Computability and Logic

23.2. ARITHMETICAL DEFINABILITY AND FORCING 289
then the only way to expand N to get a model of F(G)istotakeV as the denotation
of G.
23.2 Arithmetical Definability and Forcing
We retain the terminology and notation of the preceding section. This entire section
will be devoted to the proof of the following result.
23.3 Theorem (Addison’s theorem). The class of arithmetically definable sets of numbers
is not an arithmetically definable class of sets.
The first notion we need is that of a condition, by which we mean a finite, consistent
set of sentences of the language L G each either of the form Gm or ∼Gm. The empty
set ∅ is a condition. Other examples are {G17}, {G17, ∼G59}, and
{G0, G1, G2,...,G999 999, G1 000 000}.
We use p, q, r as variables for conditions. We say a condition q extends or is an
extension of a condition p if p is a subset of q. Thus every condition extends itself
and extends ∅.
Forcing is a relation between certain conditions and certain sentences of L G .We
write p ⊩ S to mean that condition p forces sentence S. The relation of forcing is
inductively defined by the following five stipulations:
(1) If S is an atomic sentence of L, then p ⊩ S if and only if N |= S.
(2) If t is a term of L and m is the denotation of t in N , then if S is the sentence Gt,
then p ⊩ S if and only if Gm is in p.
(3) If S is a disjunction (B ∨ C), then p ⊩ S if and only if either p ⊩ B or p ⊩ C.
(4) If S is an existential quantification ∃xB(x), then p ⊩ S if and only if, for some n,
p ⊩ B(n).
(5) If S is a negation ∼B, then p ⊩ S if and only if, for every q that extends p,itisnot
the case that q ⊩ S.
The last clause bears repeating: a condition forces the negation of a sentence
if and only if no extension forces the sentence. It follows that no condition forces
some sentence and its negation, and also that either a condition forces the negation
of a sentence or some extension forces the sentence. (It will soon be shown that if a
condition forces a sentence, so does every extension of it.)
It follows from (2) and (5) that p ⊩ ∼Gm if and only if ∼Gm is in p. For if ∼Gm
is not in p, then p ∪{Gm} is an extension of p that forces Gm.Soifp ⊩ ∼Gm, that
is, if no extension of p forces Gm, then ∼Gm must be in p. Conversely, if ∼Gm is
in p, then Gm is in no extension of p, and hence no extension of p forces Gm, and
so p ⊩ ∼Gm.
Thus {G3} forces neither G11 nor ∼G11, and so does not force (G11 ∨∼G11).
Thus a condition may imply a sentence without forcing it. (It will soon be seen that
the inverse is also possible; that, for example, ∅ forces ∼∼∃xGx, even though it
does not imply it, and does not force ∃xGx.)