How to prove $A\setminus(B\setminus C) = (A \setminus B) \cup (A \cap C)$ I have to prove that. While I know this is true by thinking about it I'm having a lot of trouble actually writing the proof
how can I prove $A\setminus(B\setminus C) = (A \setminus B) \cup (A \cap C)$ 
first of all it's true?
 A: It is IMV always handsome to write $A\setminus B$ as $A\cap B^\complement$ in cases like this:
$$A\setminus\left(B\setminus C\right)=A\cap\left(B\cap C^{\complement}\right)^{\complement}=A\cap\left(B^{\complement}\cup\left(C^{\complement}\right)^{\complement}\right)=$$$$A\cap\left(B^{\complement}\cup C\right)=\left(A\cap B^{\complement}\right)\cup\left(A\cap C\right)=\left(A\setminus B\right)\cup\left(A\cap C\right)$$
The fourth equality is based on distributivity.
A: You need to show 


*

*$A \setminus (B \setminus C) \subseteq (A \setminus B) \cup (A \cap C)$

*$(A \setminus B) \cup (A \cap C) \subseteq A \setminus (B \setminus C)$
$(\rightarrow)$ Take $x \in A \setminus (B \setminus C)$ . Then $x \in A$ and $x \notin B \setminus C$, which means that 
either 


*

*$x \notin B$, implying that $x \in A \setminus B$
or 


*$x \in B$ and $x \in C$ , implying that $x \in A \cap C$
Hence $x \in (A \setminus B) \cup (A \cap C)$. Since $x$ was arbitrary element, we can conclude $A \setminus (B \setminus C) \subseteq (A \setminus B) \cup (A \cap C)$
$(\leftarrow)$
Take $x \in (A \setminus B) \cup (A \cap C)$. Then $x \in A$ and either $x \notin B$ or $x \in C$, so in both cases $x \notin B \setminus C$.
And since $x \in A$ and $x \notin B \setminus C$, we have $x \in A \setminus (B \setminus C)$. $x$ was arbitrary, hence $(A \setminus B) \cup (A \cap C) \subseteq A \setminus (B \setminus C)$ 
$\Box$
A: Sets $S,\,T$ are equal if $x\in S$ is equivalent to $x\in T$. In this case,$$\begin{align}x\in A\setminus(B\setminus C)&\iff x\in A\land x\notin B\setminus C\\&\iff x\in A\land x\notin B\lor x\in C\\&\iff (x\in A\land x\notin B)\lor(x\in A\land x\in C)\\&\iff x\in A\setminus B\lor x\in A\cap C\\&\iff x\in (A\setminus B)\cup(A\cap C).\end{align}$$
