Re: Infinite set of HHH/DDD pairs --- truisms

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Sujet : Re: Infinite set of HHH/DDD pairs --- truisms
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory
Date : 22. Jul 2024, 16:08:03
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v7lskj$luh0$6@dont-email.me>
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User-Agent : Mozilla Thunderbird
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.
It is also a truism that any input that must be aborted
is a non-halting input.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

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