Reducing boolean algebra

40 Views Asked by At

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?

2

There are 2 best solutions below

0
On

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*}

0
On

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.