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Computability and Logic

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300 DECIDABILITY OF ARITHMETIC WITHOUT MULTIPLICATION<br />

This in turn is coextensive with<br />

(iii)<br />

∃y(D jkm (y − jki)&ks < y & y < jt)<br />

which is the sentence on the right, except for relettering the variable. For if x is as<br />

in (ii), then y = jkx will be as in (iii); <strong>and</strong> conversely, if y is as in (iii), then since<br />

jk divides y − jki, jk must divide y, which is to say that y will be of the form jkx<br />

for some x, which x will then be as in (ii).<br />

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

Replace this by the disjunction of<br />

∃x(D m (x − i)&s < x & x < t).<br />

D m (s + j − i)&s + j < t<br />

for all j with 1 ≤ j ≤ m. This step is justified because, given two integers a <strong>and</strong> b,<br />

there will be an integer strictly between them that leaves the same remainder as i<br />

when divided by m if <strong>and</strong> only if one of a + 1,...,a + m is such an integer.<br />

We now have eliminated x altogether, <strong>and</strong> have obtained a quantifier-free formula<br />

coextensive with our original formula <strong>and</strong> involving no additional free variables, <strong>and</strong><br />

we are done.<br />

Problems<br />

24.1 Consider monadic logic without identity, <strong>and</strong> add to it a new quantifier<br />

(Mx)(A(x) > B(x)), which is to be true if <strong>and</strong> only if there are more x such<br />

that A(x) than there are x such that B(x). Call the result comparative logic.<br />

Show how to define in terms of M:<br />

(a) ∀ <strong>and</strong> ∃ (so that these can be officially dropped <strong>and</strong> treated as mere<br />

abbreviations)<br />

(b) ‘most x such that A(x) are such that B(x)’<br />

24.2 Define a comparison to be a formula of the form (Mx)(A(x) >B(x)) where<br />

A(x) <strong>and</strong> B(x) are quantifier-free. Show that any sentence is equivalent to a<br />

truth-functional compound of comparisons (which then by relettering may be<br />

taken all to involve the same variable x).<br />

24.3 As with sets of sentences of first-order logic, a set of sentences of logic with<br />

the quantifier M is (finitely) satisfiable if there is an interpretation (with a finite<br />

domain) in which all sentences in the set come out true. Show that finite satisfiability<br />

for finite sets of sentences of logic with the quantifier M is decidable.<br />

(The same is true for satisfiability, but this involves more set theory than we<br />

wish to presuppose.)<br />

24.4 For present purposes, by an inequality is meant an expression of the form<br />

a 1 x 1 +···+a m x m § b<br />

where the x i are variables, the a i <strong>and</strong> b are (numerals for) specific rational<br />

numbers, <strong>and</strong> § may be any of ,≥. A finite set of inequalities is coherent

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