[EM] Condorcet and Reversal Symmetry are incompatible with DMTBR

[Kristofer Munsterhjelm]

Here's a proof that we can't have all of Condorcet, reversal symmetry, 
and dominant mutual third burial resistance.

It's a proof by exhaustion, and I see what Wikipedia means when it says 
"A proof with a large number of cases leaves an impression that the 
theorem is only true by coincidence, and not because of some underlying 
principle or connection. "

I handle ties by saying that "X wins with certainty" if there is no tie 
and "X wins with positive probability" if the probability of X being 
winning after whatever tiebreaker is greater than zero.

-

DMTBR is incompatible with Reversal Symmetry and Condorcet:

Start with (A):

4: A>B>C
4: B>A>C
1: C>A>B

A is the CW and by Condorcet, wins with certainty.
Bury A on the 4 BACs to get (B):

4: A>B>C
4: B>C>A
1: C>A>B

If B wins with positive probability, we're done: (A)->(B) is a DMTBR 
failure. Otherwise, either C wins with certainty or A wins with positive 
probability. See the end for the case where C wins with certainty.

Suppose A wins with positive probability on (B). Reverse the ballots to 
get (D):

4: A>C>B
1: B>A>C
4: C>B>A

By reversal symmetry, at least one of B and C must win with positive
probability on (D).
Suppose C wins with positive probability. Then flipping 4 CBA to CAB 
gives (E):

4: A>C>B
1: B>A>C
4: C>A>B

where A is the CW, and so (E)->(D) is a DMTBR failure.

On the other hand, if B wins, then flipping 1 BAC to BCA gives us (F)

4: A>C>B
1: B>C>A
4: C>B>A

where C is the CW, so (F)->(D) is a DMTBR failure.

==================================
The case where C wins in (B):

Suppose that C wins with certainty. Consider the reversed ballots (D) again:

4: A>C>B
1: B>A>C
4: C>B>A

By reversal symmetry and our assumption that C won on (B), C can't win.
First suppose A wins with positive probability.
Relabel the candidates according to ACB -> abc. We get:

4: b>c>a
4: a>b>c
1: c>a>b

By relabeling and assumption that A won, a must win with positive 
probability.
But this is just (B), where c must win with certainty by our initial 
assumption, so we have a contradiction.

Hence B must win with certainty, and (F)->(D) gives us a DMTBR failure 
as above.