Proof that $(A-B)\cup(B-C) = (A\cup B)\cap (A\cup C^c) \cap (B\cap C)^c$

I drew some Venn diagram's and constructed a couple of sets and this claim seems to hold (correct me if I am wrong). I have difficulties proving this however. I have tried to rewrite the lefthandside using the set identities, but no dice so far. Can anyone give me a hint?

Start by simplifying LHS: \begin{align} (A-B) \cup (B-C) &= (A \cap B') \cup (B \cap C') \\ &= ((A \cap B') \cup B) \cap((A \cap B') \cup C') \\ &= ((A\cup B) \cap (B' \cup B)) \cap((A \cup C') \cap (B' \cup C')) \\ &= (A\cup B) \cap (A \cup C') \cap (B\cap C)' \end{align}
• $(A - B) = (A \cap B')$
• Distributive property of $\cap$ and $\cup$
• Demorgon's Law: $(B' \cup C') = (B\cap C)'$