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Re: Request for comments [voting amendment]


Raul Miller:
> > > 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?
> > Yes.
> 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.
You're right. For simplification-of-rules-to-keep-track-of-in-my-head
reasons, I had "x==y is equivalent to x>>y AND y>>x" in the back of my
head, which, while useful, is not the generally-understood definition.

Matthias Urlichs     |     noris network AG     |     http://smurf.noris.de/

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