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 mistaken belief.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. 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 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 G sentenceso 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:`

[](p <-> ~[]p) -> [](p <-> ~[]f) Gödel fixed point [](p <=> [] ¬p) <=> [](p<=> []⊥)

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 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 atheorem of G, and useful, unless Bruno shoots the cowboy, because hecannot 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).`

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 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 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 G1st 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 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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