Computation in Logic and Logic in ComputationThis paper is dedicated to Alan Turing, to commemorate the Turing Centenary Year 2012 – his 100{}^{\rm th} birthyear.

Computation in Logic and Logic in Computation1

\AddToShipoutPicture

*\BackgroundPic

{cicisabstract}

The theory of addition in the domains of natural (), integer (), rational (), real () and complex () numbers is decidable; so is the theory of multiplication in all those domains. By Gödel’s Incompleteness Theorem the theory of addition and multiplication is undecidable in the domains of , and ; though Tarski proved that this theory is decidable in the domains of and . The theory of multiplication and order behaves differently in the above mentioned domains of numbers. By a theorem of Robinson, addition is definable by multiplication and order in the domain of natural numbers; thus the theory is undecidable. By a classical theorem in mathematical logic, addition is not definable in terms of multiplication and order in . In this paper, we extend Robinson’s theorem to the domain of integers () by showing the definability of addition in ; this implies that is undecidable. We also show the decidability of by the method of quantifier elimination. Whence, addition is not definable in .

{ciciskeywords}

Decidability; First-Order Logic; Gödel’s Incompleteness Theorems; Church’s Theorem; Presburger Arithmetic; Skolem Arithmetic; Quantifier Elimination.

{cicismain}

1 Introduction

The question of the decidability of logical inference has triggered the beginning of computer science. Propositional Logic is decidable, since truth tables provide a finite semantics for it. Aristotle’s Syllogism, or in modern terminology the first-order logic of unary predicates, is decidable, since it has the finite model property. The notion of a Turing Machine was a successful outcome of the struggle to settle the question of the decidability of full First-Order Logic. It is now known that the first-order logic is undecidable if it has a binary relation symbol or a binary function symbol ([cdp]). The additive theory of natural numbers was shown to be decidable by Presburger in 1929 (and by Skolem in 1930; see [lnt]). The additive theories of integer, rational, real and complex numbers (, , and ) are decidable as well. The multiplicative theory of the natural numbers is also shown to be decidable by Skolem in 1930; the theories , , and are also decidable.

Then it was expected that the theory of addition and multiplication of natural numbers would be decidable too; confirming Hilbert’s Program. But the world was shocked in 1931 by Gödel’s Incompleteness Theorem who showed that the theory is undecidable (see [lnt]). The theory is undecidable too, since is definable in this structure: by Lagrange’s Theorem . So is the theory by Robinson’s result [robinson] which shows that is definable in this structure too. However, Tarski showed that the theories and are decidable ([marker]). It is worth mentioning that the order relation is definable by means of addition and multiplication in all the above domains of numbers. For example, the formulas and define the relation in the structures and respectively. The theory of addition and order is somehow weak, in all the above number domains, since it cannot define multiplication. The theory of multiplication and order has not been extensively studied; one reason is that addition is not definable in , since the bijection of preserves multiplication and order but does not preserve addition. Also it is known that addition is definable in by Tarski’s identity ([robinson]):


where is the successor of , which is definable by the order relation: . The symbol is a shorthand for .

The question of the decidability or undecidability of the structures and are missing in the literature. In this paper, by modifying Tarski’s identity we show that addition is definable in the structure ; this implies the undecidability of . On the contrary, addition is not definable in ; here we show a stronger result by the method of quantifier elimination: the theory is decidable. Whence, by Robinson’s above-mentioned result [robinson], addition cannot be defined in this structure. An interesting outlook of our results is that though puts the domains , and on the undecidable side, and the domains and on the decidable side, the language puts the domains and on the undecidable side, but and on the decidable side.

2 Multiplication and Order in

Tarski’s identity can define the formula in when . The case was easily settled in natural numbers: for any we have . But this does not hold in , and so we have to treat this case differently. Our trick is to define the relation in terms of multiplication and successor (which is definable by order): . Thus, the following formula defines addition in terms of multiplication and order in :


So, the theories and are interdefinable, and hence is undecidable.

3 Multiplication and Order in

Unlike the case of , addition is not definable in the structure . In fact, the theory of this structure is decidable. For showing that we use the method of quantifier elimination. First let us note that the language does not allow quantifier elimination for , since e.g. the formula is not equivalent to a quantifier-free formula. So, we restrict our attention to and extend the language to , where is interpreted as “being the th power of a rational”; or in other words .

Theorem. The structure admits quantifier elimination.

We note that the above main theorem implies that the structure admits quantifier elimination as well. It is enough to distinguish the signs: for any , either or or ; so eliminating the quantifiers in each case, will eliminate all of the quantifiers. Let us also note that the quantifier-free formulas of are decidable: for any given rational number and any natural one can decide if is an th power of (an-)other rational number or not. Thus, quantifier elimination in implies the decidability of the structure , and hence .

The rest of the paper is devoted to proving the main theorem. The folklore technique of quantifier elimination starts from characterizing the terms and atomic formulas, also eliminating negations, implications and universal quantifiers, and then removing the disjunctions from the scopes of existential quantifiers, which leaves the final case to be the existential quantifier with the conjunction of some atomic (or negated atomic) formulas. Removing this one existential quantifier implies the ability to eliminate all the other quantifiers by induction. Let us summarize the first steps:

For a variable and parameter , all terms are equal to for some . Atomic formulas are in the form or or for some terms and . Negated atomic formulas are thus , and ; the formulas and are equivalent to and respectively. By de Morgan’s laws we can assume that the negation appears only behind the atomic formulas of the form , and by the equivalences and , we can assume that the implication symbol and universal quantifier do not appear in the formula (whose quantifiers are to eliminated). Finally, the equivalence leaves us with the elementary formulas of the form where each is in the form or or or or for some and terms . Whence, it suffices to show that the formula is equivalent to another formula in which (and so ) does not appear. This will finish the proof.

Here comes the next steps of quantifier elimination. The powers of can be unified: let be the least common multiplier of the ’s, ’s, ’s, ’s and ’s. From the equivalences , and , we infer that the above formula can be re-written equivalently as


for possibly new ’s, ’s, ’s, ’s, ’s, ’s and ’s. This formula is in turn equivalent to


(with the substitution ). Thus it suffices to show that the following formula


is equivalent to a quantifier-free formula. If the conjunction is not empty, then the above formula is equivalent to the quantifier-free formula


for some term . So, let us assume that the conjunction is empty, and thus we are to eliminate the quantifier of the formula
.

The formula is equivalent to (the quantifier-free formula) (that is ), since is dense.

For the formula , let p be a prime number, and put be the greatest number such that divides ; similarly is the greatest number such that divides . Then is equivalent to . By a generalized form of the Chinese Remainder Theorem ([lnt]) the existence of such an is equivalent to ; here is the greatest common divisor of and . That is equivalent to . We further note that in case of there are infinitely many solutions for which are in the form for some fixed integers and ’s; is arbitrary. In fact is the least common multiplier of ’s, and ’s are where ; the existence of ’s follows from the fact that the greatest common divisor of ’s is . Moreover, the solution is unique up to the module . So, if there exists some which satisfies for some , then it must be of the form for some (arbitrary) .

Thus, the formula is equivalent to (the quantifier-free formula) , since the solution for can be chosen to satisfy : choose a rational number between the positive real numbers and . Since the set is dense in , there exists such a rational number . Then is the desired solution.

Finally, we show that the formula


is equivalent to the following quantifier-free formula

,
where is the least common multiplier of ’s, and in which ’s satisfy .

If for some , holds, then clearly is true, and it can be easily seen that we also have . Assume ; we show that . Note that there exists some such that . Now if , then , and so by we have which contradicts the assumption . Whence, holds.

Conversely, if we have , then by the above arguments there exist some positive real numbers such that for any rational with , the number satisfies the formula where and are as above. Let P be a sufficiently large prime number which does not divide any of the numerators or denominators of (the reduced fractions of) ’s or ’s. Let and let be a positive rational number such that . We show that satisfies . Note that since we already have . For showing we distinguish two cases. (1) If then implies contradicting the assumption ; thus . (2) If , then or equivalently implies since . Since P does not divide any of the numerators or denominators of (the reduced fractions of) ’s or (’s), then we must have which holds if and only if ; this contradicts our assumption . Thus . Whence, all in all we showed that holds. Q.E.D

Acknowledgements

This research was partially supported by grant No. 90030053 of the Institute for Research in Fundamental Sciences (IPM), Tehran.

References

{biblist}\bib

cdpbook title=The Classical Decision Problem, author=Börger,E., author=Grädel,E., author=Gurevich,Y., date=2001, publisher=Springer-Verlag, Berlin, address=

\bib

robinsonarticle title=Definability and Decision Problems in Arithmetic, subtitle=, author=Robinson,J., author= , author= , journal=The Journal of Symbolic Logic, volume=14, date=1949, pages=98–114

\bib

markerbook title=Model Theory: An Introduction, author=Marker,D., date=2002, publisher=Springer-Verlag, Berlin, address=

\bib

lntbook title=Logical Number Theory I: An Introduction, author=Smoryński,C., date=1991, publisher=Springer-Verlag, Berlin, address=

Saeed Salehi, “Computation in Logic and Logic in Computation”, Invited Paper in:
Bahram B. Sadeghi (editor), Proceedings of the Third International Conference on
Contemporary Issues in Computer and Information Sciences (CICIS 2012),
Institute for Advanced Studies in Basic Sciences, Gavazangh, Zanjan, Iran,
Brown Walker Press 2012, USA, ISBN 9781612336237 (624 pages), pp. 580–583.
http://www.universal-publishers.com/book.php?method=ISBN&book=161233623X

Footnotes

  1. This paper is dedicated to Alan Turing, to commemorate the Turing Centenary Year 2012 – his 100 birthyear.
Comments 0
Request Comment
You are adding the first comment!
How to quickly get a good reply:
  • Give credit where it’s due by listing out the positive aspects of a paper before getting into which changes should be made.
  • Be specific in your critique, and provide supporting evidence with appropriate references to substantiate general statements.
  • Your comment should inspire ideas to flow and help the author improves the paper.

The better we are at sharing our knowledge with each other, the faster we move forward.
""
The feedback must be of minumum 40 characters
Add comment
Cancel
Loading ...
103261
This is a comment super asjknd jkasnjk adsnkj
Upvote
Downvote
""
The feedback must be of minumum 40 characters
The feedback must be of minumum 40 characters
Submit
Cancel

You are asking your first question!
How to quickly get a good answer:
  • Keep your question short and to the point
  • Check for grammar or spelling errors.
  • Phrase it like a question
Test
Test description