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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: People are still trying to get away with disagreeing with the
 semantics of the x86 language
Date: Wed, 3 Jul 2024 20:25:18 +0200
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Op 03.jul.2024 om 19:58 schreef olcott:
> On 7/3/2024 12:51 PM, Fred. Zwarts wrote:
>> Op 03.jul.2024 om 18:03 schreef olcott:
>>>
> 
> _DDD()
> [00002172] 55               push ebp      ; housekeeping
> [00002173] 8bec             mov ebp,esp   ; housekeeping
> [00002175] 6872210000       push 00002172 ; push DDD
> [0000217a] e853f4ffff       call 000015d2 ; call HHH(DDD)
> [0000217f] 83c404           add esp,+04
> [00002182] 5d               pop ebp
> [00002183] c3               ret
> Size in bytes:(0018) [00002183]
> 
>>> DDD is correctly emulated by HHH which calls an emulated HHH(DDD)
>>> to repeat this process an endless number of times until aborted
>>> or out-of-memory error.
>> Anyone knowing the x86 language knows that a program cannot be 
>> programmed to do two different things
>> It cannot do both run out of memory *and* abort.
> 
> DDD correctly emulated by any element of the infinite
> set of every pure function HHH cannot possibly reach
> its own ret instruction and halt. 

Exactly! Well done! This proves that HHH cannot possibly correctly 
simulate itself. If it aborts, it does so one cycle too soon.

> That HHH aborts its
> emulation at some point or never aborts its emulation
> cannot possibly change this.

Indeed, an abort does not make the simulation correct and an infinite 
simulation is not correct either.

> 
>> So make up your mind. What does it do?
>> And what does "endless number of times until aborted" mean? Does it 
>> abort after an infinite number of steps?
>> Make clear what you mean. After how many cycles is the simulation 
>> aborted?
>