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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.

4 Answers 4

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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.2010-11-05
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    Don't forget to mark the answer as correct by clicking the ✓ next to the question.2010-11-05
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    That last comment was for @noname.2010-11-05
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    @Hamster: the last one can be decomposed as `A'B'+A ≡ (A'+A)(B'+A) ≡ B'+A`, using the distributivity of disjunction over conjunction.2010-11-05
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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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First use the distributive law to pull out the C', then work on the As and Bs

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A'B'C'+AB'C'+ABC'

= B'C'+ABC' according to absorption law [ A'B'C'+AB'C' = B'C' ]

= AC'+B'C' according to absorption law [ B'C'+ABC' = AC'+B'C' ]

= (A+B')C'