On 8/11/21 10:28 AM, olcott wrote:
> On 8/10/2021 9:26 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 8/10/2021 8:41 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 8/10/2021 7:42 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 8/7/2021 7:34 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> On 8/5/2021 9:36 PM, Ben Bacarisse wrote:
>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>
>>>>>>>>>>> On 8/5/2021 5:14 PM, Ben Bacarisse wrote:
>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>      
>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>>>>
>>>>>>>>>>>>> The question is not: Does Ĥ halt on its input?
>>>>>>>>>>>> Yes it is.
>>>>>>>>>>>
>>>>>>>>>>> The question is:
>>>>>>>>>>> Does the Ĥ specified by the first ⟨Ĥ⟩ halt on its input ⟨Ĥ⟩ ?
>>>>>>>>>>> The ansswer to this question is provably no!
>>>>>>>>>> The question is: does Ĥ applied to ⟨Ĥ⟩ halt.  It does:
>>>>>>>>>>
>>>>>>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn  THIS IS NOT A CONTRADICTION
>>>>>>>>>> Indeed.  There is no contradiction.  Just an Ĥ that does not
>>>>>>>>>> meet Linz
>>>>>>>>>> spec.
>>>>>>>>>
>>>>>>>>> Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
>>>>>>>>> Because it is correct it meets the Linz spec.
>>>>>>>> I find it startling that you think that, but then it seems you
>>>>>>>> don't yet
>>>>>>>> know what the key words mean:
>>>>>>>>
>>>>>>>>> if M applied to wM does not halt
>>>>>>>>> means if the execution of the machine of the first ⟨Ĥ⟩ on its
>>>>>>>>> input of
>>>>>>>>> the seocond ⟨Ĥ⟩ does not halt then ⊢* Ĥ.qn
>>>>>>>> No.  Would you like to know "what M applied to wM does not halt"
>>>>>>>> means?
>>>>>>>> Do you need help to see that "Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn" is clearly a
>>>>>>>> case of "M
>>>>>>>> applied to wM halts"?
>>>>>>>
>>>>>>>         the Turing machine halting problem. Simply stated, the
>>>>>>> problem
>>>>>>>         is: given the description of a Turing machine M and an
>>>>>>> input w,
>>>>>>>         does M, when started in the initial configuration q0w,
>>>>>>> perform a
>>>>>>>         computation that eventually halts? (Linz:1990:317).
>>>>>> Yes.  I was offering to help you understand the key words in that
>>>>>> text.
>>>>>>
>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>>> You've missed off the key lines yet again.  Is that deliberate?  They
>>>>>> are the lines that show you are wrong so I am suspicious that you
>>>>>> keep
>>>>>> omitting them.
>>>>>>
>>>>>>> When Ĥ is applied to ⟨Ĥ⟩ the description of the Turing Machine
>>>>>>> and its
>>>>>>> input are specified as: ⟨Ĥ⟩ ⟨Ĥ⟩ for the embedded halt decider at
>>>>>>> Ĥ.qx.
>>>>>> Ungrammatical.
>>>>>>
>>>>>>> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn this is not a final state of the simulated
>>>>>>> input it is a final state of the executed Ĥ.
>>>>>> Yes.  You don't seem to know why that's wrong.
>>>>>
>>>>> What is your basis for believing that is wrong?
>>>> Ah, a question about what I'm saying.  I can help there.  The basis is
>>>> what Linz says about Ĥ.  He says that (translating to your notation)
>>>>     Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>> should be the case "if Ĥ applied to ⟨Ĥ⟩ does not halt".  But, as you
>>>> can
>>>> see, your Ĥ does halt when applied to ⟨Ĥ⟩ (qn is a halting or final
>>>> state).  Your Ĥ is not doing what it should in this one crucial case.
>>>>
>>>
>>>     the Turing machine halting problem. Simply stated, the problem
>>>     is: given the description of a Turing machine M and an input w,
>>>     does M, when started in the initial configuration q0w, perform a
>>>     computation that eventually halts? (Linz:1990:317).
>>
>> and so on.  Same old stuff.
>>
>
> When the challenge to support one's assertion with reasoning is simply
> ignored as you are ignoring it right now one can reasonably construe a
> deceptive intent.
>
> -- the Turing machine halting problem. Simply stated, the problem
> -- is: given the description of a Turing machine M and an input w,
> -- does M, when started in the initial configuration q0w, perform a
> -- computation that eventually halts? (Linz:1990:317).
>
> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>
> The input to H will be the description (encoded in some form) of M, say
> WM, as well as the input w. (Linz:1990:318)
>
> H.q0 WM w ⊢* H.qn
> if M applied to W does not halt.
>
>   becomes
>
> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
> if Ĥ applied to ⟨Ĥ⟩ does not halt.
>
>
> Pages of the Linz text to verify the above quotes in their full context:
> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>
> M STILL REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO Ĥ.qx
> M STILL REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO Ĥ.qx
> M STILL REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO Ĥ.qx
> M STILL REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO Ĥ.qx
>
> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
> if M applied to wM does not halt
>
> When we know that M refers to the Turing machine specified by the first
> wM then when Ĥ transitions to its final state of Ĥ.qn there is no direct
> contradiction formed.
>
> Can you admit when you are wrong when you really are wrong?
> Can you admit when you are wrong when you really are wrong?
> Can you admit when you are wrong when you really are wrong?
> Can you admit when you are wrong when you really are wrong?
>
> if M applied to wM does not halt (see above for definition of M)
> means when the Turing machine of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt.
>
> Ĥ.q0 ⟨Ĥ⟩  ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> Ĥ.qx correctly transitions to its final state when the Ĥ.qx acts as a
> UTM and simulates ⟨Ĥ⟩ ⟨Ĥ⟩ and determines that this input never halts.
>
> https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
>
>

But that input is also supposed to be the representation of H^, as that
is basic question being asked in the proof, can H give the right answer
for predicting what the machine H^(<H^>) does.

Since from the run of H^, we see that starting at H^.q0 <H^> we end up
at the halting state of H^.qn, we KNOW that H^(<H^>) is a Halting
computation, and thus that the input given to H represents a Halting
computation, thus the fact that H.q0 <H^> <H^> ends up at H.qn is a
wrong answer.

The fact that H's simulation did not reach the halting state is
irrelevant, since H did not complete its simulation but aborted it. It
is a FACT that if we continue simulating that exact same computation, it
WILL stop, as was show by the actual running of the actual machine
represented by that input.

Either the input to H represents that same machine, and thus we know it
will act the same, or the machines were setup wrong an the proof is void.

You don't seem to understand that the fact that H doesn't reach the
final halting state before it aborts doesn't actually prove anything.

Only a truely pure simulator that never reaches a halting state proves
that, but a truely pure simulator is one that NEVER stops until it does
reach the halting state of the machine it is simulating. A machine tht
doesn't stop its simulation until it does stop its simulation doesn't
meet that definition. If you think so, please play your local athorites
for all the red lights you didn't stop at until you did, as obviously
you never did.