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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 11:30:20 -0400
Organization: i2pn2 (i2pn.org)
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On 7/19/24 10:56 AM, olcott wrote:
> On 7/19/2024 1:40 AM, Fred. Zwarts wrote:
>> Op 18.jul.2024 om 16:18 schreef olcott:
>>> 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.
>>
>> No, but HHH would have halted when not aborted, because that is how it 
>> is programmed. That is the semantics of its x86 code.
>>
> 
> int main { DDD(); } calls HHH(DDD) that must abort the
> emulation of its input or
> HHH, emulated DDD and executed DDD never stop running.
> 
> 

And since it DOES abort that simulation (or HHH fails to be the decider 
you claim it is) that means that DDD DOES stop rubnning.

This means you logic is nothing but a LIE, and that you are proven to be 
totally ignorant of how programming works.