Re: Request for comments [voting amendment]
On Thu, Nov 14, 2002 at 01:39:39PM -0500, Buddha Buck wrote:
> > Is it accurate to say that if there is no y such that x>>y (i.e., x
> > defeats nothing), then x is NOT in the set?
On Thu, Nov 14, 2002 at 10:11:04PM +0100, Matthias Urlichs wrote:
> Yes.
Be careful here.
It should be true for the rules I proposed, because we'd require that
options beat the default option before being considered for the Schwartz
set.
However, it's not true for the general case. Imagine you have two
options and they're tied. Then, both options would be in the schwartz
set, and yet no option beats any other option.
--
Raul
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