Proof of $A \setminus B= A$ st $A \subset B$

Question

Problem statement - $A \setminus B= A$ st $A \subset B$

I think this statement is wrong as by definition of difference of sets $A \setminus B$ should contain all the elements of set $A$ which are not in $B$.But if $A$ is a subset of $B$ then all the elements of set $A$ are in set $B$ by default so shouldn't the answer be null set?

Answer 1

Sure thing, if $A = \emptyset$ then

$$A \setminus B = \emptyset \setminus B = \emptyset \subset B$$

Answer 2

Since $A\subset B$ we have that $$x\in A \implies x\in B$$

but $$\forall y\in A \setminus B\implies y\in A \quad\land \quad y\not\in B$$

than $A=\emptyset$