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devel / comp.theory / Concise refutation of halting problem proofs V61 [ Linz Proof ]

SubjectAuthor
* Concise refutation of halting problem proofs V61 [ Linz Proof ]olcott
+- Concise refutation of halting problem proofs V61 [ Linz Proof ]Richard Damon
`* Concise refutation of halting problem proofs V61 [ Linz Proof ]Mikko
 `* Concise refutation of halting problem proofs V61 [ Linz Proof ]olcott
  +- Concise refutation of halting problem proofs V61 [ Linz Proof ]Richard Damon
  `* Concise refutation of halting problem proofs V61 [ Linz Proof ]Mikko
   `- Concise refutation of halting problem proofs V61 [ Linz Proof ]olcott

1
Concise refutation of halting problem proofs V61 [ Linz Proof ]

<stm8r9$5co$1@dont-email.me>

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From: polco...@gmail.com (olcott)
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Subject: Concise refutation of halting problem proofs V61 [ Linz Proof ]
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 by: olcott - Sat, 5 Feb 2022 16:32 UTC

Halting problem undecidability and infinitely nested simulation (V3)

Linz H is defined as simulating halt decider that bases its halt status
decision on whether or not its correct simulation of its input could
ever reach the final state of this simulated input. H determines this on
the basis of matching infinite behavior patterns. When an infinite
behavior pattern is matched H aborts its simulation and transitions to
its final reject state. Otherwise H transitions to its accept state when
its simulation ends.

The following simplifies the syntax for the definition of the Linz
Turing machine Ĥ, it is now a single machine with a single start state.
A copy of Linz H is embedded at Ĥ.qx.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Can the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H possibly transition
to ⟨Ĥ⟩.qn ?

When Ĥ is applied to ⟨Ĥ⟩
Ĥ copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then embedded_H simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩

Then these steps would keep repeating:
Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩
Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩ ⟨Ĥ5⟩...

The above shows that the correctly simulated (as if Ĥ.qx was a UTM)
input to embedded_H would never reach its final state of ⟨Ĥ⟩.qn
conclusively proving that this simulated input never halts. This enables
embedded_H to abort its simulation and correctly transition to Ĥ.qn.

Because all simulating halt deciders are deciders they are only
accountable for computing the mapping from their input finite strings to
an accept or reject state on the basis of whether or not their correctly
simulated input could ever reach its final state.

embedded_H is only accountable for the behavior of its input ⟨Ĥ⟩ applied
to ⟨Ĥ⟩. embedded_H is not accountable for the behavior of the
computation that it is contained within: Ĥ applied to ⟨Ĥ⟩.

Halting problem undecidability and infinitely nested simulation (V3)

https://www.researchgate.net/publication/358009319_Halting_problem_undecidability_and_infinitely_nested_simulation_V3

--
Copyright 2021 Pete Olcott "Talent hits a target no one else can hit;
Genius hits a target no one else can see." Arthur Schopenhauer

Re: Concise refutation of halting problem proofs V61 [ Linz Proof ]

<4TxLJ.39116$%uX7.16718@fx38.iad>

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 by: Richard Damon - Sat, 5 Feb 2022 16:48 UTC

On 2/5/22 11:32 AM, olcott wrote:
> Halting problem undecidability and infinitely nested simulation (V3)
>
> Linz H is defined as simulating halt decider that bases its halt status
> decision on whether or not its correct simulation of its input could
> ever reach the final state of this simulated input. H determines this on
> the basis of matching infinite behavior patterns. When an infinite
> behavior pattern is matched H aborts its simulation and transitions to
> its final reject state. Otherwise H transitions to its accept state when
> its simulation ends.
>
> The following simplifies the syntax for the definition of the Linz
> Turing machine Ĥ, it is now a single machine with a single start state.
> A copy of Linz H is embedded at Ĥ.qx.
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>
> Can the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H possibly transition
> to ⟨Ĥ⟩.qn ?
>
> When Ĥ is applied to ⟨Ĥ⟩
>   Ĥ copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then embedded_H simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩
>
> Then these steps would keep repeating:
>   Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
>   Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩
>   Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩ ⟨Ĥ5⟩...

But only if H never aborts its simulation and goes to H.Qn to indicate
that it thinks its input is not halting.

Remember, if H does it, then the copy of H in H^, H^1, H^2, H^3 ... will
all do the exact same thing, and thus NONE of them will support the
infinite loop just indicated.

>
> The above shows that the correctly simulated (as if Ĥ.qx was a UTM)
> input to embedded_H would never reach its final state of ⟨Ĥ⟩.qn
> conclusively proving that this simulated input never halts. This enables
> embedded_H to abort its simulation and correctly transition to Ĥ.qn.
>

There either IS or there IS NOT a 'UTM' as H^.Qx, as a copy of what H
is. If there is, the BY DEFINITION, H will run forever if its input
represents a non-halting computation, and thus H fails to be able to
transition to H.Qn to indicate non-halting.

If H has modified the UTM it was based on to do that, then it is no
longer actually a UTM anymore, and thus its simulation is no longer the
defining operation of Halting.

> Because all simulating halt deciders are deciders they are only
> accountable for computing the mapping from their input finite strings to
> an accept or reject state on the basis of whether or not their correctly
> simulated input could ever reach its final state.
>
> embedded_H is only accountable for the behavior of its input ⟨Ĥ⟩ applied
> to ⟨Ĥ⟩. embedded_H is not accountable for the behavior of the
> computation that it is contained within: Ĥ applied to ⟨Ĥ⟩.

And the behavior it is accountable, if it claims to be working on the
halting problem, is the actual behavior of the computation the input
represents, which is identical to the simulation of a ACTUAL UTM, which
for <H^> <H^> has the property that if H <H^> <H^> -> H.Qn, then H^ <H^>
wlll go to H^.Qn and Halt.

Any claims that this is not true are just evidence that you are NOT
working on the halting problem unless you can point to an actual defect
in the logic I just used.

This means that H FAILS to be a correct Halting Decider, as either H
fails to go to H.Qn ever, or it goes to H.Qn when its input represents a
Halting Computation.

FAIL.

>
>
> Halting problem undecidability and infinitely nested simulation (V3)
>
> https://www.researchgate.net/publication/358009319_Halting_problem_undecidability_and_infinitely_nested_simulation_V3
>
>

We have done this before.

This has been proven wrong many times before.

YOU are just repeating your incorrect claims and seem to be stuck in a loop.

If you can not point out an actual logical error in this rebuttal, just
repeating your claims in nothing more than spreading known LIES, and
just proves that either you are too stupid to understand logic or are
just a pathological liar.

FAIL.

Re: Concise refutation of halting problem proofs V61 [ Linz Proof ]

<stolrh$9k4$1@dont-email.me>

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Subject: Re: Concise refutation of halting problem proofs V61 [ Linz Proof ]
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 by: Mikko - Sun, 6 Feb 2022 14:26 UTC

On 2022-02-05 16:32:41 +0000, olcott said:

> Halting problem undecidability and infinitely nested simulation (V3)
>
> Linz H is defined as simulating halt decider that bases its halt status
> decision on whether or not its correct simulation of its input could
> ever
...

What was wrong in V60 ?

Mikko

Re: Concise refutation of halting problem proofs V61 [ Linz Proof ]

<stomq9$rpa$1@dont-email.me>

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Subject: Re: Concise refutation of halting problem proofs V61 [ Linz Proof ]
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 by: olcott - Sun, 6 Feb 2022 14:43 UTC

On 2/6/2022 8:26 AM, Mikko wrote:
> On 2022-02-05 16:32:41 +0000, olcott said:
>
>> Halting problem undecidability and infinitely nested simulation (V3)
>>
>> Linz H is defined as simulating halt decider that bases its halt
>> status decision on whether or not its correct simulation of its input
>> could ever
>  ...
>
> What was wrong in V60 ?
>
> Mikko
>

I keep making my posts increasingly more clear.
I wish that making them clear enough to be understood worked.

I actually have to make them clear enough that any rebuttals look
foolish because most of my reviewers don't really give a rats ass for
truth they only want to show that I am wrong even if I am not wrong.
This comes from mutual animosity that has been established over the years.

Halting problem undecidability and infinitely nested simulation (V3)
https://www.researchgate.net/publication/358009319_Halting_problem_undecidability_and_infinitely_nested_simulation_V3

--
Copyright 2021 Pete Olcott "Talent hits a target no one else can hit;
Genius hits a target no one else can see." Arthur Schopenhauer

Re: Concise refutation of halting problem proofs V61 [ Linz Proof ]

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 by: Richard Damon - Sun, 6 Feb 2022 18:22 UTC

On 2/6/22 9:43 AM, olcott wrote:
> On 2/6/2022 8:26 AM, Mikko wrote:
>> On 2022-02-05 16:32:41 +0000, olcott said:
>>
>>> Halting problem undecidability and infinitely nested simulation (V3)
>>>
>>> Linz H is defined as simulating halt decider that bases its halt
>>> status decision on whether or not its correct simulation of its input
>>> could ever
>>   ...
>>
>> What was wrong in V60 ?
>>
>> Mikko
>>
>
> I keep making my posts increasingly more clear.
> I wish that making them clear enough to be understood worked.
>
> I actually have to make them clear enough that any rebuttals look
> foolish because most of my reviewers don't really give a rats ass for
> truth they only want to show that I am wrong even if I am not wrong.
> This comes from mutual animosity that has been established over the years.
>
> Halting problem undecidability and infinitely nested simulation (V3)
> https://www.researchgate.net/publication/358009319_Halting_problem_undecidability_and_infinitely_nested_simulation_V3
>
>

Except that by just ignoring the rebuttals and posting basically the
same argument, YOU are the one who looks stupid, and the one that
doesn't care for finding the actual truth.

Re: Concise refutation of halting problem proofs V61 [ Linz Proof ]

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 by: Mikko - Mon, 7 Feb 2022 09:24 UTC

On 2022-02-06 14:43:20 +0000, olcott said:

> On 2/6/2022 8:26 AM, Mikko wrote:
>> What was wrong in V60 ?

> I keep making my posts increasingly more clear.
> I wish that making them clear enough to be understood worked.

It does not work for those who already read the old version. You should
identify the unclear point and its clarification. Otherwise one just thinks
that the new version means what one thought the old version ment.

Mikko

Re: Concise refutation of halting problem proofs V61 [ Linz Proof ]

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 by: olcott - Mon, 7 Feb 2022 14:55 UTC

On 2/7/2022 3:24 AM, Mikko wrote:
> On 2022-02-06 14:43:20 +0000, olcott said:
>
>> On 2/6/2022 8:26 AM, Mikko wrote:
>>> What was wrong in V60 ?
>
>> I keep making my posts increasingly more clear.
>> I wish that making them clear enough to be understood worked.
>
> It does not work for those who already read the old version. You should
> identify the unclear point and its clarification. Otherwise one just thinks
> that the new version means what one thought the old version ment.
>
> Mikko
>

In other words you are saying that some people are deliberately taking
my words to means something besides what my words say because their main
goal is to be disagreeable?

This is the gist of my whole proof:

The following simplifies the syntax for the definition of the Linz
Turing machine Ĥ, it is now a single machine with a single start state.
A copy of Linz H is embedded at Ĥ.qx.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Can the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H possibly transition
to ⟨Ĥ⟩.qn ?

Halting problem undecidability and infinitely nested simulation (V3)

https://www.researchgate.net/publication/358009319_Halting_problem_undecidability_and_infinitely_nested_simulation_V3

--
Copyright 2021 Pete Olcott

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

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