Computability and Logic

316 NONSTANDARD MODELS
But while the quantification ‘there exists a finite sequence of sets’ can by suitable
coding be replaced by a quantification ‘there exists a set’, in general the latter quantification
cannot be eliminated. The inductive definition implicitly involves—what
the corresponding direct definition explicitly involves—an impredicative specification
of a set. In general, one cannot ‘formalize’ in P + arguments involving this kind
of inductive specification of sets, even if the conditions F 0 and F ′ involve no bound
upper variables.
Also, one cannot ‘formalize’ in P + the proof of the consistency sentence for P
indicated in the proof of the preceding proposition. [One can indeed introduce the
abbreviation True(x) for ∃X(F(X)&x ∈ X), but one cannot in P prove the existence
of {x: True(x)}, and so cannot apply the induction axiom to prove assertions involving
the abbreviation True(x).] So the proof indicated for the preceding proposition fails
for P + in place of P*. In fact, not only is the consistency sentence for P not an example
of a sentence of L* that is a theorem of P + and not of P, but actually there can be no
example of such sentence: P + is a conservative extension of P.
Our last result is a proposition immediately implying the fact just stated.
25.11 Proposition. Every model of P can be expanded to a model of P + .
Proof: Let M be a model of P. Call a subset S of the domain |M| parametrically
definable over M if there exist a formula F(x, y 1 ,...,y m )ofL* and elements
a 1 ,...,a m of |M| such that
S ={b: M |= F[b, a 1 ,...,a m ]}.
Expand M to an interpretation of L** by taking as upper domain the class of all
parametrically definable subsets of M, and interpreting ∈ as ∈.We claim the expanded
model M + is a model of P + . The axioms that need checking are induction (1) and
comprehension (3) (with F having no bound upper variables). Leaving the former to
the reader, we consider an instance of the latter:
∀u 1 ∀u 2 ∀U 1 ∀U 2 ∃X∀x(x ∈ X ↔ F(x, u 1 , u 2 , U 1 , U 2 )).
(In general, there could be more than two us and more than two Us, but the proof
would be no different.) To show the displayed axiom is true in M + , we need to show
that for any elements s 1 , s 2 of |M| and any parametrically definable subsets S 1 , S 2 of
|M| there is a parametrically definable subset T of |M| such that
M + |= ∀x(x ∈ X ↔ F(x, u 1 , u 2 , U 1 , U 2 ))[s 1 , s 2 , S 1 , S 2 , T ].
Equivalently, what we must show is that for any such s 1 , s 2 , S 1 , S 2 , the set
T ={b: M + |= F(x, u 1 , u 2 , U 1 , U 2 )[s 1 , s 2 , S 1 , S 2 , b]}
is parametrically definable. To this end, consider parametric definitions of U 1 , U 2 :
U 1 ={b: M |= G 1 [b, a 11 , a 12 ]}
U 2 ={b: M |= G 2 [b, a 21 , a 22 ]}.