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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: This function proves that only the outermost HHH examines the
 execution trace
Date: Sat, 27 Jul 2024 16:23:48 +0200
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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.