I told why CC is incompatible with FBC. Why the continuing debate?

I told why Condorcet's Criterion is incompatible with FBC. In case that 
posting wasn't noticed, I'm re-posting that demonstation here. I'm not 
saying that it wasn't already demonstrated by Kevin. But I've posted my 
demonstration too, and here is is again:

Say there are two voters whose preferences are B>A>everyone else.

Say that, with the configuration of all the other voters' votes (which may 
or may not be sincere), and sincere ballots by those two voters, there is a 
circular tie.

Since the method is anonymous and neutral, then, whatever method is used for 
solving circular ties, it's possible that someone other than A or B could 
win. Someone who is, for those two voters, worse than A. So let's say that 
that is so.

Say that, with that votes configuration for the other voters, and with 
sincere voting by those two voters, A beats everyone but B.

Say that results in a cycle in which the completion method in use, the 
circular tiebreaker, chooses someone other than A and B, someone whom those 
two voters like less than A.

But B doesn't beat A by very much. He only beats him by 3 votes.

The two voters, by changing their voting,  could change that pairwise 
result. Let's say that that is the only pairwise result that is close enough 
that they could influence it.

If those two voters ranked A and B equal 1st place, B would only beat A by 
one vote.

Then, if they reversed their preferences between A and B, and voted 
A>B>everyone else, then they make A into BeatsAll winner. Because the method 
is a Condorcet Criterion method, A wins.

They've improved the results for themselves by voting someone over their 
favorite.

The method therefore fails FBC. And I didn't say exactly what the method is. 
All that was assumed was that the method is one that meets the Condorcet 
Criterion.

Therefore the Condorcet Criterion is incompatible with FBC.

Mike Ossipoff

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