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 (for`

`some p, by incompleteness) and thus (rational formal) provability`

`behaves like believability.`

`A mathematician told me that I was dead mad by saying this, but that`

`is standard in mathematical logic (ignored by most mathematicians). It`

`is counter-intuitive. Most people believes that formal proof`

`guaranties truth, when starting from true theorem (and that is true`

`for the ideally correct machine, but no machine can know she is`

`correct, and her probability does behave like a believability, indeed`

`one on Which the application of Theaetus' definition leads to the`

`classical knowledge logic (the modal logic S4).`

`The hypostases will work for any correct machine whose beliefs extend`

`soundly the beliefs in elementary arithmetic.`

`The key is in the mathematical trick to limit us to correct machine,`

`with enough beliefs (about machines or numbers) so that they are under`

`the spell of the second incompleteness theorem, or Löb, and can prove`

`that.`

`In the literature such machine/theories are qualified as being`

`"sufficiently rich", but "Löbian" is shorter, (and then the Löb`

`formula also characterize their provability logic).`

Bruno

BrentObviously, 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) (derivable in 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. 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).--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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