using set algebra laws prove the following:
1) (A-B) U (B-A) = (A U B) - (A ∩ B)
2) ((A ∩ B)' U B)' = 0
Hello, beast!
Here is the second one.
I'll let you supply the reasons.
$\displaystyle [2]\;\;\big[(A \cap B)' \cup B\big]' \:=\:\emptyset$
$\displaystyle \big[(A \cap B)' \cup B\big]' $
. . $\displaystyle =\;\big[(A' \cup B') \cup B\big]'$
. . $\displaystyle =\;\big[A' \cup (B' \cup B)\big]'$
. . $\displaystyle =\; \big[A' \cup U\big]'$
. . $\displaystyle =\;U'$
. . $\displaystyle =\; \emptyset$