Computability and Logic

21.3. DYADIC LOGIC 275
to a quantified formula ∃xB(x) in terms of the associate (B(x)) © of the subformula
B(x), and analogously for ∀.
∃xB(x) will of course be equivalent to ∃x(B(x)) © . And (B(x)) © will be a truthfunctional
compound of clear formulas A 1 ,...,A n , each of which either is atomic or
begins with a quantifier. Consider a formula equivalent to (B(x)) © that is in disjunctive
normal form in the A i . It will be a disjunction B 1 ∨···∨B r of formulas B j , each of
which is a conjunction of some of the A i and their negations. We may assume each
B j has the form
C j,1 & ... & C j,r & D j,1 & ... & D j,s
where the Cs are the conjuncts in which the variable x does occur and the Ds those in
which it does not; by clarity, the Ds will include all conjuncts that begin with or are
the negations of formulas beginning with quantifiers, and the Cs will all be atomic.
Then as ∃x(B(x)) © we may take the disjunction B ′ 1 ∨···∨B′ r , where B′ j is
∃x(C j,1 & ··· & C j,r )&D j,1 & ... & D j,s .
(In the degenerate case where r = 0, B ′ j is thus the same as B j.)
Again we go straight to work.
21.3 Dyadic Logic
Proof of Lemma 21.2: Lemma 21.1 tells us the satisfiability problem is unsolvable
for predicate logic, and we want to show it is unsolvable for dyadic logic. It will be
enough to show how one can effectively associate to any sentence of predicate logic a
sentence of dyadic logic such that the former will be satisfiable if and only if the latter
is. What we are going to do is to show how to eliminate one three-place predicate
(at the cost of introducing new two- and one-place predicates). The same method
will work for k-place predicates for any k ≥ 3, and applying it over and over we
can eliminate all but two- and one-place predicates. The one-place ones can also be
eliminated one at a time, since given a sentence S containing a one-place predicate
P, introducing a new two-place predicate P* and replacing each atomic subformula
Px by P*xx clearly produces a sentence S* that is satisfiable if and only if S is. Thus
we can eliminate all but two-place predicates.
To indicate the method for eliminating a three-place predicate, let S be a sentence
containing such a predicate P. Let P* be a new one-place predicate, and Q i for
i = 1, 2, 3 a trio of new two-place predicates. Let w be a variable not appearing in
S, and let S* be the result of replacing each atomic subformula of form Px 1 x 2 x 3
in S by
∃w(Q 1 wx 1 & Q 2 wx 2 & Q 3 wx 3 & P ∗ w).
We claim S is satisfiable if and only if S* is satisfiable. The ‘if’ direction is easy.
For if S is unsatisfiable, then ∼S is valid, and substitution (of a formula with the
appropriate free variables for a predicate) preserves validity, so ∼S* is valid, and S*
is unsatisfiable.