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: $$\begin{align} (A-B) \cup (B-C) &= (A \cap B') \cup (B \cap C') \\ &= ((A \cap B') \cup B) \cap((A \cap B') \cup C') \\ &= ((A\cup B) \cap (B' \cup B)) \cap((A \cup C') \cap (B' \cup C')) \\ &= (A\cup B) \cap (A \cup C') \cap (B\cap C)' \end{align}$$

Here we have used

  • $(A - B) = (A \cap B')$

  • Distributive property of $\cap$ and $\cup$

  • Demorgon's Law: $(B' \cup C') = (B\cap C)'$

  • $\begingroup$ Thanks! For the record you also used the complement's law and the identity law in the third step (sorry for nitpicking) $\endgroup$ – Raymond Timmermans Oct 22 '17 at 10:10
  • $\begingroup$ Oh I didn't know its name :) $\endgroup$ – samjoe Oct 22 '17 at 10:12

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