Re: ChatGPT agrees that I have refuted the conventional Halting Problem proof technique --- Full 38 page analysis

On 2025-06-29 14:39:58 +0000, olcott said:

On 6/29/2025 4:29 AM, Mikko wrote:

On 2025-06-28 21:52:06 +0000, olcott said:
 

On 6/28/2025 12:41 PM, Richard Damon wrote:

On 6/28/25 9:54 AM, olcott wrote:

On 6/28/2025 7:04 AM, Mikko wrote:

On 2025-06-27 14:19:28 +0000, olcott said:
 

On 6/27/2025 1:55 AM, Mikko wrote:

On 2025-06-27 02:58:47 +0000, olcott said:
 

On 6/26/2025 5:16 AM, Mikko wrote:

On 2025-06-25 15:42:36 +0000, olcott said:
 

On 6/25/2025 2:38 AM, Mikko wrote:

On 2025-06-24 14:39:52 +0000, olcott said:
 

*ChatGPT and I agree that*
The directly executed DDD() is merely the first step of
otherwise infinitely recursive emulation that is terminated
at its second step.

 No matter who agrees, the directly executed DDD is mote than
merely the first step of otherwise infinitely recursive
emulation that is terminated at its second step. Not much
more but anyway. After the return of HHH(DDD) there is the
return from DDD which is the last thing DDD does before its
termination.

 *HHH(DDD) the input to HHH specifies non-terminating behavior*
The fact that DDD() itself halts does not contradict that
because the directly executing DDD() cannot possibly be an
input to HHH in the Turing machine model of computation,
thus is outside of the domain of HHH.

 The input in HHH(DDD) is the same DDD that is executed in DDD()
so the behaviour specified by the input is the behavour of
directly executed DDD, a part of which is the behaour of the
HHH that DDD calls.
 If HHH does not report about DDD but instead reports about itself
or its own actions it is not a partial halt decideer nor a partial
termination analyzer, as those are not allowed to report on their
own behavour more than "cannot determine".

 Functions computed by Turing Machines are required to compute
the mapping from their inputs and not allowed to take other
executing Turing machines as inputs.

 There is no restriction on the functions.

 counter factual.

 That is not a magic spell to create a restriction on functions.
 

A Turing machine is required
to compute the function identified in its specification and no other
function. For the halting problem the specification is that a halting
decider must compute the mapping that maps to "yes" if the computation
described by the input halts when directly executed.

 No one ever bothered to notice that because directly
executed Turing machines cannot possibly be inputs to
other Turing machines that these directly executed
Turing machines have never been in the domain of any
Turing machine.

 Irrelevant. They are the domain of the halting problem.

 That they are in the domain of the halting problem
and not in the domain of any Turing machine proves
that the requirement of the halting problem is incorrect.

 No, it just says that you don't understand the concept of representation.

 There exists no finite number of steps where N steps of
DDD are correctly simulated by HHH and this simulated DDD
reaches its simulated "return" statement final halts state.

 That is a statement about HHH that does not tell about halting of DDD.

 That this is too difficult for you to
understand count as zero rebuttal what-so-ever.

It is a sin to lie about other people. You should try to avoid sins,
especially sins that serve no purpose.
That you don't identify anything wrong in my words is sufficent to
regard them good enough at least for this discussion.

void DDD()
{
   HHH(DDD);
   return;
}
 _DDD()
[00002192] 55             push ebp
[00002193] 8bec           mov ebp,esp
[00002195] 6892210000     push 00002192  // push DDD
[0000219a] e833f4ffff     call 000015d2  // call HHH
[0000219f] 83c404         add esp,+04
[000021a2] 5d             pop ebp
[000021a3] c3             ret
Size in bytes:(0018) [000021a3]
 The x86 source code of DDD specifies that this emulated
DDD cannot possibly reach its own emulated "ret" instruction
final halt state when emulated by HHH according to the
semantics of the x86 language.

Both the source code and the x86 translation specfy that if
HHH returns then so does DDD. Whether HHH can simulate that
return is a question about HHH.
--
Mikko