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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Title: A Structural Analysis of the Standard Halting Problem
 Proof
Date: Fri, 25 Jul 2025 09:15:52 -0500
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On 7/25/2025 2:53 AM, joes wrote:
> Am Thu, 24 Jul 2025 16:41:26 -0500 schrieb olcott:
>> On 7/24/2025 4:24 PM, joes wrote:
>>> Am Thu, 24 Jul 2025 09:32:45 -0500 schrieb olcott:
> 
>>>> Aborting prematurely literally means that after N instructions of DDD
>>>> are correctly emulated by HHH that this emulated DDD would reach its
>>>> own emulated "ret" instruction final halt state.
>>>> What value of N are you proposing?
>>>
>>> Let's see: the call to HHH is #4, [waves hands], then another 4 inside
>>> the next level of simulation, and after another 4 the first simulated
>>> HHH (the one called by the input, not the outermost simulator. We are
>>> now 3 levels in) decides that enough is enough and aborts,
>>
>> Thus immediate killing its simulated DDD and everything else that HHH
>> was simulating thus no simulated DDD or simulated HHH can possibly ever
>> return no matter how many or how few X86 instructions that the executed
>> HHH correctly emulates.
>> This is the part that you fail to understand or understand that I am
>> correct and disagree anyway.

> You failed to understand I was talking about the first simulated HHH
> aborting, not the outermost simulator.

*I am trying to get you to understand that is impossible*
The only HHH that can possibly abort is the outermost
directly executed one.

>>> returning to the outermost level which takes 3 more instructions to
>>> halt, whereupon our treasured HHH returns that DDD halts. So,
>>> 4+4+4+3=15?
>>>
>>> Of course the crux is that changing "HHH" changes the input, so HHH can
>>> never do it.


-- 
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer