Computability and Logic

23.2. ARITHMETICAL DEFINABILITY AND FORCING 291
NA
G (the expansion of the standard interpretation N of the language of arithmetic
L to an interpretation of the language L G in which the new predicate G is taken to
denote A) is a model of p.
Further, say A FORCES S if some A-correct condition forces S. Note that the union
of any two A-correct conditions is still a condition and is still A-correct. It follows
that A cannot FORCE both S and ∼S, since the union of an A-correct condition forcing
S with one forcing ∼S would force both, which is impossible.
Finally, we call A generic if for every sentence S of L G , either A FORCES S or
A FORCES ∼S. If this is so at least for every sentence S with at most n occurrences
of logical operators, we call An-generic. Thus a set is generic if and only if it is
n-generic for all n.
The first fact about generic sets that we have to prove is that they exist.
23.6 Lemma. For any p, there is a generic set A such that p is A-correct.
Proof: Let S 0 , S 1 , S 2 ,... be an enumeration of all sentences of L G . Let p 0 , p 1 ,
p 2 ,... be an enumeration of all conditions. We inductively define a sequence q 0 ,
q 1 , q 2 ,...of conditions, each an extension of those that come before it, as follows:
(0) q 0 is p.
(1) If q i forces ∼S i , then q i+1 is q i .
(2) If q i does not force ∼S i , in which case there must be some q extending q i and
forcing S i , then q i+1 is the first such q (in the enumeration p 0 , p 1 , p 2 ,...).
Let A be the set of m such that Gm is in q i for some i.
We claim that p is A-correct and that A is generic. Since p = q 0 , and since for
each i, either q i+1 ⊩ S i or q i+1 ⊩ ∼S i , it will be enough to show that for each i, q i is
A-correct. And since m is in A when Gm is in q i , it is enough to show that if ∼Gm
is in q i , then m is not in A. Well, suppose it were. Then Gm would be in q j for some
j. Letting k = max(i, j), both ∼Gm and Gm would be in q k , which is impossible.
This contradiction completes the proof.
The next fact about generic sets relates FORCING and truth.
23.7 Lemma. Let S be a sentence of L G , and A a generic set. Then A FORCES S if and
only if NA G |= S.
Proof: The proof will be yet another by induction on complexity, with five cases,
one for each clause in the definition of forcing. We abbreviate ‘if and only if’ to ‘iff’.
Case 1. S is an atomic sentence of L. Then A FORCES S iff some A-correct p forces
S, iff (by Lemma 23.5) N |= S,iffNA G |= S.
Case 2. S is an atomic sentence Gt. Let m be the denotation of t in N . Then A
FORCES S iff some A-correct p forces Gt, iffGm is in some A-correct p, iffm is in
A, iffNA G |= Gt.
Case 3. S is (B ∨ C). Then A FORCES S iff some A-correct p forces (B ∨ C),
iff some A-correct p forces B or forces C, iff either some A-correct p forces B or
some A-correct p forces C, iffA FORCES B or A FORCES C, iff (by the induction
hypothesis) NA G |= B or N A G |= C,iffN A G |= (B ∨ C).