 Solution

We prove both set inclusions. If $x\in A\setminus (A\setminus B)$, then $x\in A$ and $x\notin A\setminus B$. Since $x\in A$, the latter condition forces $x\in B$, giving $x\in A\cap B$. Conversely, if $x\in A\cap B$, then $x\in A$ and $x\notin A\setminus B$, so $x\in A\setminus (A\setminus B)$. By set extensionality, the two sets are equal. This is a double inclusion proof; the implication involving the set difference uses membership in $A$ explicitly.