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From: dbush <dbush.mobile@gmail.com>
Newsgroups: comp.theory
Subject: Re: How the requirements that Professor Sipser agreed to are exactly
 met --- WDH
Date: Tue, 13 May 2025 21:54:21 -0400
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On 5/13/2025 9:48 PM, olcott wrote:
> On 5/13/2025 8:31 PM, dbush wrote:
>> On 5/13/2025 9:27 PM, olcott wrote:
>>> On 5/13/2025 8:07 PM, dbush wrote:
>>>> On 5/13/2025 5:30 PM, olcott wrote:
>>>>> On 5/13/2025 6:43 AM, Richard Damon wrote:
>>>>>> On 5/13/25 12:52 AM, olcott wrote:
>>>>>>> On 5/12/2025 11:05 PM, Richard Damon wrote:
>>>>>>>> On 5/12/25 10:53 PM, olcott wrote:
>>>>>>>>> On 5/12/2025 8:27 PM, Richard Damon wrote:
>>>>>>>>>> On 5/12/25 2:17 PM, olcott wrote:
>>>>>>>>>>> Introduction to the Theory of Computation 3rd Edition
>>>>>>>>>>> by Michael Sipser (Author)
>>>>>>>>>>> 4.4 out of 5 stars    568 rating
>>>>>>>>>>>
>>>>>>>>>>> https://www.amazon.com/Introduction-Theory-Computation- 
>>>>>>>>>>> Michael- Sipser/ dp/113318779X
>>>>>>>>>>>
>>>>>>>>>>> int DD()
>>>>>>>>>>>   {
>>>>>>>>>>>    int Halt_Status = HHH(DD);
>>>>>>>>>>>    if (Halt_Status)
>>>>>>>>>>>      HERE: goto HERE;
>>>>>>>>>>>    return Halt_Status;
>>>>>>>>>>>   }
>>>>>>>>>>>
>>>>>>>>>>> DD correctly simulated by any pure simulator
>>>>>>>>>>> named HHH cannot possibly terminate thus proving
>>>>>>>>>>> that this criteria has been met:
>>>>>>>>>>>
>>>>>>>>>>> <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
>>>>>>>>>>>
>>>>>>>>>>>      H can abort its simulation of D and correctly report that D
>>>>>>>>>>>      specifies a non-halting sequence of configurations.
>>>>>>>>>>>   </MIT Professor Sipser agreed to ONLY these verbatim words 
>>>>>>>>>>> 10/13/2022>
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Which your H doesn't do, as it can not correctly determine 
>>>>>>>>>> what doesn't happen.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Any C programmer can correctly tell what doesn't happen.
>>>>>>>>> What doesn't happen is DD reaching its "return" statement
>>>>>>>>> final halt state.
>>>>>>>>>
>>>>>>>>
>>>>>>>> Sure they can, since that is the truth, as explained.
>>>>>>>>
>>>>>>>> Since your "logic" is based on lies and equivocation, 
>>>>>>>
>>>>>>> If my logic was based on lies and equivocation
>>>>>>> then you could provide actual reasoning that
>>>>>>> corrects my errors.
>>>>>>
>>>>>> I hae.
>>>>>>
>>>>>>>
>>>>>>> It is truism that simulating termination analyzers
>>>>>>> must report on the behavior of their input as if
>>>>>>> they themselves never aborted this simulation:
>>>>>>
>>>>>> Right, of the input actually given to them, which must include all 
>>>>>> their code, and that code is what is actually there, not created 
>>>>>> by this imaginary operation.
>>>>>>
>>>>>
>>>>> In other words every single byte of HHH and DD are
>>>>> 100% totally identical except the hypothetical HHH
>>>>> has its abort code commented out.
>>>>
>>>> In other words you changed the input.
>>>>
>>>> Changing the input is not allowed.
>>>>
>>>
>>>
>>>
>>>>>
>>>>>> Thus, a HHH that aborts to return an answer, when looking at the 
>>>>>> DDD that calls it, must look at the unaborted emulation of THAT 
>>>>>> DDD, that calls the HHH that DOES abort and return an answer, as 
>>>>>> that is what the PROGRAM DDD is, If you can not create the HHH 
>>>>>> that does that without changing that input, that is a flaw in your 
>>>>>> system, not the problem.
>>>>>>
>>>>>>>
>>>>>>> *simulated D would never stop running unless aborted*
>>>>>>> or they themselves could become non-terminating.
>>>>>>
>>>>>> But you aren't simulating the same PROGRAM D that the original was 
>>>>>> given.
>>>>>>
>>>>>
>>>>> It is not supposed to be the same program.
>>>>
>>>> So you *explicitly* admit to changing the input.
>>>>
>>>
>>> The finite string of DD is specific sequence bytes.
>>
>> Which includes the specific sequence of bytes that is the finite 
>> string HHH
>>
> 
> No it does not. A function calls is not macro inclusion.
> 

Then you admit that your HHH not deciding about algorithms and therefore 
has nothing to do with the halting problem.

>>> The finite string of HHH is specific sequence bytes.
>>>
>>> The hypothetical HHH that does not abort its input
>>> cannot have input that has changed because it never
>>> comes into actual existence.
>>
>> But your HHH decides on that hypothetical non-input.
>>
> 
> The whole point here is not to critique the words
> that professor Sipser agreed to.
> 
> The whole point here is to determine whether or
> not HHH meets this spec. It is a verified fact
> that it does meet this spec.

But since you just admitted that your HHH is not deciding on algorithms, 
it doesn't matter what HHH is doing as it has nothing to do with the 
halting problem.

> 
> You try to get away with changing the subject to
> deflect attention away from this key verified fact.
> 
>>>
>>> *simulated D would never stop running unless aborted*
>>>
>>>> This proves your work has nothing to do with the halting problem.

^^^

>>>>
>>>
>>> When ZFC over-ruled naive set theory this caused
>>> Russell's Paradox to cease to exist.
>>>
>>
>> Not unless you can show how halt deciders can exist from the axioms of 
>> computation theory.
>>
>>>> If you were just honest about the fact that you're not actually 
>>>> working on the halting problem, no one would be giving you any trouble.
>>>
>>> Equally we could say that ZFC was not working
>>> on the actual Russell's Paradox.
>>>
>>> What I am doing is the same thing that ZFC did.
>>>
>>
>> Unless you can prove that the following requirements can be met from 
>> the axioms of computation theory, your argument has no basis.
>>
> 
> ZFC ruled that the equivalent arguments from set theory WERE WRONG !

 From the axioms of naive set theory.
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