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 thisdiscussion 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 alwaysthat> >>[]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 mistakenbelief.Why Isn't "mistaken belief" here merely unsound because it'spropositional variable, assuming we're speaking generally about allsystems?Yes, []p & —p, makes the machine on which [] applies, unsound. Butstill consistent. Such machines will believe (prove) an arithmeticalfalse (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 ofsystem? [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 Gsentence 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 aproposition in which p does no more occur. The main one are:Ok, that is clearer. But this is a general rule in provability logicof 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 andH on right) examples like the following:[]p corresponds to T ¬[]p corresponds to ¬[]⊥[]¬p corresponds to []⊥, which is pretty cool because you get aprovability statement that is arithmetically equivalent to its ownnon-provability iff the statement is equivalent to the statementthat 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 firstperson/third person in term of reconstitution/annihilation is clearenough for the indeterminacy purpose, and the necessity of derivingphysics from arithmetic.it is just not precise enough to get the actual technical beginningof 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 youprovethat you've proven something? How does Boolos justify that?But it nonetheless is everywhere in Boolos: []p -> [][]p IS atheorem of G, and useful, unless Bruno shoots the cowboy, becausehe cannot prove it or find his damned Boolos book.What? You don't find your sacred manual? You will have to do somepenitences 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 myattempt to remember from Boolos, for anything to be saved; "find thebook, return to earlier chapters" will suffice). I choose GL fromfuzzy 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) -> []pApply 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?

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 arelegal 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 becauseif GL can prove [][][]p etc. -> Löb formula, then its tempting tojust:[]p-> []([]p ->p) 5) Since []([]p ->p) is also ([][]p -> []p) 6) combining 4 & 5, GL: []p -> [][]pQ.E.D., but I don't believe my memory or the appropriacy to Bruno'sexercise! Too clumsy, too many levels and vertigo, lol.So, yes. Gotta get back to forever undecided and find the Boolosbook, I know. But it's fun as a kind of crossword puzzle activity,even when as confused as this partially forgotten and possiblymisunderstood thing. Scratching my head for a whole hour for thosecouple of lines of Kool-Aid crazy... PGCThis shows that the axiom 4 is redundant in G.(This is known as Sambin theorem, and is proved in Smullyan'sForever 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 firstbit.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 muchtypo!)> > Bruno >And thus you've proven that for everything you know, you can knowthatyou 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 G1st person "I" in a modal/qualitative provability sense. PGCI don't see any fixed point here.We have both the formal (and third person) []p -> [][]p and thefirst person knowledge formula [o]p -> [o][o]p, which is usuallyadmitted for "sufficiently introspective knower).BrunoThis 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 colourtheorem is true, but we cannot prove the proof is correct (at leastatthis 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 GoogleGroups "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-l...@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 GoogleGroups "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-l...@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 GoogleGroups "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 GoogleGroups "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.

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