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On 6/28/25 7:19 PM, olcott wrote:Not at all. I am the only one to totallyOn 6/28/2025 6:10 PM, Richard Damon wrote:Reallhy? By what?On 6/28/25 5:52 PM, olcott wrote:Proven to be counter-factual and over your head.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.
>
>
But there is no HHH that correctly simulates the DDD that the HHH that answers,
>
Your LIES?
Based on you not knowing what your words mean.
void Infinite_Recursion()So?
{
Infinite_Recursion();
return;
}
>
The exact same code that correctly recognizes infinite
recursion sees this non-terminating pattern after one
single recursive emulation.
>
The issue is that THAT input doesn't halt, even when correctly simulated by a UTM.
But UTM(DDD) will halt if HHH(DDD) returns an answer.
THus, you are just showing that you are just an stupid troll, that doesn't understand the basic rules of logic.
Try to actually PROVE something, which requires showing the ACTUALLY know statements that you are starting from, and then the truth preserving steps from them that reach to your final statement.*The proof is the self-evident verified fact that*
Your problem is you always stast with a strawman ststement, as it seems that is all your brain can process.When an input is deliberately designed to fool its
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