On Fri, Feb 13, 2009 at 07:31:29PM +0100, Bruno Marchal wrote:
> >
> > I'm a little confused. Did you mean Dp here? Dp = -B-p
>
>
> Fair question, given my sometimes poor random typo!
>
...
> deduce Bp) , well, if you remind the definition of the Kripke
> semantics, you can see that
>
> Bp & Dp
>
> is equivalent with
>
> Bp & Dt
>
...
> Now if you have in a world, your world if you want, Bp & Dp, you have
> at least access to a world in which p is true, and thus you have
> access to a world where t is true, given that t is true in all worlds.
> So you have Bp & Dt.
Thanks. Alles ist Klar. I think I wasn't taking seriously enough the
idea of Kripke frames before...
...
>
> A good and important exercise is to understand that with the Kripke
> semantics, ~Dt, that is B~t, that is Bf, that is "I prove 0=1", is
> automatically true in all cul-de-sac world. It is important because
> cul-de-sac worlds exists everywhere in the Kripke semantics of the
> self-reference logic G.
>
> If you interpret, if only for the fun, the worlds as state of life,
> then Bf is really "I am dead".
>
> Bruno
Yes, but I have difficulty in _simultaneously_ interpreting logic
formulae in terms of Kripke frames and B as provability. In the
former, Bp means in all successor worlds, p is true, whereas in the
latter it means I can prove that p is true.
How does one reconcile such disparate notions?
Cheers
--
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Prof Russell Standish Phone 0425 253119 (mobile)
Mathematics
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Received on Sat Feb 14 2009 - 21:41:26 PST