Re: ChatGPT refutes the key rebuttal of my work --- CORRECTION

On 10/17/2024 5:07 AM, joes wrote:

Am Wed, 16 Oct 2024 15:06:01 -0500 schrieb olcott:

On 10/16/2024 3:02 PM, joes wrote:

Am Wed, 16 Oct 2024 14:39:37 -0500 schrieb olcott:

On 10/16/2024 2:33 PM, joes wrote:

Am Wed, 16 Oct 2024 13:59:58 -0500 schrieb olcott:

On 10/16/2024 1:47 PM, joes wrote:

Am Wed, 16 Oct 2024 13:35:01 -0500 schrieb olcott:

On 10/16/2024 1:06 PM, joes wrote:

Am Wed, 16 Oct 2024 12:46:01 -0500 schrieb olcott:

On 10/16/2024 12:27 PM, joes wrote:

Am Wed, 16 Oct 2024 10:39:21 -0500 schrieb olcott:

On 10/16/2024 9:45 AM, joes wrote:

Am Wed, 16 Oct 2024 09:11:22 -0500 schrieb olcott:

On 10/16/2024 9:01 AM, joes wrote:

Am Wed, 16 Oct 2024 08:31:43 -0500 schrieb olcott:

On 10/16/2024 1:33 AM, joes wrote:

>

Terminating C functions must reach their "return"
statement.

Which DDD does.

THIS IS ALSO THE INDUSTRY STANDARD DEFINITION It is
stipulated that *correct_x86_emulation* means that a finite
string of x86 instructions is emulated according to the
semantics of the x86 language beginning with the first bytes
of this string.

You are not simulating the given program, but a version that
differs in the abort check.

HHH is correctly emulating (not simulating) the x86 language
finite string of DDD including emulating the finite string of
itself emulating the finite string of DDD up until the point
where the emulated emulated DDD would call HHH(DDD) again.

Whereupon the simulated HHH would abort, if it weren't
unnecessarily aborted.

If the first HHH to meet its abort criteria does not act on this
criteria then none of them do.

And if the first one does, all of them do.

In theory this seems true when ignoring or failing to comprehend
key details.

In practice you programmed H impurely.

Which totally does not matter to the slightest degree when you have
the discipline to stay within the precisely designated scope of the
exact words that I am saying.
When HHH is an x86 emulation based termination analyzer then each
DDD *correctly_emulated_by* any HHH that it calls cannot possibly
return no matter what this HHH does.

Exactly, because your nested HHHs do not abort.

In other words you continue to fail to understand that unless the
first one aborts then none of them can possibly abort because they all
have the exact same code.

Then HHH should report itself as halting, when they would all abort.

They would not all abort when you pay close attention to ALL of the
details. It is utterly impossible for any of them besides the outermost
one to abort because it aborts before any of the rest of them see their
abort criteria has been met.

Only because they aren't simulated. But they are still terminating

programs. An infinite loop is still non-halting even if I never run it.
 

This is just over your head.
_DDD()
[00002172] 55         push ebp      ; housekeeping
[00002173] 8bec       mov ebp,esp   ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404     add esp,+04
[00002182] 5d         pop ebp
[00002183] c3         ret
Size in bytes:(0018) [00002183]
When DDD is correctly emulated by HHH according
to the semantics of the x86 language DDD cannot
possibly reach its own machine address [00002183]
no matter what HHH does.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer