Re: DDD correctly emulated by HHH is correctly rejected as non-halting.

On 7/13/24 7:00 PM, olcott wrote:

On 7/13/2024 5:57 PM, joes wrote:

Am Fri, 12 Jul 2024 08:20:53 -0500 schrieb olcott:

On 7/12/2024 3:03 AM, Mikko wrote:

On 2024-07-11 14:10:24 +0000, olcott said:

On 7/11/2024 1:25 AM, Mikko wrote:

On 2024-07-10 17:53:38 +0000, olcott said:

On 7/10/2024 12:45 PM, Fred. Zwarts wrote:

Op 10.jul.2024 om 17:03 schreef olcott:

>

However, each of those instances has the same sequence of
instructions that the x86 language specifies the same operational
meaning.

*That is counter-factual*
When DDD is correctly emulated by HHH according to the semantics of
the x86 programming language HHH must abort its emulation of DDD or
both HHH and DDD never halt.

The assembly is not concerned with aborting or halting.
>

When DDD is correctly emulated by HHH1 according to the semantics of
the x86 programming language HHH1 need not abort its emulation of DDD
because HHH has already done this.

However, the program DDD is the same in both cases and therefore the
its behavioral meaning per x86 semantics is also the same.

HHH1(DDD) only halts because HHH(DDD) aborts its emulation thus proving
the the behaviors are different.

The simple fact remains that if the behaviour of a program depends
on what is simulating it, that simulator is faulty.
>

The behavior of DDD emulated by HHH1 is identical to the behavior of
the directly executed DDD().

Which is the behaviour of DDD accordint to the semantics of x86
language.

If you stupidly ignore that DDD does call HHH in recursive emulation it
might superfically seem that way.

 

It does that in every (non-)simulation, no change there.
>

  DDD cannot call HHH in recursive emulation WITHOUT EMULATION.
 

But, since HHH does conditional emulation, if HHH will abort its emulation it is not infinite recursive emulation.
If HHH can not handle emulating programs that call versions of it, then it is just defective or incomplete.