[EM] is the Raynaud method Smith-efficient?

[James Green-Armytage]

	A brief definition of Raynaud: 
Eliminate the candidate with the strongest defeat against him or her,
until no defeated candidates remain. (Note that when a given candidate is
eliminated, his victories over other candidates (if he
has any) are removed from consideration.)
http://userfs.cec.wustl.edu/~rhl1/rbvote/desc.html

	A brief definition of Smith-efficiency: 
Dominant set: A set of candidates such that every candidate inside the set
pairwise-beats every candidate outside the set. 
Minimal dominant set (Smith set): A dominant set that doesn't contain
smaller
dominant sets.
A Smith-efficient method always selects a member of the minimal dominant
set.
http://lists.electorama.com/pipermail/election-methods-electorama.com/2005-February/014742.html

   ***

	It seems to me that Raynaud is Smith-efficient. I'm not good at
constructing proofs, but I think a proof of this would loosely revolve
around the following line of reasoning:


Assume that some candidates are in the Smith set, and other candidates are
not in the Smith set. (If all candidates are in the Smith set, then no
method can fail Smith-efficiency in that case.)
1. If Raynaud eliminates a candidate that is not in the Smith set, then
the Smith set will not change. Why not? All of the candidates in the Smith
set still defeat all of the remaining non-Smith set candidates.
2. If Raynaud eliminates a candidate that is in the Smith set, no
candidate who was not in the initial Smith set will enter the new Smith
set. Why not? All of the remaining candidates within the initial Smith set
still defeat all of the other candidates.
3. Because initially non-Smith set candidates cannot enter subsequently
revised Smith sets as a result of candidate eliminations, the winner must
come from the initial Smith set.

	Does this make sense? Does anyone with a more academic bent know how one
would convert this into a more rigorous proof? (I could use some practice
in this, since I plan to enter an economics PhD program in the fall.)
	I recognize that Raynaud fails monotonicity, but personally I don't
consider that to be a big deal. Raynaud is arguably the most intuitively
obvious pairwise tally method. I'm not willing to argue that Raynaud is
superior to defeat-dropping pairwise methods, but I do feel that it should
continue to be part of the conversation.

my best,
James Green-Armytage
http://fc.antioch.edu/~james_green-armytage/voting.htm