Proof $(N = \emptyset) \implies (M \setminus N = M)$ and reverse could someone check if my approach is correct?

*

*if $N = \emptyset$, then $M \setminus N = M$

*if $M \setminus N = M$, then $N = \emptyset$
The second one is easy:


*Choose $M = \{1,2\}$ and $N = \{3,4\}$, then $M \setminus N = \{1,2\} \setminus \{3,4\} = \{1,2\} = M$, therefore $N \neq \emptyset$.

The first one is correct, I just don't see a way to prove this because it is so obvious.
Maybe something like this:

*

*Assume $N$ is not empty, then M and N would have to be disjoint for $M \setminus N = M$ to be true. Therefore N is empty or a disjoint set.

I'm pretty sure this isn't really a proof. This should be easy and I'm ashamed to ask this question.
 A: Just to be explicitly clear, you are not trying to prove both statements, but to prove whether each is true or false? As you establish, the second one is incorrect: if $M\setminus N = M$ then we know that $N$ contains nothing from $M$ i.e. that $N\cap M = \emptyset$. The class of counterexamples can be characterised as $M$ being any set and $N$ being any set of objects that are not elements of $M$. All we need is one counterexample of that form and you've given one: $M=\{1,2\}$, $N=\{3,4\}$.
The first statement is true and obvious, but your proof attempt starts off on the wrong foot because "assume $N\neq \emptyset$" is the wrong direction of logic (statement 1 says nothing about what might be true if $N$ is not empty). If you want the contrapositive of the first statement, it is "if $M\setminus N \neq M$ then $N\neq \emptyset$". However, we are best arguing directly from the statement itself. Without some understanding of what definition you have for $\setminus$, it is impossible to "prove" the statement.
However, you might have an informal definition like "$M\setminus N$ is the elements of $M$ that are not in $N$", which would give you the ability to write an explanation of why $M\setminus N=M$, or you might have a formal definition like "$M\setminus N:=\{x\in M: x\notin N\}$", and then you can give a slightly more formal proof that follows the same logic.
