# [EM] Problem solved (for pure ranked ballot

Abd ul-Rahman Lomax abd at lomaxdesign.com
Thu Jan 25 11:25:16 PST 2007

At 10:40 AM 1/25/2007, Kevin Venzke wrote:
I wrote:

>1 A>B>C
>1 B>C>A
>1 A>B>C (from C>A>B)
>
> > Indeed,
> > >this creates a clone. The election now has two voters for one
> > >candidate, and one for the remaining candidate. As long as the
> > >method would elect a favorite of a 2/3 majority, it violates ICC.

This is a classic error in proof. (Note that I was saying that the
procedure was defective.)

The election pattern shown above displays a clone. ICC says that the
insertion of a clone will not change the election outcome.

Okay, what is the insertion of a clone and what is it inserted into?
B and C are clones. Let's eliminate C.

So we now have
A>B
B>A
A>B

Inserting the clone C does not change the election outcome, which
remains either inside or outside the clone set, whatever it was
before the insertion. In this case A wins in both cases.

The illusion that C changes the outcome is created by comparing the
outcome of the "set with a clone in it" with that of the "set without
Favorite Betrayal," not that of the "set with no clone."

The discussion here bristles with red herrings. I remain convinced,
now, the proof is defective, it has a clear logical error in it.