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On 11/5/2024 3:01 AM, Mikko wrote:No, because HHH(Infinite_Loop) aborting its emulation and returning doesn't make Infinite_Loop halt, and HHH(Infinite_Loop) is correct that UTM(Infinite_Loop) will not halt.On 2024-11-03 15:13:56 +0000, olcott said:void Infinite_Loop()
>On 11/3/2024 7:04 AM, Mikko wrote:>On 2024-11-02 12:24:29 +0000, olcott said:>
>HHH does compute the mapping from its input DDD>
to the actual behavior that DDD specifies and this
DOES INCLUDE HHH emulating itself emulating DDD.
Yes but not the particular mapping required by the halting problem.
Yes it is the particular mapping required by the halting problem.
The exact same process occurs in the Linz proof.
The halting probelm requires that every halt decider terminates.
If HHH(DDD) terminates so does DDD. The halting problmen requires
that if DDD terminates then HHH(DDD) accepts as halting.
>
{
HERE: goto HERE;
return;
}
No that is false.
The measure is whether a C function can possibly
reach its "return" instruction final state. Your
measure determines that Infinite_Loop() halts.
Which is based on you LIES about what needs to be done, because you are just too stupid to understand what you say,If HHH(DDD) rejects as non-halting then HHH is not a halt decider.*ChatGPT's own words*
If HHH(DDD) does not terminate then HHH is not a halt deider.
>
It is easy to construct a decider that accpets DDD but that decider
is not HHH.
>
*Simplified Analogy*
Think of HHH as a "watchdog" that steps in during real execution
to stop DDD() from running forever. But when HHH simulates DDD(),
it's analyzing an "idealized" version of DDD() where nothing stops the
recursion. In the simulation, DDD() is seen as endlessly recursive, so
HHH concludes that it *would not halt* without external intervention.
https://chatgpt.com/share/67158ec6-3398-8011-98d1-41198baa29f2
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