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

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 V52 [ error or dishonesty ]

<Kk%JJ.20609$OF3.19827@fx14.iad>

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Subject: Re: Concise refutation of halting problem proofs V52 [ error or
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<875yq2h2ea.fsf@bsb.me.uk> <st62tu$f6h$1@dont-email.me>
<LCwJJ.50318$gX.12924@fx40.iad>
<UK-dnQx29oAWMmv8nZ2dnUU7-WvNnZ2d@giganews.com> <_bzJJ.7760$rU.4222@fx34.iad>
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<7FFJJ.29151$541.18496@fx35.iad> <st7a2e$oo$1@dont-email.me>
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<st9fn6$60s$2@gioia.aioe.org> <RqidnSdLIdwH2GX8nZ2dnUU7-SXNnZ2d@giganews.com>
From: Rich...@Damon-Family.org (Richard Damon)
In-Reply-To: <RqidnSdLIdwH2GX8nZ2dnUU7-SXNnZ2d@giganews.com>
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Date: Mon, 31 Jan 2022 19:41:14 -0500
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 by: Richard Damon - Tue, 1 Feb 2022 00:41 UTC

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.

H wM w -> H.Qy/H.Qn depending of if M w Halts or Not.

You don't seem to undestand this sentence, so NOTHING you have done
means anything.

Note, it DOESN'T say based on the simulation by H of its input.

It DOESN'T say if it halts in some limited number of N steps.

So, YOU FAIL.

Re: Concise refutation of halting problem proofs V52 [ error or dishonesty ]

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<Ol0JJ.27499$541.4855@fx35.iad> <875yq2h2ea.fsf@bsb.me.uk>
<st62tu$f6h$1@dont-email.me> <LCwJJ.50318$gX.12924@fx40.iad>
<UK-dnQx29oAWMmv8nZ2dnUU7-WvNnZ2d@giganews.com> <_bzJJ.7760$rU.4222@fx34.iad>
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<7FFJJ.29151$541.18496@fx35.iad> <st7a2e$oo$1@dont-email.me>
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<gIHJJ.29153$541.4042@fx35.iad> <st946r$oq0$1@dont-email.me>
From: Rich...@Damon-Family.org (Richard Damon)
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Date: Mon, 31 Jan 2022 19:46:11 -0500
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 by: Richard Damon - Tue, 1 Feb 2022 00:46 UTC

On 1/31/22 11:53 AM, olcott wrote:
> On 1/30/2022 8:20 PM, Richard Damon wrote:
>> On 1/30/22 9:05 PM, olcott wrote:
>>> On 1/30/2022 7:45 PM, Richard Damon wrote:
>>>> On 1/30/22 7:21 PM, olcott wrote:
>>>>> On 1/30/2022 6:01 PM, Richard Damon wrote:
>>>>>> On 1/30/22 6:35 PM, olcott wrote:
>>>>>>> On 1/30/2022 5:30 PM, Richard Damon wrote:
>>>>>>>> On 1/30/22 6:12 PM, olcott wrote:
>>>>>>>>> On 1/30/2022 4:39 PM, Richard Damon wrote:
>>>>>>>>>> On 1/30/22 4:21 PM, olcott wrote:
>>>>>>>>>>> On 1/30/2022 2:54 PM, Richard Damon wrote:
>>>>>>>>>>>> On 1/30/22 3:09 PM, olcott wrote:
>>>>>>>>>>>
>>>>>>>>>>>>> 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 can
>>>>>>>>>>>>> possibly reach the final state of this simulated input in
>>>>>>>>>>>>> any finite number of steps.
>>>>>>>>>>>>>
>>>>>>>>>>>>> It is like you put a guard on the front door that is
>>>>>>>>>>>>> supposed to report anyone coming in the front door (the
>>>>>>>>>>>>> actual inputs). Then someone comes in the back door (non
>>>>>>>>>>>>> inputs) and the guard does not report this. Since the guard
>>>>>>>>>>>>> is only supposed to report people coming in the front door
>>>>>>>>>>>>> it is incorrect to say that the guard made a mistake by not
>>>>>>>>>>>>> reporting people that came in the back door.
>>>>>>>>>>>>>
>>>>>>>>>>>>> embedded_H is not supposed to report on the halt status of
>>>>>>>>>>>>> the computation that it is contained within: Ĥ applied to ⟨Ĥ⟩.
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> So, you have just admitted that you aren't working on the
>>>>>>>>>>>> Halting Problem, so any claims therein are just lies.
>>>>>>>>>>>>
>>>>>>>>>>>> Since the definition of the Halting Problem refers to the
>>>>>>>>>>>> ACTUAL behavior of the machine the input represents, and NOT
>>>>>>>>>>>> the partial simulation that some simulating halt decider
>>>>>>>>>>>> might do, you are admitting that you H is NOT using the
>>>>>>>>>>>> Halting Problem definition and thus your claims that your
>>>>>>>>>>>> results apply to the Halting problem are just lies.
>>>>>>>>>>>>
>>>>>>>>>>>> For the Halting Problem, the correct results for the inputs
>>>>>>>>>>>> is based on the actual behavior of the machine, or its
>>>>>>>>>>>> equivalent the simulation of the input with a REAL UTM. Thus
>>>>>>>>>>>> the 'Front Door' to the problem s based on that, so either
>>>>>>>>>>>> you your guards are lying or, what seems more likely, you
>>>>>>>>>>>> posted them to the wrong door.
>>>>>>>>>>>>
>>>>>>>>>>>> You have basically just proved that you have totally wasted
>>>>>>>>>>>> the last years of your life, as you have been working on the
>>>>>>>>>>>> wrong problem, because you just don't understand what the
>>>>>>>>>>>> problem you wanted to solve actually was.
>>>>>>>>>>>>
>>>>>>>>>>>> FAIL.
>>>>>>>>>>>
>>>>>>>>>>> Sum(int X, int Y) { return X + Y );
>>>>>>>>>>>
>>>>>>>>>>> It is true that halt deciders must report on the actual
>>>>>>>>>>> behavior of their actual inputs in the same way that Sum(2,5)
>>>>>>>>>>> must return 7.
>>>>>>>>>>
>>>>>>>>>> Right, and the correct answer for if H(wM, w) should report
>>>>>>>>>> halting is if M x will reach a final state in a finite number
>>>>>>>>>> of steps. This is identical to if UTM(wM, w) will halt. Dosn't
>>>>>>>>>> matter what you think otherwise, that IS the definition of the
>>>>>>>>>> actual behavior.
>>>>>>>>>>
>>>>>>>>>> It is NOT something based on the partial simulation that H does.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> The you cannot understand how all kinds of infinite behavior
>>>>>>>>> patterns can be easily recognized in a finite number of steps
>>>>>>>>> is not any mistake on my part:
>>>>>>>>
>>>>>>>> Yes, MANY can, but not ALL.
>>>>>>>>
>>>>>>>> If you need to change the definition, then you are not working
>>>>>>>> on the halting problem.
>>>>>>>>
>>>>>>>>
>>>>>>>
>>>>>>> I don't have to change the definition I merely make it much more
>>>>>>> precise:
>>>>>>
>>>>>> Except that the original definition IS exactly precise. The is a
>>>>>> single WELL DEFINED answer for any instance of the question. The
>>>>>> fact that you see some abiguity just shows you don't really
>>>>>> understand the field.
>>>>>>
>>>>>>>
>>>>>>> (1) Halting is defined as reaching a final state.
>>>>>>
>>>>>> But you change the 'of what'.
>>>>>
>>>>> A directly executed TM halts when it reaches the final state of
>>>>> this directly executed TM.
>>>>>
>>>>> A simulated TM description halts when the simulated TM description
>>>>> reaches it final state.
>>>>>
>>>>
>>>> Right, but if the simulator isn't a real UTM and stops simulating,
>>>> the 'pure simulation' continues until it either halts or runs for ever.
>>>>
>>>
>>> H.q0 Wm W ⊢* H.qy
>>> iff UTM Wm W reaches its final state
>>>
>>> H.q0 Wm W ⊢* H.qn
>>> iff UTM Wm W never reaches its final state
>>
>>
>> Right, and that is the REAL UTM, not H playing one on TV and stopping
>> when it thinks it has an answer.
> As your scatterbrained mind keep repeating....
> Yet I have kept correcting. It is not necessary for a simulating halt
> decider to execute an infinite number of steps of an infinite sequence
> of configurations for it to determine that a pure simulation of its
> input would never stop running.
>
> _Infinite_Loop()
> [00000946](01)  55              push ebp
> [00000947](02)  8bec            mov ebp,esp
> [00000949](02)  ebfe            jmp 00000949
> [0000094b](01)  5d              pop ebp
> [0000094c](01)  c3              ret
> Size in bytes:(0007) [0000094c]
>
> It is self-evident that the second time the instruction at [00000949] is
> executed with no other instructions inbetween an infinite loop has been
> recognized.
>
>
>
> _Infinite_Recursion()
> [00000926](01)  55              push ebp
> [00000927](02)  8bec            mov ebp,esp
> [00000929](03)  8b4508          mov eax,[ebp+08]
> [0000092c](01)  50              push eax
> [0000092d](05)  e8f4ffffff      call 00000926
> [00000932](03)  83c404          add esp,+04
> [00000935](01)  5d              pop ebp
> [00000936](01)  c3              ret
> Size in bytes:(0017) [00000936]
>
> Begin Local Halt Decider Simulation   Execution Trace Stored at:211356
>
>  machine   stack     stack     machine    assembly
>  address   address   data      code       language
>  ========  ========  ========  =========  =============
> [00000926][00211342][00211346] 55         push ebp
> [00000927][00211342][00211346] 8bec       mov  ebp,esp
> [00000929][00211342][00211346] 8b4508     mov  eax,[ebp+08]
> [0000092c][0021133e][00000777] 50         push eax
> [0000092d][0021133a][00000932] e8f4ffffff call 00000926
> [00000926][00211336][00211342] 55         push ebp
> [00000927][00211336][00211342] 8bec       mov  ebp,esp
> [00000929][00211336][00211342] 8b4508     mov  eax,[ebp+08]
> [0000092c][00211332][00000777] 50         push eax
> [0000092d][0021132e][00000932] e8f4ffffff call 00000926
> Local Halt Decider: Infinite Recursion Detected Simulation Stopped
>
> It is self-evident that the second time the instruction at [0000092d] is
> executed with with the same input infinite recursion has been recognized.
>
>
RED HERRING.


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

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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Tue, 1 Feb 2022 00:48 UTC

On 1/31/22 11:06 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.
>
>

Right, but the BEHAVIOR of H^ applied to <H^> IS the criteria that H
needs to decide on.

H wM w -> H.Qy/H.Qn based on whether M w Halts or Not.

wM w is the input to H

The behavior of M w is what matters.

FAIL.

Re: Concise refutation of halting problem proofs V52 [ dishonesty ! ]

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Subject: Re: Concise refutation of halting problem proofs V52 [ dishonesty ! ]
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 by: olcott - Tue, 1 Feb 2022 03:40 UTC

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)?

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

embedded_H is only accountable for the actual behavior specified by its
actual inputs ⟨Ĥ⟩ ⟨Ĥ⟩.

Since the simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ at Ĥ.qx cannot possibly transition to
its final state: ⟨Ĥ⟩.qn we know that this input never halts even if it
stops running.

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

--
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 [ dishonesty ! ]

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<LCwJJ.50318$gX.12924@fx40.iad>
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Tue, 1 Feb 2022 04:17 UTC

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^>

Mayby you just don't understand what the line in Linz means when he says
that H applied to wM w must go to H.Qy if M w Halt.

Because if you did, you would understand what is required of H.

Hint, you can match M to H^, wm to <H^> and w to <H^> too, and get the
requirements for this computation.

I guess that you claim it isn't computing that requirement, you are just
admitting that you aren't working on the Halting Problem.
>
> embedded_H is only accountable for the actual behavior specified by its
> actual inputs ⟨Ĥ⟩ ⟨Ĥ⟩.

Which by the Halting Problem definition is H^ applied to <H^>

>
> Since the simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ at Ĥ.qx cannot possibly transition to
> its final state: ⟨Ĥ⟩.qn we know that this input never halts even if it
> stops running.

But it DOES if H <H^> <H^> transitions to H.Qn.

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

Right, the TURING MACHINE H^ will halt if H Halts in H.Qn.

I guess you are just pointing out that your H doesn't halt and thus
fails to be a decider.

That or you are just lying about following the requirements of the
Halting Problem.

So, your whole proof goes up in a puff of greasy black smoke as it is
shown that you aren't doing what you are claiming to be doing, so
NOTHING YOU HAVE DONE MATTERS.

If you are doing the Halting Problem, nobody cares what you prove about
your POOP.

Re: Concise refutation of halting problem proofs V52 [ ignorance? ]

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<UK-dnQx29oAWMmv8nZ2dnUU7-WvNnZ2d@giganews.com> <_bzJJ.7760$rU.4222@fx34.iad>
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<d5AJJ.57716$4C3.3626@fx13.iad>
<g6WdndvEcI0PeWv8nZ2dnUU7-VPNnZ2d@giganews.com>
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<7FFJJ.29151$541.18496@fx35.iad> <st7a2e$oo$1@dont-email.me>
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<gIHJJ.29153$541.4042@fx35.iad> <st91ek$p4g$1@dont-email.me>
<st9fn6$60s$2@gioia.aioe.org> <RqidnSdLIdwH2GX8nZ2dnUU7-SXNnZ2d@giganews.com>
<Kk%JJ.20609$OF3.19827@fx14.iad> <staa42$dgq$1@dont-email.me>
<wv2KJ.25296$tW.1549@fx39.iad>
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 by: olcott - Tue, 1 Feb 2022 04:24 UTC

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).

Halt decider are deciders thus are only ever accountable for the
properties of their actual inputs.

The function that embedded_H computes is being asked about the actual
behavior of the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ at Ĥ.qx.

--
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 [ ignorance? ]

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<UK-dnQx29oAWMmv8nZ2dnUU7-WvNnZ2d@giganews.com> <_bzJJ.7760$rU.4222@fx34.iad>
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 by: Richard Damon - Tue, 1 Feb 2022 04:33 UTC

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?

That is the DEFINITION of a Halt Decider, just like your 'Sum' is asked
to provide the sum of its digits.

>
> Halt decider are deciders thus are only ever accountable for the
> properties of their actual inputs.
>
> The function that embedded_H computes is being asked about the actual
> behavior of the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ at Ĥ.qx.
>

WRONG <H^> can't be applied to anything. <H^> is just a string, it
doesn't do anything.

Note, look again at the definition of a Halt Ddcider.

H is applied to *wM w*

and that asks about *M applied to w*

Note, H is given the representation of the machine, and is being asked
about the behavior of the actual machine.

You are looking a level of 'indirection' (actual representation) in your
arguement.

The definition NEVER talks about 'simulation', only behavior of the
actual machine. You can get to simulation by the definiton of a UTM, but
that is just a secondary operation, and only works as long as you are
only thinking about REAL simulation by a UTM, which means no aborting
the simulation, since UTMs don't do that.

It is clear that you just don't understand either the Theorm or even the
Field it comes from that you are working on as everything you say is
full of errors.

FAIL.

Re: Concise refutation of halting problem proofs V52 [ ignorance? ]

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 by: olcott - Tue, 1 Feb 2022 04:42 UTC

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 ?

--
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 [ ignorance? ]

<Pu3KJ.19025$mS1.14927@fx10.iad>

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Subject: Re: Concise refutation of halting problem proofs V52 [ ignorance? ]
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Tue, 1 Feb 2022 05:25 UTC

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.

If you want to caim that your statement is 'equivalent', then it MUST
give exactly the same answer for the problem, as that is what equivalent
means.

The problem is you are trying to use the WRONG definition of
'simulation' and trying to hide it behind fancy words.

The only simulation that can be proved equivalent is the simulation by a
REAL UTM, which if H aborts its simulation to answer, it is not.

So you get into a delema, either your H can determine the right answer,
by never aborting and thus be able to use its simulation to see what
really happens, but then isn't able to give it (since it can't abort),
or it can't actually prove what it wants because it isn't the needed UTM.

So, you FAIL.

All you have proven is that you have just proven the claim that you
aren't working on the Halting Problem because you say you H isn't
responsible to give the answer REQUIRED by the definition.

You sloppy use of language has betrayed you and made you waste many
years of your life till you don't have enough left to do something with it.

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

<H7mdnTXm59-szWT8nZ2dnUU7-bvNnZ2d@giganews.com>

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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Tue, 1 Feb 2022 15:22 UTC

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.

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

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

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

--
Copyright 2021 Pete Olcott

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

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

<493efce8-cf20-4271-8f47-2858fa3812efn@googlegroups.com>

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Subject: Re: Concise refutation of halting problem proofs V59 [ key essence ]
From: wyni...@gmail.com (wij)
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 by: wij - Tue, 1 Feb 2022 16:33 UTC

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.

The Halting Problem has a definite, commonly recognized meaning. It refers to a
real machine, no ambiguity, no one can change it, not even Linz.
Your halt-problem is absolutely certain not Linz's, or of any? textbook.
Your claim contradicts experimental truth. Otherwise, show your x86utm operating
system proof. I guess you would say xxx thousands pages, I believe there are
only few lines are yours. Show your codes.

> The halting problem does not bother to mention the requirement that
> because all halt deciders are deciders they are only accountable for
> computing the mapping from their finite string inputs to an accept or
> reject state on the basis of the actual behavior specified by this input.
>
> The halting problem does not specifically examine simulating halt
> deciders, none-the-less the behavior of a correctly simulated machine
> description is known to be equivalent to the behavior of the direct
> execution of this same machine.
>
> Since a simulating halt decider is merely a UTM for simulated inputs
> that reach their final state when a simulating halt decider correctly
> determines that its simulated its input cannot possibly reach its final
> state this is complete proof that this simulated input never halts.
> --
> Copyright 2021 Pete Olcott
>
> Talent hits a target no one else can hit;
> Genius hits a target no one else can see.
> Arthur Schopenhauer

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

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https://www.novabbs.com/devel/article-flat.php?id=26167&group=comp.theory#26167

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From: polco...@gmail.com (olcott)
Newsgroups: comp.theory,comp.ai.philosophy,sci.logic,sci.math
Subject: Re: Concise refutation of halting problem proofs V59 [ key essence ]
Followup-To: comp.theory
Date: Tue, 1 Feb 2022 11:43:15 -0600
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 by: olcott - Tue, 1 Feb 2022 17:43 UTC

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.
>
> The Halting Problem has a definite, commonly recognized meaning. It refers to a
> real machine, no ambiguity, no one can change it, not even Linz.
> Your halt-problem is absolutely certain not Linz's, or of any? textbook.
> Your claim contradicts experimental truth. Otherwise, show your x86utm operating
> system proof. I guess you would say xxx thousands pages, I believe there are
> only few lines are yours. Show your codes.
>

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

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

>> The halting problem does not bother to mention the requirement that
>> because all halt deciders are deciders they are only accountable for
>> computing the mapping from their finite string inputs to an accept or
>> reject state on the basis of the actual behavior specified by this input.
>>
>> The halting problem does not specifically examine simulating halt
>> deciders, none-the-less the behavior of a correctly simulated machine
>> description is known to be equivalent to the behavior of the direct
>> execution of this same machine.
>>
>> Since a simulating halt decider is merely a UTM for simulated inputs
>> that reach their final state when a simulating halt decider correctly
>> determines that its simulated its input cannot possibly reach its final
>> state this is complete proof that this simulated input never halts.
>> --
>> Copyright 2021 Pete Olcott
>>
>> Talent hits a target no one else can hit;
>> Genius hits a target no one else can see.
>> Arthur Schopenhauer


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

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Subject: Re: Concise refutation of halting problem proofs V59 [ key essence ]
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 by: olcott - Tue, 1 Feb 2022 18:37 UTC

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

--
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 [ key essence ]

<36edc920-e3c1-41ee-8ef8-417be1e25603n@googlegroups.com>

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Subject: Re: Concise refutation of halting problem proofs V59 [ key essence ]
From: wyni...@gmail.com (wij)
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 by: wij - Tue, 1 Feb 2022 21:14 UTC

On Wednesday, 2 February 2022 at 01:43:20 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.
> >
> > The Halting Problem has a definite, commonly recognized meaning. It refers to a
> > real machine, no ambiguity, no one can change it, not even Linz.
> > Your halt-problem is absolutely certain not Linz's, or of any? textbook..
> > Your claim contradicts experimental truth. Otherwise, show your x86utm operating
> > system proof. I guess you would say xxx thousands pages, I believe there are
> > only few lines are yours. Show your codes.
> >
> You did not even read what I said before you claimed that it was
> incorrect. The general principles that I outlined below directly apply
> to the actual Linz proof:
>
> If I am incorrect in anything that I said below then the specific error
> could be pointed out.
> >> The halting problem does not bother to mention the requirement that
> >> because all halt deciders are deciders they are only accountable for
> >> computing the mapping from their finite string inputs to an accept or
> >> reject state on the basis of the actual behavior specified by this input.
> >>
> >> The halting problem does not specifically examine simulating halt
> >> deciders, none-the-less the behavior of a correctly simulated machine
> >> description is known to be equivalent to the behavior of the direct
> >> execution of this same machine.
> >>
> >> Since a simulating halt decider is merely a UTM for simulated inputs
> >> that reach their final state when a simulating halt decider correctly
> >> determines that its simulated its input cannot possibly reach its final
> >> state this is complete proof that this simulated input never halts.
> >> --
> >> Copyright 2021 Pete Olcott
> >>
> >> Talent hits a target no one else can hit;
> >> Genius hits a target no one else can see.
> >> Arthur Schopenhauer
>
>
> --
> Copyright 2021 Pete Olcott "Talent hits a target no one else can hit;
> Genius hits a target no one else can see." Arthur Schopenhauer


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

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Subject: Re: Concise refutation of halting problem proofs V59 [ key essence ]
From: wyni...@gmail.com (wij)
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 by: wij - Tue, 1 Feb 2022 21:23 UTC

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'? 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


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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>
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<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>
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Tue, 1 Feb 2022 21:28 UTC

On 2/1/2022 3:14 PM, wij wrote:
> On Wednesday, 2 February 2022 at 01:43:20 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.
>>>
>>> The Halting Problem has a definite, commonly recognized meaning. It refers to a
>>> real machine, no ambiguity, no one can change it, not even Linz.
>>> Your halt-problem is absolutely certain not Linz's, or of any? textbook.
>>> Your claim contradicts experimental truth. Otherwise, show your x86utm operating
>>> system proof. I guess you would say xxx thousands pages, I believe there are
>>> only few lines are yours. Show your codes.
>>>
>> You did not even read what I said before you claimed that it was
>> incorrect. The general principles that I outlined below directly apply
>> to the actual Linz proof:
>>
>> If I am incorrect in anything that I said below then the specific error
>> could be pointed out.
>>>> The halting problem does not bother to mention the requirement that
>>>> because all halt deciders are deciders they are only accountable for
>>>> computing the mapping from their finite string inputs to an accept or
>>>> reject state on the basis of the actual behavior specified by this input.
>>>>
>>>> The halting problem does not specifically examine simulating halt
>>>> deciders, none-the-less the behavior of a correctly simulated machine
>>>> description is known to be equivalent to the behavior of the direct
>>>> execution of this same machine.
>>>>
>>>> Since a simulating halt decider is merely a UTM for simulated inputs
>>>> that reach their final state when a simulating halt decider correctly
>>>> determines that its simulated its input cannot possibly reach its final
>>>> state this is complete proof that this simulated input never halts.
>>>> --
>>>> Copyright 2021 Pete Olcott
>>>>
>>>> Talent hits a target no one else can hit;
>>>> Genius hits a target no one else can see.
>>>> Arthur Schopenhauer
>>
>>
>> --
>> Copyright 2021 Pete Olcott "Talent hits a target no one else can hit;
>> Genius hits a target no one else can see." Arthur Schopenhauer
>
> Your words are almost meaningless,


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

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<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>
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Followup-To: comp.theory
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 by: olcott - Tue, 1 Feb 2022 21:36 UTC

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.


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$1@dont-email.me>
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<Kk%JJ.20609$OF3.19827@fx14.iad> <staa42$dgq$1@dont-email.me>
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Subject: Re: Concise refutation of halting problem proofs V59 [ key essence ]
From: wyni...@gmail.com (wij)
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 by: wij - Tue, 1 Feb 2022 22:12 UTC

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


Click here to read the complete article
Re: Concise refutation of halting problem proofs V59 [ ignorance about halt deciders ]

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https://www.novabbs.com/devel/article-flat.php?id=26178&group=comp.theory#26178

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<RqidnSdLIdwH2GX8nZ2dnUU7-SXNnZ2d@giganews.com> <Kk%JJ.20609$OF3.19827@fx14.iad>
<staa42$dgq$1@dont-email.me> <wv2KJ.25296$tW.1549@fx39.iad>
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<rv2dnc__PYfMJ2X8nZ2dnUU7-WHNnZ2d@giganews.com> <Pu3KJ.19025$mS1.14927@fx10.iad>
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<1eqdnSCcgMlI4GT8nZ2dnUU7-LPNnZ2d@giganews.com> <4e7f9c66-8852-4bb7-b913-e94a1a174120n@googlegroups.com>
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Subject: Re: Concise refutation of halting problem proofs V59 [ ignorance
about halt deciders ]
From: wyni...@gmail.com (wij)
Injection-Date: Tue, 01 Feb 2022 22:25:57 +0000
Content-Type: text/plain; charset="UTF-8"
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 by: wij - Tue, 1 Feb 2022 22:25 UTC

On Wednesday, 2 February 2022 at 06:19:04 UTC+8, olcott wrote:
> On 2/1/2022 4:12 PM, wij wrote:
> > On Wednesday, 2 February 2022 at 05:36:39 UTC+8, olcott wrote:
> >> 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.
> --
> Copyright 2021 Pete Olcott
>
> Talent hits a target no one else can hit;
> Genius hits a target no one else can see.
> Arthur Schopenhauer


Click here to read the complete article
Re: Concise refutation of halting problem proofs V59 [ ignorance about halt deciders ]

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From: polco...@gmail.com (olcott)
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Subject: Re: Concise refutation of halting problem proofs V59 [ ignorance
about halt deciders ]
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 by: olcott - Tue, 1 Feb 2022 23:37 UTC

On 2/1/2022 4:25 PM, wij wrote:
> On Wednesday, 2 February 2022 at 06:19:04 UTC+8, olcott wrote:
>> On 2/1/2022 4:12 PM, wij wrote:
>>> On Wednesday, 2 February 2022 at 05:36:39 UTC+8, olcott wrote:
>>>> 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.
>> --
>> Copyright 2021 Pete Olcott
>>
>> Talent hits a target no one else can hit;
>> Genius hits a target no one else can see.
>> Arthur Schopenhauer
>
> There is no 'actual TM' until you it into a TM simulator,
> otherwise all empty talks.
> (I would expect to see you 'reinterpret' again)


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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)
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 by: Richard Damon - Tue, 1 Feb 2022 23:57 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.
>

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.

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.

> 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 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.

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

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

PERIOD.

FAIL.

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

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


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

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<a6adneLIPaTubWv8nZ2dnUU7-anNnZ2d@giganews.com>
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<7FFJJ.29151$541.18496@fx35.iad> <st7a2e$oo$1@dont-email.me>
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<st9fn6$60s$2@gioia.aioe.org> <RqidnSdLIdwH2GX8nZ2dnUU7-SXNnZ2d@giganews.com>
<Kk%JJ.20609$OF3.19827@fx14.iad> <staa42$dgq$1@dont-email.me>
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Wed, 2 Feb 2022 00:14 UTC

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.


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

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References: <ssh8vu$4c0$1@dont-email.me> <mWCJJ.57596$zV.23696@fx43.iad>
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<ibHJJ.56320$u41.55552@fx41.iad>
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<gIHJJ.29153$541.4042@fx35.iad> <st91ek$p4g$1@dont-email.me>
<st9fn6$60s$2@gioia.aioe.org> <RqidnSdLIdwH2GX8nZ2dnUU7-SXNnZ2d@giganews.com>
<Kk%JJ.20609$OF3.19827@fx14.iad> <staa42$dgq$1@dont-email.me>
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<1eqdnSCcgMlI4GT8nZ2dnUU7-LPNnZ2d@giganews.com>
From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Wed, 2 Feb 2022 00:20 UTC

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.


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 by: olcott - Wed, 2 Feb 2022 00:23 UTC

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.


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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)
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 by: Richard Damon - Wed, 2 Feb 2022 00:24 UTC

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).


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