Boolean Algebra Simplification with steps 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
 A: Use:
Adjacency
$xy+xy'=x$
and 
Idempotence
$x+x=x$
So:
$$a'c + b c'+a b\overset{Adjacency \ \times \ 3}{=}$$
$$a'bc + a'b'c+ ab c'+a'bc'+a bc+abc'\overset{Idempotence}{=}$$
$$a'bc + a'b'c+ ab c'+a'bc'+a bc\overset{Idempotence}{=}$$
$$a'bc + a'bc+ a'b'c+ ab c'+a'bc'+a bc\overset{Adjacency \ \times \ 2}{=}$$
$$a'bc + a'b+ a'b'c+ ab \overset{Adjacency}{=}$$
$$a'bc + b+ a'b'c \overset{Adjacency}{=}$$
$$b + a'c$$
A: 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,
$$a'c+bc'+a'c+bc'+a'b'c+ab=a'c+bc'+a'b'c'+ab=a'c+bc'+ab.$$
A: 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$2)}\\\end{align}$
