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.
 A: $(A\setminus B)\cap C = (A\cap \bar B)\cap C = A \cap (\bar B\cap C)$.
But $C\subseteq B$ and so $C\cap \bar B=\emptyset$, which implies that $A \cap (\bar B\cap C) = A\cap\emptyset = \emptyset$.
A: 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.
A: 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?
A: $C\subseteq B \implies A\setminus B\subseteq A\setminus C$,
$\therefore (A\setminus C) \cap C=\emptyset\implies (A\setminus B) \cap C=\emptyset$.
A: This can be proved using : 

  
*
  
*domination law :  $X\cap\emptyset = \emptyset $
  
*associativity of $\cap$ : $ (X \cap Y)\cap Z  = X \cap (Y\cap Z)$ 
  
*commutativity of $\cap$ : $(X \cap Y) = (Y\cap X) $

$(1)$ Supposse that $C\subseteq B $
$C\subseteq B $
$ \implies \forall (x) \space ( x\in C \rightarrow x \in B)  $
$ \implies \forall (x) \neg \space(x\in C \land x \notin B)  $
$ \implies \forall (x) \neg \space(x\in C \land x \in \overline{B}) $
$ \implies \forall (x) \neg x\in \space C\cap \overline{B} \space  $ ( by definition of $\cap$)
$ \implies \neg \space \exists x\in \space C\cap \overline{B}  $
$ \implies C\cap \overline{B} = \emptyset   $
$(2)$ Now , still under our hypothesis , 
$(A\setminus B)\cap C $
$= (A\cap \overline{B})\cap C$ (by definition of \ operation )
$ = A\cap (\overline{B}\cap C)\space \space$ (by $\cap \space $associativity )
$ = A\cap (C \cap \overline{B})\space \space$ (by $\cap \space $commutativity)
$=  A \cap \emptyset\space  $ ( in virtue of our hypothesis, substituting $\emptyset$ for $C\cap\overline {B}$)
$ = \emptyset\space $ (by domination law) 
$(3)$ Since we have derived $(A\setminus B)\cap C = \emptyset $ under the hypothesis $C\subseteq B $ , the Conditional Proof Rule allows us to conclude that : 
if $C\subseteq B $ then $(A\setminus B)\cap C = \emptyset$. 
