Re: Maybe Schulze is decisive.

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• Dear Norman, the aim of this e-mail is to prove, that Schulze wins cannot be cyclic. It is sufficient to prove that if Candidate A defeats candidate B via
Message 1 of 2 , Aug 31, 1998
Dear Norman,

the aim of this e-mail is to prove, that Schulze wins
cannot be cyclic. It is sufficient to prove that if
Candidate A defeats candidate B via beat-paths and
candidate B defeats candidate C via beat-paths, then
candidate A defeats candidate C via beat-paths.

**********

First: Suppose, that candidate A defeats candidate B
via beat-paths ab:ba. Suppose, that candidate B defeats
candidate C via beat-paths bc:cb.

( 1) ab > ba
( 2) bc > cb

Furthermore, I suppose, that the strength of the
strongest beat-path from candidate A to candidate C
is ac and that the strength of the strongest beat-path
from candidate C to candidate A is ca.

*****

Second: I get the following equations:

( 6) min{ab;bc} <= ac
( 7) min{ac;cb} <= ab
( 8) min{ba;ac} <= bc
( 9) min{bc;ca} <= ba
(10) min{ca;ab} <= cb
(11) min{cb;ba} <= ca

Example: If min{ab;bc} > ac, then I could find a
stronger beat-path from candidate A to candidate C
by going from candidate A to candidate B and then
from candidate B to candidate C. This would be a
the strength of the strongest beat-path from
candidate A to candidate C.

*****

Third: Case X: (12a) ab >= bc.

With (12a) ab >= bc and (2) bc > cb I get:

(13a) ab > cb.

When I insert (13a) into (10), I get:

(14a) ca <= cb.

When I insert (12a) into (6), I get:

(15a) bc <= ac.

With (14a), (2), and (15a) I get:

(16a) ca <= cb < bc <= ac.

Thus: ca < ac. Candidate A defeats candidate C
via beat-paths.

*****

Fourth: Case Y: (12b) bc > ab.

With (12b) bc > ab and (1) ab > ba I get:

(13b) bc > ba.

When I insert (13b) into (9), I get:

(14b) ca <= ba.

When I insert (12b) into (6), I get:

(15b) ab <= ac.

With (14b), (1), and (15b) I get:

(16b) ca <= ba < ab <= ac.

Thus: ca < ac. Candidate A defeats candidate C
via beat-paths.

**********

I have demonstrated, that if candidate A defeats candidate B
via beat-paths and candidate B defeats candidate C via
beat-paths, then candidate A defeats candidate C via
beat-paths. Thus: Schulze wins cannot be cyclic.

Markus Schulze
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