Computability and Logic

308 NONSTANDARD MODELS
Before giving the details of the proof of Lemma 25.7a, let us indicate Tennenbaum’s
main idea. Lemma 25.6a can be regarded as saying that for every z there is a y that
encodes the answers to all questions A(x)? for x less than z. Apply this to a nonstandard
element d in M. Then there is a b that encodes the answers to all questions
M |= A(x)[i]? for all i LESS THAN d. But since d is nonstandard, the denotations
of all numerals are LESS THAN d. Sob codes the answers to all the infinitely many
questions M |= A(n)? for n a natural number.
Proof of Lemma 25.7a: Let d be as in Lemma 25.6a, sowehave
Let b be such that
M |= ∃ y∀x < z (∃w((x,w)&w | y) ↔ A(x))[d].
M |= ∀x < z (∃w((x,w)&w | y) ↔ A(x))[b, d].
Since d is nonstandard, for every n, M |= n < z[d]. Thus we have
Since represents π,wehave
Thus for every n,
That is,
It follows that, for every n,
M |= (∃w((n,w)&w | y) ↔ A(n))[b].
M |= ∀w((n,w) ↔ w = p n ).
M |= p n |y ↔ A(n)[b].
M |= ∃x(p n · x = y) ↔ A(n)[b].
M |= A(n) if and only if for some a, M |= p n · x = y[a, b].
By Lemma 25.5a this means
M |= A(n) if and only if for some a, b = a ⊕···⊕a [π(n) as]
as required to complete the proof.
Proof of Theorem 25.4a: Suppose ⊕ is arithmetical. Then, since it is obtainable
from ⊕ by primitive recursion, the function f taking a to a ⊕···⊕a(n as) is arithmetical;
and then, since it is obtainable from f and π by composition, the function g
taking a to a ⊕···⊕a [π(n) as] is arithmetical. (The proof in section 16.1 that recursive
functions are arithmetical shows that processes of composition and primitive
recursion applied to arithmetical functions yield arithmetical functions.) Hence the
relation H given by
or in other words
Hbn if and only if for some a, b = a ⊕···⊕a [π(n) as]
Hbn if and only if ∃ab= g(a, n)