I need to prove that

$coNP \neq NTIME(n^2)$ using time hierarchy theorem.

as we know $DTIME(n^4)\subseteq P\subseteq coNP$

from time hierarchy theorem we can derive that $DTIME(n^4) \not \subseteq DTIME(n^2)$

therefore $DTIME(n^2) \neq coNP$


as we know $NTIME(n^4)\subseteq NP$

from time hierarchy theorem we can derive that $NTIME(n^4) \not \subseteq NTIME(n^2)$

therefore $NTIME(n^2) \neq NP$

But anyway, I could not derive $coNP \neq NTIME(n^2)$ from any combination. Can someone give me a clue?


1 Answer 1


You should use the fact that $\textsf{NTIME}(n^2)\subsetneq \textsf{NP}$.

If $\text{co}\mathsf{NP} = \textsf{NTIME}(n^2)$, then $\text{co}\mathsf{NP} \subsetneq \mathsf{NP}$ and you can reach a contradiction.

  • $\begingroup$ why is $coNP \subsetneq NP$ a contradiction?, I mean, isn't the relationship between coNP and NP one of the greatest questions in complexity theory? $\endgroup$
    – Vahan
    Jan 14, 2023 at 14:07
  • $\begingroup$ Please note that there is a difference between $\text{co}\mathsf{NP}\not \subseteq \mathsf{NP}$ and $\text{co}\mathsf{NP}\subsetneq \mathsf{NP}$. What would be problematic is a strict inclusion. Note that for $A$ a problem, $\text{coco}A = A$. $\endgroup$
    – Nathaniel
    Jan 14, 2023 at 14:11
  • $\begingroup$ I got it, thanks)) $\endgroup$
    – Vahan
    Jan 14, 2023 at 14:19

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