Proving the following statement using Sets Identities The statement goes as follow: 
$ B ∩ C ⊆ A ⇒ (C − A) ∩ (B − A) = ∅. $
First, the sign "=>" represents a tautology, no? ( apparently I get it confuse with the 3 bar sign, if you know what I mean).
Second, the fact that it equals to no solution, how do I prove that? Seems to contradict itself, at least to me.
EDIT: apparently necessary edit... Couldn't start the problem because I did not understand the notation. No need to down vote for that, the more people understand a concept properly, the better....
 A: The symbol $\Rightarrow$ means that for any sets $A,B,C$, if $B \cap C \subseteq A$ then $(C-A)\cap(B-A)=\emptyset$
You want to prove that if $B \cap C$ is a subset of $A$, then the set $(C-A)\cap (B-A)$ is empty. Drawing a Venn diagram to represent the situation might aid your intuition. 
For a formal proof you need to prove that $(C-A)\cap(B-A)$ is contained in $\emptyset$, and that $\emptyset$ is contained in $(C-A)\cap(B-A)$.
I give you the following hint: For any sets $E,F$: $E-F=E\cap F^c$. 
Added: To prove that the sets are equal you need to prove that each set is contained in the other one.
First we prove that $\emptyset \subseteq (C-A)\cap (B-A)$. This is trivial since the implication $p \rightarrow q$ is true when $p$ is false, in particular the proposition:
$$\forall x, x\in \emptyset \rightarrow x \in (C-A)\cap(B-A)$$
is true, hence $\emptyset \subseteq (C-A)\cap (B-A)$
Now we prove the other inclusion. Take $x \in (C-A) \cap (B-A)$, this means $(x \in C \wedge x \not \in A)\wedge(x \in B \wedge x \not \in A)$, which gives, by De Morgan laws: $x \in (B\cap C) \wedge x \not \in A$, and since $B \cap C \subseteq A$ we deduce $x \in A \wedge x \not \in A$, this is $x \in \emptyset$. This last step is the definition of the empty set in some books. You can also view it like this: given a proposition $p$ which is false (in this case $x \in A \wedge x \not \in A$), then the conditional $p \rightarrow q$ is true.
A: As I said in my comment above, the $"\implies"$ symbol means "implies." If you have two statements $P,Q$ and know that $P \implies Q$, then that is to say "If $P$, then $Q$". 
The $"\emptyset"$ symbol is the emptyset. It is the set that contains no elements, and is one of the most important sets in all of set theory. The empty set is somewhat analogous to zero, in as far as $S \cup \emptyset = S$ and $S \cap \emptyset = \emptyset$ for any set $S$, just as $x+0 = x$ and $x\cdot 0 = 0$ for any number $x$. The analogy is not perfect since addition and union are not quite the same, nor are multiplication and intersection. 
To start this proof, I recommend stating the following:

Let $A,B,C$ be sets such that $B\cap C\subseteq A$. 

We now assume the above is true. Next, let $x \in C-A$. (The $"\in"$ symbol means "in", or "be an element of " just in case you haven't seen it yet.) By definition, $x \in C$ and $x \notin A$. Since $x \notin A$, then $x \notin B\cap C$. Can you proceed from here to show that $x \notin B-A$?
Now let $y \in B-A$ and use similar reasoning to prove that $y \notin C-A$. Once you do that, you will have shown that $B-A$ and $C-A$ have no elements in common. We then say that their intersection is empty. This is expressed mathematically by stating $(B-A)\cap (C-A) = \emptyset$. 
Good luck!
