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I got this question:

Let $A \oplus B = (A\cap \bar{B})\cup(\bar{A}\cap B)$.

Proof that $NP = coNP$ if and only if $A,B\in NP$ and $A \oplus B\in NP$.

But I don't know how to proof the direction that if $A,B\in NP$ and $A \oplus B\in NP$ then $NP = coNP$.

I have tried to build the non-deterministic machines for $\bar{A}$ and $\bar B$, but I don't know how.

What that I tried: Say I want to build the non-deterministic machine for $\bar{A}$. I know that if $w\in A \oplus B$ and $w\in B$, then for sure $w \in \bar A$. My problem is with the condition if $w\notin A \oplus B$ and $w\notin B$ then $w \in \bar A$, because I don't know if $w\notin A \oplus B$ and $w\notin B$ since I don't know that $NP = coNP$.

So how can I show non deterministic machines for $\bar{A}$ and $\bar B$ when I don't know how to decide if "word is not in ..."?

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    $\begingroup$ Hint: it's clear that NP is closed under intersection (proof?). The task is to show that NP being closed under symmetric difference is equivalent to it being closed under complementation. $\endgroup$ – G. Bach Jan 26 '16 at 10:07
  • $\begingroup$ @G.Bach I know that NP is closed under intersection. I don't know how it can help me to solve the problem... $\endgroup$ – nrofis Jan 26 '16 at 10:37
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    $\begingroup$ Further hint: $A \oplus \Sigma^* = \bar{A}$. $\endgroup$ – Yuval Filmus Jan 26 '16 at 11:00
  • $\begingroup$ Thank you! Didn't think about using specific $B$. Now it's easy :) $\endgroup$ – nrofis Jan 26 '16 at 11:04
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    $\begingroup$ Can you answer your own question now? $\endgroup$ – Yuval Filmus Jan 26 '16 at 12:09
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Thanks to G. Bach and Yuval Filmus.

If every $A, B \in NP$ and $A \oplus B \in NP$, then let $E$ to be some language $E\in NP$. Since $\Sigma^*\in NP$ then $E \oplus \Sigma^* \in NP$. But $ E \oplus \Sigma^* = \bar E$, so $\bar E \in NP$, and we got that $NP \subseteq coNP$.

Let $D$ to be some language $D\in coNP$ ($\bar D \in NP$). Since $\Sigma^*\in NP$ then $\bar D \oplus \Sigma^* \in NP$. But $ \bar D \oplus \Sigma^* = D$, so $\bar D \in NP$, and we got that $coNP \subseteq NP$.

And then $NP = coNP$.

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