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The Majority Function is

1: A.B.¬C + A.¬B.C + ¬A.B.C + A.B.C

I can see intuitively that it can be simplified to

2: A.B + A.C + B.C

and thus A.(B + C) + B.C

but how can I use boolean algebra to get from one to the other?

This is where I get too so far

A.B.¬C + A.¬B.C + ¬A.B.C + A.B.C

= A.B.¬C + A.¬B.C + B.C.(¬A + A)

= A.B.¬C + A.¬B.C + B.C

= A(B.¬C + ¬B.C) + B.C

but I can't figure out how to rearrange the B.¬C + ¬B.C part to get the B + C that I need. Can anyone help?

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closed as too localized by Scott Morrison Dec 18 '09 at 19:58

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Closing as "too localized". I think for our target audience this question is uninteresting. –  Scott Morrison Dec 18 '09 at 19:59

1 Answer 1

up vote 7 down vote accepted

Your argument has the right idea. If you do the same thing that you did for the pair of $A$ and $\neg A$ for $B$ and $C$ too, you get $$\neg A . B . C + A . B . C = B . C$$ $$A . \neg B . C + A . B . C = A . C$$ $$A . B . \neg C + A . B . C = A . B$$ Sum the three equations and use that $X + X = X$ for all $X$ in your Boolean algebra. Voilá!

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Ah, I see then the three (A.B.C) terms simplify to one and I'm left with my #1 version. Thanks very much for your help! –  Martin Dec 18 '09 at 19:03

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