Prove (A$\Delta$B)$\Delta$(B$\Delta$C) = A$\Delta$C where A$\Delta$B = (A\B) ∪ (B\A)

Question

It looked like a smple reduction problem, but it seems not to be the way once you beging proving one side to be subset of the other, it turns on a big mess, I´m hoping someone can see a more elegant solution like a transivity property or something.

Answer 1

If you have associativity of $\Delta$; and if you know that for any set $S$, $S\Delta S=\emptyset$; and $S\Delta\emptyset=S$, the result follows quickly.

Answer 2

Use indicator function.

Recall the symmetric difference has $1_{A\triangle B}(x)=1_A(x)+1_B(x)$ for all $x$, where the sum is taken mod 2. So $$ \begin{align*} 1_{(A\triangle B)\triangle(B\triangle C)} &=1_{A\triangle B}+1_{B\triangle C}\\ &=1_A+1_B+1_B+1_C\\ &=1_A+1_C\\ &=1_{A\triangle C} \end{align*} $$