I've just realized that I didn't show, today, that Condorcet's Criterion is incompatible with Participation & Consistency. For Participation it shouldn't be so difficult, but for Consistency it could be more work. For Participation, take an example in which there's no BeatsAll winner. Whatever the circular tiebreaker is that chooses the winner, X, take some loser, Y, and modify the rankings so that he only has one defeat, by a small margin, of, say, 5. Z is the name of the candidate who defeats him. Change the example so that X's defeat(s) are by candidates other than Y & Z, but X still wins the circular tiebreaker. Then add 6 voters who vote X over Y, and Y over Z, but who don't rank X over his defeater(s). I don't call that a proof, but only a suggested outline for writing a demonstration that CC is incompatible with Participation. I merely post it to show why it looks as if such a demonstration could be written. Mike Ossipoff _________________________________________________________________ Send and receive Hotmail on your mobile device: http://mobile.msn.com
