Solution
= Solution
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.