Dear Russell:
The idea I am trying to exploit in my latest posts is that a deterministic
cascade means one that uses every ounce of inference in the FAS in every
possible way on all the current data at every single step of the
cascade. Each step is individually elegant and since deterministic is
taken to mean no other proof of any string of the cascade exists then I
called it everywhere elegant. Thus the cascade is known to be elegant by
the FAS.

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Known elegant proofs can not be more complex than the FAS itself plus a
constant.
Cascades can not stop.
Thus a contradiction.
The UD as I understand it as described by various sources I have read it is
applying all of its inference in every possible way to the current data of
each step. It is a single track machine. See below:
At 4/20/01, you wrote:
>But if P(A)=B and P'(B)=C are elegant proofs, it is very unlikely for
>P'(P(A))=C to be an elegant proof.
That is not as the UD is described as I have read it. It executes one sub
component at each step in a determined sequence. It has no choice in this
sequence. The fact that it may produce C more than once is not an issue
since this is not its goal. Its goal is to produce some collection of
strings and that just might contain C more than once but the overall proof
of that collection is still the only one available thus it is elegant.
snip
>But whilst it is necessary for each step of
>an elegant proof to be elegant, it is not sufficient
I add the constraint of "deterministic" cascade.
Hal