Prove that $A \subset B$ if and only if $A \setminus B = \emptyset$

Question

Prove that $A \subset B$ if and only if $A \setminus B = \emptyset$.

What is the correct and mathematically strict way to prove the above? (slightly different than Prove that if $A \setminus B = \emptyset$, then $A \subseteq B$ )

Answer 1

\begin{align} A \subset B \ \ :\Leftrightarrow \ \ & \forall x \ : \ (x \in A) \rightarrow (x \in B) \\ \Leftrightarrow \ \ & \forall x \ : \ \lnot (x \in A) \lor (x \in B) \end{align}

\begin{align} A \backslash B = \emptyset \ \ :\Leftrightarrow \ \ & \forall x \ : \ x \notin A \backslash B \\ \Leftrightarrow \ \ & \forall x \ : \ \lnot (x \in A \backslash B) \\ \Leftrightarrow \ \ & \forall x \ : \ \lnot ((x \in A) \land \lnot (x \in B)) \\ \Leftrightarrow \ \ & \forall x \ : \ \lnot (x \in A) \lor \lnot(\lnot (x \in B)) \\ \Leftrightarrow \ \ & \forall x \ : \ \lnot (x \in A) \lor (x \in B) \\ \Leftrightarrow \ \ & A \subset B \ \ \ \end{align}

Answer 2

If $A\subset B$, $\forall x\in A\Rightarrow x\in B$, so $A\setminus B=\emptyset$. I think this trivial.

Now, suppose $A\setminus B=\emptyset$. If $\exists x\in A$ but $x\notin B$, $x\in A\setminus B$ by the definition of set minus, which is a contradiction. So $x\in B$ and $A\subset B$.