proof that $A \triangle (B \triangle C) = (A \triangle B) \triangle C $ $ \triangle $ is symmetric difference.
I want to proof that $A \triangle (B \triangle C) = (A \triangle B) \triangle C $
It seems like join's rule. I tried to assume that $ x \in Left $ and then move to $ (...) $ so $ x \in Right $ but when I use definition of $ \triangle $ I get something like that:
$x \in A \land x \in B \land x \in C \land x \notin A \notin x \in B \notin x \in C \land (...) $
what does not have sense...
I have seen link but there is well explained algebraic way but I am trying to do this with normal set theory (By this I mean something as I presented on the begining).
 A: I think the best way to prove this is by using indicator functions:
$$1_{A\triangle (B\triangle C)}  = 1_A + 1_{B\Delta C} - 1_{A\cap (B\Delta C)}=1_A+1_B+1_C-1_{B}1_{C}-1_{A}1_{B}-1_{A}1_{C}+1_{A}1_{B}1_{C}.$$
Write the expression for $1_{(A\triangle B)\triangle C}$ and observe that they both are the same. 
A: I am assuming you mean the symmetric difference, which is given by $X\ominus Y=(X\setminus Y) \cup (Y\setminus X)$. Which is sometimes symbolized as a half-division sign, that is, dash with a dot over it.
Suppose: $$x\in[(A \ominus B)\ominus C)]$$
$$x\in[(A \ominus B)\ominus C)] \implies x\in [(A\ominus B)\setminus C] \lor x\in [C\setminus (A \ominus B)]$$
$$[x\in (A\ominus B)\land x\notin C] \lor [x\in C \land x\notin (A\ominus B)]$$
Case 1:
$$[x\in (A\ominus B)\land x\notin C] \implies x\in (A\setminus B \lor B\setminus A) \land x\notin C$$
Sub-case 1: 
$$(x\in A \land x\notin B) \land x \notin C$$
Sub-case 2: 
$$(x\in B \land x \notin A) \land x \notin C$$
From sub-case 1: 
$$x\in A \setminus B \land x\notin C$$
From sub-case 2:
$$x\in B\setminus A \land x\notin C$$
From sub-cases 1 and 2:
$$x\in [(A\setminus B)\setminus C] \lor x\in [(B\setminus A)\setminus C]$$
Therefore from subcase 1: 
$$x\in [A\setminus(B\setminus C \lor C\setminus B)]\implies x\in [A\setminus (B\setminus C \cup C\setminus B)]$$
Notice, if given any proposition $P$, it is valid to deduce $P\lor Q$. Which is what I did. Given $B \setminus C$, I deduced: $B\setminus C \lor C\setminus B$
Similarly from sub-case 2:
$$x\in [B\setminus(A\setminus C \lor C\setminus A)]\implies x\in [B\setminus (A\setminus C \cup C\setminus A)]$$
Therefore: 
$$x \in [A\ominus (B\ominus C)]$$
Following similar techniques you can show the same for case 2: 
Suppose, $[x\in C \land x\notin (A\ominus B)]$ show what I have shown for case 1, then that will complete your subset proof from right-to-left. Hopefully you can do that.
