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On 8/16/2024 1:42 AM, Fred. Zwarts wrote:Op 16.aug.2024 om 04:24 schreef olcott:On 8/15/2024 8:57 PM, Richard Damon wrote:On 8/15/24 8:12 AM, olcott wrote:On 8/15/2024 2:00 AM, joes wrote:Am Wed, 14 Aug 2024 16:07:43 +0100 schrieb Mike Terry:On 14/08/2024 08:43, joes wrote:Am Tue, 13 Aug 2024 21:38:07 -0500 schrieb olcott:On 8/13/2024 9:29 PM, Richard Damon wrote:On 8/13/24 8:52 PM, olcott wrote:
DDD would halt. HHH doesn't simulate that.DDD emulated by HHH according to the semantics of the x86 language wouldYou do not understand what we say. I have repeated many times that the>>how *HHH* returns*Try to show exactly how DDD emulated by HHH returns to itsA correct simulation of N instructions of DDD by HHH isNope, if a HHH returns to its caller,
sufficient to correctly predict the behavior of an unlimited
simulation.
caller*DDDHHH simulates DDD enter the matrix
DDD calls HHH(DDD) Fred: could be eliminated HHH
simulatesvoilasecond level
DDD calls HHH(DDD) recursion detected
HHH aborts, returns outside interference DDD haltsHHH halts>
You're misunderstanding the scenario? If your simulated HHH
aborts its simulation [line 5 above],
then the outer level H would have aborted its identical simulation
earlier. You know that, right?
That is the part that Joes and Fred do not understand.
>
simulated HHH is aborted before it would halt by itself.
never halt by itself.
DDD's return is reachable from the simulated HHH's abort.The simulating HHH fails to reach that point, proving an incompletevoid Infinite_Recursion()
simulation.
{
Infinite_Recursion();
OutputString("I never make it here!\n");
}
Unreachable code is unreachable code.
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