Re: AUDA and pronouns

```
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 always that
```[]p -> [][]p)

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

```
Consistency is []p & ~[]~p. I was saying []p & ~p, ie mistaken belief.
```
```
ISTM that Bruno equivocates and [] sometimes means "believes" and sometimes "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

```
```
Brent

```
```

```
```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.

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 you prove
```that 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

```
```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 much typo!)
```
Bruno

```
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
```
theorem is true, but we cannot prove the proof is correct (at least at
```this stage, proving program correctness is practically impossible).

```
```
--
```
You received this message because you are subscribed to the Google Groups "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.
```
```
http://iridia.ulb.ac.be/~marchal/

--
You received this message because you are subscribed to the Google Groups
"Everything List" group.
To unsubscribe from this group and stop receiving emails from it, send an email