Dear John:
Below I have slightly rewritten my argument along the lines I had
originally intended - the first effort was too rushed. The rewrite may
answer your question.
At 2/12/01, you wrote:
>Hal, you wrote (among lots of other things):
> > 2) But the other universe also has to stop given the fixed FAS complexity
> > i.e. another new running contradiction.
>Can you discern - after the split - which is "the other"?
>(excuse me for this question for a situation which I do not condone at all).
>John Mikes
---
In the context that a particular evolving universe is a recursively
enumerable theorem cascade of a particular finite consistent FAS can a
cascade branch?
Each theorem in a cascade has an AIT complexity greater than its preimage
because the cascade itself is the shortest program for computing any
theorem in it.
If the FAS has a finite AIT complexity and is fixed then the cascade must
eventually stop when the complexity of its individual theorems becomes
equal to the max complexity supported by the fixed FAS i.e. the complexity
of the FAS itself plus a constant.
The endpoint theorem must be one that has no successor under the rules of
the cascade.
Such a theorem has a low complexity since I just described it.
Thus we seem to have a contradiction - a cascade of ever more complex
theorems must stop and can only stop on a very simple theorem.
This contradiction can be resolved if the FAS is actually not constant but
can be added to by an external random oracle thus increasing its
complexity. This allows the cascade to continue.
Can the random oracle event leave behind a cascade still running in the old
FAS? No because each event of this sort would create a new running
contradiction.
The only way out seems to be that there is no branching and the FAS must
become more complex as the cascade continues.
An increase in complexity of the FAS has a possible interpretation: The
cascade is reinitialized in the new FAS with the new theorem acting as the
initiating axiom.
That is a better way of putting it and eliminates the need to distinguish
between the two universes since the running contradiction one is not allowed.
Hal
