Jobst Heitzig heitzig-j at web.de
Thu Mar 1 13:58:48 PST 2007

Dear Forest,

you wrote:
> So practically speaking, elimination of Pareto dominated alternatives
> is extremely unlikely to have any effect, although UncAAO technically
> fails the criterion.

I believe so, too...

> Note that if approval is measured in relation to a virtual "approval
> cutoff candidate," or if the approval order is automatically refined
> by the rank order on each ballot to enforce a "sincere approval"
> requirement, then the alternative Y could tie Y' but could not beat
> it. If all such ties were broken by random ballot, then Y' would beat
> Y, and that would make UncAAO Independent from Pareto Dominated
> Alternatives.

I'm not sure I grasped that but will give it a try tomorrow .-)

> But if the approval winner A is not a member of Smith, then it is
> possible (however unlikely) that the first member of the sequence A,
> f(A), f(f(A)), ... that resides in Smith is not the highest approval
> member of Smith, and that this sequence would lead to a different
> UncAAO winner if the highest approval Smith member were taken as the
> initial point.

So we could consider taking the most approved Smith set member as the 
starting point, couldn't we? That would still be monotonic, right?

> Is TACC monotone?  It seems to me that the winner W could improve in
> approval enough to overtake and surpass some W' in approval without
> defeating W' pairwise, though W' covers W.

Recalling my proof from March 3, 2005, I believe it is:
> Let us assume that the actual winner X is raised on some individual
> ballots by moving either the approval cutoff or another candidate from
> directly above X to directly below X. Then what can happen is twofold:
> First, the order in step 1 either does not change or does only change
> in that X gets moved up one position. Second, the defeats do not
> change or do only change in that X now defeats some candidate Y which
> she was defeated by before. In either case, X still must win: It was
> the last candidate added to the chain. The new chain developes exactly
> as before: As the order did not change left of X, the chain evolves
> just as before until the original position of X. If X did not change
> position, it still defeats all candidates in the chain and so gets
> added. If X did change position with Y, then Y was beaten by X or some
> other candidate already in the chain, since it was not added to the
> chain originally. If it is added now, it must hence be beaten by X, so
> that X still gets added after Y was added. In either case, when X was
> considered, the resulting chain is the old one except that perhaps Y
> is added. As no later candidate beat was added originally and X beats
> everything it beat before, also now no further candidate can be added,
> so that X is again the winner. QED.    

Still seems correct to me...

I wonder whether the two methods are more closely related than we think 
since UncAAO also constructs a chain of options in which all defeats 
are transitive. Perhaps the winner is not only uncovered but even in 
the Banks set, just like the TACC winner?

Yours, Jobst
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