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How would you simplify the following boolean expression $(!A B)+(B !C)+(BC)+(A !B !C)$?

I factorised B and managed to get $B(!A+!C+C)+(A !B !C) = B+(A !B !C)$, but I do not know how to continue.

Using a K-map, I managed to get the result of $B+A!C$ and I am trying to achieve the same result using regular identities and laws of boolean algebra. By the way, sorry for poor formatting, but I do not know how I could paste an expression from word to make it look better and easier to read.

YuiTo Cheng
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1 Answers1

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Actually, there is a non-intuitive (doesn't hold in ordinary algebra) Boolean algebra law that can be applied here: distributivity of disjunction over conjunction expressed as follows $$x+yz=(x+y)(x+z).$$

Using your notation $x=B$, $y=!B$, $z=A!C$. Hence we have $$B+!BA!C=(B+!B)(B+A!C)=1\cdot(B+A!C)=B+A!C.$$

Theta
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