Symmetric difference of symmetric differences I was looking around wikipedia for the symmetric difference when I stumbled across this fact.

From the property of the inverses in a Boolean group, it follows that the symmetric difference of two repeated symmetric differences is equivalent to the repeated symmetric difference of the join of the two multisets, where for each double set both can be removed. In particular: $$A△C = (A△B)△(B△C)$$


where $A△C$ is the symmetric difference of A and C.
Is $A△C = (A△B)△(B△C)$ true in the general case?
I tried proving it, but the amount of terms grew unwieldly for me when I tried reducing everything to unions, complements, and intersections. Is there any proofs available at hand that anyone knows for this identity?
 A: The proof is not difficult once you have proved that taking the symmetric difference of sets is associative. Using this fact, for all sets $A, B$ and $C$ we have
\begin{align} 
(A \bigtriangleup B) \bigtriangleup (B \bigtriangleup C) &= A \bigtriangleup (B \bigtriangleup (B \bigtriangleup C)) \\ &= A \bigtriangleup ((B \bigtriangleup B) \bigtriangleup C) \\ &= A \bigtriangleup(\emptyset\bigtriangleup C) \\
&= A \bigtriangleup C.
\end{align}
I am not sure at this stage whether or not this identity holds for multisets.
A: The key to this is to establish how to expand set difference (and its complement) into CNF and DNF. And, of course, make clear the form you need to convert back to a triangle.
$$\def\c{^{\small\complement}}\def\symdiff{\mathop{\triangle}}~~~~ X\symdiff Y ~{= (X\cap Y\c)\cup(X\c\cap Y)\\=(X\cup Y)\cap (X\c\cup Y\c)} \\(X\symdiff Y)\c~{=(X\cap Y)\cup(X\c\cap Y\c)\\=(X\cup Y\c)\cap(X\c\cup Y)}$$
Now to expand: first convert the root $\symdiff$ into DNF, then each disjunct into a CNF. (Or you might do it the other way). The aim of course is to distribute out all the $B$ from the $A$ and $C$.
$\quad(A\symdiff B)\symdiff (B\symdiff C)
\\=((A\symdiff B)\cap(B\symdiff C)\c)~\cup~((A\symdiff B)\c\cap(B\symdiff C))\\={((A\cup B)\cap(A\c\cup B\c)\cap(B\cup C\c)\cap(B\c\cup C))~\cup~((A\cup B\c)\cap(A\c\cup B)\cap(B\cup C)\cap(B\c\cup C\c))}\\={(((A\cap C\c)\cup B)\cap((A\c\cap C)\cup B\c))~\cup~((A\cap C\c)\cup B\c)\cap((A\c\cap C)\cup B))}$
To make it easier on the eyes, let $U=(A\cap C\c)$ and $V=(A\c\cap C)$, and we shall note that $A\symdiff C=U\cup V$.
$\quad(A\symdiff B)\symdiff (B\symdiff C)
\\=((U\cup B)\cap(V\cup B\c))~\cup~((U\cup B\c)\cap(V\cup B))$
You can take it from here.
