Simple set exercise seems not so simple Exercise about sets from Birkhoff's "Modern Applied Algebra".

Prove that for operation $\ \Delta $ , defined as

$\ R \Delta S = (R \cap S^c) \cup (R^c \cap S) $

following is true:

$\ R \Delta ( S \Delta T ) = ( R \Delta S ) \Delta T $


($\ S^c $ is complement of  $\ S $)
It's meant to be very simple, being placed in the first excercise block of the book. When I started to expand both sides of equations in order to prove that they're equal, I got this monster just for the left side:

$\ R \Delta ( S \Delta T ) = \Bigl( R \cap \bigl( (S \cap T^c) \cup (S^c \cap T) \bigr)^c \Bigr) \cup \Bigl(R^c \cap \bigl( (S \cap T^c) \cup (S^c \cap T) \bigr) \Bigr) $

For the right:

$\ ( R \Delta S ) \Delta T = \Bigl(\bigl( (R \cap S^c) \cup (R^c \cap S) \bigr) \cap T^c  \Bigr) \cup \Bigl( \bigl( (R \cap S^c) \cup (R^c \cap S) \bigr)^c  \cap T \Bigr)   $

I've tried to simplify this expression, tried to somehow rearrange it, but no luck. Am I going the wrong way? Or what should I do with what I have?

joriki's answer it the best way to understand this conceptually. Still, it's perfectly possible to continue formally the way you started: simply repeatedly apply De Morgan's laws:
$(A \cup B)' = A' \cap B'$ and $(A \cap B)' = A' \cup B'$. 
