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).
In that case we can prove Bf. From this we still cannot prove f (Bf ->
f = ~Bf = "I am consistent" (3p self-reference).
Obviously, one cannot prove []p & p, for very many statements, ie
[]p & p does not entail []([o]p)
[]p -> [][]p OK?
(and []p -> []p, and p -> p) + ([](p & p) <-> []p & []q) (derivable in
G)
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
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Prof Russell Standish Phone 0425 253119 (mobile)
Principal, High Performance Coders
Visiting Professor of Mathematics hpco...@hpcoders.com.au
University of New South Wales http://www.hpcoders.com.au
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