Boolean Simplification using Algebra Having trouble showing the following relation: 
$$A\cdot B + A'\cdot B' + B\cdot C = A\cdot B + A'\cdot B' + A'\cdot C 
$$
using Boolean Algebra. Any help is appreciated. 
 A: $$\begin{array}{rcl}(A+A')=1 &\mbox{ and } &(B+B')=1 \\
(A+A')\cdot(B+B')=1& \implies &(A\cdot B + A\cdot B' + A'\cdot B + A'\cdot B') =1\\ 
A\cdot B + A'\cdot B' +B\cdot C &=& (A\cdot B + A\cdot B' + A'\cdot B + A'\cdot B') \cdot (A\cdot B + A'\cdot B' +B\cdot C)\\
&=&(A\cdot B +0+A\cdot B\cdot C)+(0+0+0) \\
&&+ (0+0+A'\cdot B\cdot C) + (0+A'\cdot B' + 0)\\
&=& A\cdot B + A'\cdot B\cdot C  + A'\cdot B' \\
&=& A\cdot B + A'\cdot( B\cdot C + B') \\
&=& A\cdot B + A'\cdot( C+B') \\
&=& A\cdot B  +  A'\cdot C + A'\cdot B'\\
A\cdot B + A'\cdot B' +A'\cdot C &=& (A\cdot B + A\cdot B' + A'\cdot B + A'\cdot B') \cdot (A\cdot B + A'\cdot B' +A'\cdot C)\\
&=&(A\cdot B +0+0)+(0+0+0) \\
&&+ (0+0+A'\cdot B\cdot C) + (0+A'\cdot B' + A'\cdot B'\cdot C)\\
&=& A\cdot B  + A'\cdot (B'+B)\cdot C + A'\cdot B'\\
&=& A\cdot B  +  A'\cdot C + A'\cdot B'\\
\end{array}
$$
And since they are both equal to $A\cdot B  +  A'\cdot C + A'\cdot B'$ they are equal to each other.
I have used, in the first half of this proof, 
$$
B\cdot C + B' = C+B'
$$
which is much easier to prove if you don't already have it as a lemma.
A: Okay, the thing with Boolean Algebra is, that sometimes it gets messier before it gets cleaner.
We start with:
$$A\cdot B + A'\cdot B' + B\cdot C$$
Note that the following is true in general:
$$A\cdot B = A\cdot B \cdot C + A \cdot B \cdot C'$$
So, with the starting expression:
$$A\cdot B + A'\cdot B' + B\cdot C = A\cdot B \cdot C + A \cdot B \cdot C' + A'\cdot B' \cdot C + A' \cdot B' \cdot C' + A\cdot B \cdot C + A' \cdot B \cdot C$$
If you do your reduction right from here, you should be able to prove the relation.
This is because:
$$A' \cdot B' \cdot C + A' \cdot B \cdot C = A' \cdot C$$
as per the inverse property.
