preemptive vs cooperative: attempt at formalization

[Taral]

On 4/11/06, oleg at pobox.com <oleg at pobox.com> wrote:
>
> > [Rule 1]
> > * in a cooperative implementation of threading, any thread with value
> >   _|_ may cause the whole program to have value _|_. In a preemtive one,
> >   this is not true.
>
> I'm afraid that claim may need qualifications:

I was thinking that it might be more useful to express it programatically:

if preemptive then fork _|_ >> return () => ()

--
Taral <taralx at gmail.com>
"You can't prove anything."
    -- Gödel's Incompetence Theorem