Laws of logic Assertion/Reason format I am taking a Discrete Math class and we have this question.

$(B-A) \cup (C-A) = (B \cup C) -A$

Our section notes barely gloss over this, and Discrete Mathematics and Its Application, 7th Edition by Kenneth Rosen, glosses over it also. I would just like some guidance on how to solve this.
My course notes, page 104, example 1.11.2 has an example but not much else.
 A: To prove the identity
$$
(B - A) \cup (C - A) = (B \cup C) - A \tag{=}
$$
prove that each side is a subset of the other:
$$
\begin{align}
(B - A) \cup (C - A) &\subseteq (B \cup C) - A \tag{⊆} \\
(B - A) \cup (C - A)  &\supseteq (B \cup C) - A  \tag{⊇} \\
\end{align}
$$
You can do this by reasoning about elements. Membership in a term involving Boolean operations turns into a statement involving "and", "or" and "not", which you can manipulate as a statement of propositional logic using the analogous rules (De Morgan's laws, and so on.)
To show (⊆): suppose $x \in (B - A) \cup (C - A)$. Then $x$ is in $B-A$ or $x$ is in $C-A$; equivalently, ($x \in B$ and $x \notin A$) or ($x \in C$ and $x \notin A$); equivalently, ($x \in B$ or $x \in C$) and $x \notin A$; this last is equivalent to: $x \in (B \cup C) - A $.
Similarly, to show (⊇): if $x \in (B \cup C) - A$, then $x \in B \cup C$ and $x \notin A$; equivalently, ($x \in B$ or $x \in C$) and $x \notin A$; etc. It's basically reversing the steps of the other inclusion just proved.
A: I'll do my best to help, but I have never taken a discrete math course, so don't blame me if it isn't of much help :P
Consider $B-A$. By definition, $B-A = \{b\ |\ b \in B, b \notin A\}$. Now consider $C-A$. By definition, $C-A = \{c\ |\ c \in C, c \notin A\}$. Applying the definition of the union, this would mean that $(B-A) \cup (C-A)=\{x\ |\ x \in (B \cup C), x \notin A\}$. 
By definition, $(B∪C)−A = \{x\ |\ x \in (B \cup C), x \notin A\}$. This means that $(B-A) \cup (C-A) = (B∪C)−A$.
$QED$
