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From: dbush <dbush.mobile@gmail.com>
Newsgroups: comp.theory
Subject: Re: DD correctly emulated by HHH --- Totally ignoring invalid
 rebuttals ---PSR---
Date: Fri, 7 Mar 2025 21:23:21 -0500
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On 3/7/2025 9:19 PM, olcott wrote:
> On 3/7/2025 7:52 PM, dbush wrote:
>> On 3/7/2025 8:49 PM, olcott wrote:
>>> On 3/7/2025 10:25 AM, Fred. Zwarts wrote:
>>>> Op 07.mrt.2025 om 16:17 schreef olcott:
>>>>> On 3/7/2025 2:59 AM, Fred. Zwarts wrote:
>>>>>> Op 06.mrt.2025 om 21:13 schreef olcott:
>>>>>>> On 3/6/2025 3:13 AM, Fred. Zwarts wrote:
>>>>>>>> Op 06.mrt.2025 om 04:53 schreef olcott:
>>>>>>>>> On 3/5/2025 9:31 PM, dbush wrote:
>>>>>>>>>> On 3/5/2025 10:17 PM, olcott wrote:
>>>>>>>>>>> On 3/5/2025 7:10 PM, dbush wrote:
>>>>>>>>>>>>
>>>>>>>>>>>> In other words, you know that what you're working on has 
>>>>>>>>>>>> nothing to do with the halting problem, but you don't care.
>>>>>>>>>>>
>>>>>>>>>>> In other words I WILL NOT TOLERATE ANY BULLSHIT DEFLECTION.
>>>>>>>>>>> You have proven that you know these things pretty well SO 
>>>>>>>>>>> QUIT THE SHIT!
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> You want people to accept that HHH(DD) does in fact report 
>>>>>>>>>> that changing the code of HHH to an unconditional simulator 
>>>>>>>>>> and running HHH(DD) will not halt.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> DD correctly emulated by HHH cannot possibly
>>>>>>>>> reach its own "ret" instruction and terminate normally.
>>>>>>>>
>>>>>>>> Yes, we agree that HHH fails to reach the 'ret' instruction, 
>>>>>>>
>>>>>>> Despicably dishonest attempt at the straw-man deception.
>>>>>>>
>>>>>>
>>>>>> No rebuttal. So, we agree that HHH fails to reach the 'ret' 
>>>>>> instruction. 
>>>>>
>>>>> Not at all. Trying to get away with changing the subject
>>>>> WILL NOT BE TOLERATED.
>>>>>
>>>> If you do not agree that HHH fails to reach the 'ret' instruction 
>>>> (that world-class simulators do reach, just as the direct execution 
>>>> does), show how it reaches the 'ret' instruction.
>>>
>>> *set X*
>>> When-so-ever any input to any simulating termination
>>> analyzer calls the simulator that is simulating itself
>>
>> Not an issue, since termination analyzers don't exist.
> 
> I thought that you demonstrated knowledge of these things.
> Maybe I was wrong.
> 

We know termination analyzers don't exist because no algorithm exists 
that maps the halting function:

(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly