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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: Tue, 16 Jul 2024 21:12:36 -0400
Organization: i2pn2 (i2pn.org)
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On 7/16/24 9:41 AM, olcott wrote:
> On 7/16/2024 2:28 AM, Fred. Zwarts wrote:
>> Op 15.jul.2024 om 21:40 schreef olcott:
>>> On 7/15/2024 2:30 PM, Fred. Zwarts wrote:
>>>> Op 15.jul.2024 om 04:33 schreef olcott:
>>>>> On 7/14/2024 9:04 PM, Richard Damon wrote:
>>>>>> On 7/14/24 9:27 PM, olcott wrote:
>>>>>>>
>>>>>>> Any input that must be aborted to prevent the non termination
>>>>>>> of simulating termination analyzer HHH necessarily specifies
>>>>>>> non-halting behavior or it would never need to be aborted.
>>>>>>
>>>>>> Excpet, as I have shown, it doesn't.
>>>>>>
>>>>>> Your problem is you keep on ILEGALLY changing the input in your 
>>>>>> argument because you have misdefined what the input is.
>>>>>>
>>>>>
>>>>> _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]
>>>>>
>>>>> The input *is* the machine address of this finite
>>>>> string of bytes: 558bec6863210000e853f4ffff83c4045dc3
>>>>>
>>>>
>>>> It seems that you do not understand x86 language. The input is not a 
>>>> string of bytes, but an address (00002163). This points to the 
>>>> starting of the code of DDD. But a simulation needs a program, not a 
>>>> function calling undefined other functions. Therefore, all functions 
>>>> called by DDD (such as HHH) are included in the code to simulate.
>>>
>>> *The input is the machine address of this finite*
>>> *string of bytes: 558bec6863210000e853f4ffff83c4045dc3*
>>
>> It seems that olcott does not understand the x86 language.
>> The input for HHH is an address, not a finite string. 
> 
> This is such a terribly incorrect reply that I am ignoring it. Try again
> 

No, it was a VERY RIGHT reply.

You DON'T know the x86 language, particularly how a call instruction works.