= Solution
For the first <set inclusion>, take $x\in(A\cap B)\setminus C$. By the definitions of <set intersection> and <set difference>, $x\in A$, $x\in B$ and $x\notin C$. Consequently $x\in A\setminus C$ and $x\in B\setminus C$, so $x\in(A\setminus C)\cap(B\setminus C)$.
For the reverse <set inclusion>, take $x\in(A\setminus C)\cap(B\setminus C)$. Membership in the first <set difference> gives $x\in A$ and $x\notin C$, while membership in the second gives $x\in B$ and $x\notin C$. Thus $x\in A\cap B$ and $x\notin C$, proving $x\in(A\cap B)\setminus C$. The two inclusions establish \b[the required equality].
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