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From: dbush <dbush.mobile@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 16:02:39 -0400
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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.

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.