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Op 13.mei.2025 om 06:52 schreef olcott:A simulating termination analyzer according to this exact specOn 5/12/2025 11:05 PM, Richard Damon wrote:That has been done hundreths of times.On 5/12/25 10:53 PM, olcott wrote:>On 5/12/2025 8:27 PM, Richard Damon wrote:>On 5/12/25 2:17 PM, olcott wrote:>Introduction to the Theory of Computation 3rd Edition>
by Michael Sipser (Author)
4.4 out of 5 stars 568 rating
>
https://www.amazon.com/Introduction-Theory-Computation-Michael- Sipser/ dp/113318779X
>
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
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DD correctly simulated by any pure simulator
named HHH cannot possibly terminate thus proving
that this criteria has been met:
>
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider H correctly simulates its
input D until H correctly determines that its simulated D
would never stop running unless aborted then
>
H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
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Which your H doesn't do, as it can not correctly determine what doesn't happen.
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Any C programmer can correctly tell what doesn't happen.
What doesn't happen is DD reaching its "return" statement
final halt state.
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Sure they can, since that is the truth, as explained.
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Since your "logic" is based on lies and equivocation,
If my logic was based on lies and equivocation
then you could provide actual reasoning that
corrects my errors.
>One of the things that have been shown to you to be wrong.
It is truism that simulating termination analyzers
must report on the behavior of their input as if
they themselves never aborted this simulation:
Repeating it can be interpreted as lying.
The report must be about the behaviour specified in the input, including the code to abort, not about the behaviour specified in a hypothetical other HHH that does not abort
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