Prove: $\forall A,B(A-B=\emptyset \implies A\subset B)$ Lemma: $A-B=\emptyset\implies A\subset A\cap B$ where $A$ and $B$ are sets.
Proof (lemma). Let $A$ and $B$ be arbitrary sets and assume $A-B=\emptyset$. Pursuant to the axiom of extensionality our premise means
$$\forall x\in A-B(x\in \emptyset)\wedge \forall x'\in\emptyset(x'\in A-B)$$
Since there's no $y$ for which $y\in \emptyset$, we'll point out that in accordance with the foregoing statement there must be some $x$ such that $x\notin A \vee x\in B$.
With a view to show $A\subset (A\cap B)$ let $a\in A$. By choosing $x=a$, $$a\notin A \vee a\in B$$
From this foregoing statement certainly $a\in B$. Then, thus far $\forall a\in A(a\in A\cap B)$ or simply $$A\subset(A\cap B)$$
Proof.  Let $A$ and $B$ be sets. Subsequently,$$A\subset (A\cap B)\subset B$$ provided $A-B=\emptyset$.
Question: Is my proof correct?

Using Reductio ad absurdum.
The hypothesis means ( since " equality amounts to reciprocal inclusion")  : for all x, in case  x is in A and not in B, then x belongs to the empty set.
Now, suppose ( for refutation) that A is not included in B.
It means there is some object $a$ that belongs to A, but not to B.
The universal rule admitted as hypothesis applies to  this object $a$.
So $a$ belongs to the empty set, which is absurd.
So, A is included in B.
