Prove that the symmetric difference is associative using only the properties of $\cup$ and $\cap$
I don’t know how to prove this without doing a lot of nasty steps. The symmetric difference is obviously commutative, but proving that it is associative seems to involve a bunch of nasty steps. I was wondering if there was a clean way to do this?
Well, it takes just a wee bit of work to show:
$$\def\comp{^\complement}\begin{align}(A\triangle B)\triangle C&=(((A\cap B\comp)\cup(A\comp\cap B))\cap C\comp)\cup(((A\cap B\comp)\cup(A\comp\cap B))\comp\cap C)\\&~~\vdots\\&=(A\cap B\comp\cap C\comp)\cup(A\comp\cap B\cap C\comp)\cup(A\comp\cap B\comp\cap C)\cup(A\cap B\cap C)\end{align}$$
Once you've done that, make an argument from symmetry that $A\triangle(B\triangle C)$ equals the same.