Relationship between the symmetric difference of two sets and their intersection How do we prove that the union of symmetric difference of two sets and their intersection is the same as the union of the two sets?
i.e. $(A\backslash B) \cup (A\cap B) \cup (B\backslash A) = A\cup B$
where $A, B$ are two sets?
I know how it is obvious. But I want a rigorous proof from definitions of difference and union and intersection.

Maybe you were looking for a "computational" proof:
$$\underbrace{(A\backslash B)\cup(A\cap B)}_{A}\cup(B\backslash A)=A\cup (B\backslash A)=A\cup B.$$
In the first equality, I am using the fact that for any sets $A$ and $B$, we have
$$A=(A\backslash B)\cup (A\cap B).$$
Intuitively, this means that everything in $A$ either belongs to $B$ or it doesn't. For the second equality, it's just
$$A\cup (B\backslash A)=A\cup (B\cap A^c)=(A\cup B)\cap (A\cup A^c)=(A\cup B).$$
