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 ..."?