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devel / comp.theory / Re: Concise refutation of halting problem proofs V52 [ ignorance about deciders ]

SubjectAuthor
* Concise refutation of halting problem proofs V52 [ Linz Proof ]olcott
+* Concise refutation of halting problem proofs V52 [ Linz Proof ]Richard Damon
|`* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
| `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|  `* Concise refutation of halting problem proofs V52 [ Ignorant or Dishonest ]olcott
|   `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    +* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |`* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    | `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |  `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |   `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |    `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |     `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |      `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |       `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |        `* Concise refutation of halting problem proofs V52 [ Ignorant or Dishonest ]Richard Damon
|    |         `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |          `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |           +* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |           |`* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |           | `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |           |  `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |           |   `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |           |    `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |           |     `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |           |      `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |           |       `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |           |        `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |           |         `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |           |          `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |           |           `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |           |            `* Concise refutation of halting problem proofs V52 [ Ignorant or Dishonest ]Richard Damon
|    |           |             `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |           |              `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |           |               `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |           |                `- Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |           `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |            `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |             `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |              `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |               +* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |               |`- Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |               `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |                `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |                 `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |                  `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |                   `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |                    `* Concise refutation of halting problem proofs V52 [ Ignorant or Dishonest ](typo)Richard Damon
|    |                     `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |                      `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |                       `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |                        `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |                         `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |                          `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |                           `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |                            `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |                             `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |                              `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |                               `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |                                `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |                                 `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |                                  `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |                                   `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
|    |                                    `* Concise refutation of halting problem proofs V52 [ Ignorant orRichard Damon
|    |                                     `* Concise refutation of halting problem proofs V52 [ Honestolcott
|    |                                      `* Concise refutation of halting problem proofs V52 [ HonestRichard Damon
|    |                                       `* Concise refutation of halting problem proofs V52 [ Honestolcott
|    |                                        `* Concise refutation of halting problem proofs V52 [ HonestRichard Damon
|    |                                         `* Concise refutation of halting problem proofs V52 [ Honestolcott
|    |                                          `* Concise refutation of halting problem proofs V52 [ HonestRichard Damon
|    |                                           `* Concise refutation of halting problem proofs V52 [ Honestolcott
|    |                                            `* Concise refutation of halting problem proofs V52 [ HonestRichard Damon
|    |                                             `* Concise refutation of halting problem proofs V52 [ Honestolcott
|    |                                              `* Concise refutation of halting problem proofs V52 [ HonestRichard Damon
|    |                                               `* Concise refutation of halting problem proofs V52 [ Honestolcott
|    |                                                `* Concise refutation of halting problem proofs V52 [ HonestRichard Damon
|    |                                                 `* Concise refutation of halting problem proofs V52 [ Honestolcott
|    |                                                  `* Concise refutation of halting problem proofs V52 [ HonestRichard Damon
|    |                                                   `* Concise refutation of halting problem proofs V52 [ Honestolcott
|    |                                                    `* Concise refutation of halting problem proofs V52 [ HonestRichard Damon
|    |                                                     `* Concise refutation of halting problem proofs V52 [ Honestolcott
|    |                                                      `* Concise refutation of halting problem proofs V52 [ HonestRichard Damon
|    |                                                       `* Concise refutation of halting problem proofs V52 [ Honestolcott
|    |                                                        `* Concise refutation of halting problem proofs V52 [ HonestRichard Damon
|    |                                                         `* Concise refutation of halting problem proofs V52 [ Honestolcott
|    |                                                          `* Concise refutation of halting problem proofs V52 [ HonestRichard Damon
|    |                                                           `* Concise refutation of halting problem proofs V52 [ Honestolcott
|    |                                                            `* Concise refutation of halting problem proofs V52 [ HonestRichard Damon
|    |                                                             `* Concise refutation of halting problem proofs V52 [ Honestolcott
|    |                                                              `* Concise refutation of halting problem proofs V52 [ HonestRichard Damon
|    |                                                               `* Concise refutation of halting problem proofs V52 [ Honestolcott
|    |                                                                `* Concise refutation of halting problem proofs V52 [ HonestRichard Damon
|    |                                                                 `* Concise refutation of halting problem proofs V52 [ Honestolcott
|    |                                                                  `* Concise refutation of halting problem proofs V52 [ HonestRichard Damon
|    |                                                                   `* Concise refutation of halting problem proofs V52 [ Honestolcott
|    |                                                                    `* Concise refutation of halting problem proofs V52 [ HonestRichard Damon
|    |                                                                     `* Concise refutation of halting problem proofs V52 [ Honestolcott
|    |                                                                      `* Concise refutation of halting problem proofs V52 [ HonestRichard Damon
|    |                                                                       `* Concise refutation of halting problem proofs V52 [ Honestolcott
|    |                                                                        `* Concise refutation of halting problem proofs V52 [ HonestRichard Damon
|    |                                                                         `* Concise refutation of halting problem proofs V52 [ Honestolcott
|    `* Concise refutation of halting problem proofs V52 [ Ignorant orolcott
`- Concise refutation of halting problem proofs V52 [ Linz Proof ]Steve

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Re: Concise refutation of halting problem proofs V59 [ ignorance about halt deciders ]

<KbkKJ.672$Tr18.91@fx42.iad>

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Subject: Re: Concise refutation of halting problem proofs V59 [ ignorance
about halt deciders ]
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References: <ssh8vu$4c0$1@dont-email.me>
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From: Rich...@Damon-Family.org (Richard Damon)
In-Reply-To: <AJidnaw87NhNLGT8nZ2dnUU7-U3NnZ2d@giganews.com>
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 by: Richard Damon - Wed, 2 Feb 2022 00:25 UTC

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 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:
>>>>>>> On 1/31/2022 11:25 PM, Richard Damon wrote:
>>>>>>>>
>>>>>>>> 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:
>>>>>>>>>>>>> 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^>
>>>>>>>>>>> No that is not it. That is like saying "by definition"
>>>>>>>>>>> Sum(3,5) is
>>>>>>>>>>> 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.
>>>>>>> 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.
>>>>>>
>>>>>> Sounds like a NDTM.
>>>>> 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)
>>>>
>>>> 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
>>
>> 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.


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

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From: Rich...@Damon-Family.org (Richard Damon)
In-Reply-To: <jrWdnZmv2oZ9UWT8nZ2dnUU7-QvNnZ2d@giganews.com>
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Date: Tue, 1 Feb 2022 19:27:53 -0500
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 by: Richard Damon - Wed, 2 Feb 2022 00:27 UTC

On 2/1/22 7:14 PM, olcott wrote:
> On 2/1/2022 5:57 PM, Richard Damon wrote:
>> 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 1/31/2022 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:
>>>>>>>>> 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^>
>>>>>>> No that is not it. That is like saying "by definition" Sum(3,5)
>>>>>>> is 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.
>>>
>>> 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).


Click here to read the complete article
Re: Concise refutation of halting problem proofs V59 [ key essence ]

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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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From: Rich...@Damon-Family.org (Richard Damon)
In-Reply-To: <H7mdnTXm59-szWT8nZ2dnUU7-bvNnZ2d@giganews.com>
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 by: Richard Damon - Wed, 2 Feb 2022 00:33 UTC

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 1/31/2022 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:
>>>>>>> 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^>
>>>>> No that is not it. That is like saying "by definition" Sum(3,5) is
>>>>> 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.
>
> 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.

It is NOT vague. ALL computations by its definition will either be
HALTING or NON-HALTING.

Care to show one that isn't

>
> 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.

And the actual behavior specified by this input is EXACTLY the behavior
of the machine represented applied to the second part of the 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.

If you mean that UTM(wM, w) behaves exactly the same as M applied to w,
yes. But note that is actually the definition of the UTM, and any
simulation that doesn't match that definition is NOT any accurate
'simulation'

>
> 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.
>

Exceept that if the simulation the Halt Decider does doesn't EXACT match
the behavior of the machine the input represents. and that means
running forever it the input doesn't halt, it is NOT a 'UTM'.

DEFINITION.

FAIL.

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

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 by: olcott - Wed, 2 Feb 2022 00:33 UTC

On 2/1/2022 6:24 PM, Richard Damon wrote:
> 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:
>>>>>> On 1/31/2022 11:25 PM, Richard Damon wrote:
>>>>>>>
>>>>>>> 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:
>>>>>>>>>>>> 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^>
>>>>>>>>>> No that is not it. That is like saying "by definition"
>>>>>>>>>> Sum(3,5) is
>>>>>>>>>> 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.
>>>>>> 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.
>>>>>
>>>>> Sounds like a NDTM.
>>>> 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)
>>>
>>> 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.
>


Click here to read the complete article
Re: Concise refutation of halting problem proofs V52 [ error or dishonesty ]

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From: agis...@gm.invalid (André G. Isaak)
Newsgroups: comp.theory
Subject: Re: Concise refutation of halting problem proofs V52 [ error or
dishonesty ]
Date: Tue, 1 Feb 2022 17:58:22 -0700
Organization: Christians and Atheists United Against Creeping Agnosticism
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 by: André G. Isaak - Wed, 2 Feb 2022 00:58 UTC

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.

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

André

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

Re: Concise refutation of halting problem proofs V59 [ ignorance about halt deciders ]

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 by: olcott - Wed, 2 Feb 2022 01:03 UTC

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 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:
>>>>>>>> On 1/31/2022 11:25 PM, Richard Damon wrote:
>>>>>>>>>
>>>>>>>>> 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:
>>>>>>>>>>>>>> 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^>
>>>>>>>>>>>> No that is not it. That is like saying "by definition"
>>>>>>>>>>>> Sum(3,5) is
>>>>>>>>>>>> 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.
>>>>>>>> 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.
>>>>>>>
>>>>>>> Sounds like a NDTM.
>>>>>> 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)
>>>>>
>>>>> 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
>>>
>>> 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.


Click here to read the complete article
Re: Concise refutation of halting problem proofs V52 [ ignorance about deciders ]

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 by: olcott - Wed, 2 Feb 2022 01:33 UTC

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.

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

Re: Concise refutation of halting problem proofs V59 [ ignorance about halt deciders ]

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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Wed, 2 Feb 2022 01:40 UTC

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 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:
>>>>>>>>> On 1/31/2022 11:25 PM, Richard Damon wrote:
>>>>>>>>>>
>>>>>>>>>> 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:
>>>>>>>>>>>>>>> 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^>
>>>>>>>>>>>>> No that is not it. That is like saying "by definition"
>>>>>>>>>>>>> Sum(3,5) is
>>>>>>>>>>>>> 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.
>>>>>>>>> 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.
>>>>>>>>
>>>>>>>> Sounds like a NDTM.
>>>>>>> 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)
>>>>>>
>>>>>> 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
>>>>
>>>> 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 ?


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

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References: <ssh8vu$4c0$1@dont-email.me> <ecFJJ.19021$mS1.7877@fx10.iad>
<sMCdnTPlr-FDvWr8nZ2dnUU7-KXNnZ2d@giganews.com>
<7FFJJ.29151$541.18496@fx35.iad> <st7a2e$oo$1@dont-email.me>
<ibHJJ.56320$u41.55552@fx41.iad>
<hK-dnaKCNvKd2Wr8nZ2dnUU7-fPNnZ2d@giganews.com>
<gIHJJ.29153$541.4042@fx35.iad> <st91ek$p4g$1@dont-email.me>
<st9fn6$60s$2@gioia.aioe.org> <RqidnSdLIdwH2GX8nZ2dnUU7-SXNnZ2d@giganews.com>
<Kk%JJ.20609$OF3.19827@fx14.iad> <staa42$dgq$1@dont-email.me>
<wv2KJ.25296$tW.1549@fx39.iad>
<b_SdnVRGB-GdK2X8nZ2dnUU7-YPNnZ2d@giganews.com>
<4L2KJ.23685$jb1.8458@fx46.iad>
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<H7mdnTXm59-szWT8nZ2dnUU7-bvNnZ2d@giganews.com>
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<1eqdnSCcgMlI4GT8nZ2dnUU7-LPNnZ2d@giganews.com>
<4e7f9c66-8852-4bb7-b913-e94a1a174120n@googlegroups.com>
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<k5Kdndk6xuzpTGT8nZ2dnUU7-R_NnZ2d@giganews.com>
From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Wed, 2 Feb 2022 01:42 UTC

On 2/1/22 7:33 PM, olcott wrote:
> On 2/1/2022 6:24 PM, Richard Damon wrote:
>> 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:
>>>>>>> On 1/31/2022 11:25 PM, Richard Damon wrote:
>>>>>>>>
>>>>>>>> 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:
>>>>>>>>>>>>> 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^>
>>>>>>>>>>> No that is not it. That is like saying "by definition"
>>>>>>>>>>> Sum(3,5) is
>>>>>>>>>>> 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.
>>>>>>> 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.
>>>>>>
>>>>>> Sounds like a NDTM.
>>>>> 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)
>>>>
>>>> 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:


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

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References: <ssh8vu$4c0$1@dont-email.me> <osEJJ.11004$uP.10312@fx16.iad>
<9P6dnTtqj-DZhmr8nZ2dnUU7-VnNnZ2d@giganews.com>
<ecFJJ.19021$mS1.7877@fx10.iad>
<sMCdnTPlr-FDvWr8nZ2dnUU7-KXNnZ2d@giganews.com>
<7FFJJ.29151$541.18496@fx35.iad> <st7a2e$oo$1@dont-email.me>
<ibHJJ.56320$u41.55552@fx41.iad>
<hK-dnaKCNvKd2Wr8nZ2dnUU7-fPNnZ2d@giganews.com>
<gIHJJ.29153$541.4042@fx35.iad> <st91ek$p4g$1@dont-email.me>
<st9fn6$60s$2@gioia.aioe.org> <RqidnSdLIdwH2GX8nZ2dnUU7-SXNnZ2d@giganews.com>
<Kk%JJ.20609$OF3.19827@fx14.iad> <staa42$dgq$1@dont-email.me>
<wv2KJ.25296$tW.1549@fx39.iad>
<b_SdnVRGB-GdK2X8nZ2dnUU7-YPNnZ2d@giganews.com>
<4L2KJ.23685$jb1.8458@fx46.iad>
<rv2dnc__PYfMJ2X8nZ2dnUU7-WHNnZ2d@giganews.com>
<Pu3KJ.19025$mS1.14927@fx10.iad>
<H7mdnTXm59-szWT8nZ2dnUU7-bvNnZ2d@giganews.com>
<493efce8-cf20-4271-8f47-2858fa3812efn@googlegroups.com>
<1eqdnSCcgMlI4GT8nZ2dnUU7-LPNnZ2d@giganews.com> <57kKJ.670$Tr18.460@fx42.iad>
<XqCdnXK1E9aPUmT8nZ2dnUU7-fGdnZ2d@giganews.com>
From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Wed, 2 Feb 2022 01:45 UTC

On 2/1/22 7:23 PM, olcott wrote:
> On 2/1/2022 6:20 PM, Richard Damon wrote:
>> 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:
>>>>> On 1/31/2022 11:25 PM, Richard Damon wrote:
>>>>>>
>>>>>> 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:
>>>>>>>>>>> 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^>
>>>>>>>>> No that is not it. That is like saying "by definition" Sum(3,5) is
>>>>>>>>> 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.
>>>>> 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.
>>>> Sounds like a NDTM.
>>>
>>> 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).
>
>


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Re: Concise refutation of halting problem proofs V59 [ self-evident truth ]

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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Wed, 2 Feb 2022 01:47 UTC

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 Wednesday, 2 February 2022 at 05:36:39 UTC+8, 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:
>>>>>>>>>> On 1/31/2022 11:25 PM, Richard Damon wrote:
>>>>>>>>>>>
>>>>>>>>>>> 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:
>>>>>>>>>>>>>>>> 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^>
>>>>>>>>>>>>>> No that is not it. That is like saying "by definition"
>>>>>>>>>>>>>> Sum(3,5) is
>>>>>>>>>>>>>> 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.
>>>>>>>>>> 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.
>>>>>>>>>
>>>>>>>>> Sounds like a NDTM.
>>>>>>>> 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)
>>>>>>>
>>>>>>> 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
>>>>>
>>>>> 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.
>


Click here to read the complete article
Re: Concise refutation of halting problem proofs V52 [ ignorance about deciders ]

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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Wed, 2 Feb 2022 01:51 UTC

On 2/1/22 8:33 PM, 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.

And you are also wrong about how deciders wrok, because to be a correct
decider, the answer it gives must match the mapping that it is supposed
to be computing.

H <H^> <H^> BY DEFINITION is asking if H^ applied to <H^> Halts.

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

Doesn't matter. embedded_H isn't a REAL UTM.

UTM(<H^>,<H^>) will see H^ -> to H^.Qn if H(<H^>,<H^>) -> H.Qn, and if H
doesn't then it doesn't matter as H can't be 'correct' with an answwer
it doesn't give.

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

WRONG. You LIE, since you claim H -> H.Qn, that means it is NOT a real
UTM, and thus its beavior of NOT the definition of Halting.

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

WRONG, not 'Their' correct simulation, but by THE correct simulation,
which is done by a REAL UTM, which they will not be.

FAIL.

>
> 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⟩.
>

<H^> is not a machine, so does not have 'behavior'

The behavior that H(<H^>,<H^>) is accountable to is the behavior of
UTM(<H^>,H^>) BY DEFINITION.

FAIL

>
>
> 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
>>
>
>

Re: Concise refutation of halting problem proofs V59 [ self-evident truth ]

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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Wed, 2 Feb 2022 01:55 UTC

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 Wednesday, 2 February 2022 at 05:36:39 UTC+8, 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:
>>>>>>>>>>> On 1/31/2022 11:25 PM, Richard Damon wrote:
>>>>>>>>>>>>
>>>>>>>>>>>> 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:
>>>>>>>>>>>>>>>>> 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^>
>>>>>>>>>>>>>>> No that is not it. That is like saying "by definition"
>>>>>>>>>>>>>>> Sum(3,5) is
>>>>>>>>>>>>>>> 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.
>>>>>>>>>>> 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.
>>>>>>>>>>
>>>>>>>>>> Sounds like a NDTM.
>>>>>>>>> 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)
>>>>>>>>
>>>>>>>> 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
>>>>>>
>>>>>> 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.
>


Click here to read the complete article
Re: Concise refutation of halting problem proofs V52 [ ignorance about deciders ]

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From: agis...@gm.invalid (André G. Isaak)
Newsgroups: comp.theory
Subject: Re: Concise refutation of halting problem proofs V52 [ ignorance
about deciders ]
Date: Tue, 1 Feb 2022 19:08:49 -0700
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 by: André G. Isaak - Wed, 2 Feb 2022 02:08 UTC

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.

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

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

And the specification for a halt decider states that given inputs ⟨M⟩
and I it accepts the input if and only if M applied to I halts. It
doesn't make any mention of simulation by embedded_H transitioning to
⟨Ĥ⟩.qn. That's your nonsense, not the actual definition of 'halt decider'.

> 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⟩.

Its input *represents* the computation that it is contained within.
Therefore that is the computation it must answer about.

A halt decider accepts inputs ⟨M⟩ and I if and only if M applied to I
halts. That's a *DEFINITION*, and a completely noncontroversial one. The
absence of brackets around M in the second part is intentional, and you
will not find any definition of halt decider anywhere that states otherwise.

You can't "solve" the halting problem by redefining 'halt decider'.

André

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

Re: Concise refutation of halting problem proofs V59 [ self-evident truth ]

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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Wed, 2 Feb 2022 02:14 UTC

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 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 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:
>>>>>>>>>>>> On 1/31/2022 11:25 PM, Richard Damon wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>> 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:
>>>>>>>>>>>>>>>>>> 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^>
>>>>>>>>>>>>>>>> No that is not it. That is like saying "by definition"
>>>>>>>>>>>>>>>> Sum(3,5) is
>>>>>>>>>>>>>>>> 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.
>>>>>>>>>>>> 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.
>>>>>>>>>>>
>>>>>>>>>>> Sounds like a NDTM.
>>>>>>>>>> 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)
>>>>>>>>>
>>>>>>>>> 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
>>>>>>>
>>>>>>> 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.
>


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Re: Concise refutation of halting problem proofs V59 [ self-evident truth ]

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From: Rich...@Damon-Family.org (Richard Damon)
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Date: Tue, 1 Feb 2022 21:21:45 -0500
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 by: Richard Damon - Wed, 2 Feb 2022 02:21 UTC

On 2/1/22 9:14 PM, olcott wrote:
> 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 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 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:
>>>>>>>>>>>>> On 1/31/2022 11:25 PM, Richard Damon wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> 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:
>>>>>>>>>>>>>>>>>>> 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^>
>>>>>>>>>>>>>>>>> No that is not it. That is like saying "by definition"
>>>>>>>>>>>>>>>>> Sum(3,5) is
>>>>>>>>>>>>>>>>> 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.
>>>>>>>>>>>>> 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.
>>>>>>>>>>>>
>>>>>>>>>>>> Sounds like a NDTM.
>>>>>>>>>>> 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)
>>>>>>>>>>
>>>>>>>>>> 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
>>>>>>>>
>>>>>>>> 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.
>


Click here to read the complete article
Re: Concise refutation of halting problem proofs V52 [ dodgy double talk ]

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 by: olcott - Wed, 2 Feb 2022 02:37 UTC

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

Re: Concise refutation of halting problem proofs V59 [ self-evident truth ]

<vsudneA4J4TAcmT8nZ2dnUU7-IHNnZ2d@giganews.com>

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 by: olcott - Wed, 2 Feb 2022 02:41 UTC

On 2/1/2022 8:21 PM, Richard Damon wrote:
> On 2/1/22 9:14 PM, olcott wrote:
>> 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 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 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:
>>>>>>>>>>>>>> On 1/31/2022 11:25 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> 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:
>>>>>>>>>>>>>>>>>>>> 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^>
>>>>>>>>>>>>>>>>>> No that is not it. That is like saying "by definition"
>>>>>>>>>>>>>>>>>> Sum(3,5) is
>>>>>>>>>>>>>>>>>> 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.
>>>>>>>>>>>>>> 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.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Sounds like a NDTM.
>>>>>>>>>>>> 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)
>>>>>>>>>>>
>>>>>>>>>>> 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
>>>>>>>>>
>>>>>>>>> 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
You must be a liar.


Click here to read the complete article
Re: Concise refutation of halting problem proofs V52 [ dodgy double talk ]

<stcre2$o38$1@dont-email.me>

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From: agis...@gm.invalid (André G. Isaak)
Newsgroups: comp.theory
Subject: Re: Concise refutation of halting problem proofs V52 [ dodgy double
talk ]
Date: Tue, 1 Feb 2022 19:48:32 -0700
Organization: Christians and Atheists United Against Creeping Agnosticism
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 by: André G. Isaak - Wed, 2 Feb 2022 02:48 UTC

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⟩.

And your embedded_H is *not* a simulator.

If embedded_H were a simulator than it would accept ⟨M⟩ I iff M accepts
I, it would reject ⟨M⟩ I iff M rejects I, and it would fail to halt iff
M neither accepts nor rejects I (i.e. if M applied to I fails to halt).

Those are not the conditions under which embedded_H is supposed to
accept or reject inputs.

André

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

Re: Concise refutation of halting problem proofs V59 [ self-evident truth ]

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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Wed, 2 Feb 2022 02:48 UTC

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:
>>> 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 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 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:
>>>>>>>>>>>>>>> On 1/31/2022 11:25 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> 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:
>>>>>>>>>>>>>>>>>>>>> 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^>
>>>>>>>>>>>>>>>>>>> No that is not it. That is like saying "by
>>>>>>>>>>>>>>>>>>> definition" Sum(3,5) is
>>>>>>>>>>>>>>>>>>> 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.
>>>>>>>>>>>>>>> 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.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Sounds like a NDTM.
>>>>>>>>>>>>> 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)
>>>>>>>>>>>>
>>>>>>>>>>>> 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
>>>>>>>>>>
>>>>>>>>>> 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
> You must be a liar.
>
> 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⟩...


Click here to read the complete article
Re: Concise refutation of halting problem proofs V52 [ dodgy double talk ]

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References: <ssh8vu$4c0$1@dont-email.me> <Ol0JJ.27499$541.4855@fx35.iad>
<875yq2h2ea.fsf@bsb.me.uk> <st62tu$f6h$1@dont-email.me>
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Wed, 2 Feb 2022 02:54 UTC

On 2/1/22 9:37 PM, olcott wrote:
> On 2/1/2022 8:08 PM, André G. Isaak wrote:
>> 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.
>

You don't understand the problem do you. The issue is that <H^> is NOT a
machine that can be applied to anything, but just a description of a
machine.

That is like pointing to a blueprint of a grocery store and trying to go
into it to buy a loaf of bread.

Also, the DEFINITION of the behavior of the input to H, if you want to
be talking about Halting Theory isn't the simulation of that input by H
itself, but by the actual exectution of the machine that the input
represents, which by the definiton of a UTM, means that simulation of
that input by a ACTUAL UTM. Note, a actual UTM will NOT stop it
simulation until it reachs a halting state, no matter how 'sure' it is
that the input won't halt, as the UTM of an input that represents a
non-halting machine will be non-halting by definition.

Thus, if H-> H.Qn when it thinks its input is non-halting, that itself
becomes proof that H is not a UTM, and thus it isn't H's simulation that
matters.

FAIL.

Re: Concise refutation of halting problem proofs V52 [ dodgy double talk ]

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References: <ssh8vu$4c0$1@dont-email.me> <875yq2h2ea.fsf@bsb.me.uk>
<st62tu$f6h$1@dont-email.me> <LCwJJ.50318$gX.12924@fx40.iad>
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Wed, 2 Feb 2022 02:57 UTC

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.

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

Re: Concise refutation of halting problem proofs V52 [ dodgy double talk ]

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 by: Richard Damon - Wed, 2 Feb 2022 03:04 UTC

On 2/1/22 9:57 PM, 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.
>
> 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.
>

Except that the definition of the Halting Problem (combined with the
definition of a UTM) says that it is only the simulation of the input to
H by an actual UTM that matters to determine if the input represents a
Halting Computation or not.

If H aborts its simulation to answer, then H isn't a UTM, so its
simulation doesn't actually determine if the computation halts.

And, it has been shown that if H is actually just a UTM and doesn't
abort its simulation, then yes, WE can see that H^ applied to <H^> will
be non-halting, but we also see that because H doesn't abort its
simulation, it can't transition to H.Qn, so fails to be a decider in
this case.

You need to find a machine that runs forever but also halts in finite
time at H.Qn.

This is a Fairy Dust Powered Unicorn, it just doesn't exist.

You FAIL by the actual meaning of the words.

Re: Concise refutation of halting problem proofs V52 [ dodgy double talk ]

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 by: olcott - Wed, 2 Feb 2022 03:08 UTC

On 2/1/2022 9:04 PM, Richard Damon wrote:
> On 2/1/22 9:57 PM, 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.
>>
>> 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.
>>
>
> Except that the definition of the Halting Problem (combined with the
> definition of a UTM) says that it is only the simulation of the input to
> H by an actual UTM that matters to determine if the input represents a
> Halting Computation or not.
embedded_H is exactly a UTM with 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

Re: Concise refutation of halting problem proofs V59 [ self-evident truth ]

<KNqdnfmECvWLZGT8nZ2dnUU7-dfNnZ2d@giganews.com>

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 by: olcott - Wed, 2 Feb 2022 03:23 UTC

On 2/1/2022 8:48 PM, Richard Damon wrote:
> 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:
>>>> 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 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 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:
>>>>>>>>>>>>>>>> On 1/31/2022 11:25 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> 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:
>>>>>>>>>>>>>>>>>>>>>> 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^>
>>>>>>>>>>>>>>>>>>>> No that is not it. That is like saying "by
>>>>>>>>>>>>>>>>>>>> definition" Sum(3,5) is
>>>>>>>>>>>>>>>>>>>> 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.
>>>>>>>>>>>>>>>> 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.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Sounds like a NDTM.
>>>>>>>>>>>>>> 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)
>>>>>>>>>>>>>
>>>>>>>>>>>>> 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
>>>>>>>>>>>
>>>>>>>>>>> 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
>> You must be a liar.
>>
>> 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.


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