Dear Juergen:
In case what I tried to say was not clear the idea is that there are no
more than 2^(N + c) shortest possible unique proofs in an N-bit FAS. How
can number theory if it is a finite FAS contain an infinite number of
unique theorems?
Hal
Dear Jacques:
At 4/12/01, you wrote:
Maybe Hal, Russel and Jurgen should take this discussion to email and
just let us know how it turns out, because I get enough junk mail already.
I have run into those who do not like the side bar approach. I tend to
agree that it cuts all the others
Dear Juergen:
You demonstrate my point.
At 4/12/01, you wrote:
Hal, here is an infinite chain of provable unique theorems:
1+1=2, 2+1=3, 3+1=4, 4+1=5, ...
First these are not theorems they are proof chains ending in theorems.
For example:
4 + 1 = is a proof chain and the theorem proved is: 5
Dear Russell:
Yes we did indeed have a similar debate some time ago.
At that time I was still trying to express this point of view correctly and
admittedly made a number of mistakes back then [and still do].
Our debate helped me considerably and I thank you.
In response:
I just posted a
Dear Russell:
You wrote:
Why bound the proof?
It was not my idea. Chaitin equated complexity with a computing program's
length and a proof chain is a computing program according to Turing.
[rearranging your post]
1+1=2, 2+1=3, 3+1=4 ...
are all distinct theorems.
My view:
Again as in my
Dear Russell:
At 4/13/01, you wrote:
Bounded complexity does not imply bounded length. Examples include an
infinite sting of '0's, and the string '1234...9101112...'
That was part of the old debate and one of my initial mistakes. I am not
now talking about the length of theorems but the
Hal Ruhl wrote:
Dear Russell:
At 4/13/01, you wrote:
Bounded complexity does not imply bounded length. Examples include an
infinite sting of '0's, and the string '1234...9101112...'
That was part of the old debate and one of my initial mistakes. I am not
now talking about the length
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