Proving symmetric difference is associative I'm trying to prove that for $A,B,C \subset X$, we have
$$
(A \Delta B) \Delta C = A \Delta (B \Delta C). 
$$
I tried to prove it by brute force by taking an argument in the left-hand side, but I couldn't do it in a way that every step is reversible. There has to be a "clever" way to do this that I'm just not thinking of, but I'm not able to gain any kind of graphical intuition by plotting venn diagrams to see exactly how to do that.
EDIT: My attempt at brute force:
\begin{align*}
x \in (A \Delta B) \Delta C & \iff x \in ((A \Delta B) - C) \cup (C - (A \Delta B)) \\
& \iff x \in (((A - B) \cup (B -  A)) - C) \cup (C - ((A - B) \cup (B - A)) 
\end{align*}
 A: $\def\sd{\mathop{\Delta}}\def\sm{\mathop\smallsetminus}$
Prove, if you haven't been given, that: $~~~~~X\sd Y= (X\cap Y^\complement)\cup(X^\complement\cap Y)\\(X\sd Y)^\complement = (X^\complement\cap Y^\complement)\cup(X\cap Y)$
Then apply.
$${(A\sd B)\sd C\\=((A\sd B)\cap C^\complement)\cup((A\sd B)^\complement\cap C)\\\vdots\\= ((A\cap B^\complement\cap C^\complement)\cup(A^\complement\cap B\cap C^\complement))\cup((A^\complement\cap B^\complement\cap C)\cup (A\cap B\cap C))\\\vdots\\=A\sd (B\sd C)}$$
A: So to save some space, I'm going to write $a$ for the proposition $x\in A$; similarly $b$ and $c$. Then saying that $x\in A\Delta B$ is saying $(a\wedge\neg b)\vee(b\wedge\neg a)$. Seen this way, $x\in(A\Delta B)\Delta C$ can be written as $\biggl(\bigl((a\wedge\neg b)\vee(b\wedge\neg a)\bigr)\wedge\neg c\biggr)\vee\biggl(c\wedge\neg\bigl((a\wedge\neg b)\vee(b\wedge\neg a)\bigr)\biggr)$. Now, this is a mess, but if you expand it out into an or of individual 'atoms' of the form $a\wedge(\neg b)\wedge(\neg c)$, etc, using the deMorgan rules, then you should see a much more symmetric expression — in particular, one that's invariant under permutations of $a,b,c$ (or $A,B,C$). Can you see why that's enough to conclude?
