Prove that if $A$ and $B \backslash C$ are disjoint, then $A \cap B \subset C$ I am new to mathematical proofs, I wonder if my evaluation is correct.
I have to prove that 
1 - $A \cap B \backslash C = \emptyset \to A \cap B \subset C$ 
i - Break the implication, and add the antecedent to the hypothesis 
2 - Hyp: $A \cap B \backslash C = \emptyset$ 
3 - Goal: $A \cap B \subset C$ 
ii - Rewrite the goal in line 3 the following way: $ \forall x (x \in A \cap B \to x \in C)$ 
iii - Now, make x arbitrary: $x \in A \cap B \to x \in C$ 
iv - Since there's another conditional, break it and bring the antecedent as a hypothesis.
4 - Hyp: $x \in A \cap B$ - That means $x \in A \land x \in B$ 
5 - Goal: $x \in C$
vi - From hypothesis in line 2 we have the equivalent: $\forall x(x \in A \to x \notin B \backslash C)$
vii - From hyp in line 4: $x \in A$, then by Modus Ponens we conclude:
6 - $x \notin B \backslash C$ - That means $\neg (x \in B \land x \notin C) $
vii - $\neg (x \in B \land x \notin C) $ $\equiv$ $x \notin B \lor x \in C $ 
vii - From hypothesis in line 4, we have that  $x \in B$. Therefore, for line vii be true:
viii - $x \in C$ which is exacly the goal in line 6.
 A: I haven't seen any issues in your process. Since you're just starting out, I think that it's good that you are being very detailed about making sure you understand what everything means.
In this case, I think the easiest way to think about it is this: break up $B$ into the component which overlaps with $C$, and the component which overlaps with $C$'s complement. This gives:
$$A \cap B = [A \cap (B \cap C)] \cup [A \cap (B \setminus C)].$$
Since you know $A \cap (B \setminus C) = \emptyset $, you then have
$$A \cap B = A \cap B \cap C,$$
which means that $A \cap B$ is a subset of $C$.
A: Here is another way to approach it.
You can apply the following result: if $X\subseteq U$ and $Y\subseteq U$, then $X\cap Y = \varnothing$ iff $X\subset Y^{c}$.
In the present case, one has that:
\begin{align*}
A\cap(B\backslash C) &  = A\cap(B\cap C^{c}) = (A\cap B)\cap C^{c} = \varnothing \Rightarrow A\cap B\subseteq (C^{c})^{c} = C
\end{align*}
and we are done.
Hopefully this helps!
A: If $A$ and $B\setminus C$ are disjoint then $$A\cap(B\cap\lnot C)=\emptyset\implies(A\cap B)\cap\lnot C=\emptyset$$
Let $x\in A\cap B$. Then, since $(A\cap B)\cap\lnot C=\emptyset$, $x\not\in\lnot C\implies x\in C$
Since this is true for all $x\in(A\cap B)$, $A\cap B\subset C$
