Re: Proof that DDD specifies non-halting behavior --- point by point

On 8/16/24 8:42 AM, olcott wrote:

On 8/16/2024 3:53 AM, Mikko wrote:

On 2024-08-15 15:25:07 +0000, olcott said:
>

On 8/15/2024 5:22 AM, Mikko wrote:

On 2024-08-14 13:06:27 +0000, olcott said:
>

On 8/14/2024 3:17 AM, Mikko wrote:

On 2024-08-14 00:52:36 +0000, olcott said:
>

void DDD()
{
   HHH(DDD);
   return;
}

>
In order to prove that the above specifies a non-halting behavour
you must prove that HHH(DDD) does not terminate.

>
Wrong.

>
At least the proof that DDD does not terminate also proves as an
intermedate result or an obvious corollary that HHH does not halt.
>
Non-halting means that an infinite number of instructions can be
executed without halting. That means that at least one instruction
is executed infinitely many times as there are only finitely many
instructions. But not instrunctions of DDD outside HHH is executed
infinitely many times.
>

>
Wrong. Non-halting only means that when DDD is emulated
according to the semantics of the x86 language and this
emulation is unlimited that DDD would never reach its
own "return" instruction.

>
If what I said is wrong then what you said is wrong, too,
as you say what I said.
>

 *You are getting the computer science incorrectly*
 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...
 The "return" instruction is the halt state of DDD.
 

Which DDD (the program) reaches if the HHH that it calls returns, which it will if any copy of that HHH given that same input does (or you are admitting you are lying about the properties of HHH).
Remember, behavior as described above it based on the actual running (or COMPLETE simulation) of it, not on the partial simulation of it that HHH does.