$A ⊂ B$ 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?

One might proceed more directly by noting that $A\subseteq B$ is equivalent to $$\forall x,(x\in A)\implies(x\in B),$$ which is equivalent to $$\forall x,\neg(x\in A)\vee(x\in B).$$ The negation of this is $$\exists x:(x\in A)\wedge\neg(x\in B),$$ which is equivalent to $$\exists x:(x\in A)\wedge(x\notin B),$$ which is equivalent to $$\exists x:(x\in A\setminus B).$$ Renegating then shows us that $A\subseteq B$ is equivalent to $$\forall x,\neg(x\in A\setminus B),$$ at which point we're basically done.
