Showing $A \subset B \Leftrightarrow A-B=\emptyset$ How can one justify this proposition.
Propostion
$A \subset B \Leftrightarrow A-B=\emptyset$
My proof
$A-B$ means all element that appear in A but not in B. So $x\in A,x \notin B$. 
However if $A-B$ is the empty set then all element of $A$ are in $B$. And this is the definition of
$A\subset B$
Second part
If $A \subset B$ then there will no element that is in $A$ but not in $B$.
 A: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
$Here is an alternative proof.  Starting at the most complex side, so $\;A - B = \emptyset\;$, let's just unwrap the definitions, and then try to simplify.
$$\calc
A - B = \emptyset
\calcop{\equiv}{basic property of $\;\emptyset\;$}
\langle \forall x :: \lnot(x \in A - B) \rangle
\calcop{\equiv}{definition of $\;-\;$}
\langle \forall x :: \lnot(x \in A \;\land\; x \not\in B) \rangle
\calcop{\equiv}{logic: DeMorgan -- to simplify}
\langle \forall x :: x \not\in A \;\lor\; x \in B \rangle
\calcop{\equiv}{logic: $\;\lnot P \lor Q\;$ is a different way of writing $\;P \Rightarrow Q\;$}
\langle \forall x :: x \in A \;\Rightarrow\; x \in B \rangle
\calcop{\equiv}{definition of $\;\subseteq\;$}
A \subseteq B
\endcalc$$
This completes the proof.
A: That's very good. Just wrap up the second proof (i.e. state the desired conclusion): 
...That is, $A\subset B \implies A - B = \emptyset$.
A: 
Let's suppose $A \subset B$.
If we had $x\in A-B$, that would mean $x\in A$ and $x\notin B$.
However, $A\subset B$ means that $x\in A\Rightarrow x\in B$
Therefore there exists no $x\in A$ that's not in $B$.
(there's the part you need to complete your proof)
Therefore $A \subset B\Rightarrow A-B=\emptyset$

Let's suppose $A-B=\emptyset$.
That means that there exists no $x\in A$ such as $x\notin B$
Therefore $x\in A\Rightarrow x\in B$
Thus $A-B=\emptyset\Rightarrow A\subset B$

We have shown that $A-B=\emptyset\Leftrightarrow A\subset B$
