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

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.

If $x\in A\oplus B$, the definition of symmetric difference puts it in exactly one of $A,B$. It is therefore in their set union and outside their set intersection, giving $x\in (A\cup B)\setminus (A\cap B)$.

Conversely, an element of $(A\cup B)\setminus (A\cap B)$ lies in at least one of the two sets but not both. It consequently belongs to $A\setminus B$ or to $B\setminus A$, and hence to $A\oplus B$. Both set inclusions hold.