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

On 7/22/24 2:31 PM, olcott wrote:

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.
  

And each HHH creates a diffferent program DDD, and if HHH aborts, it is wrong because that DDD halts, and if HHH doesn't abort, it is wrong because it doesn't answer.
*NO* HHH is correct, NONE of them to ANY of the problems given to that one.
You are just too stupid to see that the PROGRAM DDD depends on the HHH that it is paired with, and thus you can't use one HHH's DDD to show the behavior of a different HHH's DDD.
That you do, just shows that you are nothing but a stupid LIAR.