$A ⊂ B$ if and only if $A − B = ∅$ I need to prove that $A ⊂ B$ if and only if $A − B = ∅$.
I have the following "proof":
$$ A \subset B \iff A - B = \emptyset$$
proof for $\implies:$
$$\forall x \in A, x \in B$$
Therefore,
$$A - B = \emptyset$$
proof for $\impliedby$:
If $$A - B = \emptyset$$
then
$$\forall x \in B, x \in A$$
since $\forall x \in B, x \in A$, 
$$ A \subset B $$
However the whole thing seems to be incredibly "fragile" and relies on circular logic (see how I just switched the sets in the for all statements)
Is this a valid proof? Is there a better way to write it?

Your proof is valid if one can follow the steps, but the step from $A-B=\emptyset$ to $\forall x\in B, x\in A$ looks a bit hasty and might need some clarification. For example using the definition of $A-B$ being the set of $x: x\in A \land x \notin B$, which in turn is equivalent to $x: x\in B\rightarrow x\in A$ ($\phi\rightarrow\psi$ being the same as $\neg\phi\lor\psi$).
