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Suppose that $P^{SAT} \not\subseteq coNP$. Prove that $P\ne NP$.

What I did:

Suppose that $P=NP$. Then, $P = coP = NP = coNP$.

We know that $P^P = P$.

Then, by assumption: $P^{NP} = NP = coNP$

Since $SAT$ is $NP$-complete we get: $P^{SAT} = coNP$ in contradiction to the assumption.

So $P\ne NP$.

I've been told that my proof is false, but I don't understand why.

Raphael
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blueplusgreen
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1 Answers1

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Here is a more direct proof. Suppose that $P=NP$. In particular, SAT can be solved in polynomial time. Therefore $P^{SAT} \subseteq P^P = P \subseteq coNP$.

Yuval Filmus
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