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Subject: Re: Concise refutation of halting problem proofs V59 [ key essence ]

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From: NoO...@NoWhere.com (olcott)

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Subject: Re: Concise refutation of halting problem proofs V59 [ key essence ]

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On 1/31/2022 11:25 PM, Richard Damon wrote:

The halting problem is vague on the definition of halting, it includes that a machine has stopped running and that a machine cannot reach its final state. My definition only includes the latter.

The halting problem does not bother to mention the requirement that because all halt deciders are deciders they are only accountable for computing the mapping from their finite string inputs to an accept or reject state on the basis of the actual behavior specified by this input.

The halting problem does not specifically examine simulating halt deciders, none-the-less the behavior of a correctly simulated machine description is known to be equivalent to the behavior of the direct execution of this same machine.

Since a simulating halt decider is merely a UTM for simulated inputs that reach their final state when a simulating halt decider correctly determines that its simulated its input cannot possibly reach its final state this is complete proof that this simulated input never halts.

--

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 1/31/22 11:42 PM, olcott wrote:

On 1/31/2022 10:33 PM, Richard Damon wrote:

On 1/31/22 11:24 PM, olcott wrote:

On 1/31/2022 10:17 PM, Richard Damon wrote:

On 1/31/22 10:40 PM, olcott wrote:No that is not it. That is like saying "by definition" Sum(3,5) is being asked about Sum(7,8).

On 1/31/2022 6:41 PM, Richard Damon wrote:

On 1/31/22 3:24 PM, olcott wrote:

On 1/31/2022 2:10 PM, Ben wrote:

On 1/31/2022 8:06 AM, olcott wrote:

On 1/30/2022 8:20 PM, Richard Damon wrote:

On 1/30/22 9:05 PM, olcott wrote:

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

These statements need the conditions, that H^ goes to H^.Qy/H^.Qn iff H goes to that corresponding state.

⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H in the same way that (5,3) is syntactically specified as an input to Sum(5,3)

Right, and the

Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to embedded_H in the same way that (1,2) is NOT syntactically specified as an input to Sum(5,3)

Right, but perhaps you don't understand that from you above statement the right answer is based on if UTM(<H^>,<H^>) Halts which by the definition of a UTM means if H^ applied to <H^> Halts.

The biggest reason for your huge mistakes is that you cannot stay sharply focused on a single point. It is as if you either have attention deficit disorder ADD or are addicted to methamphetamine.

>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The single point is that ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H and

Ĥ ⟨Ĥ⟩ is the NOT the input to embedded_H.

After we have mutual agreement on this point we will move on to the points that logically follow from this one.

Holy shit try to post something that makes sense.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Richard does not accept that the input to the copy of Linz H embedded at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩. He keeps insisting that it is Ĥ ⟨Ĥ⟩.

No, but apparently you can't understand actual English words.

The INPUT to H is <H^> <H^> but the CORRECT ANSWER that H must give is based on the behavior of H^ applied to <H^> BECAUSE OF THE DEFINITION of H.

In other words Sum(3,5) must return the value of Sum(7,8)?

Don't know how you get that from what I said.

Any moron knows that a function is only accountable for its actual inputs.

And the actual input to H is <H^> <H^> which MEANS by the DEFINITION of the Halting Problem that H is being asked to decide on the Halting Status of H^ applied to <H^>

Again your RED HERRING.

H is being asked EXACTLY what it being asked

H wM w -> H.Qy if M applied to w Halts, and H.Qn if it doesn't

AGREED?

No that is wrong. embedded_H is being asked:

Can the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ possibly transition to ⟨Ĥ⟩.qn ?

If you say 'No', then you aren't doing the halting problem, as the requirement I stated is EXACTLY the requirement of the Halting Problem.

The halting problem is vague on the definition of halting, it includes that a machine has stopped running and that a machine cannot reach its final state. My definition only includes the latter.

The halting problem does not bother to mention the requirement that because all halt deciders are deciders they are only accountable for computing the mapping from their finite string inputs to an accept or reject state on the basis of the actual behavior specified by this input.

The halting problem does not specifically examine simulating halt deciders, none-the-less the behavior of a correctly simulated machine description is known to be equivalent to the behavior of the direct execution of this same machine.

Since a simulating halt decider is merely a UTM for simulated inputs that reach their final state when a simulating halt decider correctly determines that its simulated its input cannot possibly reach its final state this is complete proof that this simulated input never halts.

--

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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From: polco...@gmail.com (olcott)

Newsgroups: comp.theory,comp.ai.philosophy,sci.logic,sci.math

Subject: Re: Concise refutation of halting problem proofs V59 [ key essence ]

Followup-To: comp.theory

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View all headersFrom: polco...@gmail.com (olcott)

Newsgroups: comp.theory,comp.ai.philosophy,sci.logic,sci.math

Subject: Re: Concise refutation of halting problem proofs V59 [ key essence ]

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Date: Tue, 1 Feb 2022 11:43:15 -0600

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On 2/1/2022 10:33 AM, wij wrote:

You did not even read what I said before you claimed that it was incorrect. The general principles that I outlined below directly apply to the actual Linz proof:

If I am incorrect in anything that I said below then the specific error could be pointed out.

--

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 Tuesday, 1 February 2022 at 23:22:32 UTC+8, olcott wrote:

On 1/31/2022 11:25 PM, Richard Damon wrote:Sounds like a NDTM.

The halting problem is vague on the definition of halting, it includes

On 1/31/22 11:42 PM, olcott wrote:

On 1/31/2022 10:33 PM, Richard Damon wrote:

On 1/31/22 11:24 PM, olcott wrote:

On 1/31/2022 10:17 PM, Richard Damon wrote:

On 1/31/22 10:40 PM, olcott wrote:No that is not it. That is like saying "by definition" Sum(3,5) is

On 1/31/2022 6:41 PM, Richard Damon wrote:

On 1/31/22 3:24 PM, olcott wrote:

On 1/31/2022 2:10 PM, Ben wrote:

On 1/31/2022 8:06 AM, olcott wrote:

On 1/30/2022 8:20 PM, Richard Damon wrote:

On 1/30/22 9:05 PM, olcott wrote:

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

These statements need the conditions, that H^ goes to

H^.Qy/H^.Qn iff H goes to that corresponding state.

⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H

in the same way that (5,3) is syntactically specified as an

input to Sum(5,3)

Right, and the

Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to

embedded_H in the same way that (1,2) is NOT syntactically

specified as an input to Sum(5,3)

Right, but perhaps you don't understand that from you above

statement the right answer is based on if UTM(<H^>,<H^>)

Halts which by the definition of a UTM means if H^ applied to

<H^> Halts.

The biggest reason for your huge mistakes is that you cannot

stay sharply focused on a single point. It is as if you either

have attention deficit disorder ADD or are addicted to

methamphetamine.

>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The single point is that ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H and

Ĥ ⟨Ĥ⟩ is the NOT the input to embedded_H.

After we have mutual agreement on this point we will move on

to the points that logically follow from this one.

Holy shit try to post something that makes sense.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Richard does not accept that the input to the copy of Linz H

embedded at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩. He keeps insisting that it is Ĥ ⟨Ĥ⟩.

No, but apparently you can't understand actual English words.

The INPUT to H is <H^> <H^> but the CORRECT ANSWER that H must

give is based on the behavior of H^ applied to <H^> BECAUSE OF

THE DEFINITION of H.

In other words Sum(3,5) must return the value of Sum(7,8)?

Don't know how you get that from what I said.

Any moron knows that a function is only accountable for its actual

inputs.

And the actual input to H is <H^> <H^> which MEANS by the

DEFINITION of the Halting Problem that H is being asked to decide

on the Halting Status of H^ applied to <H^>

being asked about Sum(7,8).

Again your RED HERRING.

H is being asked EXACTLY what it being asked

H wM w -> H.Qy if M applied to w Halts, and H.Qn if it doesn't

AGREED?

No that is wrong. embedded_H is being asked:

Can the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ possibly transition to ⟨Ĥ⟩.qn ?

If you say 'No', then you aren't doing the halting problem, as the

requirement I stated is EXACTLY the requirement of the Halting Problem.

that a machine has stopped running and that a machine cannot reach its

final state. My definition only includes the latter.

The Halting Problem has a definite, commonly recognized meaning. It refers to a

real machine, no ambiguity, no one can change it, not even Linz.

Your halt-problem is absolutely certain not Linz's, or of any? textbook.

Your claim contradicts experimental truth. Otherwise, show your x86utm operating

system proof. I guess you would say xxx thousands pages, I believe there are

only few lines are yours. Show your codes.

You did not even read what I said before you claimed that it was incorrect. The general principles that I outlined below directly apply to the actual Linz proof:

If I am incorrect in anything that I said below then the specific error could be pointed out.

The halting problem does not bother to mention the requirement that

because all halt deciders are deciders they are only accountable for

computing the mapping from their finite string inputs to an accept or

reject state on the basis of the actual behavior specified by this input.

The halting problem does not specifically examine simulating halt

deciders, none-the-less the behavior of a correctly simulated machine

description is known to be equivalent to the behavior of the direct

execution of this same machine.

Since a simulating halt decider is merely a UTM for simulated inputs

that reach their final state when a simulating halt decider correctly

determines that its simulated its input cannot possibly reach its final

state this is complete proof that this simulated input never halts.

-- Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;

Genius hits a target no one else can see.

Arthur Schopenhauer

--

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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<ibHJJ.56320$u41.55552@fx41.iad>

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<gIHJJ.29153$541.4042@fx35.iad> <st91ek$p4g$1@dont-email.me>

<st9fn6$60s$2@gioia.aioe.org> <RqidnSdLIdwH2GX8nZ2dnUU7-SXNnZ2d@giganews.com>

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<gIHJJ.29153$541.4042@fx35.iad> <st91ek$p4g$1@dont-email.me>

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On 2/1/2022 10:33 AM, wij wrote:

https://en.wikipedia.org/wiki/Nondeterministic_Turing_machine

It is not a NDTM, a Turing Machine only actually halts when it reaches its own final state. People not very familiar with this material may get confused and believe that a TM halts when its stops running because its simulation has been aborted. This key distinction is not typically specified in most halting problem proofs.

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

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

On Tuesday, 1 February 2022 at 23:22:32 UTC+8, olcott wrote:

On 1/31/2022 11:25 PM, Richard Damon wrote:Sounds like a NDTM.

The halting problem is vague on the definition of halting, it includes

On 1/31/22 11:42 PM, olcott wrote:

On 1/31/2022 10:33 PM, Richard Damon wrote:

On 1/31/22 11:24 PM, olcott wrote:

On 1/31/2022 10:17 PM, Richard Damon wrote:

On 1/31/22 10:40 PM, olcott wrote:No that is not it. That is like saying "by definition" Sum(3,5) is

On 1/31/2022 6:41 PM, Richard Damon wrote:

On 1/31/22 3:24 PM, olcott wrote:

On 1/31/2022 2:10 PM, Ben wrote:

On 1/31/2022 8:06 AM, olcott wrote:

On 1/30/2022 8:20 PM, Richard Damon wrote:

On 1/30/22 9:05 PM, olcott wrote:

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

These statements need the conditions, that H^ goes to

H^.Qy/H^.Qn iff H goes to that corresponding state.

⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H

in the same way that (5,3) is syntactically specified as an

input to Sum(5,3)

Right, and the

Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to

embedded_H in the same way that (1,2) is NOT syntactically

specified as an input to Sum(5,3)

Right, but perhaps you don't understand that from you above

statement the right answer is based on if UTM(<H^>,<H^>)

Halts which by the definition of a UTM means if H^ applied to

<H^> Halts.

The biggest reason for your huge mistakes is that you cannot

stay sharply focused on a single point. It is as if you either

have attention deficit disorder ADD or are addicted to

methamphetamine.

>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The single point is that ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H and

Ĥ ⟨Ĥ⟩ is the NOT the input to embedded_H.

After we have mutual agreement on this point we will move on

to the points that logically follow from this one.

Holy shit try to post something that makes sense.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Richard does not accept that the input to the copy of Linz H

embedded at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩. He keeps insisting that it is Ĥ ⟨Ĥ⟩.

No, but apparently you can't understand actual English words.

The INPUT to H is <H^> <H^> but the CORRECT ANSWER that H must

give is based on the behavior of H^ applied to <H^> BECAUSE OF

THE DEFINITION of H.

In other words Sum(3,5) must return the value of Sum(7,8)?

Don't know how you get that from what I said.

Any moron knows that a function is only accountable for its actual

inputs.

And the actual input to H is <H^> <H^> which MEANS by the

DEFINITION of the Halting Problem that H is being asked to decide

on the Halting Status of H^ applied to <H^>

being asked about Sum(7,8).

Again your RED HERRING.

H is being asked EXACTLY what it being asked

H wM w -> H.Qy if M applied to w Halts, and H.Qn if it doesn't

AGREED?

No that is wrong. embedded_H is being asked:

Can the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ possibly transition to ⟨Ĥ⟩.qn ?

If you say 'No', then you aren't doing the halting problem, as the

requirement I stated is EXACTLY the requirement of the Halting Problem.

that a machine has stopped running and that a machine cannot reach its

final state. My definition only includes the latter.

https://en.wikipedia.org/wiki/Nondeterministic_Turing_machine

It is not a NDTM, a Turing Machine only actually halts when it reaches its own final state. People not very familiar with this material may get confused and believe that a TM halts when its stops running because its simulation has been aborted. This key distinction is not typically specified in most halting problem proofs.

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

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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On 2/1/2022 3:23 PM, wij wrote:

I have shown how my system directly applies to the actual halting problem and it can be understood as correct by anyone that understands the halting problem at a much deeper level than rote memorization.

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 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ? (No means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt).

--

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 Wednesday, 2 February 2022 at 02:37:17 UTC+8, olcott wrote:

On 2/1/2022 10:33 AM, wij wrote:

On Tuesday, 1 February 2022 at 23:22:32 UTC+8, olcott wrote:https://en.wikipedia.org/wiki/Nondeterministic_Turing_machine

On 1/31/2022 11:25 PM, Richard Damon wrote:

The halting problem is vague on the definition of halting, it includes

On 1/31/22 11:42 PM, olcott wrote:

On 1/31/2022 10:33 PM, Richard Damon wrote:

On 1/31/22 11:24 PM, olcott wrote:

On 1/31/2022 10:17 PM, Richard Damon wrote:

On 1/31/22 10:40 PM, olcott wrote:No that is not it. That is like saying "by definition" Sum(3,5) is

On 1/31/2022 6:41 PM, Richard Damon wrote:

On 1/31/22 3:24 PM, olcott wrote:

On 1/31/2022 2:10 PM, Ben wrote:

On 1/31/2022 8:06 AM, olcott wrote:

On 1/30/2022 8:20 PM, Richard Damon wrote:

On 1/30/22 9:05 PM, olcott wrote:

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

These statements need the conditions, that H^ goes to

H^.Qy/H^.Qn iff H goes to that corresponding state.

⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H

in the same way that (5,3) is syntactically specified as an

input to Sum(5,3)

Right, and the

Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to

embedded_H in the same way that (1,2) is NOT syntactically

specified as an input to Sum(5,3)

Right, but perhaps you don't understand that from you above

statement the right answer is based on if UTM(<H^>,<H^>)

Halts which by the definition of a UTM means if H^ applied to

<H^> Halts.

The biggest reason for your huge mistakes is that you cannot

stay sharply focused on a single point. It is as if you either

have attention deficit disorder ADD or are addicted to

methamphetamine.

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The single point is that ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H and

Ĥ ⟨Ĥ⟩ is the NOT the input to embedded_H.

After we have mutual agreement on this point we will move on

to the points that logically follow from this one.

Holy shit try to post something that makes sense.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Richard does not accept that the input to the copy of Linz H

embedded at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩. He keeps insisting that it is Ĥ ⟨Ĥ⟩.

No, but apparently you can't understand actual English words.

The INPUT to H is <H^> <H^> but the CORRECT ANSWER that H must

give is based on the behavior of H^ applied to <H^> BECAUSE OF

THE DEFINITION of H.

In other words Sum(3,5) must return the value of Sum(7,8)?

Don't know how you get that from what I said.

Any moron knows that a function is only accountable for its actual

inputs.

And the actual input to H is <H^> <H^> which MEANS by the

DEFINITION of the Halting Problem that H is being asked to decide

on the Halting Status of H^ applied to <H^>

being asked about Sum(7,8).

Again your RED HERRING.

H is being asked EXACTLY what it being asked

H wM w -> H.Qy if M applied to w Halts, and H.Qn if it doesn't

AGREED?

No that is wrong. embedded_H is being asked:

Can the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ possibly transition to ⟨Ĥ⟩.qn ?

If you say 'No', then you aren't doing the halting problem, as the

requirement I stated is EXACTLY the requirement of the Halting Problem.

that a machine has stopped running and that a machine cannot reach its

final state. My definition only includes the latter.

Sounds like a NDTM.

It is not a NDTM, a Turing Machine only actually halts when it reaches

its own final state. People not very familiar with this material may get

confused and believe that a TM halts when its stops running because its

simulation has been aborted. This key distinction is not typically

specified in most halting problem proofs.

computation that halts … the Turing machine will halt whenever it enters

a final state. (Linz:1990:234)

Where did Linz mention 'simulation' and 'abort'?

I have shown how my system directly applies to the actual halting problem and it can be understood as correct by anyone that understands the halting problem at a much deeper level than rote memorization.

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 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ? (No means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt).

You are defining POOP [Richard Damon]

André had recommended many online sites for you to learn or test, I forget which posts it is.

But I think C program is more simpler.

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

--

Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;

Genius hits a target no one else can see.

Arthur Schopenhauer

Path: i2pn2.org!i2pn.org!eternal-september.org!reader02.eternal-september.org!.POSTED!not-for-mail

From: polco...@gmail.com (olcott)

Newsgroups: comp.theory,comp.ai.philosophy,sci.logic,sci.math

Subject: Re: Concise refutation of halting problem proofs V59 [ ignorance

about halt deciders ]

Followup-To: comp.theory

Date: Tue, 1 Feb 2022 17:37:19 -0600

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View all headersFrom: polco...@gmail.com (olcott)

Newsgroups: comp.theory,comp.ai.philosophy,sci.logic,sci.math

Subject: Re: Concise refutation of halting problem proofs V59 [ ignorance

about halt deciders ]

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Date: Tue, 1 Feb 2022 17:37:19 -0600

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On 2/1/2022 4:25 PM, wij wrote:

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 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ? (An answer of "no" means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt).

Key elements rewritten today:

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

On Wednesday, 2 February 2022 at 06:19:04 UTC+8, olcott wrote:

On 2/1/2022 4:12 PM, wij wrote:

On Wednesday, 2 February 2022 at 05:36:39 UTC+8, olcott wrote:I have already proved that I know one key fact about halt deciders that

On 2/1/2022 3:23 PM, wij wrote:

On Wednesday, 2 February 2022 at 02:37:17 UTC+8, olcott wrote:I have shown how my system directly applies to the actual halting

On 2/1/2022 10:33 AM, wij wrote:

On Tuesday, 1 February 2022 at 23:22:32 UTC+8, olcott wrote:https://en.wikipedia.org/wiki/Nondeterministic_Turing_machine

On 1/31/2022 11:25 PM, Richard Damon wrote:

The halting problem is vague on the definition of halting, it includes

On 1/31/22 11:42 PM, olcott wrote:

On 1/31/2022 10:33 PM, Richard Damon wrote:

On 1/31/22 11:24 PM, olcott wrote:

On 1/31/2022 10:17 PM, Richard Damon wrote:

On 1/31/22 10:40 PM, olcott wrote:No that is not it. That is like saying "by definition" Sum(3,5) is

On 1/31/2022 6:41 PM, Richard Damon wrote:

On 1/31/22 3:24 PM, olcott wrote:

On 1/31/2022 2:10 PM, Ben wrote:

On 1/31/2022 8:06 AM, olcott wrote:

On 1/30/2022 8:20 PM, Richard Damon wrote:

On 1/30/22 9:05 PM, olcott wrote:

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

These statements need the conditions, that H^ goes to

H^.Qy/H^.Qn iff H goes to that corresponding state.

⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H

in the same way that (5,3) is syntactically specified as an

input to Sum(5,3)

Right, and the

Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to

embedded_H in the same way that (1,2) is NOT syntactically

specified as an input to Sum(5,3)

Right, but perhaps you don't understand that from you above

statement the right answer is based on if UTM(<H^>,<H^>)

Halts which by the definition of a UTM means if H^ applied to

<H^> Halts.

The biggest reason for your huge mistakes is that you cannot

stay sharply focused on a single point. It is as if you either

have attention deficit disorder ADD or are addicted to

methamphetamine.

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The single point is that ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H and

Ĥ ⟨Ĥ⟩ is the NOT the input to embedded_H.

After we have mutual agreement on this point we will move on

to the points that logically follow from this one.

Holy shit try to post something that makes sense.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Richard does not accept that the input to the copy of Linz H

embedded at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩. He keeps insisting that it is Ĥ ⟨Ĥ⟩.

No, but apparently you can't understand actual English words.

The INPUT to H is <H^> <H^> but the CORRECT ANSWER that H must

give is based on the behavior of H^ applied to <H^> BECAUSE OF

THE DEFINITION of H.

In other words Sum(3,5) must return the value of Sum(7,8)?

Don't know how you get that from what I said.

Any moron knows that a function is only accountable for its actual

inputs.

And the actual input to H is <H^> <H^> which MEANS by the

DEFINITION of the Halting Problem that H is being asked to decide

on the Halting Status of H^ applied to <H^>

being asked about Sum(7,8).

Again your RED HERRING.

H is being asked EXACTLY what it being asked

H wM w -> H.Qy if M applied to w Halts, and H.Qn if it doesn't

AGREED?

No that is wrong. embedded_H is being asked:

Can the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ possibly transition to ⟨Ĥ⟩.qn ?

If you say 'No', then you aren't doing the halting problem, as the

requirement I stated is EXACTLY the requirement of the Halting Problem.

that a machine has stopped running and that a machine cannot reach its

final state. My definition only includes the latter.

Sounds like a NDTM.

It is not a NDTM, a Turing Machine only actually halts when it reaches

its own final state. People not very familiar with this material may get

confused and believe that a TM halts when its stops running because its

simulation has been aborted. This key distinction is not typically

specified in most halting problem proofs.

computation that halts … the Turing machine will halt whenever it enters

a final state. (Linz:1990:234)

Where did Linz mention 'simulation' and 'abort'?

problem and it can be understood as correct by anyone that understands

the halting problem at a much deeper level than rote memorization.

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 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to

⟨Ĥ⟩.qn ? (No means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt).

You are defining POOP [Richard Damon]

André had recommended many online sites for you to learn or test, I forget which posts it is.

But I think C program is more simpler.

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

-- Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;

Genius hits a target no one else can see.

Arthur Schopenhauer

André had recommended many online sites for you to learn or test, I forget which posts it is.

Type it into a TM simulator and prove your claim, your words are meaningless.

no one else here seems to know.

No one here understands that because a halt decider is a decider that it

must compute the mapping from its inputs to an accept of reject state on

the basis of the actual behavior specified by these inputs.

-- Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;

Genius hits a target no one else can see.

Arthur Schopenhauer

There is no 'actual TM' until you it into a TM simulator, otherwise all empty talks.

(I would expect to see you 'reinterpret' again)

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 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ? (An answer of "no" means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt).

Key elements rewritten today:

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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Subject: Re: Concise refutation of halting problem proofs V59 [ key essence ]

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From: NoO...@NoWhere.com (olcott)

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On 2/1/2022 5:57 PM, Richard Damon wrote:

A simulated machine description that specifies an infinite sequence of configurations stops running yet never halts when its simulation has been aborted.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ? (An answer of "no" means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never halts).

You keep erroneously believing that embedded_H computes the mapping from ⟨Ĥ⟩ ⟨Ĥ⟩ to an accept or reject state on the basis of the behavior of Ĥ applied to ⟨Ĥ⟩ rather than the actual behavior of its actual input.

I changed nothing. You simply do not know enough of the computer science of deciders.

In computability theory, the halting problem is the problem of determining, from a description of an arbitrary computer program and an input, whether the program will finish running, or continue to run forever. https://en.wikipedia.org/wiki/Halting_problem

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ? (An answer of "no" means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never halts).

I rewrote this. embedded_H <is> a UTM for all inputs that eventually transition to their final state.

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.

--

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 2/1/22 10:22 AM, olcott wrote:

On 1/31/2022 11:25 PM, Richard Damon wrote:

On 1/31/22 11:42 PM, olcott wrote:

On 1/31/22 11:24 PM, olcott wrote:

On 1/30/22 9:05 PM, olcott wrote:

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

These statements need the conditions, that H^ goes to H^.Qy/H^.Qn iff H goes to that corresponding state.

⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H in the same way that (5,3) is syntactically specified as an input to Sum(5,3)

Right, and the

Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to embedded_H in the same way that (1,2) is NOT syntactically specified as an input to Sum(5,3)

Right, but perhaps you don't understand that from you above statement the right answer is based on if UTM(<H^>,<H^>) Halts which by the definition of a UTM means if H^ applied to <H^> Halts.

The biggest reason for your huge mistakes is that you cannot stay sharply focused on a single point. It is as if you either have attention deficit disorder ADD or are addicted to methamphetamine.

>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The single point is that ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H and

Ĥ ⟨Ĥ⟩ is the NOT the input to embedded_H.

After we have mutual agreement on this point we will move on to the points that logically follow from this one.

Holy shit try to post something that makes sense.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Richard does not accept that the input to the copy of Linz H embedded at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩. He keeps insisting that it is Ĥ ⟨Ĥ⟩.

No, but apparently you can't understand actual English words.

The INPUT to H is <H^> <H^> but the CORRECT ANSWER that H must give is based on the behavior of H^ applied to <H^> BECAUSE OF THE DEFINITION of H.

In other words Sum(3,5) must return the value of Sum(7,8)?

Don't know how you get that from what I said.

Any moron knows that a function is only accountable for its actual inputs.

And the actual input to H is <H^> <H^> which MEANS by the DEFINITION of the Halting Problem that H is being asked to decide on the Halting Status of H^ applied to <H^>

Again your RED HERRING.

H is being asked EXACTLY what it being asked

H wM w -> H.Qy if M applied to w Halts, and H.Qn if it doesn't

AGREED?

No that is wrong. embedded_H is being asked:

Can the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ possibly transition to ⟨Ĥ⟩.qn ?

If you say 'No', then you aren't doing the halting problem, as the requirement I stated is EXACTLY the requirement of the Halting Problem.

The halting problem is vague on the definition of halting, it includes that a machine has stopped running and that a machine cannot reach its final state. My definition only includes the latter.

No, it is NOT 'Vague', a machine will EITHER stop running because it will reach a final state, or it can NEVER reach such a state.

Please show a machine that doesn't reach its final state but also doesn't run forever?

You seem to think that it is possible for a machine to be in some middle state.

Please provide an example of such a machine.

A simulated machine description that specifies an infinite sequence of configurations stops running yet never halts when its simulation has been aborted.

Note, the definition is stated the way it is because a simulator that aborts its simulation does NOT indicate either of the cases and does not provide evidence of the Halting state of a computation.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ? (An answer of "no" means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never halts).

The halting problem does not bother to mention the requirement that because all halt deciders are deciders they are only accountable for computing the mapping from their finite string inputs to an accept or reject state on the basis of the actual behavior specified by this input.

But if they do not compute the mapping per the definition, they are NOT

'Halt Deciders', that is your problem, what you are doing is trying to define a POOP decider can call it a Halt Decider.

You keep erroneously believing that embedded_H computes the mapping from ⟨Ĥ⟩ ⟨Ĥ⟩ to an accept or reject state on the basis of the behavior of Ĥ applied to ⟨Ĥ⟩ rather than the actual behavior of its actual input.

You are not ALLOWED to change the definiton of Halting, when you try, it just means you logic is unsound and doesn't prove anything, because it si based on a false premise.

PERIOD.

I changed nothing. You simply do not know enough of the computer science of deciders.

In computability theory, the halting problem is the problem of determining, from a description of an arbitrary computer program and an input, whether the program will finish running, or continue to run forever. https://en.wikipedia.org/wiki/Halting_problem

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ? (An answer of "no" means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never halts).

The halting problem does not specifically examine simulating halt deciders, none-the-less the behavior of a correctly simulated machine description is known to be equivalent to the behavior of the direct execution of this same machine.

Right, NON-ABORTED simuluation, and UNSTOPPED execution. So an H that aborts or 'debug steps' does NOT prove hon-halting.

PERIOD.

FAIL.

Since a simulating halt decider is merely a UTM for simulated inputs that reach their final state when a simulating halt decider correctly determines that its simulated its input cannot possibly reach its final state this is complete proof that this simulated input never halts.

No, it is NOT a UTM if it aborts its simulation for ANY reason other than the machine reached a final statee.

Even if it correctly predicts that its input is non-halting, a UTM does not abort its simulation, as BY DEFINITION, it behaves the same as the machine it is simulating, so if that is non-halting, then the UTM must be too.

I rewrote this. embedded_H <is> a UTM for all inputs that eventually transition to their final state.

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.

--

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 2/1/2022 6:20 PM, Richard Damon wrote:

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ? (An answer of "no" means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never halts).

--

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 2/1/22 1:37 PM, olcott wrote:

On 2/1/2022 10:33 AM, wij wrote:

On Tuesday, 1 February 2022 at 23:22:32 UTC+8, olcott wrote:

The halting problem is vague on the definition of halting, it includes

On 1/31/22 11:42 PM, olcott wrote:

On 1/31/22 11:24 PM, olcott wrote:

On 1/30/22 9:05 PM, olcott wrote:

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

These statements need the conditions, that H^ goes to

H^.Qy/H^.Qn iff H goes to that corresponding state.

⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H

in the same way that (5,3) is syntactically specified as an

input to Sum(5,3)

Right, and the

Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to

embedded_H in the same way that (1,2) is NOT syntactically

specified as an input to Sum(5,3)

Right, but perhaps you don't understand that from you above

statement the right answer is based on if UTM(<H^>,<H^>)

Halts which by the definition of a UTM means if H^ applied to

<H^> Halts.

The biggest reason for your huge mistakes is that you cannot

stay sharply focused on a single point. It is as if you either

have attention deficit disorder ADD or are addicted to

methamphetamine.

>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The single point is that ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H and

Ĥ ⟨Ĥ⟩ is the NOT the input to embedded_H.

After we have mutual agreement on this point we will move on

to the points that logically follow from this one.

Holy shit try to post something that makes sense.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Richard does not accept that the input to the copy of Linz H

embedded at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩. He keeps insisting that it is Ĥ ⟨Ĥ⟩.

No, but apparently you can't understand actual English words.

The INPUT to H is <H^> <H^> but the CORRECT ANSWER that H must

give is based on the behavior of H^ applied to <H^> BECAUSE OF

THE DEFINITION of H.

In other words Sum(3,5) must return the value of Sum(7,8)?

Don't know how you get that from what I said.

Any moron knows that a function is only accountable for its actual

inputs.

And the actual input to H is <H^> <H^> which MEANS by the

DEFINITION of the Halting Problem that H is being asked to decide

on the Halting Status of H^ applied to <H^>

being asked about Sum(7,8).

Again your RED HERRING.

H is being asked EXACTLY what it being asked

H wM w -> H.Qy if M applied to w Halts, and H.Qn if it doesn't

AGREED?

No that is wrong. embedded_H is being asked:

Can the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ possibly transition to ⟨Ĥ⟩.qn ?

If you say 'No', then you aren't doing the halting problem, as the

requirement I stated is EXACTLY the requirement of the Halting Problem.

that a machine has stopped running and that a machine cannot reach its

final state. My definition only includes the latter.

https://en.wikipedia.org/wiki/Nondeterministic_Turing_machine

It is not a NDTM, a Turing Machine only actually halts when it reaches its own final state. People not very familiar with this material may get confused and believe that a TM halts when its stops running because its simulation has been aborted. This key distinction is not typically specified in most halting problem proofs.

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

Halting problem undecidability and infinitely nested simulation (V3)

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

And the point that you seem to miss is that the Turing Machine doesn't stop just because some simulation of its representation gave up on simulating it.

And actual Turing machine will continue to run until it his a final state or els it will continue to run for an unbounded number of steps.

Non-Halting can only be show by showing that the actual running of the machine will continue for an unbounded number of steps, not just that there is some N that it doesn't stop in.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ? (An answer of "no" means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never halts).

--

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 V59 [ key essence ]

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Subject: Re: Concise refutation of halting problem proofs V59 [ key essence ]

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On 2/1/2022 6:24 PM, Richard Damon wrote:

The following is necessarily true on the basis of the meaning of its words:

When embedded_H correctly recognizes this infinitely repeating behavior pattern:

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⟩...

Then it is necessary correct for it to abort its simulation and transition to Ĥ.qn.

--

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 2/1/22 4:36 PM, olcott wrote:

On 2/1/2022 3:23 PM, wij wrote:

On Wednesday, 2 February 2022 at 02:37:17 UTC+8, olcott wrote:

On 2/1/2022 10:33 AM, wij wrote:

On Tuesday, 1 February 2022 at 23:22:32 UTC+8, olcott wrote:https://en.wikipedia.org/wiki/Nondeterministic_Turing_machine

On 1/31/2022 11:25 PM, Richard Damon wrote:

The halting problem is vague on the definition of halting, it includes

On 1/31/22 11:42 PM, olcott wrote:

On 1/31/2022 10:33 PM, Richard Damon wrote:

On 1/31/22 11:24 PM, olcott wrote:

On 1/31/2022 10:17 PM, Richard Damon wrote:

On 1/31/22 10:40 PM, olcott wrote:No that is not it. That is like saying "by definition" Sum(3,5) is

On 1/31/2022 6:41 PM, Richard Damon wrote:

On 1/31/22 3:24 PM, olcott wrote:

On 1/31/2022 2:10 PM, Ben wrote:

On 1/31/2022 8:06 AM, olcott wrote:

On 1/30/2022 8:20 PM, Richard Damon wrote:

On 1/30/22 9:05 PM, olcott wrote:

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

These statements need the conditions, that H^ goes to

H^.Qy/H^.Qn iff H goes to that corresponding state.

⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H

in the same way that (5,3) is syntactically specified as an

input to Sum(5,3)

Right, and the

Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to

embedded_H in the same way that (1,2) is NOT syntactically

specified as an input to Sum(5,3)

Right, but perhaps you don't understand that from you above

statement the right answer is based on if UTM(<H^>,<H^>)

Halts which by the definition of a UTM means if H^ applied to

<H^> Halts.

The biggest reason for your huge mistakes is that you cannot

stay sharply focused on a single point. It is as if you either

have attention deficit disorder ADD or are addicted to

methamphetamine.

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The single point is that ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H and

Ĥ ⟨Ĥ⟩ is the NOT the input to embedded_H.

After we have mutual agreement on this point we will move on

to the points that logically follow from this one.

Holy shit try to post something that makes sense.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Richard does not accept that the input to the copy of Linz H

embedded at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩. He keeps insisting that it is Ĥ ⟨Ĥ⟩.

No, but apparently you can't understand actual English words.

The INPUT to H is <H^> <H^> but the CORRECT ANSWER that H must

give is based on the behavior of H^ applied to <H^> BECAUSE OF

THE DEFINITION of H.

In other words Sum(3,5) must return the value of Sum(7,8)?

Don't know how you get that from what I said.

Any moron knows that a function is only accountable for its actual

inputs.

And the actual input to H is <H^> <H^> which MEANS by the

DEFINITION of the Halting Problem that H is being asked to decide

on the Halting Status of H^ applied to <H^>

being asked about Sum(7,8).

Again your RED HERRING.

H is being asked EXACTLY what it being asked

H wM w -> H.Qy if M applied to w Halts, and H.Qn if it doesn't

AGREED?

No that is wrong. embedded_H is being asked:

Can the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ possibly transition to ⟨Ĥ⟩.qn ?

If you say 'No', then you aren't doing the halting problem, as the

requirement I stated is EXACTLY the requirement of the Halting Problem.

that a machine has stopped running and that a machine cannot reach its

final state. My definition only includes the latter.

Sounds like a NDTM.

It is not a NDTM, a Turing Machine only actually halts when it reaches

its own final state. People not very familiar with this material may get

confused and believe that a TM halts when its stops running because its

simulation has been aborted. This key distinction is not typically

specified in most halting problem proofs.

computation that halts … the Turing machine will halt whenever it enters

a final state. (Linz:1990:234)

Where did Linz mention 'simulation' and 'abort'?

I have shown how my system directly applies to the actual halting problem and it can be understood as correct by anyone that understands the halting problem at a much deeper level than rote memorization.

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 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ? (No means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt).

But unless embedded_H actually IS a real UTM, that doesn't matter.

The following is necessarily true on the basis of the meaning of its words:

When embedded_H correctly recognizes this infinitely repeating behavior pattern:

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⟩...

Then it is necessary correct for it to abort its simulation and transition to Ĥ.qn.

--

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 V59 [ ignorance

about halt deciders ]

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Subject: Re: Concise refutation of halting problem proofs V59 [ ignorance

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On 2/1/2022 6:25 PM, Richard Damon wrote:

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ?

An answer of "no" means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never halts, and nothing in the universe can possibly overcome this.

If your dog bites your leg, then the answer to the question: Did your dog bite your leg? is Yes even if everyone else in the universe disagrees.

--

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 2/1/22 5:18 PM, olcott wrote:

On 2/1/2022 4:12 PM, wij wrote:

On Wednesday, 2 February 2022 at 05:36:39 UTC+8, olcott wrote:

On 2/1/2022 3:23 PM, wij wrote:

On 1/31/22 11:42 PM, olcott wrote:

On 1/31/22 11:24 PM, olcott wrote:

On 1/30/22 9:05 PM, olcott wrote:

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

These statements need the conditions, that H^ goes to

H^.Qy/H^.Qn iff H goes to that corresponding state.

⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H

in the same way that (5,3) is syntactically specified as an

input to Sum(5,3)

Right, and the

Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to

embedded_H in the same way that (1,2) is NOT syntactically

specified as an input to Sum(5,3)

Right, but perhaps you don't understand that from you above

statement the right answer is based on if UTM(<H^>,<H^>)

Halts which by the definition of a UTM means if H^ applied to

<H^> Halts.

The biggest reason for your huge mistakes is that you cannot

stay sharply focused on a single point. It is as if you either

have attention deficit disorder ADD or are addicted to

methamphetamine.

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The single point is that ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H and

Ĥ ⟨Ĥ⟩ is the NOT the input to embedded_H.

After we have mutual agreement on this point we will move on

to the points that logically follow from this one.

Holy shit try to post something that makes sense.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Richard does not accept that the input to the copy of Linz H

embedded at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩. He keeps insisting that it is Ĥ ⟨Ĥ⟩.

No, but apparently you can't understand actual English words.

The INPUT to H is <H^> <H^> but the CORRECT ANSWER that H must

give is based on the behavior of H^ applied to <H^> BECAUSE OF

THE DEFINITION of H.

In other words Sum(3,5) must return the value of Sum(7,8)?

Don't know how you get that from what I said.

Any moron knows that a function is only accountable for its actual

inputs.

And the actual input to H is <H^> <H^> which MEANS by the

DEFINITION of the Halting Problem that H is being asked to decide

on the Halting Status of H^ applied to <H^>

being asked about Sum(7,8).

Again your RED HERRING.

H is being asked EXACTLY what it being asked

H wM w -> H.Qy if M applied to w Halts, and H.Qn if it doesn't

AGREED?

No that is wrong. embedded_H is being asked:

Can the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ possibly transition to ⟨Ĥ⟩.qn ?

If you say 'No', then you aren't doing the halting problem, as the

requirement I stated is EXACTLY the requirement of the Halting Problem.

that a machine has stopped running and that a machine cannot reach its

final state. My definition only includes the latter.

Sounds like a NDTM.

It is not a NDTM, a Turing Machine only actually halts when it reaches

its own final state. People not very familiar with this material may get

confused and believe that a TM halts when its stops running because its

simulation has been aborted. This key distinction is not typically

specified in most halting problem proofs.

computation that halts … the Turing machine will halt whenever it enters

a final state. (Linz:1990:234)

Where did Linz mention 'simulation' and 'abort'?

problem and it can be understood as correct by anyone that understands

the halting problem at a much deeper level than rote memorization.

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 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to

⟨Ĥ⟩.qn ? (No means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt).

You are defining POOP [Richard Damon]

André had recommended many online sites for you to learn or test, I forget which posts it is.

But I think C program is more simpler.

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

-- Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;

Genius hits a target no one else can see.

Arthur Schopenhauer

André had recommended many online sites for you to learn or test, I forget which posts it is.

Type it into a TM simulator and prove your claim, your words are meaningless.

I have already proved that I know one key fact about halt deciders that no one else here seems to know.

No one here understands that because a halt decider is a decider that it must compute the mapping from its inputs to an accept of reject state on the basis of the actual behavior specified by these inputs.

And the ACTUAL BEHAVIOR of the input <H^> <H^> is EXACTLY the behavior of H^ applied to <H^> which does Halt if H goes to H.Qn.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ?

An answer of "no" means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never halts, and nothing in the universe can possibly overcome this.

If your dog bites your leg, then the answer to the question: Did your dog bite your leg? is Yes even if everyone else in the universe disagrees.

--

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 2/1/2022 6:58 PM, André G. Isaak wrote:

We are on the same page so far. (acknowledging when there is agreement is an essential part of an honest dialogue).

No you are flat out wrong about this. You are wrong because of your ignorance of how deciders work. Deciders compute the mapping from their finite string inputs to an accept or reject state on the basis of the actual properties of these actual inputs.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ?

An answer of "no" means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never halts, and nothing in the universe can possibly overcome this.

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 correct simulation of this 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

On 2022-01-30 19:05, olcott wrote:

On 1/30/2022 7:45 PM, Richard Damon wrote:

On 1/30/22 7:21 PM, olcott wrote:

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

These statements need the conditions, that H^ goes to H^.Qy/H^.Qn iff H goes to that corresponding state.

⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H in the same way that (5,3) is syntactically specified as an input to Sum(5,3)

Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to embedded_H in the same way that (1,2) is NOT syntactically specified as an input to Sum(5,3)

I promised myself I wouldn't involve myself in your nonsense any further, but here you've made such a terribly inaccurate analogy that I thought I had to comment.

The inputs to a function such as SUM(X, Y) are two REPRESENTATIONS of integers. If SUM were a Turing Machine, these would be two strings in the alphabet of the TM. if this were a C function, X and X would be strings of bits which form the twos complement representation of some integer. In neither case would the inputs be actual, mathematical integers. C might use the term 'integer' as one of its built in types, but C integers are NOT elements of ℤ. They are REPRESENTATIONS of the supported subset of ℤ.

So ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H in the same sense that ⟨5⟩ ⟨3⟩ are the inputs to SUM.

Ĥ ⟨Ĥ⟩ is not the input to embedded_H in the same sense that the actual mathematical integers 3 and 5 are not inputs to SUM.

We are on the same page so far. (acknowledging when there is agreement is an essential part of an honest dialogue).

If your going to make analogies, at least make ones that are accurate.

SUM takes REPRESENTATIONS of integers as its inputs, but it answers about the ACTUAL integers described by those representations. To talk about the sum of two representations is meaningless. Only actual integers have sums.

In exactly the same way, embedded_H takes a REPRESENTATION of some TM ⟨Ĥ⟩ as part of its input but it answers about the ACTUAL TM described by that input, Ĥ.

To talk about whether a representation of a TM halts is meaningless since only actual TMs, not representations of TMs, can halt. The conditions which Richard indicates above (following Linz) are therefore the correct ones.

In a previous post which I can't be botherered to find, you claimed that when the input to embedded_H is ⟨Ĥ⟩ ⟨Ĥ⟩ that embedded_H can only be expected to answer about its actual inputs and not its 'enclosing TM'.

Yes, it must answer about its input, but if its input is ⟨Ĥ⟩ ⟨Ĥ⟩, then BY THE DEFINITION OF A HALT DECIDER is must determine whether Ĥ applied to ⟨Ĥ⟩ halts.

No you are flat out wrong about this. You are wrong because of your ignorance of how deciders work. Deciders compute the mapping from their finite string inputs to an accept or reject state on the basis of the actual properties of these actual inputs.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ?

An answer of "no" means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never halts, and nothing in the universe can possibly overcome this.

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 correct simulation of this 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

And that computation happens to be the EXACT SAME computation as its 'enclosing TM'. So it is answering about *both*.

André

--

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 2/1/2022 7:40 PM, Richard Damon wrote:

When Ĥ is applied to ⟨Ĥ⟩

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

As soon as embedded_H correctly recognizes this as an infinite behavior pattern:

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⟩...

Then embedded_H can correctly abort the simulation of its input and correctly transition to Ĥ.qn.

The above words can be verified as completely true entirely on the basis of their meaning.

--

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 2/1/22 8:03 PM, olcott wrote:

On 2/1/2022 6:25 PM, Richard Damon wrote:

On 2/1/22 5:18 PM, olcott wrote:

On 2/1/2022 4:12 PM, wij wrote:

On Wednesday, 2 February 2022 at 05:36:39 UTC+8, olcott wrote:

On 1/31/22 11:42 PM, olcott wrote:

On 1/31/22 11:24 PM, olcott wrote:

On 1/30/22 9:05 PM, olcott wrote:

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

These statements need the conditions, that H^ goes to

H^.Qy/H^.Qn iff H goes to that corresponding state.

⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H

in the same way that (5,3) is syntactically specified as an

input to Sum(5,3)

Right, and the

Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to

embedded_H in the same way that (1,2) is NOT syntactically

specified as an input to Sum(5,3)

Right, but perhaps you don't understand that from you above

statement the right answer is based on if UTM(<H^>,<H^>)

Halts which by the definition of a UTM means if H^ applied to

<H^> Halts.

The biggest reason for your huge mistakes is that you cannot

stay sharply focused on a single point. It is as if you either

have attention deficit disorder ADD or are addicted to

methamphetamine.

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The single point is that ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H and

Ĥ ⟨Ĥ⟩ is the NOT the input to embedded_H.

After we have mutual agreement on this point we will move on

to the points that logically follow from this one.

Holy shit try to post something that makes sense.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Richard does not accept that the input to the copy of Linz H

embedded at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩. He keeps insisting that it is Ĥ ⟨Ĥ⟩.

No, but apparently you can't understand actual English words.

The INPUT to H is <H^> <H^> but the CORRECT ANSWER that H must

give is based on the behavior of H^ applied to <H^> BECAUSE OF

THE DEFINITION of H.

In other words Sum(3,5) must return the value of Sum(7,8)?

Don't know how you get that from what I said.

Any moron knows that a function is only accountable for its actual

inputs.

And the actual input to H is <H^> <H^> which MEANS by the

DEFINITION of the Halting Problem that H is being asked to decide

on the Halting Status of H^ applied to <H^>

being asked about Sum(7,8).

Again your RED HERRING.

H is being asked EXACTLY what it being asked

H wM w -> H.Qy if M applied to w Halts, and H.Qn if it doesn't

AGREED?

No that is wrong. embedded_H is being asked:

Can the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ possibly transition to ⟨Ĥ⟩.qn ?

If you say 'No', then you aren't doing the halting problem, as the

requirement I stated is EXACTLY the requirement of the Halting Problem.

that a machine has stopped running and that a machine cannot reach its

final state. My definition only includes the latter.

Sounds like a NDTM.

It is not a NDTM, a Turing Machine only actually halts when it reaches

its own final state. People not very familiar with this material may get

confused and believe that a TM halts when its stops running because its

simulation has been aborted. This key distinction is not typically

specified in most halting problem proofs.

computation that halts … the Turing machine will halt whenever it enters

a final state. (Linz:1990:234)

Where did Linz mention 'simulation' and 'abort'?

problem and it can be understood as correct by anyone that understands

the halting problem at a much deeper level than rote memorization.

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 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to

⟨Ĥ⟩.qn ? (No means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt).

André had recommended many online sites for you to learn or test, I forget which posts it is.

But I think C program is more simpler.

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

-- Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;

Genius hits a target no one else can see.

Arthur Schopenhauer

André had recommended many online sites for you to learn or test, I forget which posts it is.

Type it into a TM simulator and prove your claim, your words are meaningless.

I have already proved that I know one key fact about halt deciders that no one else here seems to know.

No one here understands that because a halt decider is a decider that it must compute the mapping from its inputs to an accept of reject state on the basis of the actual behavior specified by these inputs.

And the ACTUAL BEHAVIOR of the input <H^> <H^> is EXACTLY the behavior of H^ applied to <H^> which does Halt if H goes to H.Qn.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ?

Doesn't matter if embedded_H is not a ACTUAL UTM.

When Ĥ is applied to ⟨Ĥ⟩

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

As soon as embedded_H correctly recognizes this as an infinite behavior pattern:

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⟩...

Then embedded_H can correctly abort the simulation of its input and correctly transition to Ĥ.qn.

The above words can be verified as completely true entirely on the basis of their meaning.

--

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 2/1/2022 7:55 PM, Richard Damon wrote:

What I said above is true by logical necessity and you simply aren't bright enough to understand this.

If X then Y and if Y then Z and X then Z. There is no way around this.

If embedded_H correctly recognizes that its input specifies non halting behavior then it is necessarily correct for embedded_H to report this

non halting behavior.

If I see a cat then I can say that I saw a cat and be correct. I can't imagine how this is over your head, thus dishonestly is my only plausible explanation.

--

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 2/1/22 8:47 PM, olcott wrote:

On 2/1/2022 7:40 PM, Richard Damon wrote:

On 2/1/22 8:03 PM, olcott wrote:

On 2/1/2022 6:25 PM, Richard Damon wrote:

On 2/1/22 5:18 PM, olcott wrote:

On 2/1/2022 4:12 PM, wij wrote:

On 1/31/22 11:42 PM, olcott wrote:

On 1/31/22 11:24 PM, olcott wrote:

On 1/30/22 9:05 PM, olcott wrote:

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

These statements need the conditions, that H^ goes to

H^.Qy/H^.Qn iff H goes to that corresponding state.

⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H

in the same way that (5,3) is syntactically specified as an

input to Sum(5,3)

Right, and the

Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to

embedded_H in the same way that (1,2) is NOT syntactically

specified as an input to Sum(5,3)

Right, but perhaps you don't understand that from you above

statement the right answer is based on if UTM(<H^>,<H^>)

Halts which by the definition of a UTM means if H^ applied to

<H^> Halts.

The biggest reason for your huge mistakes is that you cannot

stay sharply focused on a single point. It is as if you either

have attention deficit disorder ADD or are addicted to

methamphetamine.

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The single point is that ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H and

Ĥ ⟨Ĥ⟩ is the NOT the input to embedded_H.

After we have mutual agreement on this point we will move on

to the points that logically follow from this one.

Holy shit try to post something that makes sense.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Richard does not accept that the input to the copy of Linz H

embedded at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩. He keeps insisting that it is Ĥ ⟨Ĥ⟩.

No, but apparently you can't understand actual English words.

The INPUT to H is <H^> <H^> but the CORRECT ANSWER that H must

give is based on the behavior of H^ applied to <H^> BECAUSE OF

THE DEFINITION of H.

In other words Sum(3,5) must return the value of Sum(7,8)?

Don't know how you get that from what I said.

Any moron knows that a function is only accountable for its actual

inputs.

And the actual input to H is <H^> <H^> which MEANS by the

DEFINITION of the Halting Problem that H is being asked to decide

on the Halting Status of H^ applied to <H^>

being asked about Sum(7,8).

Again your RED HERRING.

H is being asked EXACTLY what it being asked

H wM w -> H.Qy if M applied to w Halts, and H.Qn if it doesn't

AGREED?

No that is wrong. embedded_H is being asked:

Can the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ possibly transition to ⟨Ĥ⟩.qn ?

If you say 'No', then you aren't doing the halting problem, as the

requirement I stated is EXACTLY the requirement of the Halting Problem.

that a machine has stopped running and that a machine cannot reach its

final state. My definition only includes the latter.

Sounds like a NDTM.

It is not a NDTM, a Turing Machine only actually halts when it reaches

its own final state. People not very familiar with this material may get

confused and believe that a TM halts when its stops running because its

simulation has been aborted. This key distinction is not typically

specified in most halting problem proofs.

computation that halts … the Turing machine will halt whenever it enters

a final state. (Linz:1990:234)

Where did Linz mention 'simulation' and 'abort'?

problem and it can be understood as correct by anyone that understands

the halting problem at a much deeper level than rote memorization.

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 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to

⟨Ĥ⟩.qn ? (No means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt).

André had recommended many online sites for you to learn or test, I forget which posts it is.

But I think C program is more simpler.

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

-- Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;

Genius hits a target no one else can see.

Arthur Schopenhauer

André had recommended many online sites for you to learn or test, I forget which posts it is.

Type it into a TM simulator and prove your claim, your words are meaningless.

I have already proved that I know one key fact about halt deciders that no one else here seems to know.

No one here understands that because a halt decider is a decider that it must compute the mapping from its inputs to an accept of reject state on the basis of the actual behavior specified by these inputs.

And the ACTUAL BEHAVIOR of the input <H^> <H^> is EXACTLY the behavior of H^ applied to <H^> which does Halt if H goes to H.Qn.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ?

Doesn't matter if embedded_H is not a ACTUAL UTM.

When Ĥ is applied to ⟨Ĥ⟩

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

As soon as embedded_H correctly recognizes this as an infinite behavior pattern:

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⟩...

Then embedded_H can correctly abort the simulation of its input and correctly transition to Ĥ.qn.

The above words can be verified as completely true entirely on the basis of their meaning.

Nope, proven otherwise.

What I said above is true by logical necessity and you simply aren't bright enough to understand this.

If X then Y and if Y then Z and X then Z. There is no way around this.

If embedded_H correctly recognizes that its input specifies non halting behavior then it is necessarily correct for embedded_H to report this

non halting behavior.

If I see a cat then I can say that I saw a cat and be correct. I can't imagine how this is over your head, thus dishonestly is my only plausible explanation.

--

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 2/1/2022 8:08 PM, André G. Isaak wrote:

We still seem to agree. (points of mutual agreement are required for honest dialogues).

Sure and so is the "I am going to go to the" part of

"I am going to go to the store to buy some ice cream."

When you don't cut off what I said in the middle of the sentence then it makes much more sense.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ?

I said that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is being simulated by embedded_H. That you cut me off in the middle of the sentence to form you rebuttal seems ridiculously dishonest. It is like you don't even care that everyone reading this will know that you are being deliberately deceptive.

It is the case that when ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is correctly simulated by embedded_H and cannot possibly reach ⟨Ĥ⟩.qn that embedded_H is correct to report that its input does not halt.

There is no dodgy double talk way around this.

--

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 2022-02-01 18:33, olcott wrote:

On 2/1/2022 6:58 PM, André G. Isaak wrote:

On 2022-01-30 19:05, olcott wrote:

On 1/30/2022 7:45 PM, Richard Damon wrote:

On 1/30/22 7:21 PM, olcott wrote:

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

These statements need the conditions, that H^ goes to H^.Qy/H^.Qn iff H goes to that corresponding state.

⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H in the same way that (5,3) is syntactically specified as an input to Sum(5,3)

Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to embedded_H in the same way that (1,2) is NOT syntactically specified as an input to Sum(5,3)

I promised myself I wouldn't involve myself in your nonsense any further, but here you've made such a terribly inaccurate analogy that I thought I had to comment.

The inputs to a function such as SUM(X, Y) are two REPRESENTATIONS of integers. If SUM were a Turing Machine, these would be two strings in the alphabet of the TM. if this were a C function, X and X would be strings of bits which form the twos complement representation of some integer. In neither case would the inputs be actual, mathematical integers. C might use the term 'integer' as one of its built in types, but C integers are NOT elements of ℤ. They are REPRESENTATIONS of the supported subset of ℤ.

So ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H in the same sense that ⟨5⟩ ⟨3⟩ are the inputs to SUM.

Ĥ ⟨Ĥ⟩ is not the input to embedded_H in the same sense that the actual mathematical integers 3 and 5 are not inputs to SUM.

We are on the same page so far. (acknowledging when there is agreement is an essential part of an honest dialogue).

If your going to make analogies, at least make ones that are accurate.

SUM takes REPRESENTATIONS of integers as its inputs, but it answers about the ACTUAL integers described by those representations. To talk about the sum of two representations is meaningless. Only actual integers have sums.

In exactly the same way, embedded_H takes a REPRESENTATION of some TM ⟨Ĥ⟩ as part of its input but it answers about the ACTUAL TM described by that input, Ĥ.

To talk about whether a representation of a TM halts is meaningless since only actual TMs, not representations of TMs, can halt. The conditions which Richard indicates above (following Linz) are therefore the correct ones.

In a previous post which I can't be botherered to find, you claimed that when the input to embedded_H is ⟨Ĥ⟩ ⟨Ĥ⟩ that embedded_H can only be expected to answer about its actual inputs and not its 'enclosing TM'.

Yes, it must answer about its input, but if its input is ⟨Ĥ⟩ ⟨Ĥ⟩, then BY THE DEFINITION OF A HALT DECIDER is must determine whether Ĥ applied to ⟨Ĥ⟩ halts.

No you are flat out wrong about this. You are wrong because of your ignorance of how deciders work. Deciders compute the mapping from their finite string inputs to an accept or reject state on the basis of the actual properties of these actual inputs.

I am perfectly aware of how deciders work and an actual property of ⟨Ĥ⟩ is that it represents the Turing Machine Ĥ. And a halt decider is required to accept ⟨Ĥ⟩ ⟨Ĥ⟩ if and only if Ĥ ⟨Ĥ⟩ halts.

In much the same way a TM which performs SUMS might take two input strings ⟨x⟩ and ⟨y⟩ and output some third string ⟨z⟩, but specification of such a machine would be that it maps ⟨x⟩ ⟨y⟩ to ⟨z⟩ such that x + y = z. There are ⟨brackets⟩ around the inputs, but not around the entities in the specification.

It *IS* mapping from its inputs to its output, but the mapping is based on an operation over the entities which the inputs and outputs represent. That's how *all* Turing Machines work.

We still seem to agree. (points of mutual agreement are required for honest dialogues).

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ?

⟨Ĥ⟩ applied to ⟨Ĥ⟩ is completely meaningless.

Sure and so is the "I am going to go to the" part of

"I am going to go to the store to buy some ice cream."

When you don't cut off what I said in the middle of the sentence then it makes much more sense.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ?

You apply TMs to inputs. You can't apply a string to a string any more than you can add the strings "2" and "3". You can add the integers they represent, but not the strings themselves.

I said that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is being simulated by embedded_H. That you cut me off in the middle of the sentence to form you rebuttal seems ridiculously dishonest. It is like you don't even care that everyone reading this will know that you are being deliberately deceptive.

It is the case that when ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is correctly simulated by embedded_H and cannot possibly reach ⟨Ĥ⟩.qn that embedded_H is correct to report that its input does not halt.

There is no dodgy double talk way around this.

--

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 2/1/2022 8:21 PM, Richard Damon wrote:

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⟩...

--

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 2/1/22 9:14 PM, olcott wrote:You must be a liar.

On 2/1/2022 7:55 PM, Richard Damon wrote:

On 2/1/22 8:47 PM, olcott wrote:

On 2/1/2022 7:40 PM, Richard Damon wrote:

On 2/1/22 8:03 PM, olcott wrote:

On 2/1/2022 6:25 PM, Richard Damon wrote:

On 1/31/22 11:42 PM, olcott wrote:

On 1/31/22 11:24 PM, olcott wrote:

On 1/30/22 9:05 PM, olcott wrote:

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

These statements need the conditions, that H^ goes to

H^.Qy/H^.Qn iff H goes to that corresponding state.

⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H

in the same way that (5,3) is syntactically specified as an

input to Sum(5,3)

Right, and the

Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to

embedded_H in the same way that (1,2) is NOT syntactically

specified as an input to Sum(5,3)

Right, but perhaps you don't understand that from you above

statement the right answer is based on if UTM(<H^>,<H^>)

Halts which by the definition of a UTM means if H^ applied to

<H^> Halts.

The biggest reason for your huge mistakes is that you cannot

stay sharply focused on a single point. It is as if you either

have attention deficit disorder ADD or are addicted to

methamphetamine.

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The single point is that ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H and

Ĥ ⟨Ĥ⟩ is the NOT the input to embedded_H.

After we have mutual agreement on this point we will move on

to the points that logically follow from this one.

Holy shit try to post something that makes sense.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Richard does not accept that the input to the copy of Linz H

embedded at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩. He keeps insisting that it is Ĥ ⟨Ĥ⟩.

No, but apparently you can't understand actual English words.

The INPUT to H is <H^> <H^> but the CORRECT ANSWER that H must

give is based on the behavior of H^ applied to <H^> BECAUSE OF

THE DEFINITION of H.

In other words Sum(3,5) must return the value of Sum(7,8)?

Don't know how you get that from what I said.

Any moron knows that a function is only accountable for its actual

inputs.

And the actual input to H is <H^> <H^> which MEANS by the

DEFINITION of the Halting Problem that H is being asked to decide

on the Halting Status of H^ applied to <H^>

being asked about Sum(7,8).

Again your RED HERRING.

H is being asked EXACTLY what it being asked

H wM w -> H.Qy if M applied to w Halts, and H.Qn if it doesn't

AGREED?

No that is wrong. embedded_H is being asked:

Can the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ possibly transition to ⟨Ĥ⟩.qn ?

If you say 'No', then you aren't doing the halting problem, as the

requirement I stated is EXACTLY the requirement of the Halting Problem.

that a machine has stopped running and that a machine cannot reach its

final state. My definition only includes the latter.

Sounds like a NDTM.

It is not a NDTM, a Turing Machine only actually halts when it reaches

its own final state. People not very familiar with this material may get

confused and believe that a TM halts when its stops running because its

simulation has been aborted. This key distinction is not typically

specified in most halting problem proofs.

computation that halts … the Turing machine will halt whenever it enters

a final state. (Linz:1990:234)

Where did Linz mention 'simulation' and 'abort'?

problem and it can be understood as correct by anyone that understands

the halting problem at a much deeper level than rote memorization.

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 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to

⟨Ĥ⟩.qn ? (No means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt).

André had recommended many online sites for you to learn or test, I forget which posts it is.

But I think C program is more simpler.

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

-- Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;

Genius hits a target no one else can see.

Arthur Schopenhauer

André had recommended many online sites for you to learn or test, I forget which posts it is.

Type it into a TM simulator and prove your claim, your words are meaningless.

I have already proved that I know one key fact about halt deciders that no one else here seems to know.

No one here understands that because a halt decider is a decider that it must compute the mapping from its inputs to an accept of reject state on the basis of the actual behavior specified by these inputs.

And the ACTUAL BEHAVIOR of the input <H^> <H^> is EXACTLY the behavior of H^ applied to <H^> which does Halt if H goes to H.Qn.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ?

Doesn't matter if embedded_H is not a ACTUAL UTM.

When Ĥ is applied to ⟨Ĥ⟩

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

As soon as embedded_H correctly recognizes this as an infinite behavior pattern:

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⟩...

Then embedded_H can correctly abort the simulation of its input and correctly transition to Ĥ.qn.

The above words can be verified as completely true entirely on the basis of their meaning.

Nope, proven otherwise.

What I said above is true by logical necessity and you simply aren't bright enough to understand this.

Then you can provide a step by step proof of it?

If X then Y and if Y then Z and X then Z. There is no way around this.

And what are your X, Y and Z?

If embedded_H correctly recognizes that its input specifies non halting behavior then it is necessarily correct for embedded_H to report this

non halting behavior.

*IF* it correct recognizes. Since there is no pattern in H's simulation of <H^> <H^> THAT IS a proof of non-halting

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⟩...

--

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 2/1/2022 8:48 PM, André G. Isaak wrote:

So you are simply being nit picky about my use of terminology.

When ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to simulating halt decider embedded_H and embedded_H correctly determines that its simulated input cannot possibly reach any final state then embedded_H is necessarily correct to transition to Ĥ.qn indicating that its simulated input never halts.

--

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 2022-02-01 19:37, olcott wrote:

On 2/1/2022 8:08 PM, André G. Isaak wrote:

⟨Ĥ⟩ applied to ⟨Ĥ⟩ is completely meaningless.

Sure and so is the "I am going to go to the" part of

"I am going to go to the store to buy some ice cream."

When you don't cut off what I said in the middle of the sentence then it makes much more sense.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ?

That's just as meaningless. You can simulate Ĥ applied to ⟨Ĥ⟩ or you can provide ⟨Ĥ⟩ ⟨Ĥ⟩ as the input to a simulator. You cannot simulate ⟨Ĥ⟩ applied to ⟨Ĥ⟩ anymore than you can apply ⟨Ĥ⟩ to ⟨Ĥ⟩.

So you are simply being nit picky about my use of terminology.

When ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to simulating halt decider embedded_H and embedded_H correctly determines that its simulated input cannot possibly reach any final state then embedded_H is necessarily correct to transition to Ĥ.qn indicating that its simulated input never halts.

--

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 2/1/2022 8:48 PM, Richard Damon wrote:

This is a woeful lack of basic software engineering skill on your part.

When a process is terminated by the operating system no aspect of this process continues to execute at all.

embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩. When it kills its simulation every simulation that was simulated by any level of ⟨Ĥ⟩ is also immediately killed off because its parent process has been terminated.

--

Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;

Genius hits a target no one else can see.

Click here to read the complete article

On 2/1/22 9:41 PM, olcott wrote:

On 2/1/2022 8:21 PM, Richard Damon wrote:

On 2/1/22 9:14 PM, olcott wrote:You must be a liar.

On 2/1/2022 7:55 PM, Richard Damon wrote:

On 2/1/22 8:47 PM, olcott wrote:

On 2/1/2022 7:40 PM, Richard Damon wrote:

On 2/1/22 8:03 PM, olcott wrote:

On 1/31/22 11:42 PM, olcott wrote:

On 1/31/22 11:24 PM, olcott wrote:

On 1/30/22 9:05 PM, olcott wrote:

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

These statements need the conditions, that H^ goes to

H^.Qy/H^.Qn iff H goes to that corresponding state.

⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H

in the same way that (5,3) is syntactically specified as an

input to Sum(5,3)

Right, and the

Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to

embedded_H in the same way that (1,2) is NOT syntactically

specified as an input to Sum(5,3)

Right, but perhaps you don't understand that from you above

statement the right answer is based on if UTM(<H^>,<H^>)

Halts which by the definition of a UTM means if H^ applied to

<H^> Halts.

The biggest reason for your huge mistakes is that you cannot

stay sharply focused on a single point. It is as if you either

have attention deficit disorder ADD or are addicted to

methamphetamine.

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

The single point is that ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H and

Ĥ ⟨Ĥ⟩ is the NOT the input to embedded_H.

After we have mutual agreement on this point we will move on

to the points that logically follow from this one.

Holy shit try to post something that makes sense.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

Richard does not accept that the input to the copy of Linz H

embedded at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩. He keeps insisting that it is Ĥ ⟨Ĥ⟩.

No, but apparently you can't understand actual English words.

The INPUT to H is <H^> <H^> but the CORRECT ANSWER that H must

give is based on the behavior of H^ applied to <H^> BECAUSE OF

THE DEFINITION of H.

In other words Sum(3,5) must return the value of Sum(7,8)?

Don't know how you get that from what I said.

Any moron knows that a function is only accountable for its actual

inputs.

And the actual input to H is <H^> <H^> which MEANS by the

DEFINITION of the Halting Problem that H is being asked to decide

on the Halting Status of H^ applied to <H^>

being asked about Sum(7,8).

Again your RED HERRING.

H is being asked EXACTLY what it being asked

H wM w -> H.Qy if M applied to w Halts, and H.Qn if it doesn't

AGREED?

No that is wrong. embedded_H is being asked:

Can the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ possibly transition to ⟨Ĥ⟩.qn ?

If you say 'No', then you aren't doing the halting problem, as the

requirement I stated is EXACTLY the requirement of the Halting Problem.

that a machine has stopped running and that a machine cannot reach its

final state. My definition only includes the latter.

Sounds like a NDTM.

It is not a NDTM, a Turing Machine only actually halts when it reaches

its own final state. People not very familiar with this material may get

confused and believe that a TM halts when its stops running because its

simulation has been aborted. This key distinction is not typically

specified in most halting problem proofs.

computation that halts … the Turing machine will halt whenever it enters

a final state. (Linz:1990:234)

Where did Linz mention 'simulation' and 'abort'?

problem and it can be understood as correct by anyone that understands

the halting problem at a much deeper level than rote memorization.

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 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to

⟨Ĥ⟩.qn ? (No means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt).

André had recommended many online sites for you to learn or test, I forget which posts it is.

But I think C program is more simpler.

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

-- Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;

Genius hits a target no one else can see.

Arthur Schopenhauer

André had recommended many online sites for you to learn or test, I forget which posts it is.

Type it into a TM simulator and prove your claim, your words are meaningless.

I have already proved that I know one key fact about halt deciders that no one else here seems to know.

No one here understands that because a halt decider is a decider that it must compute the mapping from its inputs to an accept of reject state on the basis of the actual behavior specified by these inputs.

And the ACTUAL BEHAVIOR of the input <H^> <H^> is EXACTLY the behavior of H^ applied to <H^> which does Halt if H goes to H.Qn.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ?

Doesn't matter if embedded_H is not a ACTUAL UTM.

When Ĥ is applied to ⟨Ĥ⟩

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

As soon as embedded_H correctly recognizes this as an infinite behavior pattern:

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⟩...

Then embedded_H can correctly abort the simulation of its input and correctly transition to Ĥ.qn.

The above words can be verified as completely true entirely on the basis of their meaning.

Nope, proven otherwise.

What I said above is true by logical necessity and you simply aren't bright enough to understand this.

Then you can provide a step by step proof of it?

If X then Y and if Y then Z and X then Z. There is no way around this.

And what are your X, Y and Z?

If embedded_H correctly recognizes that its input specifies non halting behavior then it is necessarily correct for embedded_H to report this

non halting behavior.

*IF* it correct recognizes. Since there is no pattern in H's simulation of <H^> <H^> THAT IS a proof of non-halting

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 if H <H^> <H^> -> H.Qn aften N steps, then it is also true that the computation H1 <H^> <H^> -> H.Qn after N steps and the pattern ends.

This is a woeful lack of basic software engineering skill on your part.

When a process is terminated by the operating system no aspect of this process continues to execute at all.

embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩. When it kills its simulation every simulation that was simulated by any level of ⟨Ĥ⟩ is also immediately killed off because its parent process has been terminated.

--

Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;

Genius hits a target no one else can see.

Click here to read the complete article

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From: polco...@gmail.com (olcott)

Newsgroups: comp.theory,comp.ai.philosophy,sci.logic,sci.math

Subject: Re: Concise refutation of halting problem proofs V52 [ dodgy double

talk ]

Followup-To: comp.theory

Date: Tue, 1 Feb 2022 21:44:22 -0600

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View all headersFrom: polco...@gmail.com (olcott)

Newsgroups: comp.theory,comp.ai.philosophy,sci.logic,sci.math

Subject: Re: Concise refutation of halting problem proofs V52 [ dodgy double

talk ]

Followup-To: comp.theory

Date: Tue, 1 Feb 2022 21:44:22 -0600

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On 2/1/2022 9:24 PM, André G. Isaak wrote:

It is fine and good that you help correct my terminology.

What is not fine and good is for you to reject the essence of the gist of what I am saying entirely on the basis that I did not say it exactly according to conventions. The is what Ben always did. He never paid any attention to the actual substance of what I was saying.

Not at all. A simulator simulates a finite string and the actual behavior of this simulated finite string is the ultimate basis of whether or not it specifies a finite sequence of configurations.

If the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H cannot possibly reach ⟨Ĥ⟩.qn then it is necessarily correct for embedded_H to transition to Ĥ.qn and nothing else in the universe can possibly refute this.

In much simpler terms if the input to simulating halt decider H never halts then it is always correct for H to report that its input never halts.

In even simpler terms if you see an actual dog then you are correct to say: "I saw a dog", even if everyone else in the universe disagrees.

--

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 2022-02-01 19:57, olcott wrote:

On 2/1/2022 8:48 PM, André G. Isaak wrote:

On 2022-02-01 19:37, olcott wrote:

On 2/1/2022 8:08 PM, André G. Isaak wrote:

⟨Ĥ⟩ applied to ⟨Ĥ⟩ is completely meaningless.

Sure and so is the "I am going to go to the" part of

"I am going to go to the store to buy some ice cream."

When you don't cut off what I said in the middle of the sentence then it makes much more sense.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ?

That's just as meaningless. You can simulate Ĥ applied to ⟨Ĥ⟩ or you can provide ⟨Ĥ⟩ ⟨Ĥ⟩ as the input to a simulator. You cannot simulate ⟨Ĥ⟩ applied to ⟨Ĥ⟩ anymore than you can apply ⟨Ĥ⟩ to ⟨Ĥ⟩.

So you are simply being nit picky about my use of terminology.

Yes, I insist on terminology being used correctly. And any place where you attempt to publish your results will be equally, if not more, nit picky.

It is fine and good that you help correct my terminology.

What is not fine and good is for you to reject the essence of the gist of what I am saying entirely on the basis that I did not say it exactly according to conventions. The is what Ben always did. He never paid any attention to the actual substance of what I was saying.

When ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to simulating halt decider embedded_H and embedded_H correctly determines that its simulated input cannot possibly reach any final state then embedded_H is necessarily correct to transition to Ĥ.qn indicating that its simulated input never halts.

But now you've just hidden your meaningless terminological abuse. "Its simulated input" is only meaningful when it is construed as meaning the simulation of the computation REPRESENTED by the input, i.e. the

Not at all. A simulator simulates a finite string and the actual behavior of this simulated finite string is the ultimate basis of whether or not it specifies a finite sequence of configurations.

If the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H cannot possibly reach ⟨Ĥ⟩.qn then it is necessarily correct for embedded_H to transition to Ĥ.qn and nothing else in the universe can possibly refute this.

In much simpler terms if the input to simulating halt decider H never halts then it is always correct for H to report that its input never halts.

In even simpler terms if you see an actual dog then you are correct to say: "I saw a dog", even if everyone else in the universe disagrees.

--

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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From: polco...@gmail.com (olcott)

Newsgroups: comp.theory,comp.ai.philosophy,sci.logic,sci.math

Subject: Re: Concise refutation of halting problem proofs V52 [ dodgy double

talk ]

Followup-To: comp.theory

Date: Tue, 1 Feb 2022 21:59:54 -0600

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View all headersFrom: polco...@gmail.com (olcott)

Newsgroups: comp.theory,comp.ai.philosophy,sci.logic,sci.math

Subject: Re: Concise refutation of halting problem proofs V52 [ dodgy double

talk ]

Followup-To: comp.theory

Date: Tue, 1 Feb 2022 21:59:54 -0600

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On 2/1/2022 9:33 PM, André G. Isaak wrote:

embedded_H merely determines whether or not its input specifies a finite sequence of configurations. It does not give a rats ass about anything else in the universe.

--

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 2022-02-01 20:08, olcott wrote:

embedded_H is exactly a UTM with extra features added.

Apparently you don't know what 'exactly' means. embedded_H is not a UTM *at* *all*.

If embedded_H were a UTM, then

embedded_H would accept ⟨Ĥ⟩ ⟨Ĥ⟩ if Ĥ accepts ⟨Ĥ⟩

embedded_H merely determines whether or not its input specifies a finite sequence of configurations. It does not give a rats ass about anything else in the universe.

--

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 2/1/2022 9:58 PM, André G. Isaak wrote:

It is possible for Turing machines to have blank tapes.

The salient aspect for the Halting problem is whether or not the finite string machine description specifies a finite or infinite sequence of configurations. The ultimate basis for determining this is the actual behavior of the simulated finite string.

Since this equally applies to machines having inputs and machines not having inputs the distinction relative to inputs is moot.

The behavior of the simulated input when embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ is the ultimate measure of the halt status of this input.

Inputs to simulating halt deciders are already implicitly specified to be simulated. We could get verbose and say the simulated inputs to simulating halt decider:

In much simpler terms if the simulated input to simulating halt decider H never halts then it is always correct for H to report that its input never halts.

You and Richard are saying that there are exceptions to this logically necessary truth. This is like saying that when a dog bites you on the leg it might not have been a dog and it might not have been your leg even though it is stipulated that a dog bit you on the leg.

--

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 2022-02-01 20:44, olcott wrote:

On 2/1/2022 9:24 PM, André G. Isaak wrote:

On 2022-02-01 19:57, olcott wrote:

On 2/1/2022 8:48 PM, André G. Isaak wrote:

On 2022-02-01 19:37, olcott wrote:

On 2/1/2022 8:08 PM, André G. Isaak wrote:

⟨Ĥ⟩ applied to ⟨Ĥ⟩ is completely meaningless.

Sure and so is the "I am going to go to the" part of

"I am going to go to the store to buy some ice cream."

When you don't cut off what I said in the middle of the sentence then it makes much more sense.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ?

That's just as meaningless. You can simulate Ĥ applied to ⟨Ĥ⟩ or you can provide ⟨Ĥ⟩ ⟨Ĥ⟩ as the input to a simulator. You cannot simulate ⟨Ĥ⟩ applied to ⟨Ĥ⟩ anymore than you can apply ⟨Ĥ⟩ to ⟨Ĥ⟩.

So you are simply being nit picky about my use of terminology.

Yes, I insist on terminology being used correctly. And any place where you attempt to publish your results will be equally, if not more, nit picky.

It is fine and good that you help correct my terminology.

What is not fine and good is for you to reject the essence of the gist of what I am saying entirely on the basis that I did not say it exactly according to conventions. The is what Ben always did. He never paid any attention to the actual substance of what I was saying.

When ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to simulating halt decider embedded_H and embedded_H correctly determines that its simulated input cannot possibly reach any final state then embedded_H is necessarily correct to transition to Ĥ.qn indicating that its simulated input never halts.

But now you've just hidden your meaningless terminological abuse. "Its simulated input" is only meaningful when it is construed as meaning the simulation of the computation REPRESENTED by the input, i.e. the

Not at all. A simulator simulates a finite string and the actual behavior of this simulated finite string is the ultimate basis of whether or not it specifies a finite sequence of configurations.

No. A simulator simulates a Turing Machine applied to an input. It takes as its input a finite string which represents that Turing Machine/Input pair. It's completely meaningless to talk about simulating a finite string.

It is possible for Turing machines to have blank tapes.

The salient aspect for the Halting problem is whether or not the finite string machine description specifies a finite or infinite sequence of configurations. The ultimate basis for determining this is the actual behavior of the simulated finite string.

Since this equally applies to machines having inputs and machines not having inputs the distinction relative to inputs is moot.

If the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H cannot possibly reach ⟨Ĥ⟩.qn then it is necessarily correct for embedded_H to transition to Ĥ.qn and nothing else in the universe can possibly refute this.

Again, you're falling back on your belief that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is both meaningful (it isn't) and somehow distinct from H applied to ⟨Ĥ⟩.

The behavior of the simulated input when embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ is the ultimate measure of the halt status of this input.

What you're doing is essentially the same thing as if I produced a SUM TM which, given ⟨3⟩ and ⟨2⟩ as inputs, produced the output ⟨6⟩, and me claiming that it is correct because it isn't responsible for adding the integers, only the finite strings which SUM was given as inputs, and that my algorithm correctly determines that SUM ⟨3⟩⟨2⟩ is ⟨6⟩ despite the fact that SUM 3 2 is 5.

In much simpler terms if the input to simulating halt decider H never halts then it is always correct for H to report that its input never halts.

Inputs don't halt or not halt. Only the TM/input pair which the input *describes* can halt or not halt.

André

Inputs to simulating halt deciders are already implicitly specified to be simulated. We could get verbose and say the simulated inputs to simulating halt decider:

In much simpler terms if the simulated input to simulating halt decider H never halts then it is always correct for H to report that its input never halts.

You and Richard are saying that there are exceptions to this logically necessary truth. This is like saying that when a dog bites you on the leg it might not have been a dog and it might not have been your leg even though it is stipulated that a dog bit you on the leg.

--

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 2/1/2022 10:10 PM, André G. Isaak wrote:

You snipped in the middle of the sentence and then said that the sentence didn't make sense. I snip down to the most salient point.

When Ĥ is applied to ⟨Ĥ⟩

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

embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ determines that its simulated input would never reach its final state ⟨Ĥ⟩.qn on the basis of matching this infinite pattern:

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⟩...

It is a UTM that has extra features added.

--

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 2022-02-01 20:59, olcott wrote:

On 2/1/2022 9:33 PM, André G. Isaak wrote:

On 2022-02-01 20:08, olcott wrote:

embedded_H is exactly a UTM with extra features added.

Apparently you don't know what 'exactly' means. embedded_H is not a UTM *at* *all*.

If embedded_H were a UTM, then

embedded_H would accept ⟨Ĥ⟩ ⟨Ĥ⟩ if Ĥ accepts ⟨Ĥ⟩

And in a previous post you accused *me* of snipping part way through?

You snipped in the middle of the sentence and then said that the sentence didn't make sense. I snip down to the most salient point.

When Ĥ is applied to ⟨Ĥ⟩

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

embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ determines that its simulated input would never reach its final state ⟨Ĥ⟩.qn on the basis of matching this infinite pattern:

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⟩...

embedded_H merely determines whether or not its input specifies a finite sequence of configurations. It does not give a rats ass about anything else in the universe.

Which again means that it is not a UTM since that is *not* what a UTM determines.

It is a UTM that has extra features added.

Being a halt decider and being a UTM are mutually exclusive.For all inputs that reach their final state the behavior of embedded_H is equivalent to a UTM.

André

--

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 V52 [ dodgy double

talk ]

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Subject: Re: Concise refutation of halting problem proofs V52 [ dodgy double

talk ]

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On 2/2/2022 6:22 PM, Richard Damon wrote:

On the basis of the actual behavior specified by the actual input.

On the basis of the actual behavior specified by the actual input.

On the basis of the actual behavior specified by the actual input.

--

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 2/2/22 10:20 AM, olcott wrote:No it only proves that you and André don't understand that a halt decider computes the mapping from the inputs to an accept or reject state (here is the part that you two don't understand):

On 2/1/2022 9:58 PM, André G. Isaak wrote:

On 2022-02-01 20:44, olcott wrote:

On 2/1/2022 9:24 PM, André G. Isaak wrote:

On 2022-02-01 19:57, olcott wrote:

On 2/1/2022 8:48 PM, André G. Isaak wrote:

On 2/1/2022 8:08 PM, André G. Isaak wrote:

⟨Ĥ⟩ applied to ⟨Ĥ⟩ is completely meaningless.

Sure and so is the "I am going to go to the" part of

"I am going to go to the store to buy some ice cream."

When you don't cut off what I said in the middle of the sentence then it makes much more sense.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ?

That's just as meaningless. You can simulate Ĥ applied to ⟨Ĥ⟩ or you can provide ⟨Ĥ⟩ ⟨Ĥ⟩ as the input to a simulator. You cannot simulate ⟨Ĥ⟩ applied to ⟨Ĥ⟩ anymore than you can apply ⟨Ĥ⟩ to ⟨Ĥ⟩.

So you are simply being nit picky about my use of terminology.

Yes, I insist on terminology being used correctly. And any place where you attempt to publish your results will be equally, if not more, nit picky.

It is fine and good that you help correct my terminology.

What is not fine and good is for you to reject the essence of the gist of what I am saying entirely on the basis that I did not say it exactly according to conventions. The is what Ben always did. He never paid any attention to the actual substance of what I was saying.

When ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to simulating halt decider embedded_H and embedded_H correctly determines that its simulated input cannot possibly reach any final state then embedded_H is necessarily correct to transition to Ĥ.qn indicating that its simulated input never halts.

But now you've just hidden your meaningless terminological abuse. "Its simulated input" is only meaningful when it is construed as meaning the simulation of the computation REPRESENTED by the input, i.e. the

Not at all. A simulator simulates a finite string and the actual behavior of this simulated finite string is the ultimate basis of whether or not it specifies a finite sequence of configurations.

No. A simulator simulates a Turing Machine applied to an input. It takes as its input a finite string which represents that Turing Machine/Input pair. It's completely meaningless to talk about simulating a finite string.

It is possible for Turing machines to have blank tapes.

The salient aspect for the Halting problem is whether or not the finite string machine description specifies a finite or infinite sequence of configurations. The ultimate basis for determining this is the actual behavior of the simulated finite string.

Since this equally applies to machines having inputs and machines not having inputs the distinction relative to inputs is moot.

If the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H cannot possibly reach ⟨Ĥ⟩.qn then it is necessarily correct for embedded_H to transition to Ĥ.qn and nothing else in the universe can possibly refute this.

Again, you're falling back on your belief that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is both meaningful (it isn't) and somehow distinct from H applied to ⟨Ĥ⟩.

The behavior of the simulated input when embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ is the ultimate measure of the halt status of this input.

Which just proves you are not working on the Halting Problem,

On the basis of the actual behavior specified by the actual input.

On the basis of the actual behavior specified by the actual input.

On the basis of the actual behavior specified by the actual input.

--

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 V52 [ dodgy double

talk ]

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Subject: Re: Concise refutation of halting problem proofs V52 [ dodgy double

talk ]

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On 2/2/2022 8:08 PM, Richard Damon wrote:

These words prove themselves true on the basis of their meaning:

The actual behavior of the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H is the ultimate measure of the behavior specified by ⟨Ĥ⟩ ⟨Ĥ⟩.

--

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 2/2/22 8:41 PM, olcott wrote:

On 2/2/2022 6:22 PM, Richard Damon wrote:

On 2/2/22 10:20 AM, olcott wrote:No it only proves that you and André don't understand that a halt decider computes the mapping from the inputs to an accept or reject state (here is the part that you two don't understand):

On 2/1/2022 9:58 PM, André G. Isaak wrote:

On 2022-02-01 20:44, olcott wrote:

On 2/1/2022 9:24 PM, André G. Isaak wrote:

On 2/1/2022 8:08 PM, André G. Isaak wrote:

⟨Ĥ⟩ applied to ⟨Ĥ⟩ is completely meaningless.

Sure and so is the "I am going to go to the" part of

"I am going to go to the store to buy some ice cream."

When you don't cut off what I said in the middle of the sentence then it makes much more sense.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ?

That's just as meaningless. You can simulate Ĥ applied to ⟨Ĥ⟩ or you can provide ⟨Ĥ⟩ ⟨Ĥ⟩ as the input to a simulator. You cannot simulate ⟨Ĥ⟩ applied to ⟨Ĥ⟩ anymore than you can apply ⟨Ĥ⟩ to ⟨Ĥ⟩.

So you are simply being nit picky about my use of terminology.

Yes, I insist on terminology being used correctly. And any place where you attempt to publish your results will be equally, if not more, nit picky.

It is fine and good that you help correct my terminology.

What is not fine and good is for you to reject the essence of the gist of what I am saying entirely on the basis that I did not say it exactly according to conventions. The is what Ben always did. He never paid any attention to the actual substance of what I was saying.

But now you've just hidden your meaningless terminological abuse. "Its simulated input" is only meaningful when it is construed as meaning the simulation of the computation REPRESENTED by the input, i.e. the

Not at all. A simulator simulates a finite string and the actual behavior of this simulated finite string is the ultimate basis of whether or not it specifies a finite sequence of configurations.

No. A simulator simulates a Turing Machine applied to an input. It takes as its input a finite string which represents that Turing Machine/Input pair. It's completely meaningless to talk about simulating a finite string.

It is possible for Turing machines to have blank tapes.

The salient aspect for the Halting problem is whether or not the finite string machine description specifies a finite or infinite sequence of configurations. The ultimate basis for determining this is the actual behavior of the simulated finite string.

Since this equally applies to machines having inputs and machines not having inputs the distinction relative to inputs is moot.

If the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H cannot possibly reach ⟨Ĥ⟩.qn then it is necessarily correct for embedded_H to transition to Ĥ.qn and nothing else in the universe can possibly refute this.

Again, you're falling back on your belief that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is both meaningful (it isn't) and somehow distinct from H applied to ⟨Ĥ⟩.

The behavior of the simulated input when embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ is the ultimate measure of the halt status of this input.

Which just proves you are not working on the Halting Problem,

On the basis of the actual behavior specified by the actual input.

On the basis of the actual behavior specified by the actual input.

On the basis of the actual behavior specified by the actual input.

Which is DEFINED by what a the machine the input represents would do,

These words prove themselves true on the basis of their meaning:

The actual behavior of the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H is the ultimate measure of the behavior specified by ⟨Ĥ⟩ ⟨Ĥ⟩.

--

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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From: polco...@gmail.com (olcott)

Newsgroups: comp.theory,comp.ai.philosophy,sci.logic,sci.math

Subject: Re: Concise refutation of halting problem proofs V52 [ dodgy double

talk ]

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View all headersFrom: polco...@gmail.com (olcott)

Newsgroups: comp.theory,comp.ai.philosophy,sci.logic,sci.math

Subject: Re: Concise refutation of halting problem proofs V52 [ dodgy double

talk ]

Followup-To: comp.theory

Date: Wed, 2 Feb 2022 22:50:06 -0600

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On 2/2/2022 10:42 PM, Richard Damon wrote:

When embedded_H correctly matches the above infinite sequence this conclusively proves that its correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly reach ⟨Ĥ⟩.qn. (We don't even need to mention any UTM).

--

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 2/2/22 10:50 PM, olcott wrote:

On 2/2/2022 9:30 PM, Richard Damon wrote:

It must only do exactly what it actually does, if this does not meet expectations then expectations must be incorrect.

On 2/2/22 10:12 PM, olcott wrote:

On 2/2/2022 8:57 PM, Richard Damon wrote:

On 2/2/22 9:31 PM, olcott wrote:When you disagree that the correct simulation of a machine description of a machine is the ultimate measure of the behavior specified by this machine description it is just like saying that a black cat is not a cat.

On 2/2/2022 8:08 PM, Richard Damon wrote:

On 2/2/22 8:41 PM, olcott wrote:

On 2/2/2022 6:22 PM, Richard Damon wrote:

On 2/2/22 10:20 AM, olcott wrote:No it only proves that you and André don't understand that a halt decider computes the mapping from the inputs to an accept or reject state (here is the part that you two don't understand):

On 2/1/2022 9:58 PM, André G. Isaak wrote:

On 2/1/2022 8:08 PM, André G. Isaak wrote:

⟨Ĥ⟩ applied to ⟨Ĥ⟩ is completely meaningless.

Sure and so is the "I am going to go to the" part of

"I am going to go to the store to buy some ice cream."

When you don't cut off what I said in the middle of the sentence then it makes much more sense.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ?

That's just as meaningless. You can simulate Ĥ applied to ⟨Ĥ⟩ or you can provide ⟨Ĥ⟩ ⟨Ĥ⟩ as the input to a simulator. You cannot simulate ⟨Ĥ⟩ applied to ⟨Ĥ⟩ anymore than you can apply ⟨Ĥ⟩ to ⟨Ĥ⟩.

So you are simply being nit picky about my use of terminology.

Yes, I insist on terminology being used correctly. And any place where you attempt to publish your results will be equally, if not more, nit picky.

It is fine and good that you help correct my terminology.

What is not fine and good is for you to reject the essence of the gist of what I am saying entirely on the basis that I did not say it exactly according to conventions. The is what Ben always did. He never paid any attention to the actual substance of what I was saying.

But now you've just hidden your meaningless terminological abuse. "Its simulated input" is only meaningful when it is construed as meaning the simulation of the computation REPRESENTED by the input, i.e. the

Not at all. A simulator simulates a finite string and the actual behavior of this simulated finite string is the ultimate basis of whether or not it specifies a finite sequence of configurations.

No. A simulator simulates a Turing Machine applied to an input. It takes as its input a finite string which represents that Turing Machine/Input pair. It's completely meaningless to talk about simulating a finite string.

It is possible for Turing machines to have blank tapes.

The salient aspect for the Halting problem is whether or not the finite string machine description specifies a finite or infinite sequence of configurations. The ultimate basis for determining this is the actual behavior of the simulated finite string.

Since this equally applies to machines having inputs and machines not having inputs the distinction relative to inputs is moot.

Again, you're falling back on your belief that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is both meaningful (it isn't) and somehow distinct from H applied to ⟨Ĥ⟩.

The behavior of the simulated input when embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ is the ultimate measure of the halt status of this input.

Which just proves you are not working on the Halting Problem,

On the basis of the actual behavior specified by the actual input.

On the basis of the actual behavior specified by the actual input.

On the basis of the actual behavior specified by the actual input.

Which is DEFINED by what a the machine the input represents would do,

These words prove themselves true on the basis of their meaning:

The actual behavior of the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H is the ultimate measure of the behavior specified by ⟨Ĥ⟩ ⟨Ĥ⟩.

WRONG, which shows you do not actually know the meaning of the words.

The problem is that 'Correct Simulation of a machine description' has an actual meaning, in that the simulation must match the actual behavior of the machine whose description it is simulating, RIGHT?

Here is what it actually does:

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⟩...

And if that is what it actually does, then H NEVER aborts its simulation and thus never give an answer.

When embedded_H correctly matches the above infinite sequence this conclusively proves that its correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly reach ⟨Ĥ⟩.qn. (We don't even need to mention any UTM).

--

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 V52 [ dodgy double

talk ]

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Subject: Re: Concise refutation of halting problem proofs V52 [ dodgy double

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On 2/2/2022 11:09 PM, Richard Damon wrote:

--

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 2/2/22 11:50 PM, olcott wrote:That is not true. The pattern exists for at least any finite number of steps where it can be recognized. The three iterations shown above are plenty enough for it to be recogized.

On 2/2/2022 10:42 PM, Richard Damon wrote:

On 2/2/22 10:50 PM, olcott wrote:

On 2/2/2022 9:30 PM, Richard Damon wrote:

It must only do exactly what it actually does, if this does not meet expectations then expectations must be incorrect.

On 2/2/22 10:12 PM, olcott wrote:

On 2/2/2022 8:57 PM, Richard Damon wrote:

On 2/2/22 9:31 PM, olcott wrote:When you disagree that the correct simulation of a machine description of a machine is the ultimate measure of the behavior specified by this machine description it is just like saying that a black cat is not a cat.

On 2/2/2022 8:08 PM, Richard Damon wrote:

On 2/2/22 8:41 PM, olcott wrote:

On 2/2/2022 6:22 PM, Richard Damon wrote:

On 2/1/2022 8:08 PM, André G. Isaak wrote:

⟨Ĥ⟩ applied to ⟨Ĥ⟩ is completely meaningless.

Sure and so is the "I am going to go to the" part of

"I am going to go to the store to buy some ice cream."

When you don't cut off what I said in the middle of the sentence then it makes much more sense.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ?

That's just as meaningless. You can simulate Ĥ applied to ⟨Ĥ⟩ or you can provide ⟨Ĥ⟩ ⟨Ĥ⟩ as the input to a simulator. You cannot simulate ⟨Ĥ⟩ applied to ⟨Ĥ⟩ anymore than you can apply ⟨Ĥ⟩ to ⟨Ĥ⟩.

So you are simply being nit picky about my use of terminology.

Yes, I insist on terminology being used correctly. And any place where you attempt to publish your results will be equally, if not more, nit picky.

It is fine and good that you help correct my terminology.

What is not fine and good is for you to reject the essence of the gist of what I am saying entirely on the basis that I did not say it exactly according to conventions. The is what Ben always did. He never paid any attention to the actual substance of what I was saying.

But now you've just hidden your meaningless terminological abuse. "Its simulated input" is only meaningful when it is construed as meaning the simulation of the computation REPRESENTED by the input, i.e. the

Not at all. A simulator simulates a finite string and the actual behavior of this simulated finite string is the ultimate basis of whether or not it specifies a finite sequence of configurations.

No. A simulator simulates a Turing Machine applied to an input. It takes as its input a finite string which represents that Turing Machine/Input pair. It's completely meaningless to talk about simulating a finite string.

It is possible for Turing machines to have blank tapes.

The salient aspect for the Halting problem is whether or not the finite string machine description specifies a finite or infinite sequence of configurations. The ultimate basis for determining this is the actual behavior of the simulated finite string.

Since this equally applies to machines having inputs and machines not having inputs the distinction relative to inputs is moot.

Again, you're falling back on your belief that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is both meaningful (it isn't) and somehow distinct from H applied to ⟨Ĥ⟩.

The behavior of the simulated input when embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ is the ultimate measure of the halt status of this input.

Which just proves you are not working on the Halting Problem,

On the basis of the actual behavior specified by the actual input.

On the basis of the actual behavior specified by the actual input.

On the basis of the actual behavior specified by the actual input.

Which is DEFINED by what a the machine the input represents would do,

These words prove themselves true on the basis of their meaning:

The actual behavior of the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H is the ultimate measure of the behavior specified by ⟨Ĥ⟩ ⟨Ĥ⟩.

WRONG, which shows you do not actually know the meaning of the words.

The problem is that 'Correct Simulation of a machine description' has an actual meaning, in that the simulation must match the actual behavior of the machine whose description it is simulating, RIGHT?

Here is what it actually does:

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⟩...

And if that is what it actually does, then H NEVER aborts its simulation and thus never give an answer.

When embedded_H correctly matches the above infinite sequence this conclusively proves that its correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly reach ⟨Ĥ⟩.qn. (We don't even need to mention any UTM).

Excepts as previously said, that pattern only exists if H never aborts.

--

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 V52 [ pants on fire

]

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Subject: Re: Concise refutation of halting problem proofs V52 [ pants on fire

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On 2/3/2022 5:36 PM, Richard Damon wrote:

As long as the simulated ⟨Ĥ⟩ ⟨Ĥ⟩ never reaches ⟨Ĥ⟩.qn

then embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct.

It like like you are trying to use the fact that your cat scratched your arm as proof that your dog didn't bite your leg.

You are the one being dishonest:

I say that it is true that the simulated ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly ever reach ⟨Ĥ⟩.qn. (and you know that this is true).

and you try to refute this on the basis that embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn.

I say that Sum(5,2) is 7 and you say no its not because Sum(9,3) is 12.

--

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 2/3/22 9:15 AM, olcott wrote:

On 2/3/2022 5:50 AM, Richard Damon wrote:

On 2/3/22 12:24 AM, olcott wrote:

On 2/2/2022 11:09 PM, Richard Damon wrote:

On 2/2/22 11:50 PM, olcott wrote:That is not true. The pattern exists for at least any finite number of steps where it can be recognized. The three iterations shown above are plenty enough for it to be recogized.

On 2/2/2022 10:42 PM, Richard Damon wrote:

On 2/2/22 10:50 PM, olcott wrote:

On 2/2/2022 9:30 PM, Richard Damon wrote:

It must only do exactly what it actually does, if this does not meet expectations then expectations must be incorrect.

On 2/2/22 10:12 PM, olcott wrote:

On 2/2/2022 8:57 PM, Richard Damon wrote:

On 2/2/22 9:31 PM, olcott wrote:When you disagree that the correct simulation of a machine description of a machine is the ultimate measure of the behavior specified by this machine description it is just like saying that a black cat is not a cat.

On 2/2/2022 8:08 PM, Richard Damon wrote:

On 2/1/2022 8:08 PM, André G. Isaak wrote:

⟨Ĥ⟩ applied to ⟨Ĥ⟩ is completely meaningless.

Sure and so is the "I am going to go to the" part of

"I am going to go to the store to buy some ice cream."

When you don't cut off what I said in the middle of the sentence then it makes much more sense.

Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to ⟨Ĥ⟩.qn ?

That's just as meaningless. You can simulate Ĥ applied to ⟨Ĥ⟩ or you can provide ⟨Ĥ⟩ ⟨Ĥ⟩ as the input to a simulator. You cannot simulate ⟨Ĥ⟩ applied to ⟨Ĥ⟩ anymore than you can apply ⟨Ĥ⟩ to ⟨Ĥ⟩.

So you are simply being nit picky about my use of terminology.

Yes, I insist on terminology being used correctly. And any place where you attempt to publish your results will be equally, if not more, nit picky.

It is fine and good that you help correct my terminology.

What is not fine and good is for you to reject the essence of the gist of what I am saying entirely on the basis that I did not say it exactly according to conventions. The is what Ben always did. He never paid any attention to the actual substance of what I was saying.

But now you've just hidden your meaningless terminological abuse. "Its simulated input" is only meaningful when it is construed as meaning the simulation of the computation REPRESENTED by the input, i.e. the

Not at all. A simulator simulates a finite string and the actual behavior of this simulated finite string is the ultimate basis of whether or not it specifies a finite sequence of configurations.

No. A simulator simulates a Turing Machine applied to an input. It takes as its input a finite string which represents that Turing Machine/Input pair. It's completely meaningless to talk about simulating a finite string.

It is possible for Turing machines to have blank tapes.

The salient aspect for the Halting problem is whether or not the finite string machine description specifies a finite or infinite sequence of configurations. The ultimate basis for determining this is the actual behavior of the simulated finite string.

Since this equally applies to machines having inputs and machines not having inputs the distinction relative to inputs is moot.

Again, you're falling back on your belief that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is both meaningful (it isn't) and somehow distinct from H applied to ⟨Ĥ⟩.

The behavior of the simulated input when embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ is the ultimate measure of the halt status of this input.

Which just proves you are not working on the Halting Problem,

On the basis of the actual behavior specified by the actual input.

On the basis of the actual behavior specified by the actual input.

On the basis of the actual behavior specified by the actual input.

Which is DEFINED by what a the machine the input represents would do,

These words prove themselves true on the basis of their meaning:

The actual behavior of the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H is the ultimate measure of the behavior specified by ⟨Ĥ⟩ ⟨Ĥ⟩.

WRONG, which shows you do not actually know the meaning of the words.

The problem is that 'Correct Simulation of a machine description' has an actual meaning, in that the simulation must match the actual behavior of the machine whose description it is simulating, RIGHT?

Here is what it actually does:

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⟩...

And if that is what it actually does, then H NEVER aborts its simulation and thus never give an answer.

When embedded_H correctly matches the above infinite sequence this conclusively proves that its correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly reach ⟨Ĥ⟩.qn. (We don't even need to mention any UTM).

Excepts as previously said, that pattern only exists if H never aborts.

But if it only exists for a finite number of steps (till it is recognized)

Then embedded_H has conclusively proved that its simulated ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly ever reach ⟨Ĥ⟩.qn even in an infinite number of simulated steps thus meeting the Linz definition of a sequence of configurations that never halt.

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

No, it hasn't, because I just showed you that if H -> H.Qn then the computation H^ <H^> Halts.

As long as the simulated ⟨Ĥ⟩ ⟨Ĥ⟩ never reaches ⟨Ĥ⟩.qn

then embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct.

It like like you are trying to use the fact that your cat scratched your arm as proof that your dog didn't bite your leg.

Proven.

No attempt top refute.

So FAIL, you have just proved that all you are is a LIAR.

So, YOU LIE.

You are the one being dishonest:

I say that it is true that the simulated ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly ever reach ⟨Ĥ⟩.qn. (and you know that this is true).

and you try to refute this on the basis that embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn.

I say that Sum(5,2) is 7 and you say no its not because Sum(9,3) is 12.

--

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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