Computability and Logic

25.2. OPERATIONS IN NONSTANDARD MODELS 309
is arithmetical, being obtainable by existential quantification from the graph relation
of an arithmetical function.
So let B(x, y) be a formula arithmetically defining H. Let A(x) be the formula
∼B(x, x). Apply Lemma 25.7a to obtain a b such that for all n, M |= A(n) if and
only if Hbn. Since the same sentences are true in M and N , for all n, N |= A(n)if
and only if Hbn. In particular, N |= A(b) if and only if Hbb, that is, N |= ∼ B(b, b)
if and only if Hbb. But since B arithmetically defines H, we also have N |= B(b, b)
if and only if Hbb. Contradiction.
For the proof of Theorem 25.4b, we need extensions of the lemmas used for
Theorem 25.4a that will apply not just to models of arithmetic but to models of
P. We state these as Lemmas 25.5b through 25.7b below. As in the case of the
extension of Theorem 25.1a to Theorem 25.1b, some ‘formalizing’ of the kind done
in Chapter 16 is needed. What is needed for Lemma 25.6b, however, goes well beyond
this; so, leaving other details to the reader, we give the proof of that lemma, before
going on to give the derivation of Theorem 25.4b from the lemmas. The proof of
Lemma 25.6b itself uses an auxiliary lemma of some interest, Lemma 25.8 below.
25.5b Lemma. Let M be a nonstandard model of P. For any m > 0,
M |= ∀x m · x = x + ···+ x(m xs).
25.6b Lemma. Let M be a nonstandard model of P. Let A(x) be any formula of L.
Then there is a nonstandard element d such that
M |= ∃ y∀x < z (∃w((x,w)&w | y) ↔ A(x))[d].
25.7b Lemma. Let M be a nonstandard model of P. Let A(x) be any formula of L.
Then there exists a b such that for every n,
M |= A(n) if and only if for some a, b = a ⊕···⊕a [π(n) as].
25.8 Lemma (Overspill principle). Let M be a nonstandard model of P. Let B(x) be
any formula of L that is satisfied in M by all standard elements. Then B(x) is satisfied in
M by some nonstandard element.
Proof of Lemma 25.8: Assume not. Then for any d that satisfies B(x) inM, d is
standard, hence d † is standard, and hence d † satisfies B(x) inM. Thus
M |= ∀x(B(x) → B(x ′ ))
since O, being standard, satisfies B(x)inM, M |= B(0). But also
M |= (B(0)&∀x(B(x) → B(x ′ ))) →∀xB(x)
since this is an axiom of P.SoM |= ∀xB(x) and every element satisfies B(x)inM,
contrary to assumption.
Proof of Lemma 25.6b: It is possible to formalize the proof of
∀z∃y∀x < z (∃w((x,w)&w | y) ↔ A(x))