Prove: $(A-B)\cup(B-C)=(A\cup B)-(B\cap C)$ The first part is not that of a problem, but the second part:
$$(A\cup B)-(B\cap C)\to (A-B)\cup(B-C)$$
is giving me a hard time.
Trying to solve $(A-B)\cup(B-C)=(A\cup B)-(B\cap C)$ using set operations is equally confusing, as I reach this part:
$$(A\cup (B-C))-(B\cap C)$$
But don't know how to go from there to $(A\cup B)-(B\cap C)$.
Your help is appreciated.
Thank you
 A: We have
$$
(A-B)\cup(B-C)=(A\cup B)-(B\cap C)
\iff (A \setminus B) \cup (B \setminus C) = (A\cup B)\setminus (B\cap C)
$$
We also have
\begin{align*}
(A\cup B)\setminus (B\cap C) &= (A\cup B)\cap (\overline{B} \cup \overline{C})\\
&= (A \cap \overline{B}) \cup (A \cap \overline{C}) \cup (B \cap \overline{C})\\
&= (A \setminus B) \cup (A \setminus C) \cup (B \setminus C)
\end{align*}
We have to prove that $(A \cap \overline{C}) \subset (A \cap \overline{B}) \cup (B \cap \overline{C})$. Let $x \in A \cap \overline{C}$. If $x \in B$, we have $x\in B \cap \overline{C}$. If $x \notin B$, we have $x\in \overline{B}$, therefore $x\in (A \cap \overline{B})$.
Thus
$$
(A-B)\cup(B-C)=(A\cup B)-(B\cap C).
$$
A: Hint: use $X-Y=X\cap \overline{Y}$, for any sets $X,Y$ and Morgan rules.
A: For fun, you can show identities like this by considering one point in each combination of the sets is in either both sides or in neither. (Basically you are considering whether each region of the generic Venn diagram is contained in either both or neither sides of the stated set equality. Tedious, but always effective. So in this case, check that the left and right sides contain exactly the same ones of the eight points shown below. (There will be $2^k$ points to check if your identity involves $k$ sets.

