Lemma 23.5 q ⊩ ∼B(k). In the
latter case, q ⊩ ∼Gk while by Lemma 23.5 q ⊩ B(k). In either case, q ⊩ (*) by our
observations following Lemma 23.4, which is to say q ⊩ F(n). Contradiction.
Suppose that at the beginning of the proof of Lemma 23.6, instead of enumerating
all sentences we enumerate those sentences of complexity ≤ n (that is, having no
more than n occurrences of logical operators). Then the proof would establish the
existence of an n-generic set rather than of a generic set. Suppose that in the hypothesis
of Lemma 23.7 we only assume the set A is n-generic rather than generic. Then the
proof would establish the conclusion of Lemma 23.7 for sentences of complexity
≤ n, rather than for all sentences. But suppose that in the hypothesis of Lemma 23.8
we only assume the set A is n-generic rather than generic. Then the proof would
break down entirely. And indeed, in contrast to Lemma 23.8, we have the following.
23.9 Lemma. For any n, there is an n-generic set A that is arithmetical.
Proof: The proof will be indicated only in outline. The idea is to carry out the
construction in the proof of Lemma 23.6, starting from an enumeration of all sentences
of complexity ≤ n, and with p = ∅. It is necessary to show that, if code numbers are
assigned in a suitable way, then various relations among code numbers connected with
the construction will be arithmetical, with the result that the generic set constructed
is arithmetical as well.
First note that, since we have seen in the preceding section that the set of code
numbers of sentences of complexity ≤ n is recursive, the function enumerating the