0

As a follow up to this question already been asked here, I was wondering - if we supposed that P != NP, would then the following reasoning be correct:

In NP problems we can only verify in poly-time whether x is in L based on the certificate that is also given as input. Hence we could be inputting the wrong certificate and therefore x still is in L even though the output we receive is 0. In this case, NP is not closed under complement as we cannot invert the response of our verifying TM to check for the membership in coNP.

Meki21
  • 93
  • 7

1 Answers1

1

As A. Schulz said here, it is not known if $\operatorname{P} \neq \operatorname{NP} \implies \operatorname{NP} \neq \operatorname{coNP}$.

The fact that we don't know how to build a polytime NTM for $\bar{L}$ if we have a polytime NTM that decides $L$ doesn't mean there is no way to do it: we simply don't know.

Turambar
  • 41
  • 4