Set Theory Proof with Delta: $(A\mathrel{\triangle} B)\mathrel{\triangle} C = A\mathrel{\triangle} (B\mathrel{\triangle} C)$ $$(A\mathrel{\triangle} B)\mathrel{\triangle} C = A\mathrel{\triangle} (B\mathrel{\triangle} C)$$
Any thoughts? Would I use set theory laws to prove this?
I do not really know where to start, is there set theory laws for the delta notation or would I have to expand on everything?
 A: (Extreme) Brute Force $\downarrow$

[Sorry for posting picture; prefer to typeset in my LaTeX editor: \newcommand + autocompletion; I welcome edits to my post which replaces the picture with genuine code]
A: One way is brute force: (1) Some points are not in $A$, $B$, or $C$; (2) Some are in $A$ but not in $B$ or $C$; (3) Some are in $B$ and not in $A$ or $C$; (4) Some are in $C$ and not in $A$ or $B$; (5) Some are in $A$ and $B$ but not $C$; (6) Some are in $A$ and $C$ but not $B$; (7) Some are in $B$ and $C$ but not $A$; (8) Some in in all three.
Now figure out which of those points are in $(A\mathrel{\triangle} B)\mathrel{\triangle} C$ and which are in $A\mathrel{\triangle} (B\mathrel{\triangle} C)$, and see if you get the same ones both times.
That's brute force.  Another way is to show that the indicator function of $A\mathrel{\triangle}B$ is the mod-$2$ sum of the indicator functions of $A$ and $B$, and then show that mod-$2$ addition is associative.
PS: OK, here's a table:
$$
\begin{array}{c|c|c|c|c}
x\in A & x\in B & x\in C & x\in A\mathrel{\triangle} (B\mathrel{\triangle} C) & x\in (A\mathrel{\triangle} B) \mathrel{\triangle} C \\
\hline
f & f & f & \text{?} & \text{?} \\[12pt]
T & f & f & \text{?} & \text{?} \\
f & T & f & \text{?} & \text{?} \\
f & f & T & \text{?} & \text{?} \\[12pt]
T & T & f & \text{?} & \text{?} \\
T & f & T & \text{?} & \text{?} \\
f & T & T & \text{?} & \text{?} \\[12pt]
T & T & T & \text{?} & \text{?}
\end{array}
$$
Fill in the blanks and see if the last two columns match.
A: Define an operation $\delta$ where $p\delta q\Leftrightarrow \neg (p\Leftrightarrow q)$. Make a 0-1 table of $\delta$. It is easy to see that $x\in A\triangle B\Leftrightarrow (x\in A)\delta (x\in B)$. Finally show (for example by 0-1 method) that $\delta$ is associative.
