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On 2025-06-24 14:09:10 +0000, olcott said:It turns out that the question a halt decider must answer
On 6/24/2025 4:27 AM, joes wrote:Answering that question prevents HHH(DDD) from answering anyAm Mon, 23 Jun 2025 16:28:23 -0500 schrieb olcott:>On 6/23/2025 2:58 PM, joes wrote:>Am Mon, 23 Jun 2025 12:40:43 -0500 schrieb olcott:On 6/23/2025 10:34 AM, joes wrote:Am Mon, 23 Jun 2025 09:30:07 -0500 schrieb olcott:Sure, it simulates *into* the call, but it never returns, which isThus when HHH is simulating DDD and DDD calls HHH(DDD) the outer HHH[blah blah non sequitur]My claim is thatIf you read the 38 pages you will see how this is incorrect. ChatGPTSuch as HHH, making it not a decider (when simulated).
"understands" that any program that must be aborted at some point to
prevent its infinite execution is not a halting program.
>
Well MY claim is that HHH simulated HHH (itself) doesn't halt.
>obviousYou know what, it actually IS obvious that HHH can't simulate past the
call to HHH. Thanks for coming to my Ted talk.
>
does simulate itself simulating DDD.
precisely why you abort it.
>
[more irrelevant stuff]
>
void DDD()
{
HHH(DDD);
return;
}
>
*This is the question that HHH(DDD) correctly answers*
Can DDD correctly simulated by any termination analyzer
HHH that can possibly exist reach its own "return" statement
final halt state?
other question because it can only answer one question.
A termination analyzer is required to answer a different
question, which HHH(DDD) does not. Therefore HHH is not a
termination analyzer.
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