Re: HHH maps its input to the behavior specified by it --- partial simulation never reaches its halt state, but the actual behavior of the actual program does

On 8/9/24 3:56 PM, olcott wrote:

On 8/9/2024 2:45 PM, Fred. Zwarts wrote:

Op 09.aug.2024 om 19:02 schreef olcott:

On 8/9/2024 10:42 AM, Fred. Zwarts wrote:

Op 09.aug.2024 om 17:04 schreef olcott:

 >>>> void Infinite_Loop()

{
   HERE: goto HERE;
}
>
The correct simulation of the above never halts.

>

>
void DDD()
{
   HHH(DDD);
   return;
}
>
A correct simulation of N instructions of DDD <is>
A correct simulation of N instructions of DDD.

>
Irrelevant.
We are not interested in the simulation of only a few instructions, but in the simulation of the whole halting program, to determine its halting behaviour.
>

 That is a pretty stupid requirement for non-halting inputs.
You don't even know what halting is yet.

Except the input ISN'T a non-halting proggram, as it calls an HHH that will return to it since that HHH aborts its simulation and returns.
It isn't the non-halting version of DDD that calls the non-decider HHH that never aborts, and thus never answers when it is given its version of DDD.

 *Jeff said it best*
On 8/2/2024 11:32 PM, Jeff Barnett wrote:
 > ...In some formulations, there are specific states
 >    defined as "halting states" and the machine only
 >    halts if either the start state is a halt state...
  > ...these and many other definitions all have
 >    equivalent computing prowess...
 

And the PROGRAM DDD will halt if the HHH that it calls aborts and returns to ANY of its calls, since what it does for one, it does for all.