Consider $ \ A \cap (B-C)$ and $ \ (A\cap B) - (A \cap C)$. Question:
Consider $ \ A \cap (B-C)$ and $ \ (A\cap B) - (A \cap C)$.
Are the two sets equal or is one a subset of the other?
My attempt: 
We know that if $ \ x\in (A \cap B) - (A \cap C) \implies x \in A \cap B$ and $ \ x \notin A \cap C \implies x \in A$ and $ \ x\in B$ and $ \ x\notin C \implies x \in A $and $ \ x\in B-C \implies x \in A \cap (B-C)$
So, $ \ (A\cap B) - (A \cap C) \subseteq \ A \cap (B-C) $
Is the other way true?
 A: Take any $x\in \ A\cap (B -C)$ then $x\in A$ and $x\in B-C$ (so $x\in B$ and $x\notin C$).
Now $x\in A\cap B$ (since $x\in A$ and $x\in B$) and $x\notin A\cap C$ (since $x\notin C$). 
So $x\in \ (A\cap B) - (A \cap C)$. 
Since $x$ is arbitrary we have $A\cap (B-C)\subseteq \ (A\cap B) - (A \cap C)$.
A: Through logic it should be true using definition $\forall x: x\in (X-A) \iff (x \in X) \land (x \notin A)$  that
$$\forall x: (x \in A) \land ((x \in B) \land (x\notin C)) =\forall x:  ((x \in A) \land (x \in B)) \land ((x \notin C) \lor (x\notin A))$$
because in order to satisfy the formula, $x$ must be in $A$, therefore $(x \notin A)$ is always false and the rightmost term is false, meaning that $(x \notin C) \lor (x\notin A)$ is the same as $(x \notin C)$
A: Another way of proving the equality of the sets is:
\begin{align} (A\cap B)-(A\cap C) &= (A\cap B)\cap (A\cap C)^c \\&= (A\cap B)\cap (A^c\cup C^c)\\&=((A\cap B)\cap A^c) \cup ((A\cap B)\cap C^c) \\&= ((A\cap A^c)\cap B)\cup(A\cap (B\cap C^c))\\&=(\emptyset\cap B)\cup(A\cap(B-C)) \\&= \emptyset\cup(A\cap(B-C))\\&=A\cap(B-C) \end{align}
This way, you don't need the two directions, since all equalities go both ways.

I used the rules:


*

*Commutativity and associativity of $\cap, \cup$

*Distributivity among $\cap$ and $\cup$ (i.e., $X\cap(Y\cup Z) = (X\cap Y)\cup(X\cap Z)$

*Definition of $X-Y$ as $X-Y=X\cap Y^c$ (where $Y^c$ is the complement of $Y$.

