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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: Infinite set of HHH/DDD pairs
Date: Mon, 22 Jul 2024 14:32:20 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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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.

-- 
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.