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I am wondering why the following "proof" of $coNP = NP$ does not work:

$\subseteq:$Let $L$ be a language in $coNP$, that means there is a non-deterministic Turing Machine $M$ that decides the complement of $L$, denoted by $\overline{L}$, in polynomial time. Then the Turing Machine $N$ that decides $L$ in polynomial time by reversing the result of running $M$ on $\overline{L}$.

$\supseteq:$ Reverse the above arguments.

Remark: I know that this "proof" can not be correct in this way, since $coNP = NP$ is still an open problem and it is certainly not that easy. But I do not understand where the reasoning above goes wrong. Could you please explain this to me?

3nondatur
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1 Answers1

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Nondeterministic Turing machines are asymmetric in their treatment of accept and reject states: they accept if any path accepts, and reject if all paths reject.

If you swap the accept and reject states of an ordinary Turing machine then it will accept the complement of the language it accepted before, but that isn't generally true of nondeterministic Turing machines.

benrg
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