Prove that $(B - A) \cup (C - A) = (B \cup C) - A$ by showing that each side is a subset of the opposite side. A big problem is that I never even know where to start with proofs. Then I panic and get absolutely nowhere.
To reiterate: Prove that $(B - A) \cup (C - A) = (B \cup C) - A$ by showing that each side is a subset of the opposite side.
$$(B - A) \cup (C - A) = (B \cup C) - A$$
I thought to use the Distributive Laws, but I'm not sure if that would take me in the right direction.
I'm also supposed to prove it using membership tables, which I haven't even a basic understanding of. Blegh.
Edit: Here's a rough proof I attempted. How off the mark am I? And how can I rewrite it cleaner?
\ Let $x \in (B - A) \cup (C - A)$.  
\ Then $x \in (B - A)$ or $x \in (C - A)$.
\ Assume that $x \in (B - A)$. 
\ Thus $x \in B$ and $x \notin A$. 
\ Therefore $x \in (B \cup C)$.
\ Because $x \notin A$, $x \in (B \cup C) - A$
\ $(B - A) \cup (C - A) \subseteq (B \cup C) - A. \quad \quad$ (Distributive Law)
\ If $x \in (B \cup C) - A$, then $x \notin A$, and $x \in B$ or $x \in C$. 
\ Therefore $x \in (B - A)$ or $x \in (C-A)$
\ $x \in (B - A) \cup (C - A)$
\ $(B \cup C) - A \subseteq (B - A) \cup (C - A)$
 A: Hint: pick an element from the left hand side, which has to be in either $B-A$ or $C-A$, and show it has to be in the right hand side. Then, pick an element from the right hand side and show it has to be in the left hand side.
A: As hinted bt TooTone, you must "test" the equlity in terms of $\in$ relation, using the definitions :

$x \in B - A$ iff $x \in B$ and $x \notin A$
$x \in B \cup C$ iff $x \in B$ or $x \in C$.

So you can "see" (but write it the details) that :

$x \in (B - A) \cup (C - A)$ iff ($x \in B$ and $x \notin A$) or ($x \in C$ and $x \notin A)$ iff ...
... ($x \in B$ or $x \in C)$ and $x \notin A$ .

A: Your proof is alright. There is also another approach, as almost always with questions about set difference the following identity is helpful:
$$X - Y = X \cap Y^c.$$
Using it, we can transform
$$(B - A) \cup (C - A) = (B \cup C) - A$$
into
$$(B \cap A^c) \cup (C \cap A^c) = (B \cup C) \cap A^c$$
and it becomes the distributive law.
I hope this helps $\ddot\smile$
A: Let $A,B$ be sets. Then $A=B$ if and only if $A\subseteq B$ and $B\subseteq A$. This is what it means when each side is a subset of the opposite.
Everything seems correct although I would detail a little more this side: $(B \cup C) - A \subseteq (B - A) \cup (C - A)$
Let $x\in (B \cup C) - A$. Then $x\in B$ or $x\in C$ and $x\notin A$.
Assume $x\in B$. Then $x\in B$ and $x\notin A \implies x\in B-A \implies x\in (B-A)\cup (C-A)$
Then assume $x\in C$ and prove analogically.
A: Personally, I find a proof in the following style the cleanest: for all $\;x\;$,
\begin{align}
& x \in (B - A) \cup (C - A) \\
\equiv & \qquad \text{"definition of $\;\cup\;$"} \\
& x \in B - A \;\lor\; x \in C - A \\
\equiv & \qquad \text{"definition of $\;-\;$, twice"} \\
& (x \in B \land x \not\in A) \;\lor\; (x \in C \land x \not\in A) \\
\equiv & \qquad \text{"logic: simplify"} \\
& (x \in B \;\lor\; x \in C) \land x \not\in A \\
\equiv & \qquad \text{"definition of $\;\cup\;$; definition of $\;-\;$"} \\
& x \in (B \cup C) - A \\
\end{align}
so by set extensionality $\;(B - A) \cup (C - A) \;=\; (B \cup C) - A\;$.
Note how this proof starts at the most complex side, expands the definitions to go to the level of logic, and then simplifies using the rules of logic.
A: Your proof for the first inclusion is just fine! So you've established that
$$(B - A) \cup (C - A) \subseteq (B \cup C) - A$$
Now, to show set equality, we simply need to show that the right-hand side is a subset of the left-hand side.  (Note that Marnix shows away to show both inclusions simultaneously. But given your title question, it seems that you are to prove set equality by showing each side is a subset of the other side.)
For the reverse inclusion (proving the set given on right-hand side is a subset of the set given on left-hand side), we start with an arbitrary element in right-hand set, and show that it must also be an element of the set on the left.
$$\begin{align} x \in (B\cup C) - A &\iff x\in (B\cup C) \land x\notin A\\ \\
&\iff (x\in B \lor x\in C) \land x\notin A\\ \\
&\iff (x\in B \land x\notin A) \lor (x \in C \land x\notin A)\\ \\
& \iff [x\in (B-A)] \lor [x\in (C-A)] \\ \\
&\iff x\in [(B-A)\cup (C-A)]\\ \\
&\therefore\quad (B\cup C) - A \subseteq (B-A)\cup (C-A)\end{align}
$$
Note that it turns out that given the fact that step $(a)\iff$ step $(a+1)$ for each step of the proof, we in fact have proven set equality, since the proof is valid "forward" or "backward". But all we needed to "wrap up" the proof, given your work, was set inclusion.
