A⋂C ⊆ B and a∈C⇒a∉A\B Prove it A⋂C ⊆ B and a∈C⇒a∉A\B . We can see with Venn diagram more easily. But if I want to prove it , How Can I prove it. 
I tried but I dont know how to start.  
 A: It is (I think) easier to prove the contraposition.
Let $a\in A\setminus B$.  Then because $A\cap C\subseteq B$ we have
$$a\in A\setminus (A\cap C)$$ 
And then clearly
$$a\not\in C$$
A: Notice that if $a \in A \setminus B$, it would mean that $a \in A$ but $a \notin B$. Now, along with $a \in C$, this implies $a \in A \cap C \subseteq B$ which would be a contradiction, since an element cannot satisfy both $a \in B$ and $a \notin B$. Thus, the above implication must be true.
A: As @ThomasLesgourgues & @AniruddhaDeshmukh were shown, it can be proved by contradiction.
But direct proof for @MustafaErol request:
We have two cases< First:
$$a \notin A \overset{(A-B) \subseteq A}{===\Longrightarrow} a \notin A-B$$
And Second:
$$a \in A \overset{a \in C}{=\Longrightarrow} a \in (A \cap C) \overset{(A \cap C) \subseteq B}{===\Longrightarrow} a \in B \overset{(A-B) \cap B = \phi}{====\Longrightarrow} a \notin (A-B)$$
So in any case you have done.
A: I will give a proof that is probably not the most efficient, but good for understanding.
Firstly, suppose $A\cap B\neq \emptyset$, otherwise it is trivial. A good way to visualize what is happening is to use the fact $$C=C\cap (B\cup B^c)=(C\cap B)\cup (C\cap B^c).$$
It reads $a$ is in $C\cap B$ or in $C\cap B^c$.


*

*If $a\in C\cap B$, then $(C\cap B)\cap(A\cap B^c)=C\cap A\cap (B\cap B^c)=\emptyset$. It means $a\notin A\cap B^c$. Indeed, $a$ cannot be in $B$ and in $B^c$ at the same time.

*If $a\in C\cap B^c$, then $(C\cap B^c)\cap(A\cap B^c)=C\cap A\cap B^c\subseteq B\cap B^c=\emptyset$. 


We conclude that regardless the case,$a\notin (A\cap B)$.
I hope it gives you conditions to visualize the problem in Venn's Diagram now.
