Find the Simplified Sum of Products of a Boolean expression

Question

Just having some problems with a simple simplification. I am doing a simplification for the majority decoder with 3 inputs A, B and C. Its output Y assumes 1 if 2 or all 3 inputs assume 1. Y assumes 0 otherwise. Select its correct switching function Y=f(A,B,C).

So, after doing out a truth table I found the Canonical Sum of Products comes to

NOT(A).B.C + A.NOT(B).C + A.B.NOT(C) + A.B.C

This, simplified, apparently comes to Y = A * B + B * C + A * C

What are the steps taken to simply an expression like this? How is it done? How was this value gotten in this case?

Solution 1

First, note that for a Boolean expression:

A= A + A

Now, see that

NOT(A).B.C + A.NOT(B).C + A.B.NOT(C) + A.B.C
= NOT(A).B.C + A.NOT(B).C + A.B.NOT(C) + A.B.C + A.B.C + A.B.C
= (NOT(A)+A).B.C + A.(NOT(B)+B).C + A.B.(NOT(C)+C)
= B.C + A.C + A.B

Solution 2

Incidentally WolframAlpha is great for doing (checking) Boolean maths in which case the format for your example is:

~A && B && C || A && ~B && C || A && B && ~C || A && B && C

Also your specific expression is actually on this page as an example, done differently to the other answer given.

Comments

[dmc]

Nice suggestion about using WolframAlpha! Obvious in retrospect, but I hadn't thought of that before.