Showing that symmetric difference is associative $X,Y,Z$ are sets, $X+Y:=(X \cup Y) \setminus (X \cap Y)$.
I need to show that $(X+Y)+Z=X+(Y+Z)$.
Intuitively it is clear that $$(X+Y)+Z=(X \cup Y \cup Z) \setminus ( [X \cap Y] \cup [Y \cap Z] \cup [Z \cap X] \setminus [X \cap Y \cap Z ])$$
Then the claim would follow. Any idea how to prove the intuition in a smart way without considering 8 cases ($x \in X, x \notin Y, x \in Z$, etc.)?
 A: The symmetric difference $(a\cup b)\backslash (a\cap b)$ is usually written $a\Delta b. $ It is worthwhile to observe that $a\Delta b=(a\backslash b)\cup (b\backslash a). $ Now $p\in (a\Delta b)\Delta c$ iff (i) $p\in a$ or $p\in b$ but not both, and also $p\not \in c$, or (ii)$p\in c$ and $p$ doesn't belong to $a$ or $b$ unless it belongs to both of them. In other words, $p\in (a\Delta b)\Delta c$ iff $p$ belongs to  one of $a,b,c$  but not to the intersection of any 2 of them,unless $p$ belongs to all of them. What of $a\Delta (b\Delta c)$? We have $a\Delta (b\Delta c)=(b\Delta c)\Delta a.$  And $p\in (b\Delta c)\Delta a)$ iff $p$ belongs to  one of $b,c,a$  but not to the intersection of any 2 of them, unless $p$ belongs to all of them. $$(a\Delta b)\Delta c=a\Delta (b\Delta c)=$$  $$=(P\backslash Q)\cup R$$ where $P=a\cup b \cup c$ and $Q= (a\cap b)\cup (b\cap c)\cup (c\cap a)$ and $R=a\cap b \cap c$. It may be helpful to draw a Venn  diagram.
A: Both $A + (B + C)$ and  $(A + B) + C$ are exactly those things that are in an odd number of $(A, B, C)$. 
