Set theory problem: Prove: for all sets, $A$ and $B$, $A-B=A-(A \cap B).$ Prove: for all sets $A$ and $B$, $~~~A-B=A-(A \cap B).$
My work:
Proof.
In order to prove that two sets $X$ and $Y$ are equal, I must prove that $X \subseteq Y$ and $Y \subseteq X$.
Part 1. Prove $[A-B] \subseteq [A-(A \cap B)]$.
Part 2. Prove $[A-(A \cap B)] \subseteq [A-B]$.
Part 1. Proof
$\exists x, x \in A - B \Rightarrow x \in A \land x \notin B.$
$\exists y, y \in A-(A \cap B) \Rightarrow y \in A \land y \notin A \cap B.$
This is where I got stuck. I'm not sure if this is a good structure. All I'm sure about is the template. I'm not sure what to do with $y \notin A \cap B$ or even if that's a good starting point.
 A: Another method:
$$\begin{align}A-(A \cap B)=A\cap(A\cap B)^c=A\cap (A^c\cup B^c)&=(A\cap A^c)\cup(A\cap B^c)\\
\\&=\emptyset\cup(A\cap B^c)=(A\cap B^c)=A-B\end{align}$$
A: 
Part 1. Prove $[A-B] \subseteq [A-(A \cap B)]$. 
Part 2. Prove $[A-(A \cap B)] \subseteq [A-B]$.

Part 1. 
$x \in [A - B] \implies (x \in A) \wedge (x \not\in B) \implies $ 
$x\not\in (A \cap B), ~$ [since $(A \cap B) \subseteq B$]. 
Therefore, $(x\in A) \wedge [x \not\in (A \cap B)] \implies x \in [A - (A\cap B)].$
Part 2. 
This is trickier. 
$x \in [A - (A\cap B)] \implies (x \in A) \wedge [x \not\in (A\cap B)].$ 
Suppose that $~x \in B.$ 
Since you also have that $(x \in A)$, this implies that 
$x \in (A \cap B).$  This yields a contradiction. 
Therefore, you can conclude that $~x \not\in B.$ 
Therefore, $~ (x \in A) \wedge (x \not\in B).~$ 
This implies that $x \in (A - B).$
A: $$
x \in A \wedge \neg (x \in B \land x \in A ) \iff x \in A  \land x \not \in B
$$
is a tautology (modus ponendo tollens). Therefore, any element in the set $ A - (A \cap B)$ is in the set $ A - B $ and vice versa.
