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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory,sci.logic
Subject: Re: H(D,D) cannot even be asked about the behavior of D(D)
Date: Fri, 14 Jun 2024 19:27:40 -0400
Organization: i2pn2 (i2pn.org)
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On 6/14/24 9:15 AM, olcott wrote:
> On 6/14/2024 6:39 AM, Richard Damon wrote:
>> On 6/14/24 12:13 AM, olcott wrote:
>>> On 6/13/2024 10:44 PM, Richard Damon wrote:
>>>> On 6/13/24 11:14 PM, olcott wrote:
>>>>> On 6/13/2024 10:04 PM, Richard Damon wrote:
>>>>>> On 6/13/24 9:39 PM, olcott wrote:
>>>>>>> On 6/13/2024 8:24 PM, Richard Damon wrote:
>>>>>>>> On 6/13/24 11:32 AM, olcott wrote:
>>>>>>>>>
>>>>>>>>> It is contingent upon you to show the exact steps of how H 
>>>>>>>>> computes
>>>>>>>>> the mapping from the x86 machine language finite string input to
>>>>>>>>> H(D,D) using the finite string transformation rules specified by
>>>>>>>>> the semantics of the x86 programming language that reaches the
>>>>>>>>> behavior of the directly executed D(D)
>>>>>>>>>
>>>>>>>>
>>>>>>>> Why? I don't claim it can.
>>>>>>>>
>>>>>>>
>>>>>>> That means that H cannot even be asked the question:
>>>>>>> "Does D halt on its input?"
>>>>>>
>>>>>> WHy not? After all, H does what it does, the PERSON we ask is the 
>>>>>> programmer.
>>>>>>
>>>>>
>>>>> *When H and D have a pathological relationship to each other*
>>>>> There is no way to encode any H such that it can be asked:
>>>>> Does D(D) halt?
>>>>
>>>> Which just pproves that Halting is non-computable.
>>>>
>>>
>>> No it is more than that.
>>> H cannot even be asked the question:
>>> Does D(D) halt?
>>
>> No, you just don't understand the proper meaning of "ask" when applied 
>> to a deterministic entity.
>>
> 
> When H and D have a pathological relationship to each
> other then H(D,D) is not being asked about the behavior
> of D(D). H1(D,D) has no such pathological relationship
> thus D correctly simulated by H1 is the behavior of D(D).

OF course it is. The nature of the input doesn't affet the form of the 
question that H is supposed to answer.

> 
> If I ask you: What time is it?
> and my actual unstated question is:
> What is the outside temperature where you are?

Which just makes you the LIAR that you already showed you are.

H as ONE and ONLY one question it is supposed to answer, if it is a Halt 
decider, and that is "Does the Machine represented by your input halt 
when run?"

Anything else is just a lie,

> 
> Can a correct answer to the stated question be
> a correct answer to the unstated question?
But asking the quesiton you don't mean, OR answering the question you 
weren't asked are just forms of lies.

Of course, a liar like you should understand that, or is the pathology 
making it so you can't understand that nature?

> 
> H(D,D) is not even being asked about the behavior of D(D)
> 

I guses you are just admitting that you have been lying about what H is 
supposed to be for all thewse years.

IF it WAS a halt decider, that iis EXACTLY what it is being asked about.

>>>
>>> You already admitted the basis for this.
>>
>> No, that is something different.
>>
>>>
>>>> You keep on doing that, Making claims that show the truth of the 
>>>> statement you are trying to disprove.
>>>> The fact you don't undrstand that, just show how little you 
>>>> understand what you are saying.
>>>>
>>>>>
>>>>> You must see this from the POV of H or you won't get it.
>>>>> H cannot read your theory of computation textbooks, it
>>>>> only knows what it directly sees, its actual input.
>>>>
>>>> But H doesn't HAVE a "poimt of view".
>>>>
>>>
>>> When H is a simulating halt decider you can't even ask it
>>> about the behavior of D(D). You already said that it cannot
>>> map its input to the behavior of D(D). That means that you
>>> cannot ask H(D,D) about the behavior of D(D).
>>
>> OF course you can, becaue, BY DEFIINITION, that is the ONLY thing it 
>> does with its inputs.
>>
> 
> That definition might be in textbooks,
> yet H does not and cannot read textbooks.

But it programer is supposed to.

I guess you are admitting at being a failure as a programmer.

> 
> The only definition that H sees is the combination of
> its algorithm with the finite string of machine language
> of its input.

Which means the prograamer didn't do his job.

> 
> It is impossible to encode any algorithm such that H and D
> have a pathological relationship and have H even see the
> behavior of D(D).
> 

Which is what makes it impossible to build a decider that snwers the 
question.

Which is perfectly fine, as the big question is was it possible to do so.

> You already admitted there there is no mapping from the finite
> string of machine code of the input to H(D,D) to the behavior
> of D(D).

No, you are just lying agsin. Thers *IS* a mapping for the finite string 
input to H and the answer, it is based on the behavior of the UTM 
processing of the input. If it halts, then the mapping of that string is 
to Yes, if it doesn't then the mapping of that string is to No

> 
>>>
>>> What seems to me to be the world's leading termination
>>> analyzer symbolically executes its transformed input.
>>> https://link.springer.com/content/pdf/10.1007/978-3-030-99527-0_21.pdf
>>>
>>> It takes C programs and translates them into something like
>>> generic assembly language and then symbolically executes them
>>> to form a directed graph of their behavior. x86utm and HH do
>>> something similar in a much more limited fashion.
>>
>> And note, it only gives difinitive answers for SOME input.
>>
> 
> It is my understanding is that it does this much better than
> anyone else does. AProVE "symbolically executes the LLVM program".
> The LLVM program is essentially the C program translated into
> a generic assembly language.
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