Past exam of the year 1 of the computer science course of the University of Oxford / 2017 / Discrete mathematics, continuous mathematics and probability / 1 / b / i / Solution

For the first set inclusion, take $x\in A\setminus (A\setminus B)$. Then $x\in A$ but $x\notin A\setminus B$. Since membership in $A$ is already known, failure to lie in that relative complement forces $x\in B$. Hence $x\in A\cap B$.

Conversely, if $x\in A\cap B$, then $x\in A$ and its membership in $B$ excludes it from $A\setminus B$. Therefore $x\in A\setminus (A\setminus B)$. The two inclusions prove the stated equality with the set intersection.