Question

Boolean Algebra: Reduce literals (Eqations)

Hello, can someone please provide me with how to solve this problem? I must reduce this expression:

a'b'c' + a'bc' + a'bc + ab'c'

the solution should be b'c' + a'b

Please explain the axioms used.

Answer 1

This site is seriously lacking community help in Physics and Computer Science. This would be my second post where I have answered my own question.

a'b'c' + a'bc' + a'bc + ab'c' Use Distributive to get:

a'c'( b' + b ) + a'bc + ab'c' Identity : ( b' + b ) = 1
Group like terms and use Distributive Prop.

a'( c' + cb ) + ab'c' Use ID (a + a'b = a + b) where a=c' and b=b

to get:
a'( c' + b ) + ab'c' Redistribute the terms and group c' and repeat the process to get:

c'a' + c'b' + a'b


I hope we get more community support in these tougher topics.

Answer 2

i agree