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On 11/2/2024 10:44 AM, Richard Damon wrote:No, you just need to know the RESULTS of the emulation of the input even if you emulate it for an unlimited number of steps. You don't need to actually do it if you can prove what it would be.On 11/2/24 8:24 AM, olcott wrote:In other words you continue to perpetually insist on>>
When the main motive of people like Richard is to derail
any chance of mutual agreement I cannot proceed with all
of the steps achieving mutual agreement on each step one
at a time in their mandatory prerequisite order.
No, my "motive" is to hold cranks to the truth, or at least get them to admit that they are off in some other system, that they can define.
>
You keep on wanting to be in the system (since it provides the proof of the things you don't like) but can't hold yourself to actually be in the system.
>>>
void DDD()
{
HHH(DDD);
return;
}
>
_DDD()
[000020a2] 55 push ebp ; housekeeping
[000020a3] 8bec mov ebp,esp ; housekeeping
[000020a5] 68a2200000 push 000020a2 ; push DDD
[000020aa] e8f3f9ffff call 00001aa2 ; call H0
[000020af] 83c404 add esp,+04 ; housekeeping
[000020b2] 5d pop ebp ; housekeeping
[000020b3] c3 ret ; never gets here
Size in bytes:(0018) [000020b3]
>
DDD emulated by HHH according to the semantics of the x86
language cannot possibly reach its own "return" instruction
whether or not any HHH ever aborts its emulation of DDD.
Equivocation between looking at the behavor of DDD being the actual program (which include a particular version of HHH) and the behavior of a PARTIAL emulation of DDD by HHH, which ends up not having the property you want to show.
>
Partial doesn't lead to showing never.
>
the ridiculously stupid idea of requiring the complete
emulation of a non-terminating input.
I don't think this is: stupidity, ignorance, ADD.
I don't know what this leaves besides dishonesty with malice.
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