# Re: My proposed model - short form

From: Hal Ruhl <hjr.domain.name.hidden>
Date: Fri, 10 Nov 2000 20:58:13 -0800

Dear Bruno:

You commented on my step 9 which in my most recent post on the subject reads as

9) N-bit, ifc-FAS are composed of:

a) A finite theorem set [no more than 2^[N + c] as per AIT.]
b) Each theorem has a finite AIT complexity less than or equal to N +
c bits

and thus the FAS has

b) A finite symbol set or alphabet.
c) A finite set of axioms
d) A finite set of rules

The issue you raised has to do with "a": an ifc-FAS has,

a) A finite theorem set [no more than 2^[N + c] as per AIT.]

You then point to what I take to be [remember I am not a mathematician] a
theorem cascade that exists [seems to anyway] in a particular what I also
take to be a not too complex FAS.

In fact I made a slight slip from my model of the moment in this area on
one of my recent posts when I said.

"Chaitin showed that a consistent FAS that actually has a Decision
Procedure and if that procedure is finite then the FAS could not contain a
choice between alternates that were more complex than the Decision
Procedure except for a constant."

While this may not be false it is not what I wished to say, but perhaps I
stumbled onto something.

Getting back to my step 9a I take it to be Chaitin's result. His result
[neglecting for the moment his constant c] as I understand it is that an
N-bit FAS can not contain theorems more complex than the FAS. There is no
more than 2^N such theorems - a finite number if N is finite. The result
applies whether or not the N-bit FAS is complete.

Now looking at FAS that contain recursive enumeration theorem cascades
would seem to create a problem. Chaitin says they should stop because each
theorem in the cascade is more complex than the last so when you get to a
complexity of 2^N you must stop. In the case you give you only assume that
you could write down all those theorems but is there enough ink and paper
in the universe to actually do so?

Further the cascade can only stop on a theorem "that has no consequent
under the rules". That fully describes the theorem so it is not too
complex. This appears to me to be a contradiction. How to resolve it?

Well in my step 17 I gave my view that this is what spontaneously drives
the ifc-FAS to be perturbed until it is countably infinite.

My slip the other night now leads me to believe that even complete finite
FAS that contain a theorem cascade will be pushed until they are countably
infinite. The fact that Godel has them go incomplete along the way is not
relevant. This widens the scope of possible universes - excellent.

So I am really not at odds with your example. By the time you get to write
down all those theorems there might be enough ink and paper.

You also wrote:

> >1) Is "nothing" stable with respect to "everything" - Yes or No?
> >
> >2) Is "everything" stable with respect to "nothing" - Yes or No?
>
>
>What do you mean by "nothing" ? The empty set ? (In which theory ?),
>The null theory (In which logic), the physical void (in which frame?).
>
>What do you mean by stable ?

I just made a post the other night which might clear this up some. There I
try to show that both Nothing and Everything are unstable and go to a
middle ground.

By Nothing I really mean it. A completely empty system. No alphabet, no
axioms, no rules, no theorems.

By stable I mean the usual engineering/physics concept. Think of the
inside of a tea cup. Everything is on the smooth surface on one side of
the cup. It goes straight to the bottom. Nothing may get stuck on the side
for awhile but must eventually also go to the bottom. At the bottom lies
Something - my superverse which rattles around near the bottom with a
random dynamic "forever".

Hal
Received on Fri Nov 10 2000 - 18:06:09 PST

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