Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Discrete mathematics, continuous mathematics and probability / 1 / a / ii / 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,