Boolean Simplification using Algebra

Question

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.

Answer 1

$$\begin{array}{rcl}(A+A')=1 &\mbox{ and } &(B+B')=1 \\ (A+A')\cdot(B+B')=1& \implies &(A\cdot B + A\cdot B' + A'\cdot B + A'\cdot B') =1\\ A\cdot B + A'\cdot B' +B\cdot C &=& (A\cdot B + A\cdot B' + A'\cdot B + A'\cdot B') \cdot (A\cdot B + A'\cdot B' +B\cdot C)\\ &=&(A\cdot B +0+A\cdot B\cdot C)+(0+0+0) \\ &&+ (0+0+A'\cdot B\cdot C) + (0+A'\cdot B' + 0)\\ &=& A\cdot B + A'\cdot B\cdot C + A'\cdot B' \\ &=& A\cdot B + A'\cdot( B\cdot C + B') \\ &=& A\cdot B + A'\cdot( C+B') \\ &=& A\cdot B + A'\cdot C + A'\cdot B'\\ A\cdot B + A'\cdot B' +A'\cdot C &=& (A\cdot B + A\cdot B' + A'\cdot B + A'\cdot B') \cdot (A\cdot B + A'\cdot B' +A'\cdot C)\\ &=&(A\cdot B +0+0)+(0+0+0) \\ &&+ (0+0+A'\cdot B\cdot C) + (0+A'\cdot B' + A'\cdot B'\cdot C)\\ &=& A\cdot B + A'\cdot (B'+B)\cdot C + A'\cdot B'\\ &=& A\cdot B + A'\cdot C + A'\cdot B'\\ \end{array} $$ 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.

Answer 2

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.