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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Proof that DDD specifies non-halting behavior --- point by point
 --- in our head
Date: Wed, 14 Aug 2024 08:34:50 -0500
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On 8/14/2024 6:22 AM, Richard Damon wrote:
> On 8/14/24 12:24 AM, olcott wrote:
>> On 8/13/2024 11:04 PM, Richard Damon wrote:
>>> On 8/13/24 11:48 PM, olcott wrote:
>>>> On 8/13/2024 10:21 PM, Richard Damon wrote:
>>>>> On 8/13/24 10:38 PM, olcott wrote:
>>>>>> On 8/13/2024 9:29 PM, Richard Damon wrote:
>>>>>>> On 8/13/24 8:52 PM, olcott wrote:
>>>>>>>> void DDD()
>>>>>>>> {
>>>>>>>>    HHH(DDD);
>>>>>>>>    return;
>>>>>>>> }
>>>>>>>>
>>>>>>>> _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]
>>>>>>>>
>>>>>>>> A simulation of N instructions of DDD by HHH according to
>>>>>>>> the semantics of the x86 language is necessarily correct.
>>>>>>>>
>>>>>>>
>>>>>>> Nope, it is just the correct PARTIAL emulation of the first N 
>>>>>>> instructions of DDD, and not of all of DDD, 
>>>>>>
>>>>>> That is what I said dufuss.
>>>>>
>>>>> Nope. You didn't. I added clairifying words, pointing out why you 
>>>>> claim is incorrect.
>>>>>
>>>>> For an emulation to be "correct" it must be complete, as partial 
>>>>> emulations are only partially correct, so without the partial 
>>>>> modifier, they are not correct.
>>>>>
>>>>
>>>> A complete emulation of one instruction is
>>>> a complete emulation of one instruction
>>>
>>>
>>>
>>>>
>>>>>>
>>>>>>>
>>>>>>>> A correct simulation of N instructions of DDD by HHH is
>>>>>>>> sufficient to correctly predict the behavior of an unlimited
>>>>>>>> simulation.
>>>>>>>
>>>>>>> Nope, if a HHH returns to its caller, 
>>>>>>
>>>>>> *Try to show exactly how DDD emulated by HHH returns to its caller*
>>>>>> (the first one doesn't even have a caller)
>>>>>> Use the above machine language instructions to show this.
>>>>>>
>>>>>
>>>>> Remember how English works:
>>>>>
>>>>> When you ask "How DDD emulated by HHH returns to its callers".
>>>>
>>>> Show the exact machine code trace of how DDD emulated
>>>> by HHH (according to the semantics of the x86 language)
>>>> reaches its own machine address 00002183
>>>
>>> No. The trace is to long, 
>>
>> Show the Trace of DDD emulated by HHH
>> and show the trace of DDD emulated by HHH
>> emulated by the executed HHH
>> Just show the DDD code traces.
>>
> 
> First you need to make a DDD that meets the requirements, and that means 
> that it calls an HHH that meets the requirements.
> 

_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]

The is a hypothetical mental exercise and can be
accomplished even if the only DDD in the world
was simply typed into a word processor and never run.

HHH can be purely imaginary yet must emulate the
above code and itself according to the semantics
of the x86 language. In this case HHH is a pure
emulator.

On this basis we know that such an HHH would
emulate the first four instructions of DDD.

This includes a calls from the emulated DDD
to an emulated HHH(DDD).

This emulated HHH would emulate the first
four instructions of DDD.

We can do that all in our head never needing
any actually existing HHH.

All other points are moot and will simply be erased
until we have mutual agreement on this point.

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer