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

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.

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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.