[EM] LIIA is very demanding

[Kristofer Munsterhjelm]

(This might be obvious, but I don't know if it is, so here we go...)

LIIA is hard to satisfy. In particular, LIIA + Majority + unrestricted 
domain implies ISDA. To show this, I'll go from Condorcet to Smith and 
then to ISDA.

Suppose some election method satisfies {LIIA, Majority, unrestricted 
domain}, but not Condorcet. Then there exists an election (set of 
ballots) where the social ordering ranks the CW below first. Repeatedly 
eliminate the winner until the CW ranks second; by LIIA, this shouldn't 
alter the ranking of the outcomes. Then repeatedly eliminate the loser 
of the election until only two candidates are left. By LIIA, this 
shouldn't affect the social ordering of the remaining candidates. Now 
the CW is ranked last and some other candidate (call him x) is ranked 
first. But since the CW is a Condorcet winner, he beats everybody 
pairwise, including x. So Majority forces x to be ranked first, which is 
a contradiction. Hence the CW can't be ranked anywhere but first in that 
election, and so Majority + LIIA implies Condorcet.

Now suppose the election method passes Condorcet but not Smith. That is 
to say that there exists an election without a CW, where the method 
ranks a non-Smith candidate x ahead of everybody in the Smith set in its 
social ordering. Then we can pull the same trick. Repeatedly eliminate 
the method's loser until only one of the (former) Smith set members 
remain. By LIIA, this won't alter the ranking of the remaining 
candidates, so this Smith set member (y) must be ranked below x. But now 
y is the CW among the remaining candidates. That means that y, the CW, 
is not ranked first, which is impossible.

Finally, since the method must satisfy the Smith criterion, it clearly 
must satisfy ISDA. Otherwise, there exists some election where removing 
all non-Smith members would alter which of the Smith members rank first 
in the social ordering. And since the method passes Smith, it must rank 
every candidate in the Smith set above the candidates not in the set. 
Thus, it's possible to remove all non-Smith candidates by repeatedly 
eliminating the loser according to the method. This is an operation that 
shouldn't upset the ranking of the remaining candidates (by LIIA), but 
if the method fails ISDA, it will.

(Some results in the other direction: one can have the three without 
IPDA (Ranked Pairs is an example) or clone independence (Kemeny is an 
example). Everything incompatible with ISDA is also incompatible with 
the three for obvious reasons. Without unrestricted domain, neither LIIA 
nor LIIA+Majority need imply the rest (e.g. Range for the former and MJ 
for the latter). )