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

Question

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.

Answer 1

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?).

Answer 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

Answer 3

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