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.
Boolean Simplification of A'B'C'+AB'C'+ABC'
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boolean-algebra
4 Answers
7
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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0For some reason I just got stuck thank you very much. – 2010-11-05
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0Don't forget to mark the answer as correct by clicking the ✓ next to the question. – 2010-11-05
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0That last comment was for @noname. – 2010-11-05
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2@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'