On Mon, Sep 23, 2013 at 7:49 PM, Bruno Marchal <marc...@ulb.ac.be> wrote: > > On 23 Sep 2013, at 12:41, Telmo Menezes wrote: > > On Sat, Sep 21, 2013 at 9:43 PM, Bruno Marchal <marc...@ulb.ac.be> wrote: > > > On 21 Sep 2013, at 15:10, Telmo Menezes wrote: > > > On Fri, Sep 20, 2013 at 3:58 PM, Bruno Marchal <marc...@ulb.ac.be> wrote: > > > > On 19 Sep 2013, at 16:51, Telmo Menezes wrote: > > > On Thu, Sep 19, 2013 at 4:31 PM, Bruno Marchal <marc...@ulb.ac.be> > > > > If so, can't we say ~D~t and thus []t? > > > > Yes, []t is a theorem, of G and most modal logic, but not of Z! > > > > > > Isn't the only situation where ~Dt the one where this is no world? > > > > ~Dt, that is [] f, inconsistency, is the type of the error, dream, lie, and > > "near-death", or in-a-cul-de-sac. > > > Thus your interest in near-death experiences? > > > > Yes. And in all "extreme" altered state of consciousness. Those extreme > cases provide key information.

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Can this information be recovered? For example, is a NDE that did not result in death, was it really a cul-de-sac? > > > > We should *try* to avoid it, but we can't avoid it without loosing our > > universality. > > > The consistent machines face the dilemma between security and lack of > > freedom-universality. With <>p = ~[] ~p, here are equivalent way to write > > it: > > > <>t -> ~[]<>t > > <>t -> <> [] f > > []<>t -> [] f > > > I don't understand how you arrive at this equivalence. > > > I use only the fact that (p -> q) is equivalent with (~q -> ~p) (the > contraposition rule, which is valid in classical propositional logic), and > the definition of <> p = ~[] ~p. I use also that ~~p is equivalent with p. > > Note that []p = ~~[]~~p = ~<> ~p. And, > > ~[]p = <> ~p > and > ~<>p = [] ~p > > Like with the quantifier, a not (~) jumping above a modal sign makes it into > a diamond, if it was a bo, and a box, if it was a diamond. > > > Starting from <>t -> ~[]<>t. But where does <>t -> ~[]<>t come from? > Contraposition gives ~~[]<>t -> ~<>t, and this > gives by above, []<>t -> []~t, which gives > []<>t -> []f (as ~t = f, and ~f = t). > > OK? Ok! > For the third one, starting from the first one again: <>t -> ~[]<>t, By > contraposition []<>t -> ~<>t , but ~<>t = []~t = [] f. > > OK? Ok! Thanks Bruno. My only problem now is the above. Telmo. > Bruno > > > > http://iridia.ulb.ac.be/~marchal/ > > > > -- > You received this message because you are subscribed to the Google Groups > "Everything List" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to everything-list+unsubscr...@googlegroups.com. > To post to this group, send email to everything-list@googlegroups.com. > Visit this group at http://groups.google.com/group/everything-list. > For more options, visit https://groups.google.com/groups/opt_out. -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/groups/opt_out.