How to resolve $x \in A \wedge x \notin A $? Let A and B be two sets. Then
$A \setminus B = \{x: x\in A \wedge x\notin B\}$
$A \setminus B = \{x: x\in A \wedge x\notin A \cap B\}$
How can one prove that two logical statements are equal?
Suppose $x \in A \setminus B$. 
Then,
$x \in A \wedge x \notin A \cap B$
$x \in A \wedge x \notin A \wedge x\notin B$
How do I move forward?

Your implication $\{x \in A \wedge x \notin A \cap B\}\Rightarrow \{x \in A \wedge x \notin A \wedge x\notin B\}$ is not true; By De Morgan's laws it should be$$\{x \in A \wedge x \notin A \cap B\}\Rightarrow \{x \in A \wedge (x \notin A \lor x\notin B)\}$$  

Now proof.  
You want to prove:
$$\{x: x\in A \wedge x\notin B\}=\{x: x\in A \wedge x\notin A \cap B\}$$
or equivalently:
$$A \setminus B=A \setminus (A\cap B).$$


proof.1.  

$$A \setminus (A\cap B)= A\cap (A\cap B)^c = A\cap (A^c\cup B^c)=(A\cap A^c)\cup (A\cap B^c)=(A\cap B^c)= A \setminus B$$


proof.2.  

Since $A\cap B \subset B$ we have $A \setminus B\subset A \setminus (A\cap B)$.
on the other hand:
 Now let $x\in A \setminus (A\cap B)$. So $\underline{x\in A}$ and $x\notin A\cap B$; which means $\underline {x\notin  B} $. Therefore  $x\in A \setminus B$. (this direction is just Addem's proof).
