On 2/1/2022 6:20 PM, Richard Damon wrote:
> On 2/1/22 1:37 PM, olcott wrote:
>> On 2/1/2022 10:33 AM, wij wrote:
>>> On Tuesday, 1 February 2022 at 23:22:32 UTC+8, olcott wrote:
>>>> On 1/31/2022 11:25 PM, Richard Damon wrote:
>>>>>
>>>>> On 1/31/22 11:42 PM, olcott wrote:
>>>>>> On 1/31/2022 10:33 PM, Richard Damon wrote:
>>>>>>>
>>>>>>> On 1/31/22 11:24 PM, olcott wrote:
>>>>>>>> On 1/31/2022 10:17 PM, Richard Damon wrote:
>>>>>>>>> On 1/31/22 10:40 PM, olcott wrote:
>>>>>>>>>> On 1/31/2022 6:41 PM, Richard Damon wrote:
>>>>>>>>>>> On 1/31/22 3:24 PM, olcott wrote:
>>>>>>>>>>>> On 1/31/2022 2:10 PM, Ben wrote:
>>>>>>>>>>>>> On 1/31/2022 8:06 AM, olcott wrote:
>>>>>>>>>>>>>> On 1/30/2022 8:20 PM, Richard Damon wrote:
>>>>>>>>>>>>>>> On 1/30/22 9:05 PM, olcott wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> These statements need the conditions, that H^ goes to
>>>>>>>>>>>>>>>>> H^.Qy/H^.Qn iff H goes to that corresponding state.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to
>>>>>>>>>>>>>>>> embedded_H
>>>>>>>>>>>>>>>> in the same way that (5,3) is syntactically specified as an
>>>>>>>>>>>>>>>> input to Sum(5,3)
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Right, and the
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to
>>>>>>>>>>>>>>>> embedded_H in the same way that (1,2) is NOT syntactically
>>>>>>>>>>>>>>>> specified as an input to Sum(5,3)
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Right, but perhaps you don't understand that from you above
>>>>>>>>>>>>>>> statement the right answer is based on if UTM(<H^>,<H^>)
>>>>>>>>>>>>>>> Halts which by the definition of a UTM means if H^
>>>>>>>>>>>>>>> applied to
>>>>>>>>>>>>>>> <H^> Halts.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> The biggest reason for your huge mistakes is that you cannot
>>>>>>>>>>>>>> stay sharply focused on a single point. It is as if you
>>>>>>>>>>>>>> either
>>>>>>>>>>>>>> have attention deficit disorder ADD or are addicted to
>>>>>>>>>>>>>> methamphetamine.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>   >>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>>>>>   >>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> The single point is that ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to
>>>>>>>>>>>>>> embedded_H and
>>>>>>>>>>>>>> Ĥ ⟨Ĥ⟩ is the NOT the input to embedded_H.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> After we have mutual agreement on this point we will move on
>>>>>>>>>>>>>> to the points that logically follow from this one.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> Holy shit try to post something that makes sense.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>
>>>>>>>>>>>> Richard does not accept that the input to the copy of Linz H
>>>>>>>>>>>> embedded at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩. He keeps insisting that it is Ĥ
>>>>>>>>>>>> ⟨Ĥ⟩.
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> No, but apparently you can't understand actual English words.
>>>>>>>>>>>
>>>>>>>>>>> The INPUT to H is <H^> <H^> but the CORRECT ANSWER that H must
>>>>>>>>>>> give is based on the behavior of H^ applied to <H^> BECAUSE OF
>>>>>>>>>>> THE DEFINITION of H.
>>>>>>>>>>
>>>>>>>>>> In other words Sum(3,5) must return the value of Sum(7,8)?
>>>>>>>>>
>>>>>>>>> Don't know how you get that from what I said.
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Any moron knows that a function is only accountable for its
>>>>>>>>>> actual
>>>>>>>>>> inputs.
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> And the actual input to H is <H^> <H^> which MEANS by the
>>>>>>>>> DEFINITION of the Halting Problem that H is being asked to decide
>>>>>>>>> on the Halting Status of H^ applied to <H^>
>>>>>>>> No that is not it. That is like saying "by definition" Sum(3,5) is
>>>>>>>> being asked about Sum(7,8).
>>>>>>>
>>>>>>> Again your RED HERRING.
>>>>>>>
>>>>>>> H is being asked EXACTLY what it being asked
>>>>>>>
>>>>>>> H wM w -> H.Qy if M applied to w Halts, and H.Qn if it doesn't
>>>>>>>
>>>>>>> AGREED?
>>>>>>>
>>>>>>
>>>>>> No that is wrong. embedded_H is being asked:
>>>>>> Can the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ possibly transition to ⟨Ĥ⟩.qn ?
>>>>>>
>>>>>
>>>>> If you say 'No', then you aren't doing the halting problem, as the
>>>>> requirement I stated is EXACTLY the requirement of the Halting
>>>>> Problem.
>>>> The halting problem is vague on the definition of halting, it includes
>>>> that a machine has stopped running and that a machine cannot reach its
>>>> final state. My definition only includes the latter.
>>> Sounds like a NDTM.
>>
>> https://en.wikipedia.org/wiki/Nondeterministic_Turing_machine
>>
>> It is not a NDTM, a Turing Machine only actually halts when it reaches
>> its own final state. People not very familiar with this material may
>> get confused and believe that a TM halts when its stops running
>> because its simulation has been aborted. This key distinction is not
>> typically specified in most halting problem proofs.
>>
>> computation that halts … the Turing machine will halt whenever it
>> enters a final state. (Linz:1990:234)
>>
>>
>> Halting problem undecidability and infinitely nested simulation (V3)
>>
>> https://www.researchgate.net/publication/358009319_Halting_problem_undecidability_and_infinitely_nested_simulation_V3
>>
>>
>
> And the point that you seem to miss is that the Turing Machine doesn't
> stop just because some simulation of its representation gave up on
> simulating it.
>
> And actual Turing machine will continue to run until it his a final
> state or els it will continue to run for an unbounded number of steps.
>
> Non-Halting can only be show by showing that the actual running of the
> machine will continue for an unbounded number of steps, not just that
> there is some N that it doesn't stop in.

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

--
Copyright 2021 Pete Olcott

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