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 $(A\setminus B) \cup (A\setminus C) = B  \Rightarrow (B \cap C) = \emptyset$ ?

Since you already showed (1), we can use it to ease the proof of (2):
$$(A\setminus B) \cup (A\setminus C) \stackrel{(1)}= (B\setminus B) \cup (B\setminus C) = B \setminus C \stackrel{\text{req.}}= B$$
Now $B\setminus C = B \cap C^C$ by definition, so we have $B \cap C^C = B \Rightarrow B \subset C^C \Rightarrow B \cap C = \emptyset$ as claimed.
