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

Op 23.jul.2024 om 21:17 schreef olcott:

On 7/23/2024 1:55 PM, Fred. Zwarts wrote:

Op 23.jul.2024 om 20:18 schreef olcott:

On 7/23/2024 1:13 PM, Fred. Zwarts wrote:

Op 23.jul.2024 om 19:54 schreef olcott:

On 7/23/2024 11:46 AM, Fred. Zwarts wrote:

Op 23.jul.2024 om 17:03 schreef olcott:

On 7/23/2024 3:17 AM, Fred. Zwarts wrote:

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.

>
So you don't even know how foot races work.
>
The winner of the race is not supposed to wait
so that everyone crosses the finish line at once.

>
But do you know how foot races work? When the winner reaches the finish, the other players do not disappear. They do not halt either. They continue until they also reach the finish line.
>

>
Also with HHH(DDD) there are infinite instances
in the race. Waiting for the last one to finish
waits forever. The first one to cross the finish
line ends the race. Every other instance is
immediately incapacitated

You don't have to wait for the last one of an infinite number. It is sufficient to wait for the first simulated HHH, because that one aborts and halts

>
Since HHH is the exact same machine code then when the first one waits
for the second one the second one waits for the third on and on forever.
>

>
Indeed! Exactly! You are almost at the finish.

>
*THIS SEEMS PERMANENTLY TOO DIFFICULT FOR YOU TO UNDERSTAND*

>
Irrelevant nonsense ignores, because olcott seems to have difficulties with reading and writing English.
>

>
Unless the first one aborts none of them ever aborts and HHH
itself never stops running. Therefore is is correct and necessary
for the first HHH to abort.

>
But then it fails at another point. When it aborts, it aborts too soon,

 *This is just over your head no sense continuing the dialogue*

Irrelevant nonsense ignored.

Unless the first one aborts NONE OF THEM EVER ABORT
Unless the first one aborts NONE OF THEM EVER ABORT
Unless the first one aborts NONE OF THEM EVER ABORT
Unless the first one aborts NONE OF THEM EVER ABORT
Unless the first one aborts NONE OF THEM EVER ABORT
Unless the first one aborts NONE OF THEM EVER ABORT
Unless the first one aborts NONE OF THEM EVER ABORT
Unless the first one aborts NONE OF THEM EVER ABORT
Unless the first one aborts NONE OF THEM EVER ABORT
Unless the first one aborts NONE OF THEM EVER ABORT
Unless the first one aborts NONE OF THEM EVER ABORT
 

So, you prove that HHH cannot possibly simulate itself correctly.
When HHH does not abort, it needs a simulator that aborts, but when it simulates itself that is not the case, so the simulation is wrong.
When HHH aborts, it needs a simulator that does not abort, which is not the case when it simulates itself, so the simulation is also wrong.
Both aborting an non-aborting HHH variants are unable to simulate their own x86 code correctly.
It is a pity that you cannot even see such simple fact.