Help with proving that $( A - B ) ∩ B$ is equal to the empty set I am trying to prove $( A - B ) ∩ B = Ø$
I know I need to prove the following


*

*while $x∈ B$ that $x∉ ( A - B ) ∩ B$

*while $x∉ B$ that $x∉ ( A - B ) ∩ B$
I'm completely stuck, can anyone help?
 A: If you know what these terms mean it is trivial.
$A \setminus B$ is all the elements in $A$ that are not in $B$.
And $K \cap B$ is all the elements in both $K$ and $B$.
And $\emptyset$ is the set that has no elements.
......
If $(A\setminus B) \cap B$ has any element (call it $x$) then $x$ is an element of $ B$ and $x$ is an element of $A \setminus B$.  And if $x$ is an element of $A \setminus B$ then $x$ is an element of $A$ but is NOT an element  $B$.
So $x$ is both in $B$ and NOT in $B$.  That's impossible.
SO $(A\setminus B) \cap B$ can not have any elements.
So $(A\setminus B)\cap B = \emptyset$.
A: You really do not need to prove what it seems you believe you need to prove. Are you familiar with the complement of a set? Given a universal set $U$ and some set $A$, you would have $A\cup A^c=U$ because "$A$ complement," or $A^c$, is defined as "everything in the universal set that is not in $A$." More succinctly, we have $A^c=\{x : x\not\in A\}$, where it is assumed $x\in U$ from context. 
The point of all this is that we reasonably have $A\cup A^c=U$, $A\cap A^c=\emptyset$, and $A-B=A\cap B^c$. Then it's just a matter of some set algebra:
$$
\begin{align*}
(A-B)\cap B
&= (A\cap B^c)\cap B\\[0.5em]
&= A\cap(B^c\cap B)\\[0.5em]
&= A\cap\emptyset\\[0.5em]
&=\emptyset.
\end{align*}
$$
A: So, in general, if you want to prove two sets $X,Y$ are equal, you need to show $X \subseteq Y$ and $Y \subseteq X$.
Here, we have $(A-B) \cap B = \emptyset$. We want to show both subset relations: $(A-B)\cap B \subseteq \emptyset$ and $\emptyset \subseteq (A-B)\cap B$.
The latter is trivial since the empty set is a subset of every set, so we focus on the former.
Consider $x \in  (A-B)\cap B$. By definition of the intersection, $x\in (A-B)$ and $x \in B$, since it has to be both. By the definition of the set difference, $x \in A$ and $x \not \in B$: to be in the set difference, $x$ has to be in the first set and can't be in the second, or it would be deleted.
So we have $x \in A, x \in B, x\not \in B$.
So we ask: what can we conclude about this information?
If you think hard enough, you should be able to see some sort of contradiction, something that implies absolutely no element $x$ can possibly satisfy what we desire. Once you realize that, that means that there are no $x$ in $(A-B)\cap B$ and thus it must be a subset and thus equal to the empty set.
I'll leave formalizing the contradiction up to you.
A: A possible answer using (1) the truth table method (2) logical equivalences

