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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: =?UTF-8?Q?Re=3A_Computable_Functions_---_finite_string_transformati?=
 =?UTF-8?Q?on_rules_---_0_=E2=89=A01?=
Date: Mon, 28 Apr 2025 15:10:14 -0500
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In-Reply-To: <vuomsu$3dd6e$13@dont-email.me>

On 4/28/2025 3:02 PM, dbush wrote:
> On 4/28/2025 3:53 PM, olcott wrote:
>> On 4/28/2025 2:10 PM, dbush wrote:
>>> On 4/28/2025 3:03 PM, olcott wrote:
>>>> On 4/28/2025 1:37 PM, dbush wrote:
>>>>> On 4/28/2025 2:32 PM, olcott wrote:
>>>>>> On 4/28/2025 11:11 AM, dbush wrote:
>>>>>>> On 4/28/2025 12:10 PM, olcott wrote:
>>>>>>>> On 4/28/2025 4:05 AM, Mikko wrote:
>>>>>>>>> On 2025-04-27 18:23:03 +0000, olcott said:
>>>>>>>>>
>>>>>>>>>> On 4/27/2025 4:51 AM, Mikko wrote:
>>>>>>>>>>> On 2025-04-26 16:15:44 +0000, olcott said:
>>>>>>>>>>>
>>>>>>>>>>>> _DD()
>>>>>>>>>>>> [00002133] 55         push ebp      ; housekeeping
>>>>>>>>>>>> [00002134] 8bec       mov ebp,esp   ; housekeeping
>>>>>>>>>>>> [00002136] 51         push ecx      ; make space for local
>>>>>>>>>>>> [00002137] 6833210000 push 00002133 ; push DD
>>>>>>>>>>>> [0000213c] e882f4ffff call 000015c3 ; call HHH(DD)
>>>>>>>>>>>> [00002141] 83c404     add esp,+04
>>>>>>>>>>>> [00002144] 8945fc     mov [ebp-04],eax
>>>>>>>>>>>> [00002147] 837dfc00   cmp dword [ebp-04],+00
>>>>>>>>>>>> [0000214b] 7402       jz 0000214f
>>>>>>>>>>>> [0000214d] ebfe       jmp 0000214d
>>>>>>>>>>>> [0000214f] 8b45fc     mov eax,[ebp-04]
>>>>>>>>>>>> [00002152] 8be5       mov esp,ebp
>>>>>>>>>>>> [00002154] 5d         pop ebp
>>>>>>>>>>>> [00002155] c3         ret
>>>>>>>>>>>> Size in bytes:(0035) [00002155]
>>>>>>>>>>>>
>>>>>>>>>>>> When any HHH emulates DD according to the finite
>>>>>>>>>>>> string transformation rules specified by the x86
>>>>>>>>>>>> language (the line of demarcation between correct
>>>>>>>>>>>> and incorrect emulation) no emulated DD can possibly
>>>>>>>>>>>> reach its final halt state and halt.
>>>>>>>>>>>
>>>>>>>>>>> There is a type error above. First DD is introduced as a 
>>>>>>>>>>> proper name.
>>>>>>>>>>> But later it is used in the phrase "no emulated DD" where the 
>>>>>>>>>>> rules
>>>>>>>>>>> of the language require a generic name.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> *This of this as an axiom schema*
>>>>>>>>>> No DD correctly emulated by any HHH can possibly
>>>>>>>>>> reach its final halt state. This conclusively
>>>>>>>>>> proves that every HHH is correct to reject its
>>>>>>>>>> input DD as non-halting.
>>>>>>>>>
>>>>>>>>> That cannot be used as a schema before you specify what symbols 
>>>>>>>>> in it are
>>>>>>>>> placeholders and what replacements can be used for the 
>>>>>>>>> placeholders.
>>>>>>>>>
>>>>>>>>
>>>>>>>> I have gone over this many hundreds of times
>>>>>>>> do you not remember anything that I already said?
>>>>>>>>
>>>>>>>> int DD()
>>>>>>>> {
>>>>>>>>    int Halt_Status = EEE(DD);
>>>>>>>>    if (Halt_Status)
>>>>>>>>      HERE: goto HERE;
>>>>>>>>    return Halt_Status;
>>>>>>>> }
>>>>>>>>
>>>>>>>> When each element of the set of x86 emulators
>>>>>>>> named EEE 
>>>>>>>
>>>>>>> Changing the input is not allowed.
>>>>>>
>>>>>> Hypothetical possibilities numbskull.
>>>>>>
>>>>>
>>>>> So you're hypothesizing changing the input.
>>>>>
>>>>> Changing the input, hypothetically or otherwise, is not allowed.
>>>>
>>>> Examining the infinite set of HHH/DD pairs simultaneously
>>>> IS NOT CHANGING THE INPUT DUMBO.
>>>>
>>>
>>> When DD runs, HHH runs.  So HHH is part of the input.  So when you 
>>> hypothesize changing HHH, you're hypothesizing changing the input.
>>>
>>> Changing the input, hypothetically or otherwise, is not allowed.
>>
>>
>>
>> HHH[0] emulates zero instructions of DD
>> HHH[1] emulates one  instructions of DD
>> HHH[n] emulates n    instructions of DD
>> HHH[∞] emulates ∞    instructions of DD
> 
> FALSE:
> 
> HHH[0] emulates zero instructions of DD[0]
> HHH[1] emulates one  instructions of DD[1]
> HHH[n] emulates n    instructions of DD[n]
> HHH[∞] emulates ∞    instructions of DD[∞]
> 
> 
>>
>> Each DD has the exactly same machine code
> 
> False, because the algorithm DD consists of the machine code of the 
> function DD, the function HHH, and the machine code of everything that 
> HHH calls down to the OS level.
> 

You can look at it that way.

> So when you hypothesize changing the code of the function HHH, you're 
> hypothesizing changing the input algorithm DD.
> 
> Changing the input is not allowed.


Of the freaking infinite set of every damn HHH/DD
pair that can possibly exist where DD is emulated
by HHH according to the finite string transformation
rules of the x86 language NOT A DAMN ONE OF THE EMULATED DD HALTS.

It seems that you are trying to stupidly get away with
saying that you don't know the difference between one
element of a set and all of the elements of this set.

I know that you are not that stupid, thus are ever
playing the Troll.

-- 
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer