Proving that $(A\setminus C)\cap(B\setminus C)\cap(A\setminus B)=\emptyset$ For each $A,B,C$ how would I prove that $(A\setminus C)\cap(B\setminus C)\cap(A\setminus B)=\emptyset$ ?
My thoughts are if $x\in (A\setminus C)\cap(B\setminus C)\cap(A\setminus B)$, then $x\in (A\setminus C)$ and $x\in (B\setminus C)$ and $x\in (A\setminus B)$ so $x \in A$ and $x \not \in C$ and $x \in B$ and $x \not \in C$  and $x \in A$  and $x \not\in B$.
How would I continue from here?

You have $x\in B$ and $x\not\in B$. Notice a problem here?
