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?


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

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 5
    $\begingroup$ Closing as "too localized". I think for our target audience this question is uninteresting. $\endgroup$ – Scott Morrison Dec 18 '09 at 19:59

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!

  • $\begingroup$ 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! $\endgroup$ – Martin Dec 18 '09 at 19:03

Not the answer you're looking for? Browse other questions tagged or ask your own question.