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On 8/7/2024 1:25 PM, joes wrote:None of this was ever really required. We have completeAm Wed, 07 Aug 2024 08:40:31 -0500 schrieb olcott:Maybe I can do this. it requires a static local, yet thisOn 8/7/2024 2:22 AM, Mikko wrote:>On 2024-08-05 15:00:12 +0000, olcott said:On 8/5/2024 2:44 AM, Mikko wrote:On 2024-08-04 13:11:56 +0000, olcott said:On 8/4/2024 1:26 AM, Fred. Zwarts wrote:Op 03.aug.2024 om 17:20 schreef olcott:>>You could make it clearer by prefixing the simulation level.It does but it is too difficult to dig it out of emulations of emulators>In another message you have said that when HHH simulates itselfHHH and HH and the original H have proved that they simulate
simulating DDD is does not simulate itself simulating itself
simulating DDD. You have not told whether it makes a cup of coffee.
Neither action can be seen in the traces you have shown.
>
themselves simulating DDD, DD and P for three years now.
Your trace don't show siulation of exectuion differently from
simulation of simulation of execution.
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emulating inputs.
>
is only for reporting purposes and does not have an effect
on the computation.
We have had this complete proof for HH/DD and H/P for three--
years and everyone simply ignores it. This can only mean:
(a) dishonesty
(b) insufficient technical competence.
As soon as the first HHH sees the second DDD about to invoke a third HHH
it aborts the emulation. At this point DDD, the second HHH and the
second DDD all immediately stop running and HHH returns 0 to main.And HHH concludes that the second invocation of itself would somehowvoid DDD()
not also abort but run forever, and then returns that itself wouldn't
halt, and halts. ???
>
{
HHH(DDD);
return;
}
Any expert in C can tell that DDD simulated by HHH cannot
possibly ever reach its own "return" instruction final halt
state.
Is it possible for someone to have a PhD in computer science
and not have any clue about something as simple as this?
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