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 $(A\setminus B) \cup (A\setminus C) = B  \Rightarrow (B \cap C) = \emptyset$ ?
 A: You have that $(A/B) \cup (A/C)=B$. Then clearly, $B \subseteq A/C$ since it couldn't be in $A/B$ because you just removed all of $B$. So then if $B \subseteq A/C$, that is intuitively, saying, "none of $B$ is in $C$." i.e. $B \cap C = \emptyset$
A: 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 $$(A-B)\cup (A-C)=B$$  
A: 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.
A: 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)
