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computers / comp.theory / Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ correct halt deciding criteria ]

Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ correct halt deciding criteria ]

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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
correct halt deciding criteria ]
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Mon, 11 Apr 2022 01:55 UTC

On 4/10/2022 8:53 PM, Dennis Bush wrote:
> On Sunday, April 10, 2022 at 9:43:47 PM UTC-4, olcott wrote:
>> On 4/10/2022 8:38 PM, Dennis Bush wrote:
>>> On Sunday, April 10, 2022 at 9:36:11 PM UTC-4, olcott wrote:
>>>> On 4/10/2022 8:30 PM, Dennis Bush wrote:
>>>>> On Sunday, April 10, 2022 at 9:26:57 PM UTC-4, olcott wrote:
>>>>>> On 4/10/2022 8:17 PM, Dennis Bush wrote:
>>>>>>> On Sunday, April 10, 2022 at 9:13:23 PM UTC-4, olcott wrote:
>>>>>>>> On 4/10/2022 8:10 PM, Dennis Bush wrote:
>>>>>>>>> On Sunday, April 10, 2022 at 9:07:18 PM UTC-4, olcott wrote:
>>>>>>>>>> On 4/10/2022 8:03 PM, Dennis Bush wrote:
>>>>>>>>>>> On Sunday, April 10, 2022 at 9:00:25 PM UTC-4, olcott wrote:
>>>>>>>>>>>> On 4/10/2022 7:44 PM, Dennis Bush wrote:
>>>>>>>>>>>>> On Sunday, April 10, 2022 at 8:40:34 PM UTC-4, olcott wrote:
>>>>>>>>>>>>>> On 4/10/2022 7:28 PM, Dennis Bush wrote:
>>>>>>>>>>>>>>> On Sunday, April 10, 2022 at 7:58:55 PM UTC-4, olcott wrote:
>>>>>>>>>>>>>>>> On 4/10/2022 6:26 PM, Dennis Bush wrote:
>>>>>>>>>>>>>>>>> On Sunday, April 10, 2022 at 7:20:44 PM UTC-4, olcott wrote:
>>>>>>>>>>>>>>>>>> On 4/10/2022 6:14 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>>>>> On 2022-04-10 17:08, olcott wrote:
>>>>>>>>>>>>>>>>>>>> On 4/10/2022 5:59 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>>>>>>> On 2022-04-10 16:40, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>> On 4/10/2022 5:35 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>>>>>>>>> On 2022-04-10 15:56, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>>>> On 4/10/2022 4:49 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>>>>>>>>>>> On 2022-04-10 15:00, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/10/2022 3:15 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>>>>>>>> On 4/10/2022 3:07 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>> I'm trying to get you to write using correct and coherent
>>>>>>>>>>>>>>>>>>>>>>>>>>>> notation. That's one of the things you'll need to be able to
>>>>>>>>>>>>>>>>>>>>>>>>>>>> do if you ever hope to publish. That involves remembering to
>>>>>>>>>>>>>>>>>>>>>>>>>>>> always include conditions and using the same terms in your
>>>>>>>>>>>>>>>>>>>>>>>>>>>> 'equations' as in your text.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>> Not sure how that makes me a 'deceitful bastard'.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>> André
>>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>>> THAT you pretended to not know what I mean by embedded_H so
>>>>>>>>>>>>>>>>>>>>>>>>>>> that you could artificially contrive a fake basis for rebuttal
>>>>>>>>>>>>>>>>>>>>>>>>>>> when no actual basis for rebuttal exists makes you a deceitful
>>>>>>>>>>>>>>>>>>>>>>>>>>> bastard.
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>> IT IS THE CASE THAT the correctly simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
>>>>>>>>>>>>>>>>>>>>>>>>>> embedded_H never reaches its own final state of ⟨Ĥ0.qy⟩ or
>>>>>>>>>>>>>>>>>>>>>>>>>> ⟨Ĥ0.qn⟩ under any condition what-so-ever therefore ⟨Ĥ0⟩ ⟨Ĥ1⟩ is
>>>>>>>>>>>>>>>>>>>>>>>>>> proved to specify a non-halting sequence of configurations.
>>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
>>>>>>>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn
>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>> This is now the third reply you've made to the same post.
>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>> That post didn't make any arguments whatsoever about your claims.
>>>>>>>>>>>>>>>>>>>>>>>>> It simply pointed out that you are misusing your notation and
>>>>>>>>>>>>>>>>>>>>>>>>> urged you to correct it.
>>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> THE NOTATION IS A STIPULATIVE DEFINITION THUS DISAGREEMENT IS
>>>>>>>>>>>>>>>>>>>>>>>> INCORRECT.
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> If the notation is junk, then the definition is also junk.
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> That's like "stipulating" that
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> +×yz÷² = ±z+³
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> It's meaningless because the notation is meaningless, much like
>>>>>>>>>>>>>>>>>>>>>>> your notation above.
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> This is meaningless:
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy // what's the condition?
>>>>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qn // what's the condition?
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> With no conditions specified, the above is just nonsense.
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> André
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
>>>>>>>>>>>>>>>>>>>>>> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its
>>>>>>>>>>>>>>>>>>>>>> final state.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>>>>>>>>>>>>>>>>>>>> If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never reach
>>>>>>>>>>>>>>>>>>>>>> its final state.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> This is still nonsense.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
>>>>>>>>>>>>>>>>>>>> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would reach its
>>>>>>>>>>>>>>>>>>>> own final state.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>>>>>>>>>>>>>>>>>> If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would never
>>>>>>>>>>>>>>>>>>>> reach its own final state.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> And again you're still being inconsistent. You can either use H or use
>>>>>>>>>>>>>>>>>>> embedded_H, but you can't mix the two.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Sure I can. I just did.
>>>>>>>>>>>>>>>>>>>> This means that H pretends that it is only a UTM to see what its
>>>>>>>>>>>>>>>>>>>> simulated input would do in this case. If it would never reach its own
>>>>>>>>>>>>>>>>>>>> final state then H correctly rejects this input.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> A Turing Machine cannot "pretend" to be some different Turing Machine.
>>>>>>>>>>>>>>>>>> It can perform a pure simulation of its input until this simulated input
>>>>>>>>>>>>>>>>>> matches a repeating behavior pattern that proves this input never
>>>>>>>>>>>>>>>>>> reaches its own final state.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> If that's the case, why does an actual UTM applied to the *same* input halt?
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Hint: Because the result of an actual UTM applied to the input defines the correct answer, so H answers wrong.
>>>>>>>>>>>>>>>> Intuitively that would seem to be true, this intuition is incorrect.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> The ultimate definition of correct is the computation of the mapping of
>>>>>>>>>>>>>>>> the inputs to an accept or reject state on the basis of the behavior
>>>>>>>>>>>>>>>> that these inputs specify.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> That simulated inputs to embedded_H would never reach their own final
>>>>>>>>>>>>>>>> state under any condition what-so-ever
>>>>>>>>>>>>>>>> IS THE ULTIMATE MEASURE OF THEIR HALTING BEHAVIOR
>>>>>>>>>>>>>>>> and conclusively proves they specify a non-halting sequence of
>>>>>>>>>>>>>>>> configurations.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Since embedded_H is the same as H (as you have yet to provide any evidence to the contrary), then the above can be applied to any simulating halt decider.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>> The topic is the single point that the simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
>>>>>>>>>>>>>> embedded_h cannot possibly reach its own final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩.
>>>>>>>>>>>>>
>>>>>>>>>>>>> And that same logic gives us this:
>>>>>>>>>>>>>
>>>>>>>>>>>>> That simulated inputs <N><5> to Ha3 would never reach their own final
>>>>>>>>>>>>> state under any condition what-so-ever
>>>>>>>>>>>>> IS THE ULTIMATE MEASURE OF THEIR HALTING BEHAVIOR
>>>>>>>>>>>>> and conclusively proves they specify a non-halting sequence of
>>>>>>>>>>>>> configurations.
>>>>>>>>>>>>>
>>>>>>>>>>>>> THE SIMULATED INPUT CANNOT POSSIBLY REACH ITS OWN FINAL STATE
>>>>>>>>>>>>> THIS SINGLE FACT BY ITSELF PROVES THAT THE INPUT IS CORRECTLY REJECTED
>>>>>>>>>>>>>
>>>>>>>>>>>>> Try to find an error in the above.
>>>>>>>>>>>>
>>>>>>>>>>>> That simulated inputs <N><5> to Ha3
>>>>>>>>>>>> are pure nonsense gibberish.
>>>>>>>>>>>
>>>>>>>>>>> So you agree that Ha3 is correct to reject <N><5>?
>>>>>>>>>> It has no associated meaning
>>>>>>>>>
>>>>>>>>> Sure it does. Ha3 is a simulating halt decider whose halt status criteria is to abort after 3 steps, and N takes <n> as input and runs for exactly n steps before halting in state N.qy. And Ha3 rejects <N><5>.
>>>>>>>>>
>>>>>>>>> That simulated inputs <N><5> to Ha3 would never reach their own final
>>>>>>>> You are starting with a broken halt decider, my rules only apply to
>>>>>>>> correct halt deciders.
>>>>>>>>
>>>>>>>> It is like you are saying that no one can possibly drive their car and
>>>>>>>> to prove this you drive your car into a tree.
>>>>>>>
>>>>>>> So what criteria would you apply to Ha3 to determine that it's broken?
>>>>>> I would simply say that ridiculously stupid ideas such as Ha3 should
>>>>>> never be presented by someone with your top 0.04% reputation they
>>>>>> denigrate you. You must know how stupid this idea was before you first
>>>>>> mentioned it.
>>>>>
>>>>> If it's so obvious to you that Ha3 is broken, then you should be able to explain exactly how to determine that.
>>>> Obviously a very lame attempt at a head game by a guy with your reputation.
>>>
>>> Explain how you would determine Ha3 is broken. If you don't, I'll be forced to conclude that you agree that it's correct.
>> I will quit talking to you until you stop the head games.
>
> That you have difficultly following my logic (actually, your own logic applied to a different halt decider) doesn't mean I'm playing head games.
>
> Explain exactly how you would determine that Ha3 is broken, or admit that it is not.

You know that the question is mere denigration:
https://en.wikipedia.org/wiki/UFTP

--
Copyright 2022 Pete Olcott

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

SubjectRepliesAuthor
o Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key

By: olcott on Sun, 3 Apr 2022

978olcott
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