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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Anyone with sufficient knowledge of C knows that DD specifies
 non-terminating behavior to HHH
Date: Mon, 10 Feb 2025 07:41:25 -0500
Organization: i2pn2 (i2pn.org)
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On 2/10/25 7:04 AM, olcott wrote:
> On 2/10/2025 5:18 AM, joes wrote:
>> Am Sun, 09 Feb 2025 13:57:03 -0600 schrieb olcott:
>>> On 2/9/2025 1:39 PM, joes wrote:
>>>> Am Sun, 09 Feb 2025 10:49:51 -0600 schrieb olcott:
>>>>> On 2/9/2025 10:43 AM, Fred. Zwarts wrote:
>>>>>> Op 09.feb.2025 om 17:37 schreef olcott:
>>>>>>> On 2/9/2025 9:53 AM, Fred. Zwarts wrote:
>>>>>>>> Op 09.feb.2025 om 16:15 schreef olcott:
>>>>>>>>> On 2/9/2025 2:09 AM, Fred. Zwarts wrote:
>>>>>>>>>> Op 09.feb.2025 om 07:04 schreef olcott:
>>>>>>>>>>> On 2/8/2025 3:49 PM, Fred. Zwarts wrote:
>>>>>>>>>>>> Op 08.feb.2025 om 15:43 schreef olcott:
>>>>>>>>>>>>> On 2/8/2025 3:54 AM, Fred. Zwarts wrote:
>>>>>>>>>>>>>> Op 08.feb.2025 om 00:13 schreef olcott:
>>>>
>>>>>>>>> The input to HHH(DD) cannot possibly terminate normally. Referring
>>>>>>>>> to some other DD does not change this verfied fact.
>>>>>>>>>
>>>>>>>> That DD halts is a verified fact.
>>>>>>> The input to HHH(DD) DOES NOT HALT !!!
>>>>>>
>>>>>> It is a verified fact that the finite string describes a halting
>>>>>> program. Du to a bug, HHH does not see that, because it investigates
>>>>>> only the first few instructions of DD. HHH is unable to process the
>>>>>> call from DD to HHH correctly.
>>>>>
>>>>> DD simulated by HHH cannot possibly terminate normally. DD simulated
>>>>> by HHH does specify the behavioral basis of the Boolean termination
>>>>> value of the DD input to HHH.
>>>>
>>>> DD terminates, and HHH can’t simulate it normally.
>>>>
>>> That is not the same DD as the input to HHH(DD). That DD has an entirely
>>> different execution trace.
> 
>> That is entirely due to how HHH chooses to missimulate it, namely by not
>> calling itself, but a different version that doesn’t abort.
>> Why do you not pass the same DD as an input to HHH?
>>
> 
> The same machine address of DD is the only reference to DD that
> any termination analyzer ever sees.
> 
> 

And thus your model is incomplete.

The address alone isn't enough data to answer the question, for 
instance, how can you answer, as a PURE function, what the halting 
behavior is for the program at 0x2BAD is?

Remember, a PURE FUNCITION can only look at "its input", and not any 
other memory.

Thus, if we pass the address of an object, we treat the input as the 
objet pointed to by that pointer.

And thus, your decider needs to consider as its input ALL of the code 
used by that address pointed to by that address, and as such, the code 
for the HHH that DD calls has been FIXED as part of the input.

Thus, you can't change it to look at you concept of "all HHH", but need 
to use a copy in another address, like HHH1 is, and that example shows 
that some verision of HHH *CAN* simulate to the end of that specific 
input (which included the code of the original HHH)/

Sorry, you are just proving that you are ignorant of what you talk 
about, and are so stupid that you cant learn that material, or even 
understand that you need to learn it.

Too Stupid to see you are Stupid,