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

Am Fri, 16 Aug 2024 07:42:22 -0500 schrieb olcott:

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:

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

>

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.

It's kind of amazing to see you quote unrelated stuff.
How does HHH not simulate DDD returning, while not executing the same
instructions infinitely?

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.