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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: DDD incorrectly emulated by HHH is inCorrectly rejected as
 non-halting V2
Date: Wed, 17 Jul 2024 19:56:39 -0400
Organization: i2pn2 (i2pn.org)
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On 7/17/24 2:13 PM, olcott wrote:
> On 7/17/2024 12:18 PM, joes wrote:
>> Am Wed, 17 Jul 2024 08:43:04 -0500 schrieb olcott:
>>> 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:
>>
>>>>>>>> The trace does not show that HHH returns so there is no basis to
>>>>>>>> think that HHH is a decider.
>>>>>>> The trace shows the data of the executed program of HHH that does
>>>>>>> halt.
>>>>>> It shows some of the data, not all, and in particular, not the
>>>>>> halting.
>>>>> DDD emulated by HHH according to the semantic meaning of its x86
>>>>> instructions never stop running unless aborted.
>> Bla bla.
>>>> 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
>> Then HHH is not a decider.
>>
>>> 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
>>>
>>> DDD emulated by any pure function HHH according to the semantic meaning
>>> of its x86 instructions never stops running unless aborted.
>> DDD only calls HHH, which, being a decider, halts.
>>
> 
> _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]
> 
> You didn't bother to pay close enough attention.
> I referred to every pure function HHH that can possibly exist.
> 
> In each case DDD never makes it past it fourth instruction.
> This means that every HHH that halts is correct to reject its
> DDD as non-halting. Not every HHH halts.
> 

Of course it does, since the code at 000015c3 is part of DDD, or you are 
just admitting you don't understand what a program is or what a correcgt 
x86 emulaiton is.

I guess you are just admitting to being a LYING IDIOT.