= Solution
We use a <double inclusion proof>. First, let $x\in\overline{A\oplus B}$. If $x\in A$, then $x\in B$, since otherwise $x\in A\setminus B\subseteq A\oplus B$. Hence in this case $x\in A\cap B$. If $x\notin A$, then $x\notin B$, since otherwise $x\in B\setminus A\subseteq A\oplus B$. Hence in this case $x\in\overline{A\cup B}$. This proves the first <set inclusion>.
Conversely, an element of $A\cap B$ belongs to neither $A\setminus B$ nor $B\setminus A$. An element of $\overline{A\cup B}$ also belongs to neither <set difference>. Thus every element of $(A\cap B)\cup\overline{A\cup B}$ lies outside $A\oplus B$, proving the reverse <set inclusion>. Therefore
$$
\boxed{\overline{A\oplus B}=(A\cap B)\cup\overline{A\cup B}.}
$$
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