Prove that for sets $A,B,C$, if $C \subseteq B$, then $(A\setminus B)\cap C = \varnothing$. I just need the proof of this. How does one prove that given $A, B, C$, if $C\subseteq B$, then $(A\setminus B)\cap C$ is equal to an empty set.

Assume that $(A \setminus B) \cap C \neq \varnothing$. Let $x \in (A \setminus B) \cap C$. Then $x\in (A \setminus B)$, and hence, not in B. However, $x\in C$ which is a subset of $ B$. So $x\in B$. That's a contradiction.
