Simplify Boolean Expression A'B'C + A'BC + AB'C

Question

The K-map method of simplifying the Boolean Expression A'B'C + A'BC + AB'C gives the answer to be A'C+B'C. But I am not able to solve this algebraically. Please help me out. What I have tried is

A'B'C + A'BC + AB'C
= A'C (B'+B) + AB'C
= A'C(1) + AB'C
= A'C + AB'C

This is the most simplified I get using Boolean algebra but this is not matching the result. Can someone please help me out.

Answer 1

With the use of A+A=A we have

$$ A'B'C + A'BC + AB'C =\\ = \underbrace{\color{blue}{A'B'C} + A'BC} +\underbrace{\color{blue}{A'B'C}+ AB'C}=\\ =A'C (B'+B) + (A+A')B'C $$

Answer 2

Since $C$ is everywhere, we can just forget about it. With what you have already done, you obtained $A' + AB'$. You want to prove that $A' + AB' = A' + B'$. Of course, this is OK from the truth table. I do not know what rules you used to define the $+$ operator, but maybe the following computation will be OK for you:

A'+B'
= A' + 1 B'
= A' + (A + A') B'
= A' (1 + B') + A B'
= A' + A B'