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Suppose $P \neq NP$. Does it imply that there exists some superpolynomial time bound, such that any $NP$-complete problem, like SAT, can be used to simulate an arbitrary deterministc Turing Machine working in that time bound?

Rephrasing does $P \neq NP$ imply that there exists some class $D$ of languages solvable by a deterministic Turing Machine, such $P \subsetneq\ D \subseteq NP$ and SAT is $D$-hard?

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    $\begingroup$ $D$-hard under what kind of reduction? Since SAT is $NP$-hard, doesn't that automatically make SAT $D$-hard? Any language in $NP$ (and thus any language in $D$) can be reduced to SAT in polytime. $\endgroup$ May 4, 2015 at 14:54
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    $\begingroup$ Your rephrasing sounds like a question that Ladner's Theorem answers. Or do you mean something else? $\endgroup$
    – Kyle Jones
    May 4, 2015 at 16:14

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This post shows that the answer to your initial question is yes (even if P = NP).

On the other hand, I suspect there is no known proof that if unambiguousGC(polylog, NC) ⊈ coNP/poly then there is a superpolynomial time-constructible function $T$ and a superlogarithmic space-constructible function $S$ such that

DTIME(T(n)) ∩ DSPACE(S(n)) ⊆ NP/poly.

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    $\begingroup$ Dude. Stop formatting like that. How many times have the mods brought that up, now? $\endgroup$ May 6, 2015 at 4:35
  • $\begingroup$ I think once, on the meta thread. $\endgroup$
    – user12859
    May 6, 2015 at 5:28
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    $\begingroup$ @RickyDemer I recall multiple instances in comments (may be gone by now). Seriously, just use Unicode math... $\endgroup$
    – Raphael
    May 6, 2015 at 6:48
  • $\begingroup$ @Raphael: Ah yes, you commented on the meta thread too. (I don't remember any other instance in which a mod brought that up.) $\endgroup$
    – user12859
    May 6, 2015 at 6:55
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    $\begingroup$ @RickyDemer Oh, I don't know. Then consider this instance 2+x: please stop formatting like this. $\endgroup$
    – Raphael
    May 6, 2015 at 6:56

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