Computability and Logic

310 NONSTANDARD MODELS
in P, but to do so would be both extremely tedious and entirely unnecessary, since in
view of the preceding lemma it is enough to show that
∃y∀x < z (∃w((x,w)&w | y) ↔ A(x))
is satisfied by all standard elements, and for this it is enough to show that for every
n, the following is a theorem of P:
(1)
∃y∀x < n (∃w((x,w)&w | y) ↔ A(x)).
Let n = m + 1. First recall that the following is a theorem of P:
(2)
∀x(x < n ↔ (x = 0 ∨···∨x = m)).
Since represents π, writing p i for π(i), for all i < n the following is a theorem
of P:
(3)
∀w((i,w) ↔ w = p i ).
Using (2) and (3), (1) is provably equivalent in P to
(4)
∃y((p 0 | y ↔ A(0)) & ... &(p m | y ↔ A(m))).
For each sequence e = (e 0 ,...,e m ) of length n of 0s and 1s, let A e be the conjunction
of all (∼)A(i), where the negation sign is present if e i = 0 and absent if e i = 1. Let
B e (y) be the analogous formula with p i | y in place of A(i). Then the formula after
the initial quantifier in (4) is logically equivalent to the disjunction of all conjunctions
A e & B e (y). The existential quantifier may be distributed through the disjunction, and
in each disjunct confined to the conjuncts that involve the variable y. Thus (4) is
logically equivalent to the disjunction of all conjunctions A e & ∃yB e (y). Hence (1)
is provably equivalent in P to this disjunction. But ∃yB e (y) is a true ∃-rudimentary
sentence, and so is provable in P. Hence (1) is provably equivalent in P to the disjunction
of all A e . But this disjunction is logically valid, hence provable in P or any
theory. So (1) is provable in P.
Proof of Theorem 25.4b: We need a fact established in the problems at the end
of Chapter 8 (and in a different way in those at the end of Chapter 16): there exist
disjoint semirecursive sets A and B such that there is no recursive set containing A and
disjoint from B. Since the sets are semirecursive, there are ∃-rudimentary formulas
∃yα(x, y) and ∃yβ(x, y) defining them. Replacing these by
∃y(α(x, y)) & ∼∃z ≤ yβ(x, y)) and ∃y(β(x, y)&∼∃z ≤ y α(x, y))
we get ∃-rudimentary formulas α*(x) and β*(x) also defining A and B, and for which
∼∃x(α*(x)&β*(x)) is a theorem of P.Ifn is in A, then since α*(n)is∃-rudimentary
and true, it is a theorem of P, and hence M |= α*(n); while if n is in B, then similarly
β*(n) is a theorem of P and hence so is ∼α*(n), so that M |= ∼α*(n).
Now by Lemma 25.7b, there are elements b + and b − such that for every n,
M |= α*(n) if and only if for some a, b + = a ⊕···⊕a (π(n) as)
M |= ∼α*(n) if and only if for some a, b − = a ⊕···⊕a (π(n) as).