Computability and Logic

23.2. ARITHMETICAL DEFINABILITY AND FORCING 293
elements of that set in increasing order is recursive. That is, if we enumerate the
sentences S 0 , S 1 , S 2 , ...in order of increasing code number, then the function taking
us from i to the code number for S i will be recursive.
We also enumerate the conditions p 0 , p 1 , p 2 , ...in order of increasing code number,
where code numbers are assigned to finite sets of sentences—for that is what
conditions are—as in section 15.2. As we observed in section 15.2, the relation ‘the
sentence with code number i belongs to the set with code number s’ is recursive.
Using this fact and the fact that the function taking m to the code number for Gm—
essential the substitution function σ used in the preceding section—is recursive, it is
not hard to show that the set of code numbers of conditions is recursive, and that the
relation that holds between m and s if and only if s is the code number of a condition
containing Gm is recursive. We also observed in section 15.2 that the relation ‘the set
with code number s is a subset of the set with code number t’ is recursive. Hence the
relation that holds between s and t if and only if they are code numbers of conditions
p and q, with q an extension of p, is also recursive. Being recursive, the various
functions and relations we have mentioned are all arithmetical.
We also need one more fact: that for each n, the relation that holds between i and s
if and only if i is the code number of a sentence S of complexity ≤n and s is the code
number of a condition p, and p forces S, is arithmetical. The proof is very similar to
the proof in the preceding section that for each n the set V n is arithmetical, and will
be left to the reader.
Now returning to the construction of an n-generic set A, by the method of the
proof of Lemma 23.6, we see that m is in A if and only if there exists a sequence s
of conditions such that the following hold (for each i less than the length of the
sequence):
(0) The 0th entry of the sequence is the empty condition ∅
(1) If the ith entry of the sequence forces the negation of the ith sentence in the
enumeration of sentences, then the (i + 1)st entry is the same as the ith.
(2) Otherwise, the (i + 1)st entry is a condition that extends the ith and that forces
the ith sentence in the enumeration of sentences, and is such that no condition
earlier in the enumeration of conditions (that is, no condition of smaller code
number) does both these things.
(3) The sentence Gm belongs to the last entry of the sequence.
We can, of course, replace ‘there exists a sequence...’ by ‘there exists a code
number for a sequence...’. When everything is thus reformulated in terms of code
numbers, what we get is a logical compound of relations that we have noted in the
preceding several paragraphs to be arithmetical. It follows that A itself is arithmetical.
At last we are in a position to prove Addison’s theorem.
Proof of Theorem 23.3: Suppose the theorem fails. Then there is a sentence S of
L G such that for any set A, NA
G |= S if and only if A is arithmetical. Let S be of
complexity n. By Lemma 23.9 there exists an n-generic set A that is arithmetical. So
NA
G |= S. So by Lemma 23.7 (or rather, the version for n-generic sets and sentences
of complexity ≤n, as in our remarks following Lemma 23.8), A FORCES S. So some