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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Proof that DDD specifies non-halting behavior --- point by point
Date: Fri, 16 Aug 2024 00:02:17 -0400
Organization: i2pn2 (i2pn.org)
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On 8/15/24 10:26 PM, olcott wrote:
> 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.
> 

Then why does it say there were no conditional branches in the 
simulation of the code of the program "DDD" where there were in the 
simulation of the HHH that was called by DDD and thus part of the 
program DDD.

Or, are you going to admit that you fail to understand the definition of 
a program, and thus your last two decades of work just went up in the 
smoke of error due to self-created ignorance.