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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Four Chatbots figure out on their own without prompting that
 HHH(DDD)==0
Date: Sat, 19 Jul 2025 16:36:42 -0500
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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