Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

3 Answers 3

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

share|improve this answer
up vote 7 down vote accepted
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?).

share|improve this answer
    
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
1  
@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
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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.