Prove that if $A \setminus B = \emptyset$, then $A \subseteq B$ 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. 
 A: HINT: Just follow the definitions. In order to show that $A\subseteq B$, you should let $x$ be an arbitrary element of $A$ and somehow use the hypothesis that $A\setminus B=\varnothing$ to show that $x\in B$. What if $x$ were not in $B$? Then you’d have $x\in A$ and $x\notin B$, which would tell you that $x$ is in ... what? 
A: Draw venn diagram of $A - B = A \cap B^c$. can you see when it will be empty?
A: Just for grins, you could prove the (logically equivalent) contrapositive, viz.     

If $A$ is not a subset of $B$, then [$A$ "toss" $B$] is nonempty.

If $A$ is not a subset of $B$, then there is some element $x \in A$ that is not in $B$. Then when you "take away" all the things in $A$ that are/were in $B$, you have at least that element $x$ leftover.
So [$A$ "toss" $ B$] is not the emply set. 
A: Here is a complete proof in a calculational style: we start at the most complex side, and then just expand the definitions and simplify, and see where that leads us.
\begin{align}
& A \setminus B = \emptyset \\
\equiv & \qquad \text{"basic property of $\;\emptyset\;$"} \\
& \langle \forall x :: x \not\in A \setminus B \rangle \\
\equiv & \qquad \text{"definition of $\;\setminus\;$"} \\
& \langle \forall x :: \lnot (x \in A \land x \not\in B) \rangle \\
\equiv & \qquad \text{"logic: DeMorgan -- there is not much else we can do"} \\
& \langle \forall x :: x \not\in A \lor x \in B \rangle \\
\equiv & \qquad \text{"logic: $\;\lnot P \lor Q\;$ is one of the ways to write $\;P \Rightarrow Q\;$"} \\
& \langle \forall x :: x \in A \Rightarrow x \in B \rangle \\
\equiv & \qquad \text{"definition of $\;\subseteq\;$"} \\
& A \subseteq B \\
\end{align}
This completes the proof.  (Strictly speaking, this even proves the stronger statement $\;A \setminus B = \emptyset \;\equiv\;A \subseteq B\;$.)
A: We assume $A \subset E$, $B \subset E$ and $A \setminus B = \varnothing$
\begin{align*}% Utiliser & pour aligner en colonne. Enlever * pour numéroter chaque ligne et \nonumber en fin de ligne pour enlever la numérotation
 x \in A & \implies x \in A \wedge x \in E                                                                              \\
   & \implies x \in A \wedge x \in B \cup \complement_E B                                                         \\
   & \implies x \in A \wedge \left(x \in B \vee x \in \complement_E B\right)                           \\
   & \implies x \in A \wedge \left(x \in B \vee x \notin B\right)                                      \\
   & \implies \left(x \in A \wedge x \in B\right) \vee \left(x \in A \wedge x \notin B\right) \\
   & \implies x \in A \cap B \vee x \in A\setminus B                                                            \\
   & \implies x \in \left(A \cap B\right) \cup \left(A\setminus B\right)                                                   \\
   & \implies x \in \left(A \cap B\right) \cup \varnothing                                                                 \\
   & \implies x \in \left(A \cap B\right)                                                                                  \\
   & \implies x \in A \wedge x \in B                                                                              \\
   & \implies x \in B
\end{align*}
Then $A \subset B$
A: It seems to me you could benefit from writing a proof in multiple stages. Here's an example of my own thought processes on this problem.


*

*Translate the math symbols into English (or Spanish, as the case may be). I literally read "If $A \setminus B = \emptyset$, then $A \subseteq B$" as "If A with the set B removed is empty, then A is a subset of B."

*"Understand" the problem. For me with this problem I imagine myself being given A and told to remove all elements of B from it. I find I'm left with nothing, and I remark to myself that each element of A must have been in B, so A must have been contained in B.

*Find a proof without symbols. "each element of A must have been in B"--why? Maybe the intuition in (2) is wrong and there is some element of A that's not in B. But then I wouldn't have removed it from A when I removed all elements of B, so I wouldn't have ended up with the empty set, a contradiction.

*Translate your proof into math symbols. I'll do it sentence by sentence:


*

*"there is some element of A that's not in B."
-> "Pick $x \in A$. Suppose $x \not \in B$".

*"But then I wouldn't have removed it from A when I removed all elements of B"
-> "But then it would be in A minus B" -> "$x \in A \setminus B$"

*"so I wouldn't have ended up with the empty set"
-> "$A \setminus B \neq \emptyset$".

*"a contradiction" -> "so $x \not \in B$ was false, so $x \in B$".

*"A must have been contained in B."
-> "so $A \subset B$".



In all, the proof is now: "Pick $x \in A$. Suppose $x \not \in B$. Then $x \in A \setminus B$. But then $A \setminus B \neq \varnothing$, so $x \not \in B$ was false, so $x \in B$. So $A \subset B$".
This is pretty close to formal logic already. I'm not sure what system you're using, but here's one possible translation to more formal logic:


*

*$A \setminus B = \emptyset$

*$x \in A \Rightarrow x \in B$:

*

*$x \in A$

*If $x \not \in B$:

*

*$x \in A \setminus B$ (definition of $A \setminus B$)

*$A \setminus B \neq \emptyset$ (definition of $\emptyset$)

*Contradiction between 1 and 2.2.2.


*$\therefore x \in B$


*$A \subset B$ (definition of $\subset$)

