Re: How many digits is pi computable to?

From: |-|erc (h_at_r.c)
Date: 01/18/05 

Date: Tue, 18 Jan 2005 11:59:35 +1000

"Barb Knox" <see@sig.below> wrote in messag
> In article <5eGdnbxpt4DqiXHcRVn-vw@rcn.net>,
> "Bill Smythe" <chichess@beforeRCNafter.com> wrote:
>
> >"Michael Mendelsohn" wrote:
> >> IIRC Herc rejects any kind of diagonalization argument.
> >> (Maybe that has changed, though.)
>
> Oddly, he rejects ALL-BUT-ONE diagonalization argument. He DOES accept the
> one that shows the unsolvability of the halting problem. And he doesn't
> notice the logical difficulty with doing that.

Diagonalisation DISPROVES a specific function Halt()
That is different to PROVING antidiag exists.

constructivism : the denial of existence resulting in contradiction does
                           not prove existence.

======================
e.g. "this statement has no proof"

Assume it is FALSE. <denial of existence>
then it has a proof, so it is true
CONTRADICTION <resulting in contradiction>

therefore "this statement has no proof:" is true. <proves existence>
=======================

Now, compare that to the constructivist proof.

true(facts) = proven(facts) U provable(facts).

factX <=> true(factX) = ~proven(factX)
factX <=> proven(factX) U provable(factX) = ~proven(factX)
factX <=> t U p = ~t
factX <=> t = ~t (extension of case p={})
factX <=> CONTRADICTION

Sorry, no incompleteness theorem just because you got stuck working out
"YOU CANT PROVE ME"

Herc

Relevant Pages

  • Re: How many digits is pi computable to?
    ... > one that shows the unsolvability of the halting problem. ... the denial of existence resulting in contradiction does ... compare that to the constructivist proof. ...
    (sci.math)
  • Re: How many digits is pi computable to?
    ... > one that shows the unsolvability of the halting problem. ... the denial of existence resulting in contradiction does ... compare that to the constructivist proof. ...
    (comp.theory)