Kristofer Munsterhjelm
20180403 21:16:58 UTC
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.

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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.

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