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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Proof that DDD specifies non-halting behavior --- point by point
Date: Thu, 15 Aug 2024 21:26:49 -0500
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On 8/15/2024 8:57 PM, Richard Damon wrote:
> On 8/15/24 10:58 AM, olcott wrote:
>> On 8/15/2024 3:19 AM, Mikko wrote:
>>> On 2024-08-14 04:04:23 +0000, Richard Damon said:
>>>
>>>> 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, and since you HHH doesn't meet your 
>>>> requirements (since it isn't a pure function) you can't give me a 
>>>> compldte input to trace.
>>>
>>> The trace is regular enough that we could define a formal language for
>>> the trace and construct an analyzer program to detect deviations from
>>> x86 semnatics and hidden inputs.
>>>
>>
>> There are no deviations. The x86utm operating system is
>> built from libx86emu that has had decades of development
>> effort. HHH really does emulate itself emulating DDD.
>>
> 
> And then ignores that emulation,

counter-factual but you don't care.

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