How to prove that $\neg((A\cap B)\cup (\neg A \cap C)) = (A\cap\neg B)\cup (\neg A\cap \neg C)$ How can I prove the following statements are equivalent using laws of set theory?
$\neg((A\cap B)\cup (\neg A \cap C)) = (A\cap\neg B)\cup (\neg A\cap \neg C)$
I managed to use De Morgans laws to simplify the first statement down to:
$(\neg A\cup\neg B)\cap (A \cup \neg C)$ but I have no idea where to go from here.
 A: $\neg((A \cap B) \cup (\neg A \cap C)) = \neg(A \cap B) \cap \neg(\neg A \cap C)$ by de Morgans' law.
$= \neg(A \cap B) \cap \neg(\neg A \cap C) = (\neg A \cup \neg B) \cap (\neg\neg A \cup \neg C)$ by two applications of de Morgan again.
Now use distributivity ($D \cap (E \cup F) = (D \cap E) \cup (D \cap F)$, etc.) and $\neg\neg A = A$ to get that the latter equals:
$=((\neg A \cup \neg B) \cap A)  \cup ((\neg A \cup \neg B) \cap \neg C)$ which equals (twice distributity again):
$= (\neg A \cap A) \cup (\neg B \cap A) \cup (\neg A \cap \neg C) \cup (\neg B \cap \neg C)$, and now $A \cap \neg A = \emptyset$, and $(\neg B \cap \neg C)$ is absorbed by the other terms, after we split it using distributivity and $A \cup \neg A$ equals the whole space:  
$= (A \cap \neg B) \cup (\neg A \cap \neg C) \cup (\neg B \cap \neg C) = $
$= (A \cap \neg B) \cup (\neg A \cap \neg C) \cup (\neg B \cap \neg C \cap A) \cup (\neg B \cap \neg C \cap \neg A)$
and the third term is already a subset of the first, and the fourth is a subset of the second, so we are left with
$(A \cap \neg B) \cup (\neg A \cap \neg C) $, as required.
