[Math] Boolean Simplification of A’B’C’+AB’C’+ABC’

Question

My question is how do I reduce $\bar A\bar B\bar C+A\bar B\bar C+AB\bar C$ To get $(A+\bar B)\bar C$. 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?).