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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 ]
From: dbush.mo...@gmail.com (Dennis Bush)
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 by: Dennis Bush - Mon, 11 Apr 2022 01:53 UTC

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.

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

By: olcott on Sun, 3 Apr 2022

978olcott
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