Simplification of: AB + A'C + BC in boolean algebra

Question

I am trying to understand the simplification of the boolean expression:

AB + A'C + BC

I know it simplifies to

A'C + BC

And I understand why, but I cannot figure out how to perform the simplification through the expression using the boolean algebra identities. I was wondering if someone could show me the steps needed to do this. Thank you in advance.

Answer 1

The two expressions are not equal. The first expression is true when A and B is true and C false but the second is false in this case.

Answer 2

You can use a Karnaugh map (8 fields, no dividing lines this time).

See, for example:
https://nl.wikipedia.org/wiki/Karnaugh-diagram

In this diagram only one variable changes between two adjacent fields.

  B  B  B' B     'Fields
A   1   1  0  0     ABC'  ABC  AB'C AB'C'
A'  0   1  1  0     A'BC' A'BC A'B'C   A'B'C'
  C' C  C  C'

You can see that BC is not adding new 1's'.

So: AB + A'C + BC = AB + A'C

Answer 3

In this way , this can be simplified

                 LHS = AB+A'C+BC   
                 =  AB+A'C+BC (A+A')                  [ A+A'=1 ]
                  =  AB+A'C+ABC+A'BC
                  = AB+ABC+A'C+A'BC
                  =  AB (1+C)+A'C (1+B)
                  = AB+A'C                           [  1+C=1 ]
                                        =RHS..