Why Condorcet meets LO2E2:

[Mike Ossipoff]

By assumption the voters in M have all ranked the alternatives in S1
over the alternatives in S2. Therefore all the S2 alternatives
are beaten with a majority against them.

The M voters, by definition, consist of a majority of the voters.
They have the power, therefore, to ensure that any alternative
won't have a majority ranking anything over it.

In particular, they can achieve that by not including S2
alternatives in their rankings. That's because then no non-S2
alternative can have a majority ranking any S2 alternative over it.
The only way that a non-S2 alternative can have a majority against
it is if there's a cycle among the non-S2 alterntives, in which
every one of them has another non-S2 alternative ranked over
it by a majority. Any would-be order-reverser wanting to
create a circular tie that an S2 alternative could win would
have to somehow be able to engineer such a cycle among the
non-S2 alternatives. This would require an impossible degree
of predictive knowledge and good luck.

Because would-be order-reversers couldn't count on there being
a cycle among the non-S2 alternatives in which everything 
has a majority against it, and since they couldn't count
on being able to create such a cycle among the non-S2 alternatives,
the would-be order-reversers could only expect that S2 alternatives
can't win if they have a majority against them, as they do,
since the M voters rank the S1 alternatives over them.

That's why Condorcet meets LO2E2 for all practical purposes.

***

Why Condorcet, with the "subcycle rule" strictly meets LO2E2:

The subcycle rule:

If there are 1 or more subcycles (by which I mean cycles
whose members don't beat everything outside the cycle),
then the first thing to be done is to solve the subcycles,
in the following way: If a subcycle (A) is an element of
another subcycle (B), then subcycle B shall be solved first,
by Condorcet's scoring rule, and the winner shall take 
subcycle A's place in subcycle B.

The only way that refusal by the M voters could fail to ensure
the defeat of the S2 alternatives is if there's a cycle
among the non-S2 alternatives. If that cycle isn't a subcycle,
that means its members all beat everything outside the
set, and so that cycle contains the Smith set. If so, then
Smith Condorcet will chose from within that cycle, and won't
choose anything in S2. But if that cycle is a subcycle, then
it will be solved and replaced by the winner of Condorcet
scoring within the cycle. When the cycle has been thus eliminated,
that remaining alternative from it will no longer be beaten
by anything outside of S2. Since it can't have anything in S2
ranked over it by a majority (since the M voters refused to
rank any S2 alternatives), it has to have a better Condorcet
score than anything in S2 has. Therefore when Smith//Condorcet
uses the subcycle rule, it strictly meets LO2E2.

***

I don't know whether plain Condorcet will also strictly meet
LO2E2 if the subcycle rule is used.

I believe it would if the subcycle rule is changed so that _any_
cycle is solved as the rule specifies. That's because, then, 
a cycle of alternatives outside S2 will be solved & replaced
by the winner within that cycle, even if that cycle is the
Smith set.

It would. That isn't a conjecture. Plain Condorcet strictly meets
LO2E2 if the right form of the subcycle rule is used. In fact
the subcycle rule version described in the paragraph before last
does the job when Smith//Condorcet is the method too.

So both plain Condorcet & Smith//Condorcet meet LO2E2 when that
form of the subcycle rule is used.

***

But I wouldn't include the subcycle rule in a public proposal,
because, as I said, Condorcet, even without the subcycle rule,
meets LO2E2 for all pracatical purpposes. And, of course, 
strictly meets LO2E1.

***

Mike
 


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