On Tue, Oct 22, 2013 at 3:52 PM, Bruno Marchal <marc...@ulb.ac.be> wrote:
> > On 22 Oct 2013, at 03:21, Platonist Guitar Cowboy wrote: > > > > > 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. >> > > OK. > > > >> >> >> >> >>> >>> > >>> > > >>> > >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? > > > You can consider formula like "p <-> [] (... p ...)" in all modal logic, > but few will have solution. > > This can be said: any K4 reasonner will have the fixed point in case he > visit Knave/ Knight island. And that K4 reasonner will becaome a G > reasonner (a Lôbian entity). > > Any K4 consistent machines "rich enough" (like PA, ZF, but unlike RA) will > become Löbian too, and got the G-like fixed point. The Gödel > diagonalization lemma will somehow emulate the Knight-Knaves island. > > > > > >> >> [](p <-> ~[]p) -> [](p <-> ~[]f) Gödel fixed point >> >> [](p <-> [] ¬p) <=> [](p<-> []⊥) >> >> > Yes, that's the kind of thing I think we're talking about. > > > OK. An important one is: > > [](p <-> <>p) <-> [](p <-> f) (all machines/sentences asserting their own > consistency can prove 0=1) > > And the one related to Löb's theorem: > > [](p <-> []p) <-> [](p <-> t) > > That is very amazing: all machines/sentences asserting their own > provability are true and provable. > That's a sort arithmetical "placebo". > > > > >> >> >> 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) > > > ? > > I will read the rest later, but in the meantime you might try to explain > what you mean by "so". > The formula []([]p->p) -> ([]p->p) is not a theorem (for any p), as you > can see by substituting p by false. > > []([]f -> f) -> ([]f -> f) would entail []<>t -> "correctness", but []<>t > -> []f "inconsistency! (by Gödel's second incompleteness). > > You cannot apply "[]p -> p" in G, as self-correctness is not provable in G > (or by the correct machine). > > OK? > I think :-) But something *has* to substitute p here like []p&p here for p...hmm. This would get rid of the inconsistency in G that you point towards, I guess. Isn't that more in line with unprovable self-correctness by the machine? Like this: GL proves []([]p->p)->[]p 1) Substitute p with ([]p&p) thus: (([][]p & []p) -> ([]p&p)) -> []([]p&p) or: ([]([]p &p) -> ([]p & p)) -> []([]p&p) 2) generalizing with [], keeping Löb axiom: []([]([]p & p) -> ([]p & p)) -> []([]p &p) which is still Löb: []([]p -> p ) -> []p 3) Truth-functionally reducing without mess: []p -> []([]p&p) 4) and through [](p->q)->([]p->[]q) []([]p&p) -> [][]p 5) Combining 3) and 4) this time: []p -> [][]p Another possible wrong road I can see is this linguistic attempt: if []p is true, then necessity of p is due to its "form", which is not empirical fact, so the truth of []p is necessary. So [][]p follows. But this cowboy is running out of roads as he still can't find the book or remember! Logicians on the net all use the different abbreviations and terms, which is a) not consistent as scientific group and b) not as self-evident as their use of them, because "abbreviation" itself should be a shorter word... and for a people who use these every day, that should be evident.. ;-) PGC > > > > > 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. > 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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