The solid coalition cloneproof resistant set extension breaks the property that a voter who prefers some candidate outside the set to the winner has nothing to gain by burying the winner. Suppose that B is the current winner and there's a coalition {A1, A2, A3} that is almost a clone set. Suppose that if {A1, A2, A3} were replaced with a single candidate so that each voter would replace his highest ranked set member with that single candidate, then that candidate (call him A) would disqualify B. But because a few voters vote something like ... > A1 > B > A2 > A3 > ..., A1 does not obtain >1/k of the first preferences in every sub-election containing A1 and B (but not A2 or A3). Suppose furthermore that the method electing from the extended set elects B. But then if these almost-voters bury B so that instead we have ... > A1 > A2 > A3 > B > ... then the solid coalition and the "little more elegant" relation could both establish {A1, A2, A3} ~> B, thus kicking B off the set. The problem is that it's too weak, i.e. it doesn't consistently handle near-clone sets. It is possible to be cloneproof and elect from the resistant set (e.g. IRV). But if IRV elects from the extended resistant set induced by the solid coalition/more elegant relations, then it has some additional property that makes it never elect B in the example above. I don't know if cloneproof unburiability of the disqualification relation is incompatible with Condorcet, but I don't see a reason why it would be. Except that passing both LNHs is. The first-preference one is too strong. As a result, it is cyclical. The example from the other post: 34: A>B>C 33: B>C>A 33: C>A>B Now (A,B) ~> C since 34 + 33 > 50, {B, C} ~> A since 33 + 33 > 50 and {C, A} ~> B since 33 + 34 > 50. Now we could imagine altering the threshold to get around this problem. But for the cycle to be broken, then in the three-candidate case, it needs to be greater than v * floor(2/3) = 66, because that's the weakest link in this cycle. But if it's 2/3, then: 51: A>B 49: B>A A ~> B, but clone A into A1 and A2: 26: A1>A2>B 25: A2>A1>B 49: B>A the sum of A1 and A2's first preferences is less than 2/3 and so B wouldn't be disqualified. There's a gap that's just too large for first preferences on its own to work: it can't simultaneously be acyclical in the Condorcet cycle example and work in the 26/25/49 example. -km