Computability and Logic

256 NORMAL FORMS
in a sense to be made precise, be ‘eliminated’. (Constants will be treated as a special
sort of function symbol, namely 0-place function symbols. Whatever we say about
function symbols in this section goes for constants, too, and they will not be given
separate consideration.)
Let us take up the elimination of function symbols first. The first fact we need is
that any sentence is logically equivalent to one in which all function symbols occur
immediately to the right of the identity symbol. This means that no function symbols
occur in the blanks to the right of predicates other than the identity predicate, or
in the blanks to the right of a function symbol, or in the blank to the left of the
identity symbol, so the only occurrences of an n-place function symbol f are in
atomic subformulas of the type v = f (u 1 ,...,u n ), where v and the u i are variables
(not necessarily all distinct).
The proof is quite simple: Suppose that S is a sentence with at least one occurrence
of a function symbol f in a position other than immediately to the right of the identity
symbol. In any such occurrence, f occurs as the first symbol in some term t that
occurs (possibly as a subterm of a more complex term) in some atomic subformula
A(t). Let v be any variable not occurring in S, and let S − be the result of replacing
A(t) by the logically equivalent ∃v(v = t & A(v)). Then S is logically equivalent to
S − , and S − contains one fewer occurrence of function symbols in positions other
than immediately to the right of the identity symbol. Reducing the number of such
occurrences one at a time in this way, S is ultimately equivalent to a sentence with no
such occurrences. So for the remainder of this chapter, we consider only sentences
without such occurrences.
We show how to eliminate function symbols one at a time from such sentences.
(The process may be repeated until all function symbols, including constants, have
been eliminated.) If S is such a sentence and f an n-place function symbol occurring
in it, let R beanew(n + 1)-place predicate. Replace each subformula of the type
v = f (u 1 ,...,u n ) in which f occurs—and remember, these are the only kind of
occurrences of f in S—by R(u 1 ,...,u n , v), and call the result S ± . Let C be the
following sentence, which we call the functionality axiom:
∀x 1 ...∀x n ∃y∀z(R(x 1 ,...,x n , z) ↔ z = y).
Let D be the following sentence, which we call the auxiliary axiom:
∀x 1 ...∀x n ∀z(R(x 1 ,...,x n , z) ↔ z = f (x 1 ,...,x n )).
The precise sense in which the symbol f is ‘dispensable’ is indicated by the following
proposition (and its proof).
19.12 Proposition. S is satisfiable if and only if S ± & C is satisfiable.
Proof: Let us begin by sorting out the relationships among the various sentences
we have introduced. If we call the language to which the original sentence S belonged
L, the language obtained by adding the new predicate R to this language L + , and the
language obtained by removing the old function symbol f from the latter language
L ± , then S belongs to L, D to L + , and S ± and C to L ± . Note that D implies C, and
D implies S ↔ S ± .