$A\subseteq B\to A\setminus B = \emptyset$. Is this true? If $A\subseteq B$, is then $A\setminus B = \emptyset$?
If we take a some $x\in A$, then also $x\in B$. Right?
So, if we take all the elements of $B$ away from $A$, we have nothing left?
 A: There actually is a somewhat nice way you can write up the proof for what you already know to be intuitively true. First you must know how $A\subseteq B$ and $A\setminus B$ are defined though. We have
$$
A\subseteq B = \{x : x\in A\to x\in B\}\tag{1}
$$
and
$$
A\setminus B = \{x : x\in A\land x\not\in B\}\tag{2}
$$
Now consider the following argument.
Claim: $A\subseteq B\to A\setminus B=\emptyset$.
Proof. Suppose $A\subseteq B$ (we are given this). Then $x\in A\to x\in B$ by $(1)$. Consider the negation of $(1)$. That is, what is $\neg(x\in A\to x\in B)$? Note the following:
\begin{align} A\nsubseteq B &\equiv \neg(x\in A\to x\in B)\tag{definition}\\[0.5em] 
&\equiv \neg(x\not\in A\lor x\in B)\tag{$p\to q\equiv\neg p\lor q$}\\[0.5em]
&\equiv x\in A\land x\not\in B\tag{DeMorgan}\\[0.5em]
&\equiv A\setminus B.\tag{definition}
\end{align}
As you can see from the argument above, if we are given that $A\subseteq B$, then it must necessarily be true that $A\setminus B=\emptyset$. $\blacksquare$ 
This is just a more formal way of asserting what you noted at the beginning. Does it make sense?
A: Yes, that's correct. $\qquad{}{}$
A: Well the definition of $A/ B$ is $A/B=A \cap B^c$. Since $A \subset B$ then $A \cap B^c=\emptyset$.
