# Without using the Time Hierarchy Theorem, is there any other way to prove P /= EXP

I haven't found any way to show this was the case (that P/=EXP, without using the Time Hierarchy Theorem) If you can find a way to show this, can you give a short explanation of this,and also if possible link a paper supporting your conclusion. Thank you so much

• But I have a question Kaban: How exactly did you derive your conclusion from the post you linked. Dec 29 '18 at 4:42

There is a way, actually! Kind of.

Consider two cases:

1. $$P \not= NP$$. This case is easy: then $$P \not= EXP$$, because $$NP \subset EXP$$.
2. $$P = NP$$. It is known that $$P = NP$$ implies that there is a language from $$EXP$$ that has circuit complexity of at least $$\frac{2^n}{10n}$$, so $$EXP$$ is not a subset of $$P/\mathrm{poly}$$ in this case, let alone $$P$$. I will recall the proof here with more details if it is necessary, but that question has already been answered, for example, here: https://mathoverflow.net/questions/57828/p-np-exp-has-circuit-size-o2n-n (the answer is pretty sketchy, though, so feel free to ask for more details).
• Do we have any evidence that we have either results for the question P vs. NP: P = NP or P != NP? Since it is possible that P vs. NP are independent of axioms of mathematics (like Axiom of Choice). Scott Aaronson (may be others) tried to show for current techniques that have been used to show that a mathematical statement is an independents of mathematical axioms cannot be used for the case of P vs. NP. So it is open that P vs. NP is an independent of mathematical axioms, P = NP, or P != NP. So, I believe that your proof is based on very strong assumption that P = NP or P != NP. Jul 16 at 18:52