Simplification of: A’(BC + AB + BA’) in boolean algebra

Question

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

A’(BC + AB + BA’)

I was wondering if someone could show me the steps needed to do this. Thank you in advance

Answer 1

First distribute $A'$: $$A'BC+A'AB+A'BA'$$ By idempotence and annihilation this is equal to $$A'BC+A'B$$ Then by absorption on $A'B$ we get the final result as $$A'B$$