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From: joes <noreply@example.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 11:18:40 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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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?

-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.