Premises for $A \setminus B = \emptyset$

I'm trying to find all the possible premises for $$A \setminus B = \emptyset$$ and so far I have $$A \subseteq B$$ or $$A = \emptyset$$ or $$A=B$$ but I'm not sure if these are ALL the possible premises.

• Isn't $A \setminus B = \emptyset$ simply equivalent to $A \subseteq B$, which includes the other two? This is because $A \setminus B = \{ x \in A : x \notin B \}$, so if this is empty, it means that for all $x \in A$ one has $x \in B$. – Andreas Caranti Sep 23 '18 at 17:32
• Nvm it is. My bad. – Hai Sep 23 '18 at 17:35

$$A\subseteq B$$ is equivalent to $$A\setminus B = \emptyset$$, let's try to prove it.

Being more formal, $$A\subseteq B$$ means $$\forall x \in A$$, $$x\in B$$.

You can deduce from this that the empty set is a subset of any set. The sentence $$\forall x \in \emptyset$$, $$x\in B$$ is true since there are no elements $$x\in\emptyset$$ to begin with. This means that $$A=\emptyset$$ is just a case of $$A\subseteq B$$. It's also true to see that $$A=B$$ is just a case of $$A\subseteq B$$. All in all this means that

$$A\subseteq B\qquad \text{OR} \qquad A=\emptyset\quad \text{OR} \quad A = B$$ is equivalent to

$$A \subseteq B.$$

Now, you should be able to prove that $$A\subseteq B\quad \Leftrightarrow \quad A\setminus B=\emptyset.$$

As for the premises, well all that imply $$A\subseteq B$$ are true. So $$A=\emptyset$$ and $$A=B$$ are correct.