Why SSD is a Cycle Condorcet method
SSD is a Cycle Condorcet method if it can be shown that
any defeat among members of the current Schwartz set
(I'll just call it the Schwartz set) is in a cycle.
That's the same as saying that no member of the Schwartz set
has an unreturned beatpath to him, meaning that if A has a
beatpath to B, then B must have one to A.
Saying that if B is a member of the Schwartz set then B
doesn't have an unreturned beatpath to him is the same as
saying that if B has an unreturned beat path to him then
B isn't in the Schwartz set. That's what I'll demonstrate:
The premise, then, is that A has a beatpath to B, but B doesn't
have a beatpath to A. It's desired to show that B isn't in
the Schwartz set.
Let S1 be the set of candidates including A and the candidates
who have a beatpath to A. Let S2 be the set of all the other
candidates in the election.
By the definition of a beatpath, if a candidate beats A or someone
who has a beatpath to A, then that candidate has a beatpath to A.
That means that if a candidate _doesn't_ have a beatpath to A,
then he doesn't beat A or anyone who has a beatpath to A.
That means that no S2 member beats any S1 member. That means
that S1 is an unbeaten set. S2 is not an unbeaten set, since
A beats B.
In fact, any set containing B but not containing A is not an
unbeaten set, because A beats B.
Now, say some set contains both A & B, and is an unbeaten set.
A subset of that set that leaves out B and any other S2 members
is also an unbeaten set, since neither B nor any S2 member beats
A or any S2 member. Since that unbeaten set containing both
A & B therefore contains a smaller unbeaten set, then the set
containing A & B can't be a "small unbeaten set", and therefore
can't be part of the Schwartz set, by the Schwartz set's definition.
So B can't be in the Schwartz set. That's what we set out to
Since a candidate with an unreturned beatpath can't be in the
Schwartz set then any candidate in the Schwartz set must not
have an unreturned beatpath to him. That means that any defeat
in the Schwartz set must be in a cycle. That's what I originally
said I'd prove.
That means that SSD never drops a candidate unless that candidate
is the weakest defeat in a cycle (since SSD only drops the
weakest defeat in the Schwartz set). So SSD is a Cycle Condorcet
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