# Re: tautology

From: Russell Standish <R.Standish.domain.name.hidden>
Date: Thu, 9 Dec 1999 10:02:04 +1100 (EST)

>
> Russell Standish wrote:
>
> >None of this is in defence of QTI! It merely is to show that your
> >measure argument fails - unless you happen to be a Copenhagener :)
> >
> >As I have mentioned before, in order for QTI to work, there must also be no
> >possible "cul-de-sac" branches. In Bruno's model logic, I believe this
> >would be expressed as
> >
> >\forall \alpha \in W, \models_\alpha^W \neg \Box (\Box\top \wedge \Box\bot)
> >
> >or in slightly more English notation (\top == true, \bot == false,
> >\wedge == and)
> >
> >for every world alpha in the model W, there cannot be a successor
> >world that can only access a terminal world
> >
> >where \Box p is trivially true in a terminal world, regardless of the
> >truth table of p, but \Box true and \Box false cannot both be true at
> >the same time.
>
> I'm not sure I understand what you mean by \models_\alpha^W"
>

I'm using LaTeX symbols. In English, the above phrase reads "is a
theorem in world alpha of model W".

> But "BOX TRUE AND BOX FALSE" is indeed true in all, and only in all
> terminal worlds. (actually "BOX FALSE" is enough)

Consider the following diagram:

/------ [False]
|
[\Box False]{--- [False]
|
\------[False]

I was reading this as saying "False is a true statement", therefore
\Box False is true is the predecsessor world. Does one rule out
statements like "False is a true statement" from the picture utterly?

>
> The interesting formula to find here is a formula with one
> variable "p" which would caracterize Kripke frames with no
> terminal worlds. Solution : (BOX p -> DIAMOND p). This formula
> is verified, for all the valuation of p, iff the frame is
> ideal. There are no cul-de-sac at all.

True, but idealism is a sufficient, but not necessary condition for
absence of cul-de-sacs. One can have a model in which every terminal
world is preceded by a world that acesses non-terminal states. Such a
model is required for QTI to hold.

> I agree with you that this is needed for any kind of
> immortality" (quantum immortality, comp immortality, etc.).
> It is highly interesting to look at it as a kind of abstract
> renormalisation.
>
> >In anycase, I haven't got a clue as to how one might start proving
> >this, or working out under what conditions it might hold.
>
> With BOX being read as the Godel's BEWEISBAR
> arithmetical provability predicate, it can be show that
>
> 1) (BOX p -> DIAMOND p) is true
> 2) BOX (BOX p -> DIAMOND p) is false
>
> Roughly put : comp immortality is a true but not provable
> statement for the self-referentially correct machine!
> (remember that true but unprovable statements are still "bettable").
>
> Bruno
>
>

----------------------------------------------------------------------------
Dr. Russell Standish Director
High Performance Computing Support Unit,
University of NSW Phone 9385 6967
Sydney 2052 Fax 9385 6965
Australia R.Standish.domain.name.hidden
Room 2075, Red Centre http://parallel.hpc.unsw.edu.au/rks
----------------------------------------------------------------------------
Received on Wed Dec 08 1999 - 15:02:23 PST

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