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When $\overline{M}$ is a non-deterministic polynomial by time Turing machine that final states switched: accept to reject and vice versa. I'm thinking that this equal to $P=NP$, but I saw a solution (an example) that I disagree with: $M$ is a non-deterministic polynomial by time Turing machine that decide $SAT$, if all that paths are rejected then $L(\overline{M})=\Sigma^*\in P$

Is it a valid solution, or as I'm thinking $L(M)\in NPC$ and $L(\overline{M})\in P \Leftrightarrow P=NP$

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Let's imagine a Non Deterministic Turing Machine $M$ that decide $SAT$. If we tune this machine a bit, and add a transition on the initial state, for every letter read, that reject the entry. Let $M'$ be the new NTM. Then, $L(M') = L(M)$, since $u\in L(M) \Leftrightarrow \exists$ at least one computational path in $M$ to an accepting state (and the same thing for $M'$).

Now consider $\overline{M'}$. Since we added a rejecting transition which is possible for every entry in $M'$, that means that there is an accepting transition for every entry in $\overline{M'}$, so it means that $L(\overline{M'}) = \Sigma^* \in P$.

That does not necessarily imply that $P = NP$, the reason being that $L(\overline{M'}) \neq \overline{L(M')}$.

Generaly speaking, $L \in NPC \Leftrightarrow \overline{L} \in \text{co-}NPC$, but that is not the case if you consider a Turing Machine and its switched version.

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  • $\begingroup$ Thanks, I'm trying to understand your answer. $\endgroup$ Feb 21 at 21:21
  • $\begingroup$ The key is that to be accepted by a NTM, there must exists at least one accepting path. But if for an entry $u$ there exist both an accepting path and a rejecting path, then $u \in L(M)$ AND $u \in L(\overline{M})$. $\endgroup$
    – Nathaniel
    Feb 21 at 21:24
  • $\begingroup$ Thanks, I think I got it $\endgroup$ Feb 21 at 21:58
  • $\begingroup$ Is it possible to "add a transition on the initial state" without changing the accepting states only because it's Non Deterministic Turing Machine, or I missed something? $\endgroup$ Feb 21 at 22:23
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    $\begingroup$ Yes, because on a deterministic TM, you can't have multiple choices in a given configuration. $\endgroup$
    – Nathaniel
    Feb 21 at 23:02

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