Sujet : Re: DDD correctly emulated by HHH is correctly rejected as non-halting.
De : noreply (at) *nospam* example.org (joes)
Groupes : comp.theoryDate : 12. Jul 2024, 10:08:59
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <ce0a250902cd90698367e11a5900e68f6163eec3@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 Thu, 11 Jul 2024 15:56:09 -0500 schrieb olcott:
On 7/11/2024 3:19 PM, joes wrote:
Am Thu, 11 Jul 2024 10:05:58 -0500 schrieb olcott:
On 7/11/2024 9:25 AM, joes wrote:
Am Thu, 11 Jul 2024 09:10:24 -0500 schrieb olcott:
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:
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.
If the recursive call to HHH from DDD halts, the outer HHH doesn't
need to abort.
Do you mean that HHH doesn't halt?
This.
DDD depends totally on HHH; it halts exactly when HHH does.
Which it does, because it aborts.
What does HHH do after it aborts?
And this one.
DDD correctly simulated by HHH has provably different behavior than
DDD correctly simulated by HHH1.
Which means that HHH is not doing the simulation correctly.
When HHH simulates DDD according to the semantics of the x86 language
then HHH is simulating correctly. When people disagree with the
semantics of the x86 language THEY ARE WRONG !!!
Aborting is not a correct simulation.
Please answer the other questions above.
-- Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:Objectively I am a genius.
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