On Mon, Oct 21, 2013 at 4:15 PM, Bruno Marchal <marc...@ulb.ac.be> wrote:
> > On 21 Oct 2013, at 04:48, Platonist Guitar Cowboy wrote: > > > 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? > > > Yes, []p & —p, makes the machine on which [] applies, unsound. But still > consistent. Such machines will believe (prove) an arithmetical false (but > consistent) sentence. > > > > > >> >> > >> > > >> > >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? > > > This is a bit unclear. > A fixed point is when a proposition says something about herself (like p > <-> []p, p <-> [] ~p, etc.). The fixed point will be a proposition in which > p does no more occur. The main one are: > Ok, that is clearer. But this is a general rule in provability logic of every modal system, right? > > [](p <-> ~[]p) -> [](p <-> ~[]f) Gödel fixed point > > [](p <=> [] ¬p) <=> [](p<=> []⊥) > > Yes, that's the kind of thing I think we're talking about. > > > 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<=> []⊥) > > > OK. > > > > This is the kind of thing that clarifies the pronoun issue imho. > > > In arithmetic. But in UDA I think that the definition of first > person/third person in term of reconstitution/annihilation is clear enough > for the indeterminacy purpose, and the necessity of deriving physics from > arithmetic. > it is just not precise enough to get the actual technical beginning of the > derivation of physics. > > > > > >> > >> > []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. > > > What? You don't find your sacred manual? You will have to do some > penitences or something :) > > Let me give you a difficult exercise: derive []p -> [][]p in *any* normal > modal logic satisfying Löb's formula (that is derive []p -> [][]p from > []([]p -p) -> []p (and [](p->q)->([]p->[]q). > > Hmm, I'll try but feel free to ignore if there is too much bs in my attempt to remember from Boolos, for anything to be saved; "find the book, return to earlier chapters" will suffice). I choose GL from fuzzy memory as normal modal logic for the exercise: *GL derives []p -> [][]p in a confused cowboy's dream from ** []([]p -p) -> []p (and [](p->q)->([]p->[]q):* 1) Löb formula: []([]p ->p) -> *[]*p Apply truth reflexivity ([]p->p) to *bold* box above to isolate p as ([]p->p), so: []([]p->p) -> ([]p->p) 2) Assuming 1), GL can prove 1), so *[]*1): *[]*([]([]p->p) -> ([]p->p)) 3) Apply Löb formula again: []([]([]p->p) -> ([]p->p)) ->* []([]p ->p)* 4) Naturally or mistakenly because I don't know if these moves are legal through "[](p->q)->([]p->[]q)" [](([][]p->[]p) -> ([][]p->[]p)) -> []([]p ->p) and ([][][]p-> [][]p-> [][][]p-> [][]p) -> []([]p ->p) If that weirdness is ok then the following holds more simply because if GL can prove [][][]p etc. -> Löb formula, then its tempting to just: []p-> []([]p ->p) 5) Since []([]p ->p) is also ([][]p -> []p) 6) combining 4 & 5, GL: []p -> [][]p Q.E.D., but I don't believe my memory or the appropriacy to Bruno's exercise! Too clumsy, too many levels and vertigo, lol. So, yes. Gotta get back to forever undecided and find the Boolos book, I know. But it's fun as a kind of crossword puzzle activity, even when as confused as this partially forgotten and possibly misunderstood thing. Scratching my head for a whole hour for those couple of lines of Kool-Aid crazy... PGC > This shows that the axiom 4 is redundant in G. > > (This is known as Sambin theorem, and is proved in Smullyan's Forever > Undecided). I can give the solution, some day). > > > > > >> >> >> > (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. > > > Russell was right. I made a typo error, as predicted (!). > > > > > >> 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 > > > I don't see any fixed point here. > > We have both the formal (and third person) []p -> [][]p and the first > person knowledge formula [o]p -> [o][o]p, which is usually admitted for > "sufficiently introspective knower). > > Bruno > > > > > >> 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. > > > 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. 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