On 25 Sep 2013, at 23:11, Telmo Menezes wrote:
On Wed, Sep 25, 2013 at 4:52 PM, Bruno Marchal <marc...@ulb.ac.be>
wrote:
On 25 Sep 2013, at 15:50, Telmo Menezes wrote:
On Mon, Sep 23, 2013 at 7:49 PM, Bruno Marchal <marc...@ulb.ac.be>
wrote:
On 23 Sep 2013, at 12:41, Telmo Menezes wrote:
<snip>
Thus your interest in near-death experiences?
Yes. And in all "extreme" altered state of consciousness. Those
extreme
cases provide key information.
Can this information be recovered?
A part of it, but it is experiential and not really communicable or
rationally justifiable. The wise searcher will remain mute on this,
or be
precise that it gives only a report of an experience, or perhaps
suggests a
(meta) theory in which such experience exists and are not
communicable.
Ok.
For example, is a NDE that did not result in death, was it really a
cul-de-sac?
I would say no. But this is complex question, and the salvia
experience
confused me on this topic.
In the NDE, people come back, but with salvia, locally, people does
not come
back, only an approximative copy.
I read several reports on this sensation of having been copied. It's
very intriguing.
I agree. It is *very* intriguing. I thought even that salvia somehow
refutes comp, but I realize that it was only from the first person
perspective, and this is coherent with the fact that the "knower" (Bp
& p) is already quite comp resistant.
It is the diabolical aspect of comp: it predicts that the machine have
necessarily an hard time with comp. Somehow, like the Gödelian
sentence, comp says about itself "you can't really believe me". From
that perspective salvia plays a similar trick, just very realistically!
That's why i say that salvia is not an NDE, but a DE. That is why
it can be
quite literally "life changing", and why I would not recommend it to
anybody.
For the feeler or observer there is no cul-de-sac (thanks to the "&
Dt"
added to the Bp, by Gödel's completeness theorem (not
incompleteness!).
But the scientist part of him has cul-de-sac, perhaps the publish
or perish,
that is the fact that proofs must be finite, before publication.
But it is hard to interpret all this literally. Caution.
Sure.
We should *try* to avoid it, but we can't avoid it without
loosing our
universality.
The consistent machines face the dilemma between security and
lack of
freedom-universality. With <>p = ~[] ~p, here are equivalent way
to
write
it:
<>t -> ~[]<>t
<>t -> <> [] f
[]<>t -> [] f
I don't understand how you arrive at this equivalence.
I use only the fact that (p -> q) is equivalent with (~q -> ~p)
(the
contraposition rule, which is valid in classical propositional
logic),
and
the definition of <> p = ~[] ~p. I use also that ~~p is
equivalent with
p.
Note that []p = ~~[]~~p = ~<> ~p. And,
~[]p = <> ~p
and
~<>p = [] ~p
Like with the quantifier, a not (~) jumping above a modal sign
makes it
into
a diamond, if it was a bo, and a box, if it was a diamond.
Starting from <>t -> ~[]<>t.
But where does <>t -> ~[]<>t come from?
<>t -> ~[] <> t
is the same as
~[] f -> ~[] (~[] f)
OK?
Ok.
And that is .... the modal writing of Gödel's second incompleteness
theorem:
if the false is not provable, then that fact (that the false is not
provable) is itself not provable.
Nice, that is very satisfying.
OK. Nice.
Keep in mind that <>t (which I write also Dt) is the same as —[]
f, which
is equivalent with "I am consistent", and Gödel's second theorem
asserts: If
I am consistent then I cannot prove that I am consistent. Note that
the "I"
is a third person I. (The first person "I" considers his/her
consistency
trivial).
Contraposition gives ~~[]<>t -> ~<>t, and this
gives by above, []<>t -> []~t, which gives
[]<>t -> []f (as ~t = f, and ~f = t).
OK?
Ok!
For the third one, starting from the first one again: <>t ->
~[]<>t, By
contraposition []<>t -> ~<>t , but ~<>t = []~t = [] f.
OK?
Ok! Thanks Bruno. My only problem now is the above.
Tell me if you see that it was the modal version of Gödel's second
incompleteness theorem.
You might as an exercise show that it follows from Löb's theorem:
[] ([] p -> p) -> [] p
Two hints: 1) "~p" is the same as "p -> f", 2) replace p by f.
OK?
Löb's formula *is* the main axiom of the modal logic G.
Alright, it's simple with the hints:
[] ([] p -> p) -> [] p
repalce p by f:
[] ([] f -> f) -> [] f
[]f -> f = ~[]f:
[](~[]f) -> []f
contraposition:
~[]f -> ~(~[]f)
Thanks!
You are welcome :)
Bruno
http://iridia.ulb.ac.be/~marchal/
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