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Concise refutation of halting problem proofs V47 | olcott |

Re: Concise refutation of halting problem proofs V47, REFUTED, FAIR | olcott |

Re: Concise refutation of halting problem proofs V47, [ H is an | olcott |

Re: Concise refutation of halting problem proofs V47, [ H is an | olcott |

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Subject: Concise refutation of halting problem proofs V47

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// Simplified Linz(1990) Ĥ

// and Strachey(1965) P

void P(ptr x)

{

if (H(x, y))

HERE: goto HERE;

}

H and P are defined according to the standard HP counter-example template shown above.

H bases its halt status decision on the behavior of the simulation of its input.

Then P demonstrates an infinitely repeating pattern that cannot possibly ever reach its final state.

This conclusively proves that the input to H meets the Linz definition of non-halting:

computation that halts … the Turing machine will halt whenever it enters a final state. (Linz:1990:234)

thus the sufficiency condition for H to report that its input specifies a non-halting computation.

Halting problem undecidability and infinitely nested simulation V2

https://www.researchgate.net/publication/356105750_Halting_problem_undecidability_and_infinitely_nested_simulation_V2

--

Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;

Genius hits a target no one else can see.

Arthur Schopenhauer

// and Strachey(1965) P

void P(ptr x)

{

if (H(x, y))

HERE: goto HERE;

}

H and P are defined according to the standard HP counter-example template shown above.

H bases its halt status decision on the behavior of the simulation of its input.

Then P demonstrates an infinitely repeating pattern that cannot possibly ever reach its final state.

This conclusively proves that the input to H meets the Linz definition of non-halting:

computation that halts … the Turing machine will halt whenever it enters a final state. (Linz:1990:234)

thus the sufficiency condition for H to report that its input specifies a non-halting computation.

Halting problem undecidability and infinitely nested simulation V2

https://www.researchgate.net/publication/356105750_Halting_problem_undecidability_and_infinitely_nested_simulation_V2

--

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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Subject: Re: Concise refutation of halting problem proofs V47, REFUTED, FAIR

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On 1/8/2022 8:55 PM, Richard Damon wrote:

The problem is that you are simply too stupid to ever understand that P specifies a sequence of configurations that never reach its final state and thus is correctly determined to be a non-halting computation according to Linz.

Malcolm, Kaz and Flibble are not too stupid to understand this.

Ben, André and Mike are not interested in understanding what I say they are only interested in finding some basis for rebuttal. If there is at least one minor point that I have not proven completely they count everything that I say as incorrect on the basis of this minor point.

--

Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;

Genius hits a target no one else can see.

Arthur Schopenhauer

On 1/8/22 8:41 PM, olcott wrote:

// Simplified Linz(1990) Ĥ

// and Strachey(1965) P

void P(ptr x)

{

if (H(x, y))

HERE: goto HERE;

}

H and P are defined according to the standard HP counter-example template shown above.

H bases its halt status decision on the behavior of the simulation of its input.

Then P demonstrates an infinitely repeating pattern that cannot possibly ever reach its final state.

This conclusively proves that the input to H meets the Linz definition of non-halting:

computation that halts … the Turing machine will halt whenever it enters a final state. (Linz:1990:234)

thus the sufficiency condition for H to report that its input specifies a non-halting computation.

Halting problem undecidability and infinitely nested simulation V2

https://www.researchgate.net/publication/356105750_Halting_problem_undecidability_and_infinitely_nested_simulation_V2

Full Proof with Request for Rebuttal

We have gone around the circle of this MANY times, and you keep just rearranging things and not every answering the refutation.

The problem is that you are simply too stupid to ever understand that P specifies a sequence of configurations that never reach its final state and thus is correctly determined to be a non-halting computation according to Linz.

Malcolm, Kaz and Flibble are not too stupid to understand this.

Ben, André and Mike are not interested in understanding what I say they are only interested in finding some basis for rebuttal. If there is at least one minor point that I have not proven completely they count everything that I say as incorrect on the basis of this minor point.

--

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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On 1/8/2022 9:59 PM, Richard Damon wrote:

It is always correct for H to report on what the behavior of its input would be if H did not interfere with the behavior of this input.

H is an objective observer.

It is never correct for H to report on what the behavior of its input would be if H did interfere with the behavior of this input.

H is not an objective observer.

--

Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;

Genius hits a target no one else can see.

Arthur Schopenhauer

On 1/8/22 10:20 PM, olcott wrote:

On 1/8/2022 8:55 PM, Richard Damon wrote:

On 1/8/22 8:41 PM, olcott wrote:

// Simplified Linz(1990) Ĥ

// and Strachey(1965) P

void P(ptr x)

{

if (H(x, y))

HERE: goto HERE;

}

H and P are defined according to the standard HP counter-example template shown above.

H bases its halt status decision on the behavior of the simulation of its input.

Then P demonstrates an infinitely repeating pattern that cannot possibly ever reach its final state.

This conclusively proves that the input to H meets the Linz definition of non-halting:

computation that halts … the Turing machine will halt whenever it enters a final state. (Linz:1990:234)

thus the sufficiency condition for H to report that its input specifies a non-halting computation.

Halting problem undecidability and infinitely nested simulation V2

https://www.researchgate.net/publication/356105750_Halting_problem_undecidability_and_infinitely_nested_simulation_V2

Full Proof with Request for Rebuttal

We have gone around the circle of this MANY times, and you keep just rearranging things and not every answering the refutation.

The problem is that you are simply too stupid to ever understand that P specifies a sequence of configurations that never reach its final state and thus is correctly determined to be a non-halting computation according to Linz.

And you are too stupid to see that it doesn't if H(P,P) returns 0, as this just proved.

It is always correct for H to report on what the behavior of its input would be if H did not interfere with the behavior of this input.

H is an objective observer.

It is never correct for H to report on what the behavior of its input would be if H did interfere with the behavior of this input.

H is not an objective observer.

Your failure to point out an error will be taken as an admission that you accept that your logic is incorrect.

FAIR WARNING.

Malcolm, Kaz and Flibble are not too stupid to understand this.

Ben, André and Mike are not interested in understanding what I say they are only interested in finding some basis for rebuttal. If there is at least one minor point that I have not proven completely they count everything that I say as incorrect on the basis of this minor point.

--

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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On 1/9/2022 12:06 PM, Richard Damon wrote:

It is the job of H to determine what the behavior of the input would be if H did not interfere with this behavior.

Alternatively H could correctly recognize inputs that would never stop running if H did not interfere and then report that every input does halt when H does interfere.

Such an H could simply accept every input in that some of its inputs halt on their own and the other inputs must be aborted. In this case It would be impossible to create an input that H would not correctly decide.

--

Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;

Genius hits a target no one else can see.

Arthur Schopenhauer

On 1/9/22 12:57 PM, olcott wrote:

On 1/9/2022 11:43 AM, Richard Damon wrote:

On 1/9/22 11:40 AM, olcott wrote:

On 1/8/2022 9:59 PM, Richard Damon wrote:

On 1/8/22 10:20 PM, olcott wrote:

On 1/8/2022 8:55 PM, Richard Damon wrote:

On 1/8/22 8:41 PM, olcott wrote:

// Simplified Linz(1990) Ĥ

// and Strachey(1965) P

void P(ptr x)

{

if (H(x, y))

HERE: goto HERE;

}

H and P are defined according to the standard HP counter-example template shown above.

H bases its halt status decision on the behavior of the simulation of its input.

Then P demonstrates an infinitely repeating pattern that cannot possibly ever reach its final state.

This conclusively proves that the input to H meets the Linz definition of non-halting:

computation that halts … the Turing machine will halt whenever it enters a final state. (Linz:1990:234)

thus the sufficiency condition for H to report that its input specifies a non-halting computation.

Halting problem undecidability and infinitely nested simulation V2

https://www.researchgate.net/publication/356105750_Halting_problem_undecidability_and_infinitely_nested_simulation_V2

Full Proof with Request for Rebuttal

We have gone around the circle of this MANY times, and you keep just rearranging things and not every answering the refutation.

The problem is that you are simply too stupid to ever understand that P specifies a sequence of configurations that never reach its final state and thus is correctly determined to be a non-halting computation according to Linz.

And you are too stupid to see that it doesn't if H(P,P) returns 0, as this just proved.

It is always correct for H to report on what the behavior of its input would be if H did not interfere with the behavior of this input.

H is an objective observer.

It is never correct for H to report on what the behavior of its input would be if H did interfere with the behavior of this input.

H is not an objective observer.

IMPROPERLY PHRASED, H must report on what the machine that its input represents will do, even if that includes a copy of itself. That is not H 'interfering' with the behavior of that machine.

It is Impossible for the copy of a decider doing the deciding to 'interfere' with the behavior of a machine, as that behavior is defined independent of the decider.

Yes, the aborting of a simulation by the copy of the decider doing the deciding doesn't affect the behavior of the machine it is deciding on, a copy of it IN the machine it is trying to decide on, DOES, as it IS part of the machine it is deciding on.

FAIL.

The fact that you can't keep the different copies of H separate shows your lack of reasoning ability.

When H reports on what the behavior of its simulated input would be if H did not interfere, it is the same for P, infinite loops, or infinite recursion, H must only reject its input as non-halting.

Except it isn't 'interference' for the copy of H in the input to do what it is programmed to do.

It is the job of H to determine what the behavior of the input would be if H did not interfere with this behavior.

Alternatively H could correctly recognize inputs that would never stop running if H did not interfere and then report that every input does halt when H does interfere.

Such an H could simply accept every input in that some of its inputs halt on their own and the other inputs must be aborted. In this case It would be impossible to create an input that H would not correctly decide.

In fact, it is interference for the H that is deciding to NOT let the copy of H inside P to do that that H is programmed to do.

DEFINITIONS, you know, The correct answer for H(<X>, y) is basd on what X(y) would do when run.

The behavior of 'the input' is what that program would do when run 'without outside interference', that means that copy of H inside P does what it will do.

Since H(P,P) returns 0, ALL Copies of H(P,P) return 0, so the copy inside P does thins, and P(P) will Halt.

FAIL.

This has been PROVEN and no actual rebuttal provided, so you have conceded the point.

Your failure to point out an error will be taken as an admission that you accept that your logic is incorrect.

FAIR WARNING.

Malcolm, Kaz and Flibble are not too stupid to understand this.

Ben, André and Mike are not interested in understanding what I say they are only interested in finding some basis for rebuttal. If there is at least one minor point that I have not proven completely they count everything that I say as incorrect on the basis of this minor point.

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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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