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On 8/3/2024 9:04 AM, Fred. Zwarts wrote:If HHH aborts and halts, it did it one cycle before the simulated HHH would halt and return.Op 03.aug.2024 om 15:50 schreef olcott:HHH(DDD) simulates DDD that calls HHH(DDD) to repeat the process.On 8/3/2024 3:14 AM, Fred. Zwarts wrote:>Op 02.aug.2024 om 22:57 schreef olcott:>Who here is too stupid to know that DDD correctly simulated>
by HHH cannot possibly reach its own return instruction?
>
void DDD()
{
HHH(DDD);
return;
}
>
Which proves that the simulation is incorrect.
When are you going to understand that you are not allowed
to disagree with the semantics of the x86 language?
>
>
I do not disagree.
When are you going to understand that it is a deviation of the semantics of the x86 language to skip instructions of a halting program,
If it does this an infinite number of times the simulated DDD
never reaches its own return instruction.
If it does this a googolplex number of times the simulated DDD
never reaches its own return instruction.
If the simulation is ever aborted the simulated DDD never reachesOnly, because the simulation prevented to reach tat instruction, because it aborts prematurely.
its own return instruction.
When we construe the halt state of DDD as its "return"That would be your dream of a HHH that does not halt.
instruction then the simulated DD never halts.
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