Prove that if $A$, $B$, and $C$ are sets then $(A - B) \cup (A - C) = A - (B \cap C)$ 
Prove that if $A$, $B$, and $C$ are sets then $(A - B) \cup (A - C) = A - (B \cap C)$.

I have the proof for the first direction:
Let $x \in (A - B) \cup (A - C)$ be given. Hence, $x \in (A - B)$ or $x \in (A - C)$. Suppose $x \in (A - B)$. Hence, $x \in A$ and $x \notin B$. Since $x \notin B$ it is logically true that $x \notin (B \cap C)$. So, $x \in A$ and $x \notin (B \cap C)$. Hence $x \in A - (B \cap C)$ by definition. A similar argument works in the case where $x \in (A - C)$. So, $(A - B) \cup (A - C) = A - (B \cap C)$.
I'm confused as to how to go about proving the other direction. Would I assume that $x \in A - (B \cap C)$?
 A: Let $x \in A - (B \cap C)$. Hence, $x \in A$ and $x \notin B\cap C$.
Suppose $x \notin B\cap C$. Hence $x \notin B $ or $x\notin C$.
$x \in A$ and $x \notin B $ then $x \in (A - B)$.
$x \in A$ and $x \notin C $ then $x \in (A - C)$.
Hence $x \in (A - B) \cup (A - C)$ by definition.
A: \begin{align*} (A - B) \cup (A - C) &=(A \cap B^c) \cup (A \cap C^c) \\ & = A\cap (B^c\cup C^c) \\ &\underbrace{=}_{\text{De Morgan}}A\cap (B \cap C)^c \\ &= A - (B \cap C)\end{align*}
A: It is a lot easier if you do the next thing: Let A be universal set, so all elements which we speak of are in A. Now, you put $A-B$ as a negation of q, where q is a fact that something is i n B. Then you simply have to prove that negation of conjuction of two is disjuction of separate negations, De Morgan rule. Or you can make 2 times 2 table. 
A: 
I'm confused as to how to go about proving the other direction. Would I assume that $x∈A−(B∩C)$

Sure. 
An arbitrary element of the set difference will be an element of A that is not also an element of both B and C.   It can be an element of A and C but not B, or an element of A and B but not C, or an element of A but neither B nor C.
  Thus it is either an element of A that is not an element of B, or it is an element of A that is not an element of C. $$A\setminus(B\cap C)~\subseteq~(A\setminus B)\cup(A\setminus C)$$
