So I got this equation: (NOT A + B) x ( A + C)
When I try to reduce this I get (Not A AND C) OR (A AND B) OR (B AND C) But wolframalpha.com(here) says it ends up being A'C + AB. I also tried to input my original reduction and got this (here) whitch gave me the same result. How does (B AND C) just get reduced to nothing?
This is the Consensus Theorem:
\begin{align*} A'C + AB + BC &= A'C + AB + (1)BC \\ &= A'C + AB + (A + A')BC \\ &= A'C + AB + (ABC + A'BC) \\ &= (A'C + A'BC) + (AB + ABC) \\ &= A'C(1 + B) + AB(1 + C) \\ &= A'C(1) + AB(1) \\ &= A'C + AB \\ \end{align*}
Note by logical definitions: if either $B$ or $C$ hold, then either ($A'$ and $C$) or ($A$ and $B$) must hold.
If you want to go purely with formal addition/multiplication, we can note that $$ A'C + AB + BC= \\ A'C + AB + (A + A')BC= \\ A'C + AB + ABC + A'BC =\\ A'(C + BC) + A(B + BC) =\\ A'C(1 + B) + AB(1 + C) =\\ A'C + AB $$ If you're interested in computing such reductions, look into Karnaugh maps.
