# Re: AUDA and pronouns

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On 21 Oct 2013, at 23:34, meekerdb wrote:```
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```On 10/21/2013 7:25 AM, Bruno Marchal wrote:
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On 21 Oct 2013, at 05:09, meekerdb wrote:

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```On 10/20/2013 2:15 PM, Russell Standish wrote:
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```On Sun, Oct 20, 2013 at 06:22:15PM +0200, Bruno Marchal wrote:
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```On 20 Oct 2013, at 12:01, Russell Standish wrote:

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```On Sun, Oct 20, 2013 at 08:52:41AM +0200, Bruno Marchal wrote:
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We have always that [o]p -> [o][o]p (like we have also always that
```[]p -> [][]p)

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There may be things we can prove, but about which we are in fact
mistaken, ie
[]p & -p
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```That is consistent. (Shit happens, we became unsound).

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Consistency is []p & ~[]~p. I was saying []p & ~p, ie mistaken belief.
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ISTM that Bruno equivocates and [] sometimes means "believes" and sometimes "provable".
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But I am allowed to do that, because []p -> p is not a theorem (for some p, by incompleteness)
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Isn't incompleteness ~(p->[]p) ?
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Yes.

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And you assume the machine is consistent, so doesn't that entail []p- >p ?
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"[]p -> p" is correctness. It is trivially true for the machine I consider, because they are correct by definition/choice.
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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').
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Believing '0=1', does not make you inconsistent. Only non correct.

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```and thus (rational formal) provability behaves like believability.

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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).
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That still seems like equivocation to me. Even if they "obey the same modal" logic, that only means they have the axioms and rules of inference. It doesn't follow that the true but unprovable propositions are the same propositions as the true but not- believable ones.
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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).
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The hypostases will work for any correct machine whose beliefs extend soundly the beliefs in elementary arithmetic.
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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.
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But doesn't limiting us to a correct machine mean the []p->p ; isn't that what "correct" means?
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Yes.
The key point is that the machine itself cannot prove []p -> p.
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[]p->p is true, about the machine, but not provable in general by the machine (which can already not prove
```[]f->f).

Bruno

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Brent

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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).
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Bruno

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Brent

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```Obviously, one cannot prove []p & p, for very many statements, ie
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[]p & p does not entail []([o]p)
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```[]p -> [][]p  OK?

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```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.

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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?

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(and []p -> []p, and p -> p) + ([](p & p) <-> []p & []q) (derivable
```in G)

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```Did you mean [](p&q) <-> []p & []q? That theorem at least sounds

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```so    []p & p -> [][]p & ([]p & p)
-> []([]p & p) & ([]p & p),

thus ([]p & p) ->  [][o]p    (& [o]p : thus [o]p -> [o][o]p)

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```Therefore, it cannot be that [o]p -> [o]([o]p) ???

Something must be wrong...

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I hope I am not too short above, (and that there is not to much typo!)
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Bruno

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And thus you've proven that for everything you know, you can know that you 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 colour
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theorem is true, but we cannot prove the proof is correct (at least at
```this stage, proving program correctness is practically impossible).

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