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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Simulation vs. Execution in the Halting Problem
Date: Wed, 11 Jun 2025 18:45:46 -0500
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On 6/11/2025 6:25 PM, wij wrote:
> On Wed, 2025-06-11 at 17:33 -0500, olcott wrote:
>> On 6/11/2025 4:57 PM, wij wrote:
>>> On Wed, 2025-06-11 at 16:44 -0500, olcott wrote:
>>>> On 6/11/2025 4:23 PM, wij wrote:
>>>>> On Wed, 2025-06-11 at 16:10 -0500, olcott wrote:
>>>>>> On 6/11/2025 3:59 PM, wij wrote:
>>>>>>> On Wed, 2025-06-11 at 15:30 -0500, olcott wrote:
>>>>>>>> On 6/11/2025 2:45 PM, wij wrote:
>>>>>>>>> On Wed, 2025-06-11 at 14:39 -0500, olcott wrote:
>>>>>>>>>> On 6/11/2025 2:31 PM, wij wrote:
>>>>>>>>>>> On Wed, 2025-06-11 at 14:14 -0500, olcott wrote:
>>>>>>>>>>>> On 6/11/2025 1:25 PM, wij wrote:
>>>>>>>>>>>>> On Wed, 2025-06-11 at 12:59 -0500, olcott wrote:
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Yes all other people (especially Dennis Bush) are saying
>>>>>>>>>>>>>>>> that H(D) is required to report on the behavior of the
>>>>>>>>>>>>>>>> direct execution of D() never noticing that this stupidly
>>>>>>>>>>>>>>>> requires H(D) to report on the behavior of its caller.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> If the H above means the H that the HP refers to. The H is required to
>>>>>>>>>>>>>>> report its argument's behavior (ie. by H(D)). But NOT required by
>>>>>>>>>>>>>>> simulation.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> It turns out that no one ever noticed that simulating halt
>>>>>>>>>>>>>> deciders nullify the HP counter-example input in that this
>>>>>>>>>>>>>> input cannot possibly reach its contradictory part.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> The HP does not care what D does (simply to say).
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Everyone says that H(D) must re[port on the behavior of
>>>>>>>>>>>>>> the direct execution of D().
>>>>>>>>>>>>>
>>>>>>>>>>>>> That is what the HP asks.
>>>>>>>>>>>>>
>>>>>>>>>>>>>>> The HP only requires: H(D)==1 iff D() halts
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> int main()
>>>>>>>>>>>>>> {
>>>>>>>>>>>>>>           D(); // calls H(D)
>>>>>>>>>>>>>> }
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Which requires H(D) to report on the behavior of its
>>>>>>>>>>>>>> caller instead of reporting on the behavior that its
>>>>>>>>>>>>>> input actually specifies.
>>>>>>>>>>>>>
>>>>>>>>>>>>> That is no problem. H does not care what D does inside (simply to say).
>>>>>>>>>>>>> The HP simply asks for a H that "H(D)==1 iff D() halts".
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Which requires H to report on something that it cannot possibly see.
>>>>>>>>>>>
>>>>>>>>>>> On the contrary, what the HP proves is very useful.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> I am not talking about the halting problem, I have always
>>>>>>>>>> been talking about the conventional halting problem proof.
>>>>>>>>>> THIS PROOF IS WRONG
>>>>>>>>>
>>>>>>>>> When talking about proof, we say it is valid or not. By doing so, we have
>>>>>>>>> to unambiguously pose the problem and the derivation to the conclusion.
>>>>>>>>> The HP proof just did that.
>>>>>>>>>
>>>>>>>>
>>>>>>>> It may seem that way if you pay less than 100%
>>>>>>>> complete attention.
>>>>>>>>
>>>>>>>> The HP proof depends on an *INPUT* that does
>>>>>>>> the opposite of whatever value that H returns
>>>>>>>> and no such *INPUT* can possibly exist.
>>>>>>>
>>>>>>> That is absolutely correct. No such *INPUT* (i.e. D) can possible exit is because
>>>>>>> the H inside D does not exist at all.
>>>>>>> So, if the H is assumed to exist, then D will exist to make H undecidable.
>>>>>>>
>>>>>>
>>>>>> There is no *input* to any termination analyzer
>>>>>> that can do the opposite of whatever value that
>>>>>> this termination analyzer returns
>>>>>
>>>>> Your reinterpretation of of HP case is wrong.
>>>>> Your D or H is not the case mention in the HP proof.
>>>>>
>>>>
>>>> There cannot possibly exist any D mine or
>>>> anyone else's that is encoded to do the opposite
>>>> of whatever value that H returns.
>>>
>>> Why not? D and H are supposed to be TM (or C function).
>>> If the D cannot do the opposite of whatever value that H returns, then
>>> that D is not powerful enough to be a TM, not an interesting case.
>>>
>>
>> Can you be your biological mother's biological father?
> 
> What is the same reason? What's the relationship of 1+1=2 relates to HP?
> 
>> It is for this same reason that the function's caller
>> cannot simultaneously be its input.
> 
> D and H belong to the same set of TM equivalent stuff.

Yes and we have the exact same issue with TM's it
is merely more difficult to see.

I am not going to get into that until after you totally
understand this at the C level. I am unwilling to talk
about this endlessly in circles.

> D has to be able to perform exactly H's function (if D is a TM and if H exists).
> Otherwise, that D is not the counter-example mentioned in the HP proof.
>
I have to covered too. Unless you understand that
D cannot be both an input to H and its caller there
is no sense going there.

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