Sujet : Re: Sequence of sequence, selection and iteration matters
De : noreply (at) *nospam* example.org (joes)
Groupes : comp.theoryDate : 09. Jul 2024, 19:50:47
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <ee5f2371ef699b2907a5a3d8dc3709889b85284f@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
Am Tue, 09 Jul 2024 11:44:53 -0500 schrieb olcott:
On 7/9/2024 10:21 AM, Fred. Zwarts wrote:
Op 09.jul.2024 om 16:46 schreef olcott:
On 7/9/2024 9:38 AM, Fred. Zwarts wrote:
*When DDD is correctly emulated by any pure function*
*HHH x86 emulator that can possibly exist* which calls an emulated
HHH(DDD) to repeat this process until the emulated DDD is aborted.
And the fact *that* it aborts, makes the simulation incorrect (as
Sipser would agree with), because the X86 code does not specify an
abort at that point. Therefore, the only conclusion must be: No such
HHH exists.
HHH is fully operational in the x86utm operating system.
It does not fulfill the specification. Your HHH is not the true HHH.
It is bugged.
When DDD is correctly emulated by any pure function x86 emulator HHH
calls an emulated HHH(DDD) this call cannot possibly return. This
prevents the emulated DDD from ever reaching past its own machine
address of 0000216b and halting.
Are you saying that the called HHH(DDD) does not terminate?
HHH is required to report that it must abort the emulaton of its input.
No, why?
HHH cannot correctly report that DDD need not be aborted on the basis of
the behavior of a directly executed DDD(DDD) after HHH has already
aborted its emulated DDD.
Why did it abort DDD then?
-- Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:Objectively I am a genius.