# Boolean Simplification of A'B'C'+AB'C'+ABC'

My question is how do I reduce $\bar A\bar B\bar C+A\bar B\bar C+AB\bar C$ To get $(A+\bar B)\bar C$. I'm so lost just been trying to get it for awhile only using the 10 boolean simplification rules.

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## 3 Answers

A'B'C'+AB'C'+ABC'
=  B'C'+ABC'        by absorption [ A'B'C'+AB'C'  =  B'C' ]
=  AC'+B'C'         by absorption [ B'C'+ABC'  =  AC'+B'C' ]
=  (A+B')C'


Used tool at http://www.logicminimizer.com

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A'B'C'+AB'C'+ABC'
C'(A'B'+AB'+AB)
C'(A'B'+A(B'+B))
C'(A'B'+A)
C'(B'+A)


It's that last step that used to trip me up. A'+AB = A'+B Forget what that law is called (identity?).

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For some reason I just got stuck thank you very much. – noname Nov 5 '10 at 1:05
Don't forget to mark the answer as correct by clicking the ✓ next to the question. – Alexsander Akers Nov 5 '10 at 2:33
That last comment was for @noname. – LarsH Nov 5 '10 at 3:35
@Hamster: the last one can be decomposed as A'B'+A ≡ (A'+A)(B'+A) ≡ B'+A, using the distributivity of disjunction over conjunction. – Niel de Beaudrap Nov 5 '10 at 10:35

First use the distributive law to pull out the C', then work on the As and Bs

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