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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Infinite set of HHH/DDD pairs --- truisms
Date: Mon, 22 Jul 2024 20:01:02 -0400
Organization: i2pn2 (i2pn.org)
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On 7/22/24 2:31 PM, olcott wrote:
> On 7/22/2024 12:45 PM, Fred. Zwarts wrote:
>> Op 22.jul.2024 om 17:08 schreef olcott:
>>> On 7/22/2024 9:32 AM, joes wrote:
>>>> Am Mon, 22 Jul 2024 09:13:33 -0500 schrieb olcott:
>>>>> On 7/22/2024 3:01 AM, Mikko wrote:
>>>>>> On 2024-07-21 13:50:17 +0000, olcott said:
>>>>>>> On 7/21/2024 4:38 AM, Mikko wrote:
>>>>>>>> On 2024-07-20 13:28:36 +0000, olcott said:
>>>>>>>>> On 7/20/2024 3:54 AM, Mikko wrote:
>>>>>>>>>> On 2024-07-19 14:39:25 +0000, olcott said:
>>>>>>>>>>> On 7/19/2024 3:51 AM, Mikko wrote:
>>>>
>>>>>>>>>> Anyway you did not say that some HHHᵢ can simulate the
>>>>>>>>>> corresponding DDDᵢ to its termination. And each DDDᵢ does
>>>>>>>>>> terminate, whether simulated or not.
>>>>
>>>>
>>>>>>> Then DDD correctly simulated by any pure function HHH cannot 
>>>>>>> possibly
>>>>>>> reach its own return instruction and halt, therefore every HHH is
>>>>>>> correct to reject its DDD as non-halting.
>>>>>> That does not follow. It is never correct to reject a halting
>>>>>> comoputation as non-halting.
>>>>> In each of the above instances DDD never reaches its return 
>>>>> instruction
>>>>> and halts. This proves that HHH is correct to report that its DDD 
>>>>> never
>>>>> halts.
>>>> It can't return if the simulation of it is aborted.
>>>>
>>>>> Within the hypothetical scenario where DDD is correctly emulated by 
>>>>> its
>>>>> HHH and this HHH never aborts its simulation neither DDD nor HHH ever
>>>>> stops running.
>>>> In actuality HHH DOES abort simulating.
>>>>
>>>>> This conclusively proves that HHH is required to abort the 
>>>>> simulation of
>>>>> its corresponding DDD as required by the design spec that every 
>>>>> partial
>>>>> halt decider must halt and is otherwise not any kind of decider at 
>>>>> all.
>>>> Like Fred recognised a while ago, you are arguing as if HHH didn't 
>>>> abort.
>>>>
>>>>> That HHH is required to abort its simulation of DDD conclusively 
>>>>> proves
>>>>> that this DDD never halts.
>>>> You've got it the wrong way around.
>>>>
>>>
>>> I am talking about hypothetical possible ways that HHH could be encoded.
>>> (a) HHH(DDD) is encoded to abort its simulation.
>>> (b) HHH(DDD) is encoded to never abort its simulation.
>>>
>>> Therefore (a) is correct and (b) is incorrect according to the
>>> design requirements for HHH that it must halt.
>>
>> Both are incorrect. An HHH, when encoded to abort does not need to be 
>> aborted when simulated, because it already halts on its own.
> 
> You must have attention deficit disorder.
> (a) At least one HHH aborts.
> (b) No HHH ever aborts.
> 
> Every X has property Y or not, there is no inbetween.
> 
> 
> 

And each HHH creates a diffferent program DDD, and if HHH aborts, it is 
wrong because that DDD halts, and if HHH doesn't abort, it is wrong 
because it doesn't answer.

*NO* HHH is correct, NONE of them to ANY of the problems given to that one.

You are just too stupid to see that the PROGRAM DDD depends on the HHH 
that it is paired with, and thus you can't use one HHH's DDD to show the 
behavior of a different HHH's DDD.

That you do, just shows that you are nothing but a stupid LIAR.