How prove $(A\triangle B)\triangle C=A\triangle(B\triangle C)$ by defining function from $(A\triangle B)\triangle C$ to $A\triangle(B\triangle C)$ How do you prove $(A \triangle{}B) \triangle{}C=A \triangle{}(B \triangle{}C)$ by defining  function from $(A \triangle{}B) \triangle{}C$ to $A \triangle{}(B \triangle{}C)$?
$(A \triangle{}B:=A\cup B-(A\cap B)$) I know this proof : let $x\in(A \triangle{}B) \triangle{}C$ the show that $x \in A \triangle{}(B \triangle{}C)$ and  similarly prove $x\in A \triangle{}(B\triangle{}C)$ then  $x \in (A \triangle{}B) \triangle{}C.$ Thanks in advance.
 A: Just to expand a bit on the solution of @Did
Take any set $U$, then you want to prove that its powerset $\mathfrak{P}(U)$ is a ring with respect to symmetric difference and intersection. 
This is possibly best seen noting that there is a bijection of $\mathfrak{P}(U)$ with the set $\mathbf{Z}_2^{U}$ of functions from $U$ to $\mathbf{Z}_2$: a subset $S$ of $U$ goes into its characteristic function $\mathbf{1}_S$, which is $1$ exactly on the elements of $S$.
Now by general facts $\mathbf{Z}_2^{U}$ is a ring with respect to pointwise addition and multiplication. And these two operations correspond to symmetric difference and intersection in $\mathfrak{P}(U)$.
A: For every $F$ and $G$, $H=F\triangle G$ may be defined by the identity $\mathbf 1_H=\mathbf 1_F+\mathbf 1_G\pmod{2}$.
Hence $U=A\triangle(B\triangle C)$ is uniquely defined by the fact that $\mathbf 1_U=\mathbf 1_A+\mathbf 1_{B\triangle C}\pmod{2}$, that is, $\mathbf 1_U=\mathbf 1_A+\mathbf 1_B+\mathbf 1_C\pmod{2}$. By symmetry of the RHS, the same holds for $\mathbf 1_V$ where $V=(A\triangle B)\triangle C$, hence $U=V$.
Edit: To complete the answer above, recall that, for every $E$, the function $\mathbb 1_E$ is the indicator function of $E$, defined by $\mathbf 1_E(x)=1$ if $x$ is in $E$ and $\mathbf 1_E(x)=0$ otherwise. Thus, $\mathbf 1_F+\mathbf 1_G$ equals $2$ on $F\cap G$, $1$ on $F\triangle G$, and $0$ on the complement of $F\cup G$, that is, $1\pmod{2}$ on $F\triangle G$ and $0\pmod{2}$ on the complement of $F\triangle G$. Thus, $\mathbf 1_F+\mathbf 1_G=\mathbf 1_{F\triangle G}\pmod{2}$, as claimed above.
A: This is more or less the same solution as did's. Anyhow, here is how proofs by characteristic functions work.
Consider $X$ a set which is fixed and assumed to contain all the sets involved in the exercise you are trying to solve.
For any $A \subset X$ define $1_A : X \to \{ 0,1\}$ by
$$1_A(x) = \left\{ 
\begin{eqnarray}
\begin{split}
1 & \mbox{if } x \in A \\
0 & \mbox{if } x \notin A \\
\end{split}
\end{eqnarray}\right.$$
Now for subsets of $X$ the following are true, and easy to prove:


*

*$A \subset B \Leftrightarrow 1_A \leq 1_B$

*$A= B \Leftrightarrow 1_A = 1_B$

*$1_{A \cap B}=1_A \cdot 1_B$

*$1_{A \cup B}=1_A+1_B-1_A \cdot 1_B$

*$1_{A \backslash B}= 1_A-1_A1_B$

*$1_{A \Delta B}=1_A+1_B-2\cdot 1_A \cdot 1_B=(1_A-1_B)^2$


For your problem all you need to prove is that $1_{A \Delta (B \Delta C)}= 1_{(A \Delta B) \Delta C}$. Using the above properties you get
$$1_{A \Delta (B \Delta C)}=1_A+1_{B \Delta C}-2\cdot1_A1_{B \Delta C}$$
$$=1_A+1_{B}+1_C-2\cdot 1_B1_C-2\cdot1_A\cdot(1_C-2\cdot 1_B1_C)$$
$$=1_A+1_B+1_C-2[1_A1_B+ 1_B1_C+\cdot1_A1_C]+4\cdot 1_A1_B1_C$$
Now, you just prove that $1_{(A \Delta B) \Delta C}$ leads to the same. The computation is almost identical. 
