Prove that $A-B=A$ implies and implied by $A\cap B=\emptyset $ My work 
Let $x$ be any arbitrary element of $A-B$
$$A-B=\{x: x\in A,\; x\notin B\}$$
$$=\{x: x\in A,\;  x\in B'\}$$
$$=\{x:x\in A\cap B\}$$
How do I proceed further? 
 A: alternatively:

${\color{Red} {\text{to proof}}}$:  $A\cap B=\emptyset$:

    
*
    
*We can use Symmetric Difference of Equal Sets and $(A\setminus B)\cap B=\emptyset$ in fact: $$ \begin{align} (A\cap B)\Delta \emptyset&= ((A \cap B) \setminus \emptyset) \cup (\emptyset \setminus (A \cap B)) \\ &=((A \cap B) \setminus \emptyset) \cup \emptyset \\ &=((A \cap B) \setminus \emptyset)\\ &=(A \cap B) \setminus ((A\setminus B)\cap B) \\ &=(A \cap B) \setminus (A\cap B) \text{ (because we have that } A \setminus B= A) \\ &=\emptyset \\ &{\color{Red} \Longrightarrow} A \cap B=\emptyset \end{align}$$
    
  



    
*We can use $(A\setminus B)\cap B=\emptyset$ and
    " $A \setminus B= A\cap C \to A\cap B\cap C=\emptyset$ " in fact: $$\begin{align}A \cap B\cap C &= (A\cap C) \cap B \\ &= (A \setminus B) \cap B \text{ (because we have that }A \setminus B= A \cap C)\\ &=\emptyset \\ &{\color{Red} \Longrightarrow}A\cap B=\emptyset \text{ (because we have that } C=A) \end{align}$$
    
  

A: *

*Let $A-B=A$.


Assume for the sake of contradiction that $x\in A \cap B$. Then, in particular, $x\in A=A-B$, so $x\notin B$, which is a contradiction with  $x\in A \cap B$.


*Let $A\cap B=\emptyset$.


a) If $x\in A$ then, since $A\cap B=\emptyset$, $x\notin B$, so $x\in A-B$. Thus $A\subset A-B$.
b) If $x\in A-B$ then, by definition, $x\in A$, so $A-B\subset A$.
In conclusion, $ A-B=A$.
A: Hints:
For the forward direction.


*

*To prove a set is empty, the typical approach is by contradiction.

*Assume for contradiction that $A\cap B\not=\emptyset$ and $A\setminus B=A$, so let $x\in A\cap B$ ($x$ exists because $A\cap B$ is not empty).

*Can you prove that $x\in A$, but $x\not\in A\setminus B$?  Does this give a contradiction?
For the backwards direction, 


*

*Assume that $A\cap B=\emptyset$.

*Since $A\setminus B\subseteq A$, all you need to show is that if $x\in A$, then $x\in A\setminus B$.

*Since $x\in A$, $x\not\in B$ as $A\cap B=\emptyset$, so $x\in A\setminus B$.
You could also do the backwards direction using the contrapositive.


*

*Suppose that $A\setminus B\not=A$.  

*Since $A\setminus B\subseteq A$, there is some $x\in A$ such that $x\not\in A\setminus B$.

*Prove that $x\in B$ and get that $A\cap B\not=\emptyset$.
A: Hint
$A\backslash B = A\backslash(A\cap B)$
