= Solution
For the first <set inclusion>, take $x\in(A\cap B)\setminus C$. Membership in the <intersection> gives $x\in A$ and $x\in B$, while the <set difference> gives $x\notin C$. Consequently $x\in(A\setminus C)\cap(B\setminus C)$.
For the reverse <set inclusion>, an element of $(A\setminus C)\cap(B\setminus C)$ belongs to both $A$ and $B$ and does not belong to $C$. It therefore belongs to $(A\cap B)\setminus C$. By <double inclusion proof>,
$$
\boxed{(A\cap B)\setminus C=(A\setminus C)\cap(B\setminus C).}
$$
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