I thought <>P meant P was possible? If so wouldn't P imply <>P? Or have I
misremembered what <>P means?

On 7 March 2015 at 21:08, Bruno Marchal <[email protected]> wrote:

>
> On 07 Mar 2015, at 02:51, meekerdb wrote:
>
>  On 3/6/2015 7:24 AM, Bruno Marchal wrote:
>
> That might depend on the context. Usually, in our computationalist context
> it means true in the standard model of arithmetic, which is "this reality"
> if you want.
>
>  In the modal context, it means true in this world (which in our
> arithmetical context is NOT necessarily among the accessible world, because
> we don't have []p -> p). With the logic of provability, we cannot access
> the world we are in. p does not imply <>p
>
>
> I wonder about such definitions of modal operators.  WHY doesn't p imply
> <>p?  We *could* define <> so that it did.  Is there some good reason not
> to?
>
>
>
> The modal logic are imposed by the fact that he box (and thus the diamond)
> are the one describing the self-reference, by Solovay theorem. The box is
> Gödel's beweisbar. It is an arithmetical predicate. We really assume only
> Robinson (and Peano) arithmetic. We don't have p -> <>p, because this would
> mean in particular t -> <>t, and if that was a theorem of G, then <>t would
> be provable, contradicting Gödel's incompleteness.
>
> All modal logics are extracted from arithmetic. They are shortcuts
> provided by Solovay's completeness theorem of G and G*, and the Theaetetus'
> variants.
>
> Bruno
>
>
>
>
>
> Brent
>
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