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: Sidenote: a common way to denote $(A\cup B)-(A\cap B)$ is $A\Delta B$.
It is the symmetric difference of $A$ and $B$.

Using indicator functions we find that:$$1_{A\Delta B}=1_A+1_B-2\cdot1_A1_B$$
From: $$1_A+1_B-2\cdot1_A1_B=1_A+1_C-2\cdot1_A1_C$$it follows that:
$$(1_B-1_C)(1-2\cdot1_A)=0$$
This implies that $1_B=1_C$ (since we cannot have $1_A=\frac12$) or equivalently: $$B=C$$
A: Proof by contradiction: suppose there were some $A, B,$ and $C$ such that $(A \cup B) - (A \cap B) = (A \cup C) - (A \cap C)$ but $B \neq C$.
Because $B \neq C$, there must be some element $x$ which is in either $B$ or $C$ but not the other. Otherwise, if every element of $B$ was in $C$ and vice versa, clearly the sets must be equal by mutual containment.
Suppose that $x$ is in $B$ and not in $C$. If $x$ is also in $A$, then $x$ is in both $A \cup B$ and $A \cap B$, so $x$ is not in $(A \cup B) - (A \cap B)$. However, $x$ will be in $A \cup C$ and not in $A \cap C$, so $x$ is in $(A \cup C) - (A \cap C)$, violating the equation and causing a contradiction.
If $x$ is instead not in $A$, then $x$ is in $A \cup B$ but not $A \cap B$, so $x$ is in $(A \cup B) - (A \cap B)$, but $x$ is in neither $A \cup C$ nor $A \cap C$ so it is not in $(A \cup C) - (A \cap C)$, also violating the equation.
Because of the symmetry of the question, you can apply the same argument to show that we reach a contradiction if $x$ is in $C$ but not $B$. Therefore, our assumption has led us to a contradiction and it must be false.
A: 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.
