Sujet : Re: This function proves that only the outermost HHH examines the execution trace
De : F.Zwarts (at) *nospam* HetNet.nl (Fred. Zwarts)
Groupes : comp.theoryDate : 27. Jul 2024, 15:23:48
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v82vtl$3dq41$2@dont-email.me>
References : 1 2 3
User-Agent : Mozilla Thunderbird
Op 27.jul.2024 om 15:48 schreef olcott:
On 7/27/2024 3:36 AM, Fred. Zwarts wrote:
Op 26.jul.2024 om 17:56 schreef olcott:
This is meant for Mike, Joes and Fred don't have the technical competence to understand it.
>
I have pity with you,
I am not the one that stupidly believes that a non-terminating
input must be emulated to non-existent completion or the emulation
is wrong.
(That is something I never said, but it seems too difficult for you.)
You are the one that believes that the simulation of a halting program must be aborted to prevent non-halting. Ha ha.
That the first HHH to see the non-halting behavior pattern must
abort or none of them abort is simply too difficult for you.
That two recursions is not equal to an infinite recursion is already too difficult for you.
You keep dreaming of an infinite recursion, when HHH is encoded to abort after two recursions. It is too difficult for you to understand that dreams do not play a role in logic.
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