Re: Four Chatbots figure out on their own without prompting that HHH(DDD)==0

Liste des GroupesRevenir à s logic 
Sujet : Re: Four Chatbots figure out on their own without prompting that HHH(DDD)==0
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory
Date : 19. Jul 2025, 22:36:42
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <105h35a$2uujj$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
User-Agent : Mozilla Thunderbird
On 7/19/2025 4:26 PM, wij wrote:
On Sat, 2025-07-19 at 16:05 -0500, olcott wrote:
On 7/19/2025 3:57 PM, wij wrote:
On Sat, 2025-07-19 at 15:41 -0500, olcott wrote:
On 7/19/2025 3:14 PM, wij wrote:
>
HP is very simple: H(D)=1 if D halts, H(D)=0 if D does not halt.
>
>
The standard proof assumes a decider
H(M,x) that determines whether machine
M halts on input x.
>
But this formulation is flawed, because:
>
Whatever the 'formulation' is, the HP result is a fact that no H can decide
the halting status of any given D.
>
>
And that is wrong because H(⟨D⟩) is correctly determined.
It has always been a type mismatch error when H(D) was
assumed.
 Yes, there is type mismatch problems in nearly all discussions.
But I don't think you will understand what it is.
 
I have proven that I do and you only deny this
because you are not interested in an honest
dialogue.

Turing machines can only process finite encodings
(e.g. ⟨M⟩), not executable entities like M.
>
So the valid formulation must be
H(⟨M⟩,x), where ⟨M⟩ is a string.
>
Halting Problem::= H(D)=1 if D halts, H(D)=0 if D does not halt.
The conclusion is, no such H exists.
>
>
And that is wrong because H(⟨D⟩) is correctly determined.
It has always been a type mismatch error when H(D) was
assumed.
>
int DD()
{
    int Halt_Status = HHH(DD);
    if (Halt_Status)
      HERE: goto HERE;
    return Halt_Status;
}
>
 A type mismatch: HHH(DD) or HHH(<DDD>)?
 
DD points to the finite string machine
description of DD it does not point to
the executing process of DD.

DD correctly simulated by HHH cannot reach past
the "if" statement thus cannot reach the "return"
statement. T
 That is roughly what HP proof says.
 
Not at all. The HP proof claims that DD
correctly simulated by HHH reaches the
self-contradictory part of DD and thus
forms a contradiction.

his makes HHH(DD)==0 correct.
 How is this statement from?
You chopped up my statement in the middle of a word.

HHH(DD) above shows it cannot return to report 0.
(I guess you might say something and doing another, again)
 
Factually incorrect.

'formulation' does not really matter.
If 'formulation' matters, it is another problem.
>
>
>
 
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

Date Sujet#  Auteur
23 Jul 25 o 

Haut de la page

Les messages affichés proviennent d'usenet.

NewsPortal