Proving statement - $(A \setminus B) \cup (A \setminus C) = B\Leftrightarrow A=B , C\cap B=\varnothing$ I`m trying to prove this claim and I need some advice how to continue,
$$(A \setminus B) \cup (A \setminus C) = B \Leftrightarrow A=B , C\cap B=\varnothing$$
what I did is:
$$(A \setminus B) \cup (A \setminus C) = (A \cap B') \cup (A \cap C') = A \cup (B' \cap C')$$
thanks!
 A: Hint for $\implies$ direction:
When using the distributive law, note the equivalence between $(A\cap B') \cup (A\cap C') \iff A\cap (B' \cup C')$:
$$(A/B) \cup (A/C) = \color{blue}{(A \cap B')\cup(A\cap C')} = \color{blue}{\bf A \cap(B'\cup C')}$$
Note that by DeMorgan's $$A \cap(B'\cup C') = A\cap (B \cap C)'$$
Now recall that the premise is $$(A\setminus B) \cup (A\setminus C) = B$$
And now we're at 
$$\begin{align} (A\setminus B) \cup (A\setminus C) & = B \\ \\ 
A \cap(B'\cup C') & = B \\ \\
A\cap (B \cap C)' & = B  \\ \\
A\setminus (B\cap C) & = B
\end{align}$$
Now, what can you conclude about the relationship between $A$ and $B$, and about the intersection $B\cap C\;?$
A: One direction is trivial (if $A=B$ and $C\cap B=\emptyset$ is given). For the other direction note
$$B=(A\backslash C)\cup (A\backslash B)\subseteq A$$
and $$A\backslash B\subseteq B $$ which together imply $A=B$. Thus your equation simplifies to
$$A\backslash C=A$$
and therefore $A\cap C=\emptyset.$
A: I usually start this type of set proofs by translating the set-level expressions to logic-level expressions using set extensionality and the definitions of the set operators, and then complete the proof on the logic level.  That way I don't have to remember all the set-theoretic rules, and can just use ordinary logic.
Let's do that for both sides, and see where that leads us.  For the left hand side,
\begin{align}
& (A \setminus B) \cup (A \setminus C) = B \\
\equiv & \;\;\;\;\;\text{"extensionality"} \\
& \langle \forall x :: x \in (A \setminus B) \cup (A \setminus C) \;\equiv\; x \in B \rangle \\
\equiv & \;\;\;\;\;\text{"expand $\;\cup\;$ and then $\;\setminus\;$ twice using their definitions"} \\
& \langle \forall x :: (x \in A \land x \not\in B) \lor (x \in A \land x \not\in C) \;\equiv\; x \in B \rangle \\
\equiv & \;\;\;\;\;\text{"in the left hand side: factor out $\;x \in A\;$; apply DeMorgan"} \\
(0) \;\; \phantom\equiv & \langle \forall x :: x \in A \land \lnot(x \in B \land x \in C) \;\equiv\; x \in B \rangle \\
\end{align}
For the right hand side we similarly get the following
$$
(1) \;\; \langle \forall x :: (x \in A \equiv x\in B)
\land \lnot(x \in B \land x\in C) \rangle
$$
Seeing that the shapes of $(0)$ and $(1)$ have a lot in common, we can complete the proof by proving (for any boolean expressions $\;P\;$, $\;Q\;$, and $\;R\;$)
$$
(2) \;\; P \land \lnot(Q \land R) \:\:\equiv\;\; Q \:\:\equiv\:\: (P \equiv Q) \land \lnot(Q \land R)
$$
Since $\;\equiv\;$ is associative and symmetric, and the first and last of the three parts look similar, we can bring these parts together by rearranging this to
$$
\phantom{(2)} \;\; P \land \lnot(Q \land R) \:\:\equiv\:\: (P \equiv Q) \land \lnot(Q \land R) \:\:\equiv\:\: Q
$$
Now we calculate
\begin{align}
& P \land \lnot(Q \land R) \:\:\equiv\:\: (P \equiv Q) \land \lnot(Q \land R) \\
\equiv & \;\;\;\;\;\text{"factor out common conjunct"} \\
& \lnot(Q \land R) \;\Rightarrow\; (P \;\equiv\; P \equiv Q) \\
\equiv & \;\;\;\;\;\text{"simplify right hand side"} \\
& \lnot(Q \land R) \;\Rightarrow\; Q \\
\equiv & \;\;\;\;\;\text{"use negation of right hand side $\;Q\;$ in left hand side"} \\
& \lnot(\textrm{false} \land R) \;\Rightarrow\; Q \\
\equiv & \;\;\;\;\;\text{"simplify"} \\
& Q \\
\end{align}
This proves $(2)$ and completes the proof.
