Sujet : Re: ChatGPT refutes the key rebuttal of my work
De : noreply (at) *nospam* example.org (joes)
Groupes : comp.theoryDate : 16. Oct 2024, 19:47:22
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <8cb59c1760f051701155070c17b7828ef660aaad@i2pn2.org>
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User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
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.
When HHH is an x86 emulation based termination analyzer then each
DDD *correctly_emulated_by* any HHH that it calls never returns.
It is not a correct emulation if it has a different termination
status.
-- 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.