Proof of $(A-B) \cup (B-A) = (A-C) \cup (C-A)$ then $B=C$ I know intuitively that if $(A-B) \cup (B-A) = (A-C) \cup (C-A)$ then $B=C$. But it just seems hard to prove it without using a pictorial argument. Can anyone show how I may proceed?
 A: Assume that $x$ is an element that is in $B$ but not in $A$. Then $$x\in B-A=B\cap A^c$$ and therefore $$x \in (A-B)\cup (B-A)$$ Due to the given equality we must also have that $$x \in (A-C)\cup (C-A)$$ but since $x \notin A$ by assumption we have that $$x \in C-A$$ or $x \in C$. The same holds for every element in $C$ witht that property, so that we have that $$B\cap A^c =C \cap A^c\tag1$$ Assume know that $y$ is an element that lies in $B$ and in $A$, that is $y \in A \cap B$. Then, this element is not included in the set on LHS and therefore (due to the given equality) it is also not included in the set on the RHS. Repeating for $C$, we conclude also that $$B\cap A=C \cap A\tag2$$ Combining $(1)$ and $(2)$ with the fact that $$B=(B\cap A) \cup (B\cap A^c)$$ and similarly for $C$, we have that $B=C$ as required.
A: Here is another proof.
The problem statement contains two expressions of the form $\;(A \setminus X) \cup (X \setminus A)\;$.  Let's first try to simplify those.  Which elements $\;x\;$ are members of such a set? Let's calculate:
\begin{align}
& x \in (A \setminus X) \cup (X \setminus A) \\
\equiv & \qquad \text{"definition of $\;\cup\;$; definition of $\;\setminus\;$, twice"} \\
& (x \in A \land x \not\in X) \lor (x \in X \land x \not\in A) \\
\equiv & \qquad \text{"logic: simplify"} \\
& x \in A \;\not\equiv\; x \in X \\
\end{align}
If that second step is not immediately obvious, here is a detailed proof:
\begin{align}
& (P \land \lnot Q) \lor (Q \land \lnot P) \\
\equiv & \qquad \text{"DeMorgan -- so that we can introduce $\;\Rightarrow\;$ later"} \\
& \lnot ((\lnot P \lor Q) \land (\lnot Q \lor P)) \\
\equiv & \qquad \text{"$\;\lnot \phi \lor \psi\;$ is one way to write $\;\phi \Rightarrow \psi\;$"} \\
& \lnot ((P \Rightarrow Q) \land (Q \Rightarrow P)) \\
\equiv & \qquad \text{"simplify: mutual implication is equivalence"} \\
& \lnot (P \equiv Q) \\
\equiv & \qquad \text{"simplify"} \\
& P \not\equiv Q \\
\end{align}
Now we can simplify
\begin{align}
& (A \setminus B) \cup (B \setminus A) \;=\; (A \setminus C) \cup (C \setminus A) \\
\equiv & \qquad \text{"set extensionality; the first calculation above, twice} \\
& \qquad \phantom{\text{"}}\text{-- note that $\;\equiv\;$ and $\;\not\equiv\;$ are associative, and mutually associative,} \\
& \qquad \phantom{\text{"}}\text{therefore we can leave out the parentheses"} \\
& \langle \forall x :: x \in A \not\equiv x \in B \;\equiv\; x \in A \not\equiv x \in C \rangle \\
\equiv & \qquad \text{"logic: $\;\equiv\;$ is symmetric"} \\
& \langle \forall x :: x \in A \not\equiv x \in A \;\equiv\; x \in B \not\equiv x \in C \rangle \\
\equiv & \qquad \text{"logic: simplify using the fact that $\;\phi \not\equiv \phi\;$ is false"} \\
& \langle \forall x :: x \in B \equiv x \in C \rangle \\
\equiv & \qquad \text{"set extensionality"} \\
& B = C \\
\end{align}
That completes the proof.
A: One way to look at it: define the symmetric difference $A \oplus B = (A-B) \cup (B-A)$. This defines a (commutative) group operation on all subsets of the universe set, where every set  is its own inverse and $\emptyset$ is the neutral element. Your statement is then $A \oplus B = A \oplus C \rightarrow B = C$, which follows because we cancel by adding $A$ to both sides...
