On 21 Oct 2013, at 23:34, meekerdb wrote:

On 10/21/2013 7:25 AM, Bruno Marchal wrote:On 21 Oct 2013, at 05:09, meekerdb wrote:On 10/20/2013 2:15 PM, Russell Standish 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 & -pThat is consistent. (Shit happens, we became unsound).Consistency is []p & ~[]~p. I was saying []p & ~p, ie mistakenbelief.ISTM that Bruno equivocates and [] sometimes means "believes" andsometimes "provable".But I am allowed to do that, because []p -> p is not a theorem (forsome p, by incompleteness)Isn't incompleteness ~(p->[]p) ?

Yes.

And you assume the machine is consistent, so doesn't that entail []p->p ?

`"[]p -> p" is correctness. It is trivially true for the machine I`

`consider, because they are correct by definition/choice.`

`Consistency is correctness on the f: []f -> f. It is a very particular`

`case of correctness.`

`There are machines which are not correct, yet consistent. For example`

`Peano-Arithmetic + the axiom beweisbar('f').`

Believing '0=1', does not make you inconsistent. Only non correct.

and thus (rational formal) provability behaves like believability.A mathematician told me that I was dead mad by saying this, butthat is standard in mathematical logic (ignored by mostmathematicians). It is counter-intuitive. Most people believes thatformal proof guaranties truth, when starting from true theorem (andthat is true for the ideally correct machine, but no machine canknow she is correct, and her probability does behave like abelievability, indeed one on Which the application of Theaetus'definition leads to the classical knowledge logic (the modal logicS4).That still seems like equivocation to me. Even if they "obey thesame modal" logic, that only means they have the axioms and rules ofinference. It doesn't follow that the true but unprovablepropositions are the same propositions as the true but not-believable ones.

`It depends of your theory of belief. I agree that this concerns a form`

`of rational beliefs, extending the belief in Peano Axiom (or`

`combinators axioms, etc.) in a correct and recursively enumerable way`

`(using comp).`

The hypostases will work for any correct machine whose beliefsextend soundly the beliefs in elementary arithmetic.The key is in the mathematical trick to limit us to correctmachine, with enough beliefs (about machines or numbers) so thatthey are under the spell of the second incompleteness theorem, orLöb, and can prove that.But doesn't limiting us to a correct machine mean the []p->p ; isn'tthat what "correct" means?

Yes. The key point is that the machine itself cannot prove []p -> p.

`[]p->p is true, about the machine, but not provable in general by the`

`machine (which can already not prove`

[]f->f). Bruno

BrentIn the literature such machine/theories are qualified as being"sufficiently rich", but "Löbian" is shorter, (and then the Löbformula also characterize their provability logic).BrunoBrentObviously, one cannot prove []p & p, for very many statements, ie[]p & p does not entail []([o]p)[]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?(and []p -> []p, and p -> p) + ([](p & p) <-> []p & []q)(derivablein G)Did you mean [](p&q) <-> []p & []q? 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...I hope I am not too short above, (and that there is not to muchtypo!)BrunoAnd thus you've proven that for everything you know, you can knowthatyou know it. This seems wrong, as the 4 colour theorem indicates.Wecan prove the 4 colour theorem by means of a computer program,and itmay indeed be correct, so that we Theatetically know the 4 colourtheorem is true, but we cannot prove the proof is correct (atleast atthis stage, proving program correctness is practically impossible).--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.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.

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