Prove: $A\cap (B - C) = (A \cap B) - (A \cap C)$ Prove: $A \cap (B - C) = (A \cap B) − (A \cap C)$
I can understand this using Venn Diagrams, however I am struggling to translate this into a formal proof. 
 A: The most common strategy for proving that sets $S=T$ is to show that $S\subset T$ and $T\subset S$.  And the most common strategy for proving $S\subset T$ is to prove that $(x\in S) \implies (x\in T)$ for any element $x$ in the universe of the problem.  So let's do that!
First, let $x\in A\cap (B-C)$ be given.  This means that $x\in A$ and $x\in B-C$, meaning that $x\in B$ and $x\notin C$.  In other words, $x\in A\cap B$ but $x\notin A\cap C$, so $x\in (A\cap B)-(A\cap C)$.  
Conversely, let $x\in(A\cap B)-(A\cap C)$ be given.  Then $x\in A\cap B$ and $x\notin A\cap C$.  Since $x\in A\cap B$, $x\in A$ and $x\in B$.  But $x\notin A\cap C$, so it must be that $x\notin C$.  Since $x\in B$ and $x\notin C$, $x\in B-C$.  Therefore, $x\in A\cap (B-C)$.
A: To show that two sets are equal you normaly show the two relations $\subseteq$ and $\supseteq$.
This is how it is done, where I write it down explanatory and not how I would do so in a mathematical proof.
So we first want to show that $A\cap (B-C)\subseteq (A\cap B)-(A\cap C)$
So let $x\in A\cap (B-C)$.
By definition of $\cap$ this means that $x\in A$ and $x\in B-C$.
By definition of $-$ we have that $B-C$ implies $x\in B$ and $x\notin C$.
Now we have $x\in A$ and $x\in B$. So $x\in A\cap B$. (By definitino of $\cap$)
Since $x\in A$ and $x\notin C$, it is $x\notin A\cap C$. (By definition of $\cap$)
So $x\in (A\cap B)-(A\cap C)$ (By defintion of $-$)
Now show $A\cap (B-C)\supseteq (A\cap B)-(A\cap C)$.
Start like this: Let $x\in (A\cap B)-(A\cap C)$. Then ...
The proof is done (exactly) like above.
A: $$A \cap (B-C) = A \cap (B \cap C^\complement)$$ while
$$(A\cap B) - (A \cap C) = (A \cap B) \cap (A \cap C)^\complement = \\
(A \cap B) \cap (A^\complement \cup C^\complement) (\text{ de Morgan) } =\\
(A \cap B \cap A^\complement) \cup (A \cap B \cap C^\complement) = A \cap B \cap C^\complement$$
so we have equality (the first component of the final union is empty).
A: Here it is an approach based on the definition of difference and the De Morgan Laws:
\begin{align*}
(A\cap B) - (A\cap C) & = (A\cap B)\cap(\overline{A\cap C}) = (A\cap B)\cap(\overline{A}\cup\overline{C})\\
& = (A\cap B\cap\overline{A})\cup(A\cap B\cap\overline{C})\\
& = A\cap B\cap\overline{C} = A\cap(B - C)
\end{align*}
