Boolean Algebra Simplification with steps

Question

Need to simplify the algebra $$a'c+bc'+a'c+bc'+a'b'c+ab$$ I know the answer is $$b+a'c$$ but I can't figure it why I have been able to simplify the algebra to $$a'c + b c'+a b$$ but can't simplify it further

Answer 1

You know that Boolean operations $+$ and $\cdot$ are idempotent, right? Then $$a'c+bc'+a'c+bc'=a'c+bc'.$$ Also, $a'c+a'b'c=a'c$, which you can derive using distributivity, commutativity and the equations $x1=x$ and $x+1=1$, yielding $$a'c + a'b'c = a'1c + a'b'c = a'c(1+b') = a'c1 = a'c.$$ Then, $$a'c+bc'+a'c+bc'+a'b'c+ab=a'c+bc'+a'b'c'+ab=a'c+bc'+ab.$$

Answer 2

Use:

Adjacency

$xy+xy'=x$

and

Idempotence

$x+x=x$

So:

$$a'c + b c'+a b\overset{Adjacency \ \times \ 3}{=}$$

$$a'bc + a'b'c+ ab c'+a'bc'+a bc+abc'\overset{Idempotence}{=}$$

$$a'bc + a'b'c+ ab c'+a'bc'+a bc\overset{Idempotence}{=}$$

$$a'bc + a'bc+ a'b'c+ ab c'+a'bc'+a bc\overset{Adjacency \ \times \ 2}{=}$$

$$a'bc + a'b+ a'b'c+ ab \overset{Adjacency}{=}$$

$$a'bc + b+ a'b'c \overset{Adjacency}{=}$$

$$b + a'c$$

Answer 3

The obvious thing to do would be to distribute out the common term b with the aim of simplifying. But first we complicate things a bit, to get the terms looking similar (conjunctions of a, c, and their complements).

$\begin{align}&a'c+bc'+ab\\&a'c+(a+c')b&&\textsf{distribution}\\&a'c+(a(c+c')+(a+a')c')b&&\textsf{complementation}\\&a'c+(ac+ac'+ac'+a'c')b&&\textsf{distribution}\\&a'c+(ac+ac'+a'c')b&&\textsf{idempotence}\end{align}$

Now, how might we get that last a'c into those braces. Well $1+b=1$ by anihilation.

$\begin{align}&a'c(1)+(ac+ac'+a'c')b&&\textsf{conjunctive identity}\\&a'c(1+b)+(ac+ac'+a'c')b&&\textsf{anihilation}\\&a'c+(a'c+ac+ac'+a'c')b&&\textsf{distribution($\times$2)}\\\end{align}$