(A⊕B)⊕C = A⊕(B⊕C) how to prove I have to prove this equation. I thought about it and did the venn-diagram on the equation which shows me that the equation is true but I'm having a hard time coming up with a proper way to prove the equation. Picute of the Venn-Diagram
 A: An element $x$ is in $(A\oplus B)\oplus C$ iff it is in $A\oplus B$ or in $C$ but not in both. Iff it is either in exactly one of $A$ or $B$, or otherwise it is in $C$. Iff exactly one or all three of $A,\ B,\ C$ contains it. Iff the number of summands is odd in which it is contained.
The same conclusion applies to $A\oplus(B\oplus C)$.
A: Hint: Keep these equivalences in mind (or prove them should need be):$$\begin{align}(X\oplus Y)&=(X\cap Y^\complement)\cup(X^\complement\cap Y)\tag 1\\[2ex](X\oplus Y)^\complement&= (X\cap Y)\cup(X^\complement\cap Y^\complement)\tag 2\end{align}$$
So to begin...
$$\begin{align}(A\oplus B)\oplus C &= ((A\oplus B)\cap C^\complement)\cup((A\oplus B)^\complement\cap C)\\ &~~\vdots\end{align}$$
A: Let $x \in (A\oplus B) \oplus C$ be arbitary. We will show $x \in A\oplus (B \oplus C)$, to show one inclusion at least.
Immediately we have two cases, which we again split:
1: $x \in A \oplus B$ and $x \notin C$.
1a: $x \in A$, $x \notin B$, $x \notin C$.
1b: $x \notin A$, $x \in B$, $x \notin C$.
2: $x \notin A \oplus B$ and $x \in C$.
2a: $x \in A$, $x \in B$, $x \in C$.
2b. $x \notin A$, $x \notin B$, $x \in C$.
Note that these four cases correspond to four areas in the Venn diagram, and also follow straight from the definition of the symmetric difference.
Also note that I use $x \notin A\oplus B$ iff ($x \in A$ and $x \in B$) or ($x \notin A$ and $x \notin B$).
Finishing the cases:
1a: $x \notin B\oplus C$ and $x \in A$ so $x \in A\oplus (B \oplus C)$.
1b: $x \in B \oplus C$ and $x \notin A$ so $x \in A\oplus (B \oplus C)$.
2a: $x \notin B \oplus C$ and $x \in A$ so $x \in A\oplus (B \oplus C)$.
2b: $x \in B \oplus C$ and $x \notin A$ so $x \in A\oplus (B \oplus C)$.
The reverse inclusion is easy, read all cases backwards....
