# Re: AUDA and pronouns

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On 21 Oct 2013, at 04:48, Platonist Guitar Cowboy wrote:```
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Disclaimer: No idea if I am even on the same planet on which this discussion is taking place. So pardon my questions and confusions:
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On Sun, Oct 20, 2013 at 11:15 PM, Russell Standish <li...@hpcoders.com.au > 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:
> >>
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> >>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.

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Why Isn't "mistaken belief" here merely unsound because it's propositional variable, assuming we're speaking generally about all systems?
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Yes, []p & —p, makes the machine on which [] applies, unsound. But still consistent. Such machines will believe (prove) an arithmetical false (but consistent) sentence.
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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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Isn't that a rule for any modal sentence though, independent of system? [o]p is ([]p & p)
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Isn't []p & p = []([o]p) the definition of fixed point theorem? That, plus modalization conditionals to remove p from the G sentence so that every sentence H is a fixed point of it?
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This is a bit unclear.
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A fixed point is when a proposition says something about herself (like p <-> []p, p <-> [] ~p, etc.). The fixed point will be a proposition in which p does no more occur. The main one are:
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[](p <-> ~[]p) -> [](p <-> ~[]f)   Gödel fixed point

[](p <=> [] ¬p) <=> [](p<=> []⊥)

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So that you get that list of instances with (G sentence on left and H on right) examples like the following:
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[]p   corresponds to  T
¬[]p corresponds to  ¬[]⊥
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[]¬p corresponds to []⊥, which is pretty cool because you get a provability statement that is arithmetically equivalent to its own non-provability iff the statement is equivalent to the statement that arithmetic is inconsistent. Because G proves in this fashion:
```
[o](p <=> [] ¬p) <=> [o](p<=> []⊥)
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OK.

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This is the kind of thing that clarifies the pronoun issue imho.
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In arithmetic. But in UDA I think that the definition of first person/ third person in term of reconstitution/annihilation is clear enough for the indeterminacy purpose, and the necessity of deriving physics from arithmetic. it is just not precise enough to get the actual technical beginning of the derivation of physics.
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>
> []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?

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But it nonetheless is everywhere in Boolos: []p -> [][]p IS a theorem of G, and useful, unless Bruno shoots the cowboy, because he cannot prove it or find his damned Boolos book.
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What? You don't find your sacred manual? You will have to do some penitences or something :)
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Let me give you a difficult exercise: derive []p -> [][]p in *any* normal modal logic satisfying Löb's formula (that is derive []p -> [] []p from []([]p -p) -> []p (and [](p->q)->([]p->[]q).
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This shows that the axiom 4 is redundant in G.

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(This is known as Sambin theorem, and is proved in Smullyan's Forever Undecided). I can give the solution, some day).
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> (and []p -> []p, and p -> p) + ([](p & p) <-> []p & []q) (derivable
> in G)
>

Did you mean [](p&q) <-> []p & []q?

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I'm not sure of the q and whether you can just leave out the first bit.
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Russell was right. I made a typo error, as predicted (!).

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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...

Why?

> >
>
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> 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.

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Not sure, I'd guess we're comparing G's reasoning 3rd person "I" with reasoning of a fixed point corresponding to some sentence of G 1st person "I" in a modal/qualitative provability sense. PGC
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I don't see any fixed point here.

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We have both the formal (and third person) []p -> [][]p and the first person knowledge formula [o]p -> [o][o]p, which is usually admitted for "sufficiently introspective knower).
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Bruno

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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).

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