What is another more direct proof of this exercise? The exercise states that " If we know that $(A \cup B)-(A \cap B)=(A \cup C)-(A \cap C)$, then $B=C$ ". At my first sight, I don't know how to do that. However, when I see another exercise in the later of the book, I find a way to work out. That exercise states that we can definite $+$ and $.$, such that, $A+B:=(A \cup B)-(A \cap B), A.B=A \cap B$. Then it let me prove that :
$1.A+B=B+A$
$2.A+\emptyset=A$
$3.A+A=\emptyset$
$4.A.A=A$
$5.A+(B+C)=(A+B)+C$
These are very easy to prove. And from these property, I start my proof of the former exercise. I use property 3 to prove the former. Because now I can translate the  $(A \cup B)-(A \cap B)=(A \cup C)-(A \cap C)$ to $A+B=A+C$. By using property 3, $A+A+B=A+A+C$, using property 5, I can conclude that $(A+A)+B=(A+A)+C$, so by using property 3, $B=C$.
Though I work out the exercise finally, I think that there must be some simpler and more direct proof, since this exercise is situated before than those five properties. Can anyone help with a more direct and more intuitive proof of this?

A direct proof would be to take $x\in B$ and then to deduce $x\in C$, that is $B\subset C$.
If we can do that then we may exchange $B$ and $C$ and reach $B=C$.
So, let us assume $x\in B$.
Now either  $$x\in (A\cup B) - (A\cap B)\tag{1}$$ or
$$x\notin (A\cup B) - (A\cap B) \tag{2}$$
Our assumption $x\in B$ certainly imply $x\in A\cup B$, hence in case (1) $x\notin A$.
But by the assumption on $C$ we have
$$(A\cup C) - (A\cap C) = (A\cup B) - (A\cap B)\ni x$$
and since $x\notin A$, $x$ must belong to $C$.
The case (2) can be handled similarly, I leave the fun for you.
