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

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:

>

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.

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. In other words, some HHH abort, some HHH do not abort.
Both produce an incorrect simulation, when they try to simulate themselves.
(a) Some HHH abort and they cause an incorrect simulation when simulating themselves, because they abort too soon, making the simulation incomplete and, therefore, incorrect.
(b) Some HHH never abort and they do not finish the simulation of themselves.
Both (a) and (b) result in incorrect simulations.
HHH cannot possibly simulate itself correctly.
When analysing (a), there is no need to dream about (b). The HHH that aborts will simulate itself, a HHH that aborts. So, in that case no abort is needed for the simulated HHH, because the simulated HHH is encoded to abort. However, the simulating HHH performs the unneeded abort, because it is encoded to do so.
When analysing (b), there is no need to dream about (a). The HHH that does not abort, should abort its own simulation, but cannot do that, because it is encode to never abort.