how can i simplify this logical circuit (A'C+'A'B'C) + 'B'C

Question

                        _____|A|______|/C|_____
             ____|                                                        |___
            |           |___|/A|__|/B|__|/C|_____|     |
    ___|                                                                           |____
            |                                                                            |
            |_______|/B|_______|/C|________|

Answer 1

It can be simplified as follows:

\begin{array}{l} & AC' + A'B'C + B'C' & \text{ Given }\\ & AC' + A'B' + B'C' & \text{ Consensus 1 }\\ & AC' + A'B' & \text{ Consensus 2 }\\ \end{array}

There's two different rules that I've seen called "Consensus".

Proof of "Consensus 1":

\begin{array}{l} & ABC + A'B & \text{ Given }\\ & (ABC + A')(ABC + B) & \text{ Distributive }\\ & (BC + A')(ABC + B) & \text{ Reduction }\\ & (BC + A')(AC + 1)B & \text{ Distributive }\\ & (BC + A')1B & \text{ Identity }\\ & BC + A'B & \text{ Distributive }\\ \end{array}

Proof of "Consensus 2":

\begin{array}{l} & AB + A'C + BC & \text{ Given }\\ & AB + A'C + BC1 & \text{ Identity }\\ & AB + A'C + BC(A + A') & \text{ Identity }\\ & AB + A'C + ABC + A'BC & \text{ Distributive }\\ & AB(1 + C) + A'C(1 + B) & \text{ Distributive }\\ & AB1 + A'C'1 & \text{ Identity }\\ & AB + A'C' & \text{ Identity }\\ \end{array}