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From: joes <noreply@example.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: Two dozen people were simply wrong --- Try to prove otherwise ---
 pinned down
Date: Sat, 1 Jun 2024 19:12:18 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Sat, 01 Jun 2024 11:22:05 -0500 schrieb olcott:

> On 6/1/2024 11:06 AM, Fred. Zwarts wrote:
>> Op 01.jun.2024 om 17:51 schreef olcott:
>>> On 6/1/2024 10:32 AM, Fred. Zwarts wrote:
>>>> Op 01.jun.2024 om 17:17 schreef olcott:
>>>>> On 6/1/2024 3:57 AM, Fred. Zwarts wrote:
>>>>>> Op 01.jun.2024 om 01:57 schreef olcott:
>>>>>>> On 5/31/2024 6:33 PM, Richard Damon wrote:
>>>>>>>> On 5/31/24 6:54 PM, olcott wrote:
>>>>>>>>> On 5/31/2024 5:46 PM, Richard Damon wrote:
>>>>>>>>>> On 5/31/24 6:08 PM, olcott wrote:
>>>>>>>>>>> On 5/31/2024 4:36 PM, Richard Damon wrote:
>>>>>>>>>>>> On 5/31/24 10:10 AM, olcott wrote:
>>>>>>>>>>>>> On 5/31/2024 6:16 AM, Richard Damon wrote:
>>>>>>>>>>>>>> On 5/30/24 11:27 PM, olcott wrote:
>>>> Halting criteria are the same for all functions. If the direct
>>>> execution of HH(DD,DD) proves that HH halts, then the direct
>>>> execution of DD also proves that DD halts.
>>>
>>> *HH is required to report on the behavior that its input specifies* HH
>>> is not allowed to report on the behavior of DD(DD) {the computation
>>> that itself is contained within}.
>> 
>> The input of HH is HH as part of DD. (Remember DD calls HH.)
>> So, when HH is required to report about its input, it reports about the
>> behaviour of both DD and HH.
>> 
>> 
> HH correctly reports that because DD calls HH(DD,DD) in recursive
> simulation that DD never halts.
> 
> HHH(HH,DD,DD) would report that HH halts.
What is this now? All H behave the same, if given identical inputs.

-- 
joes