300 DECIDABILITY OF ARITHMETIC WITHOUT MULTIPLICATION

This in turn is coextensive with

(iii) y(D jkm (y − jki) & ks < y & y < jt)

which is the sentence on the right, except for relettering the variable. For if x is as in (ii), then y = jkx will be as in (iii); and conversely, if y is as in (iii), then since jk divides y − jki, jk must divide y, which is to say that y will be of the form jkx for some x, which x will then be as in (ii).

(30) At this point, the only occurrences of x are in sentences of the form

x(Dm (x − i) & s <x & x <t).

Replace this by the disjunction of

Dm (s + j − i) & s + j <t

for all j with 1 ≤ j ≤ m. This step is justified because, given two integers a and b, there will be an integer strictly between them that leaves the same remainder as i when divided by m if and only if one of a + 1, . . . , a + m is such an integer.

We now have eliminated x altogether, and have obtained a quantifier-free formula coextensive with our original formula and involving no additional free variables, and we are done.

Problems

24.1Consider monadic logic without identity, and add to it a new quantifier (Mx)(A(x) > B(x)), which is to be true if and only if there are more x such that A(x) than there are x such that B(x). Call the result comparative logic. Show how to define in terms of M:

(a)and (so that these can be officially dropped and treated as mere abbreviations)

(b)‘most x such that A(x) are such that B(x)’

24.2Define a comparison to be a formula of the form (Mx)(A(x) > B(x)) where A(x) and B(x) are quantifier-free. Show that any sentence is equivalent to a

truth-functional compound of comparisons (which then by relettering may be taken all to involve the same variable x).

24.3As with sets of sentences of first-order logic, a set of sentences of logic with the quantifier M is (finitely) satisfiable if there is an interpretation (with a finite domain) in which all sentences in the set come out true. Show that finite satisfiability for finite sets of sentences of logic with the quantifier M is decidable. (The same is true for satisfiability, but this involves more set theory than we wish to presuppose.)

24.4For present purposes, by an inequality is meant an expression of the form

a1 x1 + · · · + am xm § b

where the xi are variables, the ai and b are (numerals for) specific rational numbers, and § may be any of <, ≤, >, ≥. A finite set of inequalities is coherent

if there are rational numbers ri that if taken for the xi would make each inequality in the set come out true (with respect to the usual addition operation and order relation on rational numbers). Show that there is a decision procedure for the coherence of finite sets of inequalities.

24.5In sentential logic the only nonlogical symbols are an enumerable infinity of sentence letters, and the only logical operators are negation, conjunction, and disjunction , &, . Let A1, . . . , An be sentence letters, and consider sentences of sentential logic that contain no sentence letters, but the Ai , or equivalently, that are truth-functional compounds of the Ai . For each sequence e = (e1, . . . , en ) of 0s and 1s, let Pe be ( )A1 & . . . & ( )An , where for each i, 1 ≤ i ≤ n, the negation sign preceding Ai is present if ei = 0, and absent if ei = 1. For present purposes a probability measure μ may be defined as an assignment of a rational number μ(Pe) to each Pe in such a way that the sum of all these numbers is 1. For a truth-functional combination A of the Ai we define μ(A) to be the sum of the μ(Pe) for those Pe that imply A, or equivalently, that are disjuncts in the full disjunctive normal form of A). The conditional probability μ(A\B) is defined to be the quotient μ(A & B)/μ(A) if μ(A) = 0, and is conventionally taken to be 1 if μ(A) = 0. For present purposes, by a constraint is meant an

expression of the form μ(A) § b or μ(A\B) § b, where A and B are sentences of sentential logic, b a nonnegative rational number, and § any of <, ≤, >, ≥. A finite set of constraints is coherent if there exists a probability measure μ that makes each constraint in the set come out true. Is the set of constraints μ(A\B) = 3/4, μ(B\C) = 3/4, and μ(A\C) = 1/4 coherent?

24.6Show that there is a decision procedure for the coherence of finite sets of constraints.