Solution
= 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.