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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Who here understands that the last paragraph is Necessarily true?
Date: Fri, 19 Jul 2024 11:33:15 -0400
Organization: i2pn2 (i2pn.org)
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On 7/19/24 10:49 AM, olcott wrote:
> On 7/19/2024 4:14 AM, Mikko wrote:
>> On 2024-07-18 14:18:51 +0000, olcott said:
>>
>>> On 7/18/2024 3:41 AM, Fred. Zwarts wrote:
>>>> Op 17.jul.2024 om 16:56 schreef olcott:
>>>>> On 7/17/2024 9:32 AM, Fred. Zwarts wrote:
>>>>>> Op 17.jul.2024 om 16:20 schreef olcott:
>>>>>>> On 7/17/2024 8:54 AM, Fred. Zwarts wrote:
>>>>>>>> Op 17.jul.2024 om 15:27 schreef olcott:
>>>>>>>>>
>>>>>>>>> HHH is not allowed to report on the behavior of it actual self
>>>>>>>>> in its own directly executed process. HHH is allowed to report on
>>>>>>>>> the effect of the behavior of the simulation of itself 
>>>>>>>>> simulating DDD.
>>>>>>>>>
>>>>>>>>
>>>>>>>> But only on the effect of a correct simulation.
>>>>>>>
>>>>>>> _DDD()
>>>>>>> [00002163] 55         push ebp      ; housekeeping
>>>>>>> [00002164] 8bec       mov ebp,esp   ; housekeeping
>>>>>>> [00002166] 6863210000 push 00002163 ; push DDD
>>>>>>> [0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
>>>>>>> [00002170] 83c404     add esp,+04
>>>>>>> [00002173] 5d         pop ebp
>>>>>>> [00002174] c3         ret
>>>>>>> Size in bytes:(0018) [00002174]
>>>>>>>
>>>>>>> *THIS IS SELF EVIDENT THUS DISAGREEMENT IS INCORRECT*
>>>>>>> DDD emulated by any pure function HHH according to the
>>>>>>> semantic meaning of its x86 instructions never stops
>>>>>>> running unless aborted.
>>>>>>>
>>>>>>
>>>>>> It is self evident that a program that aborts will halt.
>>>>>> The semantics of the x86 code of a halting program is also 
>>>>>> self-evident: it halts.
>>>>>> So, the aborting HHH, when simulated correctly, stops.
>>>>>> Dreaming of a HHH that does not abort is irrelevant.
>>>>>>
>>>>>
>>>>> That is all the dishonest dodge of the strawman deception.
>>>>> HHH is required to halt by its design spec.
>>>>>
>>>>> _DDD()
>>>>> [00002163] 55         push ebp      ; housekeeping
>>>>> [00002164] 8bec       mov ebp,esp   ; housekeeping
>>>>> [00002166] 6863210000 push 00002163 ; push DDD
>>>>> [0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
>>>>> [00002170] 83c404     add esp,+04
>>>>> [00002173] 5d         pop ebp
>>>>> [00002174] c3         ret
>>>>> Size in bytes:(0018) [00002174]
>>>>>
>>>>> *THIS IS SELF EVIDENT THUS DISAGREEMENT IS INCORRECT*
>>>>> DDD emulated by any pure function HHH according to the
>>>>> semantic meaning of its x86 instructions never stops
>>>>> running unless aborted.
>>>>>
>>>>
>>>> Dreaming of a HHH that does not halt, when we are talking about a 
>>>> HHH that aborts and halts is irrelevant. Therefore, the 'unless 
>>>> aborted' is irrelevant. The semantics of the x86 instructions are 
>>>> self-evident: HHH halts.
>>>
>>> When you are hungry you remain hungry until you eat.
>>>    Before HHH(DDD) aborts its emulation the directly
>>>    executed DDD() cannot possibly halt.
>>>
>>> After you eat you are no longer hungry.
>>>    After HHH(DDD) aborts its emulation the directly
>>>    executed DDD() halts.
>>
>> If DDD does not halt it indicates that HHH is faulty. Therefore the
>> interesting question is whether DDD halts, not when DDD halts.
>>
> 
> *By your same reasoning*
> If Infinite_Loop() does not halt HHH is faulty.
> 
> In other words if Infinite_Loop()  is an actual infinite
> loop then this is all the fault of HHH.
> 

No, because Infinite_Loop just never halts when run, and doesn't depend 
on the behavior of HHH.

DDD does depend on the behavior of HHH, so HHH to CORRECT analyze its 
behavior must correct decide on its own behavior. If HHH will halt, then 
HHH needs to include that knowledge in its analysis of DDD, but it 
doesn't, so its analysis is just wrong.

HHH could have used the Flibble method, and gotten the answer, but it 
failed to do that, so it failed at a solvable problem.

> void Infinite_Loop()
> {
>    HERE: goto HERE;
> }
> 
> int main()
> {
>    HHH(Infinite_Loop);
> }
>