On 20 Oct 2013, at 23:15, 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 always that []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 mistaken belief.

`I was saying that ( []p & ~p) is consistent. ( []p & ~[]~p) is "p is`

`provable and p is consistent).`

Mistaken belief is consistent with arithmetic.

Obviously, 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.

`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.

If this were a theorem of G, then it suggests G does not capture the nature of proof.

`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.`

`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.`

`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).`

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?

`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).`

(and []p -> []p, and p -> p) + ([](p & p) <-> []p & []q) (derivable in G)Did you mean [](p&q) <-> []p & []q?

Yes.

That theorem at least sounds plausable as being about proof.

OK.

`[]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.`

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 know that you know it. This seems wrong, as the 4 colour theorem indicates.

I would not trust my intuition about this.

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).

`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.`

`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).`

`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).`

Bruno

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