Given a three-candidate Condorcet cycle, it's always possible to eliminate a candidate who didn't win so that the winner changes. Thus the Condorcet loser criterion is incompatible with independence of irrelevant alternatives.

Is this correct?

Consider two systems. System A is straight Score and System B is to eliminate the Condorcet loser (beaten-by-all-pairwise candidate) if there is one repeatedly and then apply Score to the rest.

If there are only three candidates in the election and they form a Condorcet cycle, there is no Condorcet loser so systems A and B produce the same outcome. How does this show a dependence on irrelevant alternatives? Isn't plain Score famously compliant with IAA?

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In an election with only two candidates, the Condorcet loser criterion implies the majority criterion.Given a three-candidate Condorcet cycle, it's always possible to eliminate a candidate who didn't win so that the winner changes. Thus the Condorcet loser criterion is incompatible with independence of irrelevant alternatives.

Say you've got an A>B>C>A cycle. Without loss of generality, A wins in some method that passes the Condorcet loser criterion.

Now remove B. C beats A head-to-head so must now win (A is the Condorcet loser with only A and C). Given that removing B has changed the winner from A to C, the method must fail IIA.

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