How to prove DeMorgan's law? How to prove DeMorgan's Law?
$$A - (B \cup C) = (A - B) \cap (A - C)$$
$$A - (B \cap C) = (A - B) \cup (A - C)$$
EDIT:
Here is what I have tried so far:  
Considering the first equation, assuming $x \in A - (B \cup C)$ then $x \in A$ and $x \not\in B$ and $x \not\in C$, while the right hand means ($x \in A$ and $x \not\in B$) or ($x \in A $ and $x \not\in C$) which is the same as $x \in A$ and $x \not\in B$ and $x \not\in C$. So the two set is the same.
But I do not know whether this is sufficient for a proof. Am I wrong?
 A: I'd "reduce" it to logic
$x\in A-(B\cup C) \iff x\in A \text{ and not } (x\in B \text{ or } x\in C)$
thus by DeMorgan in logic (which can be proved with truth table)
$x\in A \text{ and } x\notin B \text{ and } x\notin C$
Thus $(x \in A  \text{ and } x \notin B) \text{ and } (x \in A  \text{ and } x \notin C)$. Thus
$x\in (A- B)\cap (A- C)$
and vice versa.
A: Maybe you're familiar with the fact that if $A\subseteq B$ and $B\subseteq A$, then $A=B$?
Since you're viewing $A,B,C$ as sets, you can prove these by showing the set on the left of $=$ is a subset of the set on the right of $=$, and vice versa.
For example, suppose $x\in A-(B\cup C)$. So $x\in A$, but $x\not\in B\cup C$. In particular, $x\not\in B$, and $x\not\in C$. It follows that $x\in A-B$ and $x\in A-C$, that is, $x\in (A-B)\cap (A-C)$. Thus $A-(B\cup C)\subseteq (A-B)\cap (A-C)$. 
Try using a similar method to show the reverse containment, and you'll get your result. This works for the second law also.
A: In words
http://www.youtube.com/watch?v=bED-wffoK_g4
If you need an explanation and have noone to discuss with.
