$A ⊂ B$ 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?
 A: I like using "proof by contradiction" for this one, because it is logically and intuitively clear.


*

*(⇒) Assume $A\subseteq B$. By contradiction, suppose $A \setminus B \ne \varnothing$. Therefore there exists $x \in A \setminus B$. Therefore $x \in A$ and $x \notin B$ — a contradiction. As a result, $A \setminus B = \varnothing$.

*(⇐) Assume $A \setminus B = \varnothing$. By contradiction, suppose $A \nsubseteq B$. Therefore there exists $x \in A$ such that $x \notin B$. Therefore $x \in A \setminus B$ — a contradiction. As a result, $A \subseteq B$.
As a result, $A \subseteq B$ if and only if $A \setminus B = \varnothing$.   ◻
A: 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.
A: I think your proof is Ok. I just add more details.
$ (\Rightarrow) $; Suppose $ A\subset B $. Then $ \forall x\in A,x\in B $. Therefore if $ x\notin B $ then $ x\notin A $. That is if $ x\in B^{c} $ then $ x\in A^{c} $. Since $ A-B=A\cap B^{c} $ we have that $ A-B=\varnothing $.
$ (\Leftarrow) $; Conversely suppose $ A-B=\varnothing $. So $ A\cap B^{c}=\varnothing $. Therefore $ \forall x\in A,x\notin B^{c} $. Hence $ \forall x\in A,x\in B $. So we have that $ A\subset B $.
A: 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 $\neg\phi\lor\psi$).
A: Informally:
 $A - B$ is all elements of $A$ out of $B$.
and
$A \subseteq B$ means $A$ is surrounded by $B$.
so $\cdots$
