[EM] Condorcet and Participation

[Markus Schulze]

Dear participants,

this is Moulin's proof that participation and Condorcet
are incompatible.

Situation 1:

   3 ADBC
   3 ADCB
   4 BCAD
   5 DBCA

Situation 2:

   Suppose candidate B is elected with positive probability
   in situation 1. When we add 6 BDAC voters then candidate B
   must be elected with positive probability according to
   participation and candidate D must be elected with
   certainty according to Condorcet.

Situation 3:

   Suppose candidate C is elected with positive probability
   in situation 1. When we add 8 CBAD voters then candidate C
   must be elected with positive probability according to
   participation and candidate B must be elected with
   certainty according to Condorcet.

Situation 4:

   Suppose candidate D is elected with positive probability
   in situation 1. When we add 4 DABC voters then candidate D
   must be elected with positive probability according to
   participation and candidate A must be elected with
   certainty according to Condorcet.

Situation 5:

   Because of the considerations in Situation 2-4 we get
   to the conclusion that candidate A must be elected with
   certainty in situation 1. When we add 4 CABD voters then
   candidate B and candidate D must be elected each with
   zero probability according to participation.

Situation 6:

   Suppose candidate A is elected with positive probability
   in situation 5. When we add 6 ACBD voters then candidate A
   must be elected with positive probability according to
   participation and candidate C must be elected with
   certainty according to Condorcet.

Situation 7:

   Suppose candidate C is elected with positive probability
   in situation 5. When we add 4 CBAD voters then candidate C
   must be elected with positive probability according to
   participation and candidate B must be elected with
   certainty according to Condorcet.

Markus Schulze