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1$\begingroup$ No, the undecidability of DEP (Hilbert's tenth problem) wasn't shown until 1970, by Matiyesevich. The Entscheidungsproblem is not Hilbert's tenth problem; concerns the validity of formulae of first-order logic. And, once again, the P vs. NP problem is absolutely not a problem about whether undecidable problems are verifiable. $\endgroup$David Richerby– David Richerby2016-02-18 09:38:46 +00:00Commented Feb 18, 2016 at 9:38
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2$\begingroup$ If you want to provide more details you should edit your original post. $\endgroup$Tom van der Zanden– Tom van der Zanden2016-02-18 12:06:31 +00:00Commented Feb 18, 2016 at 12:06
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$\begingroup$ @DavidRicherby Note that the answer given by Ben : « the set of problems decidable by NTMs is identical to the set of problems decidable by TMs ». Just in this sense, I think that the definition of NP confuses P with NP, and it leads to P=NP (NDTM). If this definition needs to be questioned, then other conclusions deduced from this definition, like the equivalence of a deterministic verifier and a non-deterministic decider, need also to be questioned. $\endgroup$yu LI– yu LI2016-02-24 04:30:37 +00:00Commented Feb 24, 2016 at 4:30
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$\begingroup$ @YuLi "it leads to P=NP (NDTM)." I have no idea what you mean by that. Also, I don't see the relevance of pointing out that TMs and NTMs decide the same languages. If they didn't decide the same languages, NTMs would be a completely unreasonable model of computation and it's hard to imagine that we'd care what they can compute in polynomial time. In complexity theory, we're taking a finer-grained view and asking about the computational resources required and the definition of NP doesn't confuse that at all. $\endgroup$David Richerby– David Richerby2016-02-24 04:57:48 +00:00Commented Feb 24, 2016 at 4:57
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$\begingroup$ @DavidRicherby Thanks, I have modified my answer according to your remark in order to clarify the relation of the Entscheidungsproblem and Hilbert's tenth problem. Concerning the question about the current definition of NP, it is difficult to discuss in several words. The objective of my answer is just to evoke some reflexions about this basic topic, … $\endgroup$yu LI– yu LI2016-02-24 05:49:18 +00:00Commented Feb 24, 2016 at 5:49
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