Disclaimer: No idea if I am even on the same planet on which this discussion is taking place. So pardon my questions and confusions:

On Sun, Oct 20, 2013 at 11:15 PM, Russell Standish <li...@hpcoders.com.au>wrote: > On Sun, Oct 20, 2013 at 06:22:15PM +0200, Bruno Marchal wrote: > > > > On 20 Oct 2013, at 12:01, Russell Standish wrote: > > > > >On Sun, Oct 20, 2013 at 08:52:41AM +0200, Bruno Marchal wrote: > > >> > > >>We have always that [o]p -> [o][o]p (like we have also always that > > >>[]p -> [][]p) > > >> > > > > > > > > >There may be things we can prove, but about which we are in fact > > >mistaken, ie > > >[]p & -p > > > > That is consistent. (Shit happens, we became unsound). > > > > Consistency is []p & ~[]~p. I was saying []p & ~p, ie mistaken belief. > > Why Isn't "mistaken belief" here merely unsound because it's propositional variable, assuming we're speaking generally about all systems? > > > > > > > > >Obviously, one cannot prove []p & p, for very many statements, ie > > > > > >[]p & p does not entail []([o]p) > Isn't that a rule for any modal sentence though, independent of system? [o]p is ([]p & p) Isn't []p & p = []([o]p) the definition of fixed point theorem? That, plus modalization conditionals to remove p from the G sentence so that every sentence H is a fixed point of it? So that you get that list of instances with (G sentence on left and H on right) examples like the following: []p corresponds to T ¬[]p corresponds to ¬[]⊥ []¬p corresponds to []⊥, which is pretty cool because you get a provability statement that is arithmetically equivalent to its own non-provability iff the statement is equivalent to the statement that arithmetic is inconsistent. Because G proves in this fashion: [o](p <=> [] ¬p) <=> [o](p<=> []⊥) This is the kind of thing that clarifies the pronoun issue imho. > > > > []p -> [][]p OK? > > > > Why? This is not obvious. It translates as being able to prove that > you can prove stuff when you can prove it. > > If this were a theorem of G, then it suggests G does not capture > the nature of proof. > > Oh, I see that you are just restating axiom "4". But how can you prove > that you've proven something? How does Boolos justify that? > But it nonetheless is everywhere in Boolos: []p -> [][]p IS a theorem of G, and useful, unless Bruno shoots the cowboy, because he cannot prove it or find his damned Boolos book. > > > > (and []p -> []p, and p -> p) + ([](p & p) <-> []p & []q) (derivable > > in G) > > > > Did you mean [](p&q) <-> []p & []q? I'm not sure of the q and whether you can just leave out the first bit. > That theorem at least sounds > plausable as being about proof. > > > > so []p & p -> [][]p & ([]p & p) > > -> []([]p & p) & ([]p & p), > > > > thus ([]p & p) -> [][o]p (& [o]p : thus [o]p -> [o][o]p) > > > > > > > > >Therefore, it cannot be that [o]p -> [o]([o]p) ??? > > > > > >Something must be wrong... > Why? > > > > > > > I hope I am not too short above, (and that there is not to much typo!) > > > > Bruno > > > > And thus you've proven that for everything you know, you can know that > you know it. Not sure, I'd guess we're comparing G's reasoning 3rd person "I" with reasoning of a fixed point corresponding to some sentence of G 1st person "I" in a modal/qualitative provability sense. PGC > This seems wrong, as the 4 colour theorem indicates. We > can prove the 4 colour theorem by means of a computer program, and it > may indeed be correct, so that we Theatetically know the 4 colour > theorem is true, but we cannot prove the proof is correct (at least at > this stage, proving program correctness is practically impossible). > > > -- > > > ---------------------------------------------------------------------------- > Prof Russell Standish Phone 0425 253119 (mobile) > Principal, High Performance Coders > Visiting Professor of Mathematics hpco...@hpcoders.com.au > University of New South Wales http://www.hpcoders.com.au > > ---------------------------------------------------------------------------- > > -- > 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.