Show that $X-Y = X \cap \overline{Y} $ Very basic problem, just wanted to be sure I did this correctly.
The problem is "Show that $X-Y = X \cap \overline{Y}  $". There was hint in the problem telling one to let our universe $U=X \cup Y$. Here's my work:
Notation: $\overline{A}$ is the complement of a set, $A$, and the operator $-$ is meant to be a set difference, so that give two sets, $A,B$ we have $A-B=\{x\in A : x\not\in B\}$.

$$X \cap \overline{Y} = X \cap [(X\cup Y)-Y]$$
  $$=X \cap [(X-Y)\cup(Y-Y)]$$
  $$=X \cap [(X-Y) \cup \emptyset]$$
  $$=X \cap (X-Y)$$
  $$=X-Y$$
  $$\implies X \cap \overline{Y}=X-Y$$
Q.E.D

I think I got it. It seemed really trivial, considering the whole "let $U=X \cup Y$" bit. Just wanted to be sure I didn't overlook anything
 A: To prove set equality one needs to show that each set is a subset of the other. That is, 
$$A=B \, \, \iff\left(A\subseteq B\right)\, \land \left(B\subseteq A\right)$$
So, to prove problem presented we need to show $X \cap Y^{c} \subseteq X\setminus Y$ and $X\setminus Y \subseteq X \cap Y^{c}$
Proof:
Part 1: $X \cap Y^{c} \subseteq X\setminus Y$
Assume: $x \in X \cap Y^{c} \quad$ Show: $x \in X\setminus Y$
So by definition of set intersection, $x \in X$ and $x \in Y^{c}$
Set complement is defined as $Y^{c} = U \setminus Y$, so $x \in U$ and $x \notin Y$, where $U$ is defined to be our universe.
So, $x \in X$ and $x \notin Y$
So by the definition of set difference, $x \in X\setminus Y$
Part 2: $X\setminus Y \subseteq X \cap Y^{c}$
Assume: $x \in X\setminus Y \quad$ Show: $x \in X \cap Y^{c}$
So by the definition of set difference, $x \in X$ and $x \notin Y$
Trivially (assuming well-defined sets), $x \in U$, where $U$ is our universe.
So, $x \in U$ and $x \notin Y$, which by the definition of set difference implies $x \in U \setminus Y$
Set compliment is defined as $Y^{c} = U \setminus Y$, where $U$ is defined to be our universe.
Therefore, since $x \in U \setminus Y$, $\, \, x \in Y^{c}$
Since $x \in X$ and $x \in Y^{c}$ and by the definition of set intersection, $x \in X \cap Y^{c}$
$\Box$
