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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Who knows that DDD correctly simulated by HHH cannot possibly
 reach its own return instruction final state? BUT ONLY that DDD
Date: Sat, 3 Aug 2024 22:56:20 -0400
Organization: i2pn2 (i2pn.org)
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On 8/3/24 7:36 PM, olcott wrote:
> On 8/3/2024 5:51 PM, Richard Damon wrote:
>> On 8/3/24 6:15 PM, olcott wrote:
>>> On 8/3/2024 5:07 PM, Richard Damon wrote:
>>>>
>>>> The problem is that every one of those emulation is of a *DIFFERENT* 
>>>> input, so they don't prove anything together except that each one 
>>>> didn't go far enough. 
>>>
>>> void DDD()
>>> {
>>>    HHH(DDD);
>>>    return;
>>> }
>>>
>>> When each HHH correctly emulates 0 to infinity steps of
>>> its corresponding DDD and none of them reach the "return"
>>> halt state of DDD then even the one that emulated infinite
>>> steps of DDD did not emulate enough steps?
>>>
>>>
>>
>> Just says lying YOU.
>>
>> You got any source for that other than yourself?
>>
> 
> It is self-evident and you know it. I do have four
> people (two with masters in CS) that attest to that.
> *It is as simple as I can possibly make it*

Maybe to your mind filled with false facts, but it isn't true.

> 
> I wonder how you think that you are not swearing your
> allegiance to that father of lies?

Because, I know I speak the truth.

Why do you not think you are lying?

> 
> Anyone that truly understands infinite recursion knows
> that DDD correctly simulated by HHH cannot possibly reach
> its own "return" final state.

Right, but for every other HHH, which the ones that answer are, it isn't 
a fact.


> 
> Surpisingly (to me) Jeff Barnett set the record straight
> on exactly what halting means.
> 

No, there is one, and only one definition, it is a machine that reaches 
its final state.

Note, *a machine*, not a (partial) emulation of the machine

And it is the FULL machine, so for DDD, it includes the HHH that it was 
built on.