# Re: AUDA and pronouns

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On 21 Oct 2013, at 17:59, Platonist Guitar Cowboy wrote:```
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On Mon, Oct 21, 2013 at 4:26 PM, Bruno Marchal <marc...@ulb.ac.be> wrote:
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On 21 Oct 2013, at 08:24, Russell Standish wrote:

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On Mon, Oct 21, 2013 at 04:48:42AM +0200, Platonist Guitar Cowboy wrote:
```Disclaimer: No idea if I am even on the same planet on which this
discussion is taking place. So pardon my questions and confusions:

You and me both - we're all students here :).

I'm just rather doubtful about an axiomatisation of proof that assumes
we can prove that we can prove something, as with that we can know
that we (Theatetically) know something (since truth is usually
inherently unknowable).

It reminds me of a 3 year old's question "but why?" Ultimately, you
will not be able to answer a question like that.

It is quite possible I haven't drunk enough Kool-Aid.

Question for Bruno (raised from PGC's earlier comments):

Is axiom 4, ie  []p -> [][]p, called a fixed point theorem?

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No. PGC is a bit unclear/mysterious when referring to the fixed point theorem here.
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But I don't refer to fixed point theorem there.

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Concerning []p -> [][]p, I just stated it is a theory of G, used in all manner of proofs fruitfully.
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I remember something like "if []p -> [][]p weren't a theory of G as proven by some usual suspect, Kripke I think, then we would extend GL sufficiently until it was!"
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And that shows how often this is used; almost axiomatically in practice. Boolos at least seems addicted to it. PGC
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No problem. Minor vocabulary details. I guess you mean theorem of G (or is in the theory G).
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By Solovay theorem, G is complete and sound for the arithmetical provability, and wht is really oimprtant is that for all arithmetical proposition beweisbar('p') -> beweisbar('beeisbar('p')') is a theorem of (Peano) arithmetic. If G was not proving []p -> [][]p, it would not be arithmetically complete.
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Bruno

http://iridia.ulb.ac.be/~marchal/

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