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

On 2024-07-04 12:41:30 +0000, olcott said:
 

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.

 You haven't proven that in any of those cases. In particular, about DDD it
seems that your claim cannot be proven. The other cases might be provable.