Re: DDD incorrectly emulated by HHH is inCorrectly rejected as non-halting V2

On 7/17/24 9:43 AM, olcott wrote:

On 7/17/2024 8:35 AM, Fred. Zwarts wrote:

Op 17.jul.2024 om 15:02 schreef olcott:

On 7/17/2024 1:48 AM, Mikko wrote:

On 2024-07-16 15:57:04 +0000, olcott said:
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New slave_stack at:1038c4
Begin Local Halt Decider Simulation   Execution Trace Stored at:1138cc
[00002172][001138bc][001138c0] 55         push ebp      ; housekeeping
[00002173][001138bc][001138c0] 8bec       mov ebp,esp   ; housekeeping
[00002175][001138b8][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][001138b4][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
New slave_stack at:14e2ec
[00002172][0015e2e4][0015e2e8] 55         push ebp      ; housekeeping
[00002173][0015e2e4][0015e2e8] 8bec       mov ebp,esp   ; housekeeping
[00002175][0015e2e0][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][0015e2dc][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

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The trace does not show that HHH returns so there is no basis to
think that HHH is a decider.
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The trace shows the data of the executed program of HHH that
does halt.

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It shows some of the data, not all, and in particular, not the halting.
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_DDD()
[00002163] 55         push ebp      ; housekeeping
[00002164] 8bec       mov ebp,esp   ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404     add esp,+04
[00002173] 5d         pop ebp
[00002174] c3         ret
Size in bytes:(0018) [00002174]
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DDD emulated by HHH according to the semantic meaning of
its x86 instructions never stop running unless aborted.
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You have shown that you do not understand the semantics of the x86 language.
HHH does abort and halt after N cycles,

That is counter-factual
 When we examine the infinite set of every HHH/DDD pair such that:
HHH1  One step of DDD is correctly emulated by HHH
HHH2  Two steps of DDD are correctly emulated by HHH
HHH3  Three steps of DDD are correctly emulated by HHH
...
HHH∞  The emulation of DDD by HHH never stops

Right, but the DDD being EMULATED does (as long as the code of the HHH shows that any copy of it wll return to its called)/

 *THIS IS SELF EVIDENT THUS DISAGREEMENT IS INCORRECT*
DDD emulated by any pure function HHH according to the
semantic meaning of its x86 instructions never stops
running unless aborted.
 

And youj are just showing you INTENTIONALLY trying to be deceptive.
Yes, the PARTIAL EMULATION by HHH doesn't reach that point, but DDD it self does, and THAT is what the question to HHH asks about if we are in the area of computation theory.
You are just proving yourself to be just a damn liar.