Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability / 1 / a

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 the required equality.