Computability and Logic

19.2. SKOLEM NORMAL FORM 249
take f1
N (a 1)tobeε(B 1 ): we use ε to choose one particular element b 1 such that
a 1 and b 1 satisfy ∀x 2 ∃y 2 R(x 1 , y 1 , x 2 , y 2 ). Then since a 1 and f1
N (a 1) will satisfy
∀x 2 ∃y 2 R(x 1 , y 1 , x 2 , y 2 ), it follows that a 1 will satisfy the foregoing conditional, and
since this will be the case for any a 1 , it follows that (3.1) will be true.
We next want to assign a f2
N that will make the Skolem axiom (3.2) come out true.
We proceed in exactly the same way. For any a 1 and a 2 , consider the set B 2 of all b 2
such that a 1 , a 2 , and b 2 satisfy R(x 1 , f 1 (x 1 ), x 2 , y 2 ). If B 1 is empty, we take f2 N (a 1, a 2 )
to be ε(|M|), and otherwise take it to be ε(B 2 ). The procedure would be the same no
matter how many Skolem function symbols we needed to assign denotations to, and
how many Skolem axioms we needed to make true.
Let Ɣ be any set of sentences of any language L, and for each sentence A in Ɣ, first
associate to it a logically equivalent prenex sentence A* as in the preceding section,
and then associate to A* its Skolem form A # as above, and let Ɣ # be the set of all these
sentences A # for A in Ɣ. Then Ɣ # is a set of ∀-sentences equivalent for satisfiability
to the original set Ɣ. For if Ɣ # is satisfiable, there is an interpretation N in which
each A # in Ɣ # comes out true, and since A # implies A* and A* is equivalent to A,we
thus have an interpretation in which each A in Ɣ comes out true, so Ɣ is satisfiable.
Conversely, if Ɣ is satisfiable, there is an interpretation M of the original language
in which each A in Ɣ and hence each A* comes out true. By the preceding lemma,
M has an expansion N to an interpretation in which each A* remains true and all
Skolem axioms come out true. Since A* together with the Skolem axioms implies
A # , each A # in Ɣ # comes out true in N , and Ɣ # is satisfiable. We have thus shown
how we can associate to any set of sentences a set of ∀-sentences equivalent to it for
satisfiability. This fact, however, does not exhaust the content of the Skolemization
lemma. For it can also be used to give a proof of the Löwenheim–Skolem theorem,
and in a stronger version than that stated in chapter 12 (and proved in chapter 13).
To state the strong Löwenheim–Skolem theorem we need the notion of what it is
for one interpretation B to be a subinterpretation of another interpretation A. Where
function symbols are absent, the definition is simply that (1) the domain |B| should be
a subset of the domain |A|, (2) for any b 1 , ..., b n in |B| and any predicate R one has
(S1)
and (3) for every constant c one has
(S2)
R B (b 1 ,...,b n ) if and only if R A ((b 1 ),...,(b n ))
(c B ) = c A .
Thus, B is just like A, except that we ‘throw away’ the elements of |A| that are not
in |B|.
Where function symbols are present, we have also to require that for any b 1 , ...,
b n in |B| and any function symbol f the following should hold:
(S3)
f B (b 1 ,...,b n ) = f A (b 1 ,...,b n ).
Note that this last implies that f A (b 1 , ..., b n ) must be in |B|: Where function
symbols are absent, any nonempty subset B of A can be the domain of a subinterpretation
of A, but where function symbols are present, only those nonempty subsets B