Originally Posted by

**dunsta** This is an example solution I am reading but I can't understand the 3rd transformation.

$\displaystyle (A-B) \cup(B-A) =(A \cup B) - (A \cap B)$

I understand changing the LHS to

$\displaystyle (A-B) \cup(B-A) = (A\bigcap B)\cup(B\cap A')$ < ---a (this)

using set diff representation

(The notes I am reading use this notation where sometimes the intersection operator is "big", does it have an alternate meaning when it is big, or simply a typo?)

If it is just a typo, then I needn't worry about it.

but the next step I don't understand:

(to this) b ---> $\displaystyle (A \cup (B \cap A')) \cap (B' \cup (B \cap A'))$

the notes claim this is using distributivity,

but I can't see how they arrive at that equation using this rule

$\displaystyle A \cap (B \cup C)) = (A \cap B) \cup (A \cap C)$ <-- this I understand,

but I can;t see how they went from a to b