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".

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
plausable as being about proof.


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).



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