Prove that if $A \setminus B = \emptyset$, then $A \subseteq B$.
The Venn Diagram helped me to visualize what I'm trying to show (thanks @GA316), but the book asks for a written proof (step by step) by contradiction. Sorry if I wasn't more specific at first, is just that I've had many troubles in the past with proofs, somehow I have many ideas but I can't seem to connect them to get to the final proof.
This is what I have so far: $P \rightarrow Q$ is equivalent to $\neg Q \rightarrow \neg P$ contraposition (thanks @The Chaz 2.0)
With P: $ A \setminus B = \emptyset$ and Q: $ A \subseteq B$
so $ \neg Q \equiv A \not\subseteq B\ , \exists x \in A : x \notin B $
be $ t: t \in A \wedge t \notin B $ ...is this right?
as this is the definition for $A \setminus B \ne \emptyset$ ...is this right?
$\therefore \neg Q \rightarrow \neg P \equiv A \not\subseteq B\ \rightarrow A \setminus B \ne\emptyset$
I have many concerns regarding if I'm using the correct notation. I am trying to learn this by myself and have nobody else to ask.
Also, sorry if it took me too long to update, I just started learning about this LaTEX notation.
Thank you very much in advance, you guys are so nice and helpful. You made me feel very welcomed and sure I need to read more about the rules and instructions for using this site.