Decidability of Arithmetic without Multiplication

296 DECIDABILITY OF ARITHMETIC WITHOUT MULTIPLICATION

Given a sentence S of L without ·, replace everywhere in it by +1, and replace every quantification x or x by a relativized quantification

x((x = 0 0 <x) → · · ·) x((x = 0 0 <x) & . . .).

to obtain a sentence S* of K . Then S will be true in N if and only if S* is true in M. Thus to prove Presburger’s theorem, it will be sufficient to describe an effective procedure for determining whether or not a given sentence of K is true in M.

For the remainder of this chapter, therefore, term or formula or sentence will always mean term or formula or sentence of K, while denotation or satisfaction or truth will always mean denotation or satisfaction or truth in M. We call two terms r and s coextensive if v1 . . . vn r = s is true, where the vi are all the variables occurring in r or s. We call two formulas F and G coextensive if v1. . . vn (F ↔ G) is true, where the vi are all the free variables occurring in F or G.

Given any closed term, we can effectively calculate its denotation. Given any atomic sentence, we can effectively determine its truth value; and we can therefore do the same for any quantifier-free sentence. We are going to show how one can effectively decide whether a given sentence S is true by showing how one can effectively associate to S a coextensive quantifier-free sentence T : once T is found, its truth value, which is also the truth value of S, can be effectively determined.

The method to be used for finding T , given S, is called elimination of quantifiers. It consists in showing how one can effectively associate to a quantifier-free formula F(x), which may contain other free variables besides x, and quantifier-free G such that G is coextensive with x F(x) and G contains no additional free variables beyond the free variables in x F(x). This shown, given S, we put it in prenex form, then replace each quantification x by x , and work from the inside outward, successively replacing existential quantifications of quantifier-free formulas by coextensive quantifier-free formulas with no additional free variables, until at last a sentence with no free variables, which is to say, a quantifier-free sentence T , is obtained.

So let F(x) be a quantifier-free formula. We obtain G, coextensive with x F(x) and containing no addition free variables beyond those in x F(x), by performing, in order, a sequence of 30 operations, each of which replaces a formula by a coextensive formula with no additional free variables.

In describing the operations to be gone through, we make use of certain notational conventions. When writing of a positive integer k and a term t we allow ourselves to write

for instance. With such notation, the 30 operations are as follows:

(1)Put F into disjunctive normal form. (See section 19.1.) Thus we get a disjunction of conjunctions of atomic formulas of the forms r = s or r <s or Dm s (where r and s are terms) and negations of such.

(2)Replace each formula of form r = s by (r < s + 1 & s < r + 1).

(3)Replace each formula of form r = s by (r < s s < r).

(4)Put the result back into disjunctive normal form.

(5)Replace each formula of form r < s by s < r + 1.

(6)Replace each formula of form Dm s by the disjunction of Dm (s + i) for all i with 0 < i < m. The result is coextensive with the original, because for any number

a, m divides exactly one of a, a + 1, a + 2, . . . , a + m − 1.

(7)Put the result back into disjunctive normal form.

(8)At this point we have a disjunction of conjunctions of atomic formulas of the forms r < s and Dm s. Replace each formula of form r < s by 0 < (s --- r).

(9)We say a term is in normal form if it has one of the five forms

kx, --- kx, kx + t, --- kx + t, or t, wherein t is a term not containing the variable x. For every term one can effectively find a coextensive term in normal form by ordinary algebraic operations, such as regrouping and reordering summands. Replace each term in the formula that is not in normal form by a coextensive one that is.

(10)Replace each formula of form

as the case may be.

(11)At this point all atomic formulas with predicate < that contain the variable x have either the form t < kx or the form kx < t, where k is positive and t does not contain x. We call those of the former form lower inequalities and those of the latter form upper inequalties. Rearrange the order of conjuncts in each disjunct so that all lower inequalities occur on the left.

(12)Towards reducing the number of lower inequalities occurring in any disjunct, if a conjunction of form t1 < k1 x & t2 < k2 x occurs in a disjunct, replace it by the disjunction of the following three conjunctions:

(i)t1 < k1 x & k1t2 < k2t1

(ii)t1 < k1 x & k1t2 = k2t1

(iii)t2 < k2 x & k2t1 < k1t2.

To see that this substitution is justified (that is, to see that it produces a result coextensive with the original), note that exactly one of the second conjuncts in

(i)–(iii) must hold, and that (i) or (ii) holds, then so do k2t1 < k1k2 x and

k1t2 < k1k2 x, and hence so does t2 < k2 x; while similarly (iii) yields t1 < k1 x.

(13)Eliminate any occurrences of = introduced at the previous step by repeating steps

(2) and (4).

(14)The effect of the preceding three steps is to reduce by one the number of lower inequalities in any disjunct where there were more than one to begin with. (Note that k1t2 < k2t1, for instance, does not count as a lower inequality, since it does not

298 DECIDABILITY OF ARITHMETIC WITHOUT MULTIPLICATION

contain the variable x.) Repeat these three steps over and over until no disjunct has more than one lower inequality among its conjuncts.

(15)Carry out an analogous process for upper inequalities, until no disjunct has more than one lower or upper inequality among its conjuncts.

(16)Replace each formula of form

as the case may be. This step is justified because for any number a, m divides a if and only if m divides −a.

(17)At this point all atomic formulas with Dm and involving x have the form

Dm (kx --- t), where k is a positive integer. Replace any formula of this form by the disjunction of all conjunctions

Dm (kx --- i) & Dm (t --- i)

for 0 ≤ i < m. To see that this step is justified, note that m divides the difference of two numbers a and b if and only if a and b leave the same remainder on division by m, and that the remainder on dividing a (respectively, b) by m is the unique i with 0 ≤ i < m such that m divides a − i (respectively, b − i).

(18)Put the result back into disjunctive normal form.

(19)At this point all atomic formulas with Dm and involving x have the form

Dm (kx --- i), where k is a positive integer and 0 ≤ i < m. Replace any formula of this form with k >1 by the disjunction of the formulas Dm (x --- j) for all j with 0 ≤ j < m such that m divides k j − i. This step is justified because for any

number a, ka leaves a remainder of i on division by m if and only if kj does, where j is the remainder on dividing a by m.

(20)Put the result back into disjunctive normal form.

(21)At this point all atomic formulas with Dm and involving x have the form

Dm (x --- i), where i is a nonnegative integer. In any such case consider the prime decomposition of m; that is, write

m = p1e1 · · · pkek where p1 < p2 < · · · < pk and all ps are primes. If k >1, then let m1 = p1e1 , . . . , mk = pkek , and replace Dm (x --- i) by

Dm1 (x --- i) & . . . & Dmk (x --- i).

This step is justified because the product of two given numbers having no common factor (such as powers of distinct primes) divides a given number if and only if each of the two given numbers does.

(22)At this point all atomic formulas with Ds and involving x have the form

Dm (x − i), where i is a nonnegative integer, and m a power of a prime. If in a given disjunct there are two conjuncts Dm (x − i) and Dn (x − j) where m and n are powers of the same prime, say m = pd , n = pe, d ≤ e, then drop Dm (x --- i) in favor of Dn (i − j), which does not involve x. This step is justified because, since

m divides n, for any number a, if a leaves remainder j on division by n, a will leave remainder i on division by m if and only if j does.

(23)Repeat the preceding step until for any two conjuncts Dm (x --- i) and Dn (x --- j) in a single disjunct, m and n are powers of distinct primes, and therefore have no common factors.

(24)Replace each Dm (x --- i) by Dm (x --- i*), where i* is the remainder on dividing i by m.

(25)Rewrite each disjunct so that all atomic formulas with with Ds and involving x are on the left.

(26)At this point each disjunct has the form

Dm1 (x --- i1) & . . . & Dmk (x --- ik ) & (other conjuncts)

where 0 ≤ i1 < m1, . . . , 0 ≤ ik < mk . Let m = m1 · · · · · mk . According to the Chinese remainder theorem (see Lemma 15.5), there exists a (unique) i with 0 ≤ i < m such that i leaves remainder i1 on division by m1, . . . , i leaves

remainder ik on division by mk . Replace the conjuncts involving Ds by the single formula Dm (x --- i).

(27)At this point we have a disjunction F1 · · · Fk each of whose disjuncts is a conjunction containing at most one lower inequality, at most one upper inequality, and at most one formula of form Dm (x − i). Rewrite x(F1 · · · Fk ) as

x F1 · · · x Fk .

(28)Within each disjunct xF, rewrite the conjunction F so that any and all conjuncts involving x occur on the left, and confine the quantifier to these conjuncts, of which there are at most three; if there are none, simply omit the quantifier.

(29)At this point, the only occurrences of x are in sentences of one of the seven types listed in Table 24-1. Replace these by the sentences listed on the right.

Table 24-1. Elimination of quantifiers

This step is justified in the first five cases because in these cases the sentence on the right is automatically true. (In the fourth case this is because there are arbitrarily large integers leaving a prescribed remainder i on division by m, and similarly in the fifth case.) The sixth and seventh cases are similar to each other. We discuss the latter because it is slightly more complicated. First note that the sentence on the left,