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.

Instead of contradiction you might want to prove the contrapositive:

if $(A\setminus B)\cap C\ne\emptyset$, then $C\not\subseteq B$

Suppose $x\in (A\setminus B)\cap C$. Then $x\in A\setminus B$ and $x\in C$. Therefore


*

*$x\in A$

*$x\notin B$

*$x\in C$
What do 2 and 3 tell you?
