Associativity of symmetric difference of sets By $A\oplus B$ we denote the symmectric difference of two sets. The definition is $A\oplus B =(A\setminus B) \cup (B\setminus A)$. Now I hope to show that $A\oplus (B\oplus C) = (A\oplus B)\oplus C$. I remember that there's an elegant proof, but I forget its detail. (The first step is to show $A\oplus (A\oplus B) = B$, and maybe applying this formula, we can obtain associativity.) 
 A: $A \oplus B$ contains those elements (of the given basic set) which belong to exactly one of the sets $A$ and $B$. Thus, $A \oplus (B \oplus C)$ consists of those elements which belong to exactly one of the sets $A$ and $B \oplus C$, i.e. to exactly one or all three of the sets $A$, $B$ and $C$. This shows $A \oplus (B \oplus C) = (A \oplus B) \oplus C$.
There is also a more algebraic proof, using the isomorphism of structures $(P(X),\oplus,\cap) \cong (\mathbb{F}_2^X,+,*)$, which immediately implies that $(P(X),\oplus,\cap)$ is a ring.
A: You could use characteristic functions to make it more algebraic. Observe
$$
1_{A\cap B}=1_A\cdot 1_B \qquad\mbox{and}\qquad 1_{A\Delta B}=1_A+1_B-2\cdot 1_{A\cap B}.
$$
Then
$$
1_{A\Delta(B\Delta C)}=1_A+1_{B\Delta C}-2\cdot1_{A\cap (B\Delta C)}
$$
$$
=1_A+1_B+1_C-2\cdot1_{B\cap C}-2\cdot1_A\cdot(1_B+1_C-2\cdot1_{B\cap C})
$$
$$
=1_A+1_B+1_C-2\cdot1_{A\cap B}-2\cdot1_{B\cap C}-2\cdot1_{C\cap A}+4\cdot1_{A\cap B\cap C}.
$$
I used that $\cap $ is associative to make sense of $A\cap (B\cap C)=A\cap B\cap C$. To be consistent, this can also be proven with characteristic functions:
$$
1_{A\cap(B\cap C)}=1_A1_{B\cap C}=1_A1_B1_C=1_{A\cap B}1_C=1_{(A\cap B)\cap C}.
$$
Now starting from $1_{(A\Delta B)\Delta C}$, you end up with the same symmetric formula. Hence your two sets have the same characteristic functions, which means that they are equal.
A: Another way to prove this, is to first prove that $\;x \in A \oplus B \;\equiv\; x \in A \not\equiv x \in B\;$, and then use the analogous law of logic ($\;\not\equiv\;$ is associative, i.e. $\;P \not\equiv (Q \not\equiv R) \;\equiv\; (P \not\equiv Q) \not\equiv R\;$).
A: This can be done in a second.
$$
i_{\left( A∆B \right)}≡i_A+i_B\left( mod2 \right) .Thus\,\,i_{\left( A∆B \right) ∆C}=i_{A∆\left( B∆C \right)}≡i_A+i_B+i_C\left( mod2 \right) .
$$
