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

An systematic solution:
In the table below, every column represents some subexpression and holds truth values meaning "belongs to this set". The rows exhaust all $2^3$ combinations.
The $7^{th}$ column is the LHS of the equivalence and the $11^{th}$ is the RHS. As you can see, they are identical.
A B C A/B A/C  +  =B  A=B B.C =0  and
0 0 0  0   0   0   1   1   0   1   1
1 0 0  1   1   1   0   0   0   1   0
0 1 0  0   0   0   0   0   0   1   0
1 1 0  0   1   1   1   1   0   1   1
0 0 1  0   0   0   1   1   0   1   1
1 0 1  1   0   1   0   0   0   1   0
0 1 1  0   0   0   0   0   1   0   0
1 1 1  0   0   0   0   1   1   0   0

(+ for $\cup$, . for $\cap$)
Grouping all rows in a single hexadecimal number:
A  B  C  A/B A/C  +  =B  A=B B.C =0  and
55 33 0F 44  50  54  98  99  03  FC  98

(/ is and not, + is or, . is and, = is not xor, bitwise)
