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

Question

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.

score 8 · Accepted Answer · 2010-11-04T23:41:36.120

8

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?).

edited Nov 04 '10 at 23:41

answered Nov 04 '10 at 23:27

For some reason I just got stuck thank you very much. – Nov 05 '10 at 01:05
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. – Niel de Beaudrap Nov 05 '10 at 10:35

score 2 · Answer 2 · answered Oct 11 '12 at 14:29

2

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

answered Oct 11 '12 at 14:29

Mika

151

score 0 · Answer 3 · edited Apr 05 '18 at 17:25

0

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'

edited Apr 05 '18 at 17:25

amWhy

209,954

answered Apr 05 '18 at 14:51

unknown

1

score 0 · Answer 4 · answered Oct 23 '19 at 15:51

0

Check for similarities between terms.
Apply IDEMPOTENCY rule for letting you group 1 term with more.
Check to see if it is optimized or it can be more. Here you see C' which is a common factor.

answered Oct 23 '19 at 15:51

Dragos Makovei

25

score 0 · Answer 5 · answered Nov 04 '10 at 23:18

0

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

answered Nov 04 '10 at 23:18

Ross Millikan

374,822

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

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