Proving $A\cap (B - C) = (A\cap B) - (A\cap C)$

[Dustinsfl]

$A\cap (B - C) = (A\cap B) - (A\cap C)$For the identity, we will show $A\cap (B - C) \subseteq (A\cap B) - (A\cap C)$ and $A\cap (B - C) \supseteq (A\cap B) - (A\cap C)$.
Let $x\in A\cap (B - C)$.
Then $x\in A$ and $x\in B - C$.
So $x\in A$ and $x\in B$ and $x\notin B\cap C$.

Is this the right approach? I know $B-C = $ some other expression but I can't remember it.

 

[Amer]

B-C = B \cap C^c
C^c it is the complement of C, About you work it is right since x in A and in B so it is in the intersection and x it is not in C so it is not in the intersection of A and C

now the other direction

 

[Plato2]

$A\cap (B - C) = (A\cap B) - (A\cap C)$

$\begin{align*}(A\cap B)-(A\cap C)&= (A\cap B)\cap(A^c\cup C^c)\\&=(A\cap B\cap A^c)\cup (A\cap B\cap C^c) \\&=\emptyset \cup (A\cap B\cap C^c)\\&= (A\cap B\cap C^c)\\&=(A\cap B)-C\end{align*}$