Let A, B and C be subsets of a universal set U.
Prove by contradiction that $$A\cap B \subseteq C \to (A-C)\cap B = \emptyset$$
Suppose otherwise, $A\cap B\subseteq C \land (A-C)\cap B \neq \emptyset$. Let $n\in A$, then by definition of the subset, $n\in C$. Since $n\in A \land n\in C, n\not \in (A-C)$, by definition of the set difference, this means $(A-C)= \emptyset$. Therefor, $n\not \in (A-C)\cap B$, by definition of the intersection. Thus, by definition of the empty set, $(A-C)\cap B = \emptyset$. This is a contradiction. Thus, $A\cap B \subseteq C \to (A-C)\cap B = \emptyset$ must be true.
Can someone tell me whether I did this right. I think it makes sense, but it also seems like I made a mistake somewhere because it seems too easy and short. Thank you.