Re: DDD correctly emulated by HHH --- Correct Emulation Defined

Am Fri, 21 Mar 2025 13:58:02 -0500 schrieb olcott:

On 3/21/2025 12:56 PM, joes wrote:

Am Fri, 21 Mar 2025 08:07:38 -0500 schrieb olcott:

On 3/21/2025 2:59 AM, Mikko wrote:

On 2025-03-20 22:43:34 +0000, olcott said:

On 3/20/2025 4:16 AM, Mikko wrote:

On 2025-03-20 02:32:43 +0000, olcott said:

1,2,3...4,294,967,296 steps of DDD are correctly emulated by HHH and
DDD never reaches its "ret" instruction and terminates normally.

But your HHH does not simulate correctly more steps of DDD than your
HHH1 does.

This HHH is the hypothetical HHH the emulates an arbitrary number of
steps of DDD according to the semantics of the x86 language.

No, your HHH is real and it does not simulate arbitrarily many
instructions.

If DDD behaves the same way for every possible HHH then it behaves this
way for any particular HHH.

Yes, all of them simulate only partially.

It is always at machine address 000015d2. This same HHH also has the
ability to emulate itself emulating DDD.

No, it can't simulate itself to its definite termination.

HHH does simulate itself simulating DDD to the final state of HHH.

If it did that, the simulated HHH would return and DD would halt
instead of getting aborted.

--
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.