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On 7/19/2025 2:29 PM, wij wrote:But that isn't the correct answer, as DD() Halts since HHH(DD) returns 0.On Sat, 2025-07-19 at 14:19 -0500, olcott wrote:It is a key element of my refutation of this proofOn 7/19/2025 12:02 PM, Richard Damon wrote:>On 7/19/25 10:42 AM, olcott wrote:>On 7/18/2025 3:49 AM, joes wrote:>
>That is wrong. It is, as you say, very obvious that HHH cannot simulate>
DDD past the call to HHH. You just draw the wrong conclusion from it.
(Aside: what "seems" to you will convince no one. You can just call
everybody dishonest. Also, they are not "your reviewers".)
>
For the purposes of this discussion this is the
100% complete definition of HHH. It is the exact
same one that I give to all the chat bots.
>
Termination Analyzer HHH simulates its input until
it detects a non-terminating behavior pattern. When
HHH detects such a pattern it aborts its simulation
and returns 0.
So, the only HHH that meets your definition is the HHH that never
detects the pattern and aborts, and thus never returns.
>
All of the Chat bots conclude that HHH(DDD) is correct
to reject its input as non-halting because this input
specified recursive simulation. They figure this out
on their own without any prompting.
>
https://chatgpt.com/share/687aa4c2-b814-8011-9e7d-b85c03b291eb
It is still nothing to do with the Halting Problem proof (Because it is POOH)
>
because HHH also correctly determines that HHH(DD)==0.
DD correctly simulated by HHH cannot possibly everBut Your HHH doesn't DO a correct smulation, so your claim is just fantasy,
reach past its first statement because it specifies
recursive simulation.
int DD()And that isn't the complete definition of the PROGRAM DD, thus showing that your arguement is just a category error.
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
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