Show that $(A\setminus B) \cup (A\setminus C) = B \Leftrightarrow A=B \wedge (B \cap C) = \emptyset$ I believe there are 3 parts to this.
1) $(A\setminus B) \cup (A\setminus C) = B  \Rightarrow  A=B $
2) $(A\setminus B) \cup (A\setminus C) = B  \Rightarrow (B \cap C) = \emptyset$
3) $A=B \wedge (B \cap C) = \emptyset  \Rightarrow (A\setminus B) \cup (A\setminus C) = B$ 
I can do the the parts labelled 1 and 3 but cannot show part 2.  Anyone who can explain how you do the 2nd part ie show $(A\setminus B) \cup (A\setminus C) = B  \Rightarrow (B \cap C) = \emptyset$ ?

Hint: Let $(A-B)\cup (A-C)=B$ hold true and assume that there is $x \in B \cap C$. This means in particular that $x\in B$ and $x \in C$. Hence $$x \notin A-B$$ and for the same reason $$x\notin A-C$$ Therefore $$x\notin (A-B)\cup (A-C)$$ but on the other hand $x\in B$ which a contradiction to the assumption that $$(A-B)\cup (A-C)=B$$  
