Simplification of: AB’ + BC + C’A in boolean algebra

Question

I am trying to understand the simplification of the boolean expression:

AB’ + BC + C’A

I know it simplifies to

A + BC

And I understand why, but I cannot figure out how to perform the simplification through the expression using the boolean algebra identities. I was wondering if someone could show me the steps needed to do this. Thank you in advance.

@Henry. It was a question to check if the questioner understood or just wanted a copy of the answer. Anyway, thank you for the answer and +1 from me. — zkutch, Oct 08 '20 at 01:07

score 2 · Accepted Answer · answered Oct 08 '20 at 01:01

2

You can use absorption to say $B'=B'+B'C$ so
$$B'+BC+C' = B'+B'C+BC+C'=B'+(B'+B)C+C'= B'+C+C'=B'+1=1$$ and so $$AB'+ABC+AC' = A(B'+BC+C')=A$$

You can use absorption again to say $BC=BC+ABC$ so $$AB' +BC + AC' = AB' +BC + ABC+ AC' = A(B'+BC+C')+BC=A+BC$$

answered Oct 08 '20 at 01:01

Henry

157,058

can you use a distribution law ? – dobdobeh alias Oct 08 '20 at 01:08
@dobdobehalias I did when I said $B'C+BC=(B'+B)C$ and $AB'+ABC+AC'=A(B'+BC+C')$ – Henry Oct 08 '20 at 01:13
thanks for your answer – dobdobeh alias Oct 08 '20 at 01:23

Simplification of: AB’ + BC + C’A in boolean algebra

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