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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: DDD correctly emulated by HHH is correctly rejected as
 non-halting.
Date: Mon, 15 Jul 2024 08:21:35 -0500
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On 7/15/2024 2:52 AM, Mikko wrote:
> On 2024-07-14 14:44:27 +0000, olcott said:
> 
>> On 7/14/2024 3:48 AM, Mikko wrote:
>>> On 2024-07-13 12:19:36 +0000, olcott said:
>>>
>>>> On 7/13/2024 2:55 AM, Mikko wrote:
>>>>> On 2024-07-12 13:28:15 +0000, olcott said:
>>>>>
>>>>>> On 7/12/2024 3:27 AM, Mikko wrote:
>>>>>>> On 2024-07-11 14:02:52 +0000, olcott said:
>>>>>>>
>>>>>>>> On 7/11/2024 1:22 AM, Mikko wrote:
>>>>>>>>> On 2024-07-10 15:03:46 +0000, olcott said:
>>>>>>>>>
>>>>>>>>>> typedef void (*ptr)();
>>>>>>>>>> int HHH(ptr P);
>>>>>>>>>>
>>>>>>>>>> void DDD()
>>>>>>>>>> {
>>>>>>>>>>    HHH(DDD);
>>>>>>>>>> }
>>>>>>>>>>
>>>>>>>>>> int main()
>>>>>>>>>> {
>>>>>>>>>>    HHH(DDD);
>>>>>>>>>> }
>>>>>>>>>>
>>>>>>>>>> We stipulate that the only measure of a correct emulation
>>>>>>>>>> is the semantics of the x86 programming language. By this
>>>>>>>>>> measure when 1 to ∞ steps of DDD are correctly emulated by
>>>>>>>>>> each pure function x86 emulator HHH (of the infinite set
>>>>>>>>>> of every HHH that can possibly exist) then DDD cannot
>>>>>>>>>> possibly reach past its own machine address of 0000216b
>>>>>>>>>> and halt.
>>>>>>>>>
>>>>>>>>> For every instruction that the C compiler generates the x86 
>>>>>>>>> language
>>>>>>>>> specifies an unambiguous meaning, leaving no room for "can".
>>>>>>>>>
>>>>>>>>
>>>>>>>> then DDD cannot possibly reach past its own machine
>>>>>>>> address of 0000216b and halt.
>>>>>>>
>>>>>>> As I already said, there is not room for "can". That means there is
>>>>>>> no room for "cannot", either. The x86 semantics of the unshown code
>>>>>>> determines unambigously what happens.
>>>>>>>
>>>>>>
>>>>>> Of an infinite set behavior X exists for at least one element
>>>>>> or behavior X does not exist for at least one element.
>>>>>> Of the infinite set of HHH/DDD pairs zero DDD elements halt.
>>>>>
>>>>> That is so far from the Common Language that I can't parse.
>>>>>
>>>>
>>>> *This proves that every rebuttal is wrong somewhere*
>>>> No DDD instance of each HHH/DDD pair of the infinite set of
>>>> every HHH/DDD pair ever reaches past its own machine address of
>>>> 0000216b and halts thus proving that every HHH is correct to
>>>> reject its input DDD as non-halting.
>>>
>>> Here you attempt to use the same name for a constant programs and 
>>> univesally
>>> quantifed variable with a poorly specified range. That is a form of a 
>>> well
>>> known mistake called the "fallacy of equivocation".
>>
>> I incorporated your suggestion in my paper.
>> DDD is a fixed constant finite string that calls its
>> HHH at the same fixed constant machine address.
> 
> That does not make sense. Which HHH does that DDD call? Which HHH
> is at that fixed machine address?
> 

HHH₁ to HHH∞ forming an infinite set of HHH/DDD pairs

HHH₁/DDD₁ to HHH∞/DDD∞ is another way to specify this
infinite set of HHH/DDD pairs.

>> 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 running.
> 
> It is not possible to execute more steps than there are, so you add that
> the emulation may terminate earlier if the input halts. Unless you
> only want to prove that those programs don't halt unless the halt.
> 

Of the infinite set of HHH/DDD pairs no DDD ever halts thus
every HHH that halts is necessarily correct to reject its DDD
as non-halting.

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