I probably have a less "academic bent" than you; but to me it is just *obvious* that as we only ever eliminate
pairwise losers it is impossible for all the members of the Smith set to be eliminated.
Therefore I'm happy to believe Rob Le Grand and Blake Cretney when they tell us on their web pages that
Raynaud meets Smith.
I gave some thought to Raynaud a few months ago:
"It seems to me that three versions of Raynaud are possible ( one that meets Symmetric Completion and two that don't). The obvious one that does is Raynaud (Margins). The other two could be called Raynaud (Pairwise Opposition), which eliminates the candidate which loses the pairwise comparison in which the winner has the highest gross score (explicit winning votes); and Raynaud (Gross Loser), which eliminates the candidate with the lowest gross score in any pairwise comparison.
In this example of Kevin Venzke's:
49: A 24: B 27: C>B
A>C 49-27 (m 22) C>B 27-24 (m 3) B>A 51-49 (m 2)
The three versions each give a different winner. PO eliminates A and elects C, failing Plurality. GL eliminates B and elects A, failing Minimal Defense. Margins eliminates C and elects B."
At the time, I preferred Raynaud (Gross Loser).
Chris Benham
---- Election-methods mailing list - see http://electorama.com/em for list info
