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

Op 22.jul.2024 om 21:57 schreef 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:

>

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.

 Sure and when we start a race with a single file line of
people that are 15 feet apart and everyone goes the same
speed then everyone will reach the finish line, eventually.
 When the first HHH that reaches the finish line stops
simulating its input then no other HHH can possibly reach
the finish line because nothing is simulating them.

Exactly! That is the error in HHH. It stops simulating before the other HHH could reach the finish line. The simulated HHH and the simulating HHH have the same finish line, because both abort after N cycles.
So, it is ncorrect to abort the simulated HHH, which would have reached the finish line on its own, if not aborted.

 If the first HHH waits on the second HHH and the second
waits on the third... Then no HHH ever aborts.
 

Excatly! That is why HHH cannot possibly simulate itself correctly. Do you finally get it?