= Solution
We prove both <set inclusions>, as required by a <double inclusion proof>. If $x\in A\setminus(B\cup C)$, then $x\in A$ and $x\notin B\cup C$. Thus $x\notin B$ and $x\notin C$, so $x\in A\setminus B$ and $x\in A\setminus C$. This proves the forward <set inclusion>.
Conversely, if $x\in(A\setminus B)\cap(A\setminus C)$, then $x\in A$, $x\notin B$, and $x\notin C$. Hence $x\notin B\cup C$, giving $x\in A\setminus(B\cup C)$. Therefore
$$
\boxed{A\setminus(B\cup C)=(A\setminus B)\cap(A\setminus C).}
$$
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