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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: Infinite set of HHH/DDD pairs --- truisms
Date: Mon, 22 Jul 2024 21:19:18 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Mon, 22 Jul 2024 14:57:21 -0500 schrieb olcott:
> On 7/22/2024 2:30 PM, Fred. Zwarts wrote:
>> Op 22.jul.2024 om 20:31 schreef olcott:
>>> 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:

>>>>>>>>> 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.

>>>>> 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.
Please no ableism.
>>> (a) At least one HHH aborts.
>>> (b) No HHH ever aborts.
>>> Every X has property Y or not, there is no inbetween.
>> 
>> Do you have difficulty reading and writing English?
>> If every X has property Y or not, then it is clear that every HHH abort
>> or not.

> If the first HHH waits on the second HHH and the second waits on the
> third... Then no HHH ever aborts.
Yes, exactly. That's one half of the contradiction. The other half:
When all of them abort, all of them are wrong to do so, because
what they are simulating also aborts, making the abort unnecessary.

-- 
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.