Re: Flat out dishonest or totally ignorant? Can ADD be this severe?

On 7/4/24 8:41 AM, olcott wrote:

On 7/4/2024 1:42 AM, Mikko wrote:

On 2024-07-04 00:40:37 +0000, olcott said:
>

On 7/3/2024 6:18 PM, Richard Damon wrote:

On 7/3/24 10:19 AM, olcott wrote:

On 7/3/2024 9:11 AM, joes wrote:

Am Tue, 02 Jul 2024 22:55:12 -0500 schrieb olcott:

On 7/2/2024 10:50 PM, joes wrote:

Am Tue, 02 Jul 2024 14:46:38 -0500 schrieb olcott:

On 7/2/2024 2:17 PM, Fred. Zwarts wrote:

Op 02.jul.2024 om 21:00 schreef olcott:

On 7/2/2024 1:42 PM, Fred. Zwarts wrote:

Op 02.jul.2024 om 14:22 schreef olcott:

On 7/2/2024 3:22 AM, Fred. Zwarts wrote:

Op 02.jul.2024 om 03:25 schreef olcott:

>

HHH repeats the process twice and aborts too soon.

>
DDD is correctly emulated by any HHH that can exist which calls this
emulated HHH(DDD) to repeat the process until aborted (which may be
never).

Whatever HHH does, it does not run forever but aborts.
>

HHH halts on input DDD.
DDD correctly simulated by HHH cannot possibly halt.

WTF? It only calls HHH, which you just said halts.
>

>
An aborted simulation does not count as halting.

>
And doesn't show non-halting either.
>

Reaching it own machine address 00002183 counts as halting.
DDD correctly simulated by HHH cannot possibly do that.

>
But HHH doesn't DO a "Correct Simulation" that can show that, it only does a PARTIAL simulation.
>

>
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     If simulating halt decider H correctly simulates its input D
     until H correctly determines that its simulated D would never
     stop running unless aborted then
>
     H can abort its simulation of D and correctly report that D
     specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
>
until H correctly determines

>
Does that ever happen?
>

 Knowledge of the C programming language proves that it happens
in these three cases.
 void Infinite_Loop()
{
   HERE: goto HERE;
}
 void Infinite_Recursion()
{
   Infinite_Recursion();
}
 void DDD()
{
   HHH(DDD);
}
 int main()
{
   HHH(Infinite_Loop);
   HHH(Infinite_Recursion);
   HHH(DDD);
}
 Every C programmer that knows what an x86 emulator is knows that when HHH emulates the machine language of Infinite_Loop, Infinite_Recursion, and DDD that it must abort these emulations so that itself can terminate normally.
 

And any competent programmer knows the proofs that show that Infinite_Loop and Infinte_Recursion will never halt, and the fact that the execution of DDD will halt since your HHH(DDD) returns so it can be a decider means that, by the right definition, the correct simulation of DDD shows that it does halt, and HHH's claim that it doesn't is just a LIE.
You are, of course, excluded from that class of competent programmers, as you think that the partial simulation of DDD some how makes DDD to just stop running at the point its simulation is aborted.
You don't seem to understand that the details of the program fully define its behavior, and the simulation is just a method to observe that, and shutting your eyes by terminating the simulation doesn't stop the behavior that is there.
The behavior of just the partial simulation of DDD by HHH isn't a behavior of DDD (but shows part of it) but is part of the behavior of HHH as applied to DDD.
So, you are just using wrong (and improper) definitions in your logic and don't seem to know better.