# Re: AUDA and pronouns

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On 20 Oct 2013, at 23:15, 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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I was saying that ( []p & ~p) is consistent. ( []p & ~[]~p) is "p is provable and p is consistent).
```Mistaken belief is consistent with arithmetic.

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Obviously, one cannot prove []p & p, for very many statements, ie

[]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.
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You are right, this is not so easy to prove. It follows from the "provable sigma_1 completeness", that is the fact that if p is a sigma_1 formula, then Peano Arithmetic can prove p -> Bp. (That is not easy to prove, but it is done in Hilbert-Bernays, also in the books by Boolos). It is the hard part of the second incompleteness theorem. It presupposes some induction axioms, like in Peano Arithmetic (PA).
```Then Bp is itself a sigma_1 arithmetical proposition, so []p -> [][]p.

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If this were a theorem of G, then it suggests G does not capture
the nature of proof.
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PA, ZF, all Löbian machines prove []p -> [][]p. That is why they are "intrspective enough". When they prove something, they can prove that can prove that something.
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Robinson Arithmetic, which has no induction axiom at all, cannot. It is still sigma_1 complete, but cannot prove its own sigma_1 completeness, nor []p -> [][]p. RA is typically NOT Löbian.
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sigma_1 completeness = a provability characterization of Turing universality. Löbian machines are universal, like RA, but unlike RA, thay can prove (heir universality (in some weak sense).
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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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As I say, by showing that PA proves all instances of p -> []p, for p sigma_1. This is long and subtle to show. Boolos 1979 sums it rather well, and Boolos 193, gives all the details (but it is more messy).
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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?
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Yes.

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```That theorem at least sounds
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OK.

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[]p -> [][]p is not so astonishing, because the induction axioms provides strong provability power to the theories. Let me try to say a bit more. If PA proves p, PA will proves []p. This is much more easy to prove, and is indeed captured by the necessaitation rule p/[]p. That is almost obvious, because if PA proves p, such a proof exists, and thus has a Gödel number, and the machines will prove it (provable('p') is sigma_1, and the theories are sigma_1 complete). That is true for both RA and PA. But PA, like ZF, can internalise that rule (p/[]p), they can proof in general that []p -> [][]p. This will be proved by induction on the complexity of the formula p, and well, it is not easy at all to see that.
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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.
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```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).
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It should be easy once we have a concrete formal proof. As far as I know, we don't have this for the 4 colour theorem. But once we have such proofs, it should be trivial to prove that the proof exists, making []p -> [][]p easy to prove for the particular case of p = 4- color.
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A correct machine is automatically Löbian if she is sigma_1 complete, and has enough induction axioms to prove its own sigma_1 completeness (in the sense of proving all formula p -> []p, for p sigma_1).
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In fact "p-> []p" characterizes sigma_1 completeness (by a result by Albert Visser), and that is why to get the proba on the UD*, we use the intensional nuance []p & <>t (= proba) starting from G extended with the axiom "p-> []p" (limiting the proposition to the UD).
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Bruno

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