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From Wikipedia's article on Presburger arithmetic:

Then Fischer and Rabin (1974) proved that any decision algorithm for Presburger arithmetic has a worst-case runtime of at least $2^{2^{cn}}$, for some constant $c>0$. Hence, the decision problem for Presburger arithmetic is an example of a decision problem that has been proved to require more than exponential run time.

(The referenced paper is Michael J. Fischer and Michael O. Rabin, Super-Exponential Complexity of Presburger Arithmetic, Proceedings of the SIAM-AMS Symposium in Applied Mathematics 7 (1974), pp.27–41)

Some other source: "Quantifier-free Boolean Algebra with Presburger Arithmetic is NP-Complete" (Viktor Kuncak, Technical Report MIT-CSAIL-TR-2007-001, MIT CSAIL, January 2007.)

So an $\mathsf{NP}$ problem that is proved to require more than exponential time. Why doesn't this establish that $\mathsf{P} \neq \mathsf{NP}$.

Rodrigo de Azevedo
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Bill Cody
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    That other source is wrong. Could you link to it? – wlad Jul 06 '15 at 12:47
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    It isn't immediately clear that the decision problem for Presburger arithmetic is in NP. One first thinks of checking a proof for any statement in (the additive theory of natural numbers), which of course can be done in time polynomial in the length of the proof, but what would be required is a check that can be done in time polynomial in the length of the statement, a substantially higher bar. – hardmath Jul 06 '15 at 12:51
  • It was the first link on a google of [Presburger np]. The very first sentence in Google's summary was "Presburger Arithmetic is NP-Complete". However the actual title of the paper (which was not included in full) was, "Quantifier-Free Boolean Algebra with Presburger Arithmetic is NP-Complete" – Bill Cody Jul 06 '15 at 12:56
  • I'm not an expert in this field, but the conjunction of those two quotes immediately drew my attention. Apologies if I should have linked to the actual paper first before posting here, but frankly wasn't actually aware that any decision problem had been proven to be exponential or greater. – Bill Cody Jul 06 '15 at 13:01
  • So is there still a question? Or is the question resolved after that closer look at the "other source"? – David K Jul 06 '15 at 13:41
  • Don't know yet; From abstract: "Quantifier-Free fragment of Boolean Al- gebra with Presburger Arithmetic (QFBAPA). Previous algorithms for QFBAPA had non-deterministic exponential time complexity. In this re- port I show that QFBAPA is in NP, and is therefore NP-complete" – Bill Cody Jul 06 '15 at 13:50
  • But if I dumbly stumbled into a proof that has eluded the greatest mathematicians for decades, I'm claiming full credit ;-) – Bill Cody Jul 06 '15 at 13:52
  • I've added a link to the "other source", a brief glance at which suggests that the fragment of Presburger arithmetic which is quantifier-free is there shown to be in NP, but avowedly not all of Presburger arithmetic is included in the theory considered there. – hardmath Jul 06 '15 at 14:09
  • Seems like if Presburger Arithmetic is combined with something else, then the Time Complexity for QFBAPA will be determined by the greater of the two, meaning its time complexity is 2^2^cn. Maybe somehow the paper writer wasn't aware of the previous results from 1974, who knows. Wishful thinking in the extreme maybe. – Bill Cody Jul 06 '15 at 14:11
  • wikipedia: "A theory has quantifier elimination if for every formula there exists another formula without quantifiers that is equivalent to it (modulo the theory). Examples of theories that have been shown decidable using quantifier elimination are Presburger arithmetic" – Bill Cody Jul 06 '15 at 14:34
  • @BillCody: Have a look at the paper. The "quantifier-free" qualification applies to both the Boolean algebra and the Presburger arithmetic formulas. The author Viktor Kuncak has written previously on decision procedures for the full theory of Boolean algebra and Presburger arithmetic, and its reduction to a decision procedure for Presburger arithmetic, so it seems he would be fully aware of the literature. – hardmath Jul 06 '15 at 14:37
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    From wikipedia page on NP-complete https://en.wikipedia.org/wiki/NP-complete "Since NP-complete problems are in NP, their running time is at most exponential. However, some problems provably require more time, for example Presburger arithmetic." – TomGrubb Jul 06 '15 at 14:39
  • Thanks for that, bburGsamohT. Hardmath, My analysis at this point is at the level of text-matching via google, but that quote above by me on quantifier elimination seems to imply that in Presburger arithmetic the quantifiers are just a convenience, as there is always an equivalent expression without quantifiers. Probably just need to bow out of this discussion at this point and delve into the paper. – Bill Cody Jul 06 '15 at 15:01
  • But interestingly, he doesn't mention the 1974 paper in the footnotes, bibliography or anywhere. – Bill Cody Jul 06 '15 at 15:03
  • For a paper co-authored by Kuncak in 2005 that cites the 1974 paper by Fischer and Rabin, see here. – hardmath Jul 06 '15 at 15:17

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The doubly exponential worse case runtime for deciding Presburger arithmetic theorems occurs only when nested quantifiers are used. Since the paper describes a quantifier-free version of Presburger arithmetic, the doubly exponential lower bound does not apply. We're left with the singly-exponential runtime upper-bound known for all NP-complete problems and an unknown lower-bound, leaving the P = NP question unresolved.

Kyle Jones
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