Prove that two sets are the same I need to prove that $(A \Delta  B ) \Delta C = A \Delta (B\Delta C)$for any sets A,B,C, and $A \Delta B = (A-B)\cup(B-A)$. I tried to expand both the left hand side and the right hand side expression, and after that my expresion was composed only of sample proposition ($x \in A / x\in B/x\in C\ x\notin A\ x\notin B $ or $ x \notin C $) and then prove that using a truth table. My problem is that the truth table does not get me that the expression is a tautology.  
 A: I think it is the easiest way to prove it by using characteristic functions:
Since
$$\chi_{A \Delta B}=\chi_A+\chi_B\mod 2,$$ we have:
$$\chi_{(A \Delta B)\Delta C}=\chi_{(A \Delta B)}+\chi_C\mod 2=(\chi_A+\chi_B)+\chi_C\mod 2=\chi_A+(\chi_B+\chi_C)\mod 2=\chi_A+\chi_{B \Delta C}\mod 2=\chi_{A \Delta (B\Delta C)}.$$ 
A: By definition
$$
x\in A\bigtriangleup B\iff (x\in A \land x\notin B) \lor(x\in B \land x\notin A)\tag1
$$
$$
x\in B\bigtriangleup C\iff (x\in B \land x\notin C) \lor(x\in C \land x\notin B)\tag2
$$
Hence
\begin{align}
x\notin A\bigtriangleup B&\iff (x\notin A \lor x\in B) \land (x\notin B \lor x\in A)
\\
&\iff (x\notin A \land x\notin B) \lor (x\notin A \land x\in A) 
\\
&\quad\quad\quad\lor (x\in B \land x\notin B) \lor (x\in B \land x\in A) 
\\
&\iff (x\notin A \land x\notin B) \lor (x\in B \land x\in A)\tag3
\end{align}
And
\begin{align}
x\notin B\bigtriangleup C&\iff (x\notin B \lor x\in C) \land (x\notin C \lor x\in B)
\\
&\iff (x\notin B \land x\notin C) \lor (x\notin B \land x\in B) 
\\
&\quad\quad\quad\lor (x\in C \land x\notin C) \lor (x\in C \land x\in B) 
\\
&\iff (x\notin B \land x\notin C) \lor (x\in C \land x\in B)\tag4
\end{align}
Thus
\begin{align}
x\in (A\bigtriangleup B)\bigtriangleup C &\iff (x\in A\bigtriangleup B \land x\notin C) \lor (x\in C \land x\notin A\bigtriangleup B)
\\
&\iff (((x\in A \land x\notin B) \lor(x\in B \land x\notin A))\land x\notin C)
\\
&\quad\quad\quad\lor (x\in C \land ((x\notin A \land x\notin B) \lor (x\in B \land x\in A)))\tag{by (1),(3)}
\\
&\iff (x\in A \land x\notin B \land x\notin C) \lor (x\notin A \land x\in B \land x\notin C)
\\
&\quad\quad\quad\lor (x\notin A \land x\notin B \land x\in C) \lor (x\in A \land x\in B \land x\in C)\tag5
\end{align}
And
\begin{align}
x\in A\bigtriangleup (B\bigtriangleup C) &\iff (x\in A \land x\notin B\bigtriangleup C) \lor (x\notin A \land x\in B\bigtriangleup C)
\\
&\iff ((x\in A \land ((x\notin B \land x\notin C) \lor (x\in B \land x\in C)))
\\
&\quad\quad\quad\lor (x\notin A \land ((x\in B \land x\notin C) \lor(x\in C \land x\notin B)))\tag{by (2),(4)}
\\
&\iff (x\in A \land x\notin B \land x\notin C) \lor (x\in A \land x\in B \land x\in C)
\\
&\quad\quad\quad\lor (x\notin A \land x\in B \land x\notin C) \lor (x\notin A \land x\notin B \land x\in C)\tag6
\end{align}
By $(5)$ and $(6)$
$$
x\in (A\bigtriangleup B)\bigtriangleup C\iff x\in A\bigtriangleup (B\bigtriangleup C)
$$
Hence
$$
(A\bigtriangleup B)\bigtriangleup C=A\bigtriangleup (B\bigtriangleup C)
$$
A: A variation of hermes's proof.

For any subsets $A, B, C$ of a given set $X.$ By notation, $A^c = X\setminus A.$
$A\Delta B = (A\setminus B)\cup (B\setminus A) = (A\cap B^c)\cup (B\cap A^c) = (A\cup B)\cap (A\cap B)^c.$
In particular, $\Delta$ is commutative.
$$
\begin{align*}
&(A\Delta B)\Delta C\\
=\;&\{[(A\cup B)\cap (A\cap B)^c]\cup C\}\cap[(A\cup B)\cap (A\cap B)^c\cap C]^c\\
=\;&(A\cup B\cup C)\cap(A^c\cup B^c\cup C)\cap[(A^c\cap B^c)\cup (A\cap B)\cup C^c]\\
=\;&(A\cup B\cup C)\cap(A^c\cup B^c\cup C)\cap\{[(A^c\cup B)\cap (B^c\cup A)]\cup C^c\}\\
=\;&(A\cup B\cup C)\cap(A^c\cup B^c\cup C)\cap(A^c\cup B\cup C^c)\cap (B^c\cup A\cup C^c)
\end{align*}
$$
Then, doing permutation $A\mapsto C, B\mapsto B, C\mapsto A$ to get
$$
\begin{align*}
&A\Delta (B\Delta C)\\
=\;&(C\Delta B)\Delta A\\
=\;&(C\cup B\cup A)\cap (C^c\cup B^c\cup A)\cap (C^c\cup B\cup A^c)\cap (B^c\cup C\cup A^c)\\
=\;&(A\Delta B)\Delta C
\end{align*}
$$
