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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: HHH maps its input to the behavior specified by it
Date: Wed, 7 Aug 2024 21:03:03 -0400
Organization: i2pn2 (i2pn.org)
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On 8/7/24 2:14 PM, olcott wrote:
> On 8/7/2024 1:02 PM, joes wrote:
>> Am Wed, 07 Aug 2024 08:54:41 -0500 schrieb olcott:
>>> On 8/7/2024 2:29 AM, Mikko wrote:
>>>> On 2024-08-05 13:49:44 +0000, olcott said:
>>
>>>> I know what it means. But the inflected form "emulated" does not mean
>>>> what you apparently think it means. You seem to think that "DDD
>>>> emulated by HHH" means whatever HHH thinks DDD means but it does not.
>>>> DDD means what it means whether HHH emulates it or not.
>>>>
>>> In other words when DDD is defined to have a pathological relationship
>>> to HHH we can just close our eyes and ignore it and pretend that it
>>> doesn't exist?
>> It doesn't change anything about DDD. HHH was supposed to decide anything
>> and can't fulfill that promise. That doesn't mean that DDD is somehow
>> faulty, it's just a counterexample.
>>
> 
> void DDD()
> {
>    HHH(DDD);
>    return;
> }
> 
> *HHH is required to report on the behavior of DDD*
> Anyone that does not understand that HHH meets this criteria
> has insufficient understanding.

But it doesn't, as a correct simulation of a DDD that calls an HHH that 
returns will stop running, and thus HHH can never correctly determine 
something that doesn't happen, so it is never correct to abort its 
emulation.

The only DDD that is non-halting, is the one built on an HHH that will 
NEVER return an answer for it, and such an HHH can not abort to return, 
or it isn't what it needed to be to make that the corret answer.

> 
> <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
>      If simulating halt decider H correctly simulates its input D
>      until H correctly determines that its simulated D would never
>      stop running unless aborted then
> 
> I can't imagine that any expert in the C language can say
> that HHH does not meet this criteria without lying. All
> four of them that answered agreed that it does. Two of
> these four have MSCS.
> 
> If most everyone here hardly knows C at all that would
> be quite a shock to me.
> 

Nope, YOU don't know the C language, or Logic, or what Truth is, or it 
seems ANYTHING you talk about.