$\text{Let } A, B, \text{and } X \text{ be sets. Suppose }A \subset X. \text{If }X \setminus B \subset X \setminus A, \text{then }A \subset B.$ I am trying to prove the following: $$\text{Let } A, B, \text{and } X \text{ be sets. Suppose }A \subset X. \text{If }X \setminus B \subset X \setminus A, \text{then }A \subset B.$$
I am currently working with the very definitions of the operations used in this claim, specifically those involving set difference and subsets. I am struggling to figure out what to do after assuming the antecedent; where do I go after assuming that $X \setminus B \subset X \setminus A$?
Any help is appreciated!
 A: You want to show that $A$ is a subset of $B$, that is: every element $x$ that belongs to $A$ also belongs to $B$.
Start with $x \in A$ (then $x \in X$ because $A \subset X$). You want to show that $x\in B$. If this wasn't true, what would happen? More precisely, if you assume that $x \not \in B$, which means $x \in X \setminus B$, then what could you say?

Recall that you have the hypothesis $X \setminus B \subset X \setminus A$. Since $x \in X \setminus B$, you can say that $x \in X \setminus A$... which contradicts the fact that $x \in A$. Therefore, the assumption "$x \not \in B$" can't be true, and you've just showed that $x \in B$ as desired!

A: The argument is abit wierd but here it goes
assuming 
$$ X \setminus B \subset X \setminus A$$
so $$x\in X \wedge x \not \in B \Rightarrow  x\in X \wedge x\not \in A $$
contrapositive is $$x \not \in X \vee x\in A \Rightarrow x \not \in X \vee x\in B $$
now assume that $x\in A$ so $x\not \in X $ or $x\in B$. Since $A\subset X$ $x\in A $ and $x\in X$. It must be that $x\in B$
