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Im trying to prove $(a'+b')((b'+c)+b'c) = b'+a'c$ and I am stuck on the second half. I know I need to get it to $(a'+b')(b'+c)$ to use the Distributive Property of OR over AND, but I cant seem to workout how to transform $((b'+c)+b'c)$ to $(b'+c)$. Any ideas?

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$b'+c+b'c=b'(1+c)+c=b'+c$, because $1+c=1$.

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  • $\begingroup$ Had to apply associative first to make (b'+c) + b'c = (b' + b'c) + c $\endgroup$ – pstatix Dec 13 '17 at 0:10
  • $\begingroup$ @pstatix Yes, and also the commutative property and again the distributive property. $\endgroup$ – egreg Dec 13 '17 at 7:02

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