On 10/20/2021 11:46 AM, André G. Isaak wrote:
> On 2021-10-20 09:51, olcott wrote:
>> On 10/20/2021 10:42 AM, André G. Isaak wrote:
>>> On 2021-10-20 09:25, olcott wrote:
>>>> On 10/20/2021 10:13 AM, André G. Isaak wrote:
>>>>> On 2021-10-20 07:57, olcott wrote:
>>>>>> On 10/19/2021 8:21 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 10/19/2021 11:55 AM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> On 10/18/2021 5:59 PM, Ben Bacarisse wrote:
>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>
>>>>>>>>>>>> On 10/17/2021 7:47 PM, Ben Bacarisse wrote:
>>>>>>>>>>>>> Quotes re-ordered for clarity...
>>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>>>> On 10/17/2021 10:03 AM, Ben Bacarisse wrote:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> But you agree that
>>>>>>>>>>>>>>>         Ĥ.q0 ⟨Ĥ⟩ ⊦ Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦ Ĥ.qn
>>>>>>>>>>>>>>>         if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> is (one half of) what Linz means by Ĥ, yes?  (I don't
>>>>>>>>>>>>>>> expect an
>>>>>>>>>>>>>>> answer -- I expect you'll just repeat the waffle.)
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> No not at all, not in the least little bit
>>>>>>>>>>>>>
>>>>>>>>>>>>> OK, so you are not using Ĥ to denote an exemplar of the
>>>>>>>>>>>>> same class of
>>>>>>>>>>>>> TMs that Linz does.  If you were, you would have to agree
>>>>>>>>>>>>> that Linz's
>>>>>>>>>>>>> specification of when Ĥ transitions to qn is the correct one.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Oh, and you need to apologise for some very deceptive
>>>>>>>>>>>>> claims to the
>>>>>>>>>>>>> contrary.
>>>>>>>>>>>>
>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>>> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>>>>>>>>>>
>>>>>>>>>> https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>>>>>>>>>> page 319
>>>>>>>>>>
>>>>>>>>>> That you do not understand the most basic point that Linz has
>>>>>>>>>> adapted
>>>>>>>>>> TM H by prepending states and appending states such that the
>>>>>>>>>> original
>>>>>>>>>> H "not" path of H remains in the middle is a necessary
>>>>>>>>>> prerequisite.
>>>>>>>>> I'll leave it to others to decide which one of use us is
>>>>>>>>> lacking basic
>>>>>>>>> understanding.
>>>>>>>>> Just stop dishonestly removing the key criterion given in Linz,
>>>>>>>>> or admit
>>>>>>>>> that you are not using this notation as Linz does.
>>>>>>>>>
>>>>>>>>>> q0 Wm Wm ⊢* y1 qn y2  / The "no" path of TM H
>>>>>>>>> Linz says it better: if M applied to wM does not halt.  Your silly
>>>>>>>>> phrase does not state what it is that H is saying "no" to.
>>>>>>>>
>>>>>>>> Linz is vague about the point in the execution trace that matters.
>>>>>>>
>>>>>>> You replacement is useless.  Linz's specification is clear.
>>>>>>> Obviously
>>>>>>> it does not way what you want, but it's not vague.
>>>>>>>
>>>>>>>>>> q0 Wm Wm ⊢* y1 qn y2  / The "no" path of TM H'
>>>>>>>>>> q0 Wm ⊢* Ĥq0 Wm Wm ⊢* Ĥ y1 qn y2 // The "no" path of TM Ĥ
>>>>>>>>>
>>>>>>>>> Linz says it better: if Ĥ applied to wM does not halt.
>>>>>>>>
>>>>>>>> The only point in the execution trace that matters is where the
>>>>>>>> halt
>>>>>>>> decider makes its halting decision.
>>>>>>>
>>>>>>> What matters is that if H is as specified in Linz, the
>>>>>>> corresponding Ĥ
>>>>>>> behaves as Linz says and you keep removing.
>>>>>>>
>>>>>>>>>> This leaves the "no" path of TM H at Linz Ĥq0
>>>>>>>>>>
>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>> And Linz says exactly when this transitions sequence occurs: if Ĥ
>>>>>>>>> applied to ⟨Ĥ⟩ does not halt.
>>>>>>>>
>>>>>>>> If he is referring to something besides the input to the halt
>>>>>>>> decider
>>>>>>>> then Linz is incorrect.
>>>>>>>
>>>>>>> I am happy to explain "if Ĥ applied to wM does not halt" when
>>>>>>> attached
>>>>>>> to the line you keep leaving it off if you think it would help.
>>>>>>> What is
>>>>>>> it you don't understand?
>>>>>>>
>>>>>>
>>>>>> q0 Wm ⊢* Ĥq0 Wm Wm ⊢* Ĥ y1 qn y2 // The "no" path of TM Ĥ
>>>>>> if M applied to Wm does not halt
>>>>>>
>>>>>> Linz cannot be correctly referring to q0 applied to Wm because
>>>>>> there is no halt decider at q0. If he is referring to that then he
>>>>>> is wrong.
>>>>>
>>>>> Nor is there a halt decider at Ĥq0. Both q0 and Ĥq0 refer to single
>>>>> states in the execution of Ĥ. There isn't a halt decider anywhere
>>>>> in Ĥ.
>>>>>
>>>>
>>>> https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf It is
>>>> clear that the "no" path of the H halt decider from which Ĥ was
>>>> constructed begins at Ĥq0 and its input is Wm Wm. Pages 318 and 319
>>>> prove this.
>>>
>>> ??
>>>
>>> Ĥ always begins in state q0, so all paths ultimately begin there.
>>>
>>
>> That is only the point in the execution trace of Ĥ that copies its input
>
> So what? It's still the starting point of the computation. Ĥ is a Turing
> Machine which takes a *single* input string. The fact that it
> subsequently duplicates that is an integral part of the computation
> performed by Ĥ on a given input.
>
>> so that its halt decider at Ĥq0 has its required input pair.
>
> There IS NO halt decider at Ĥq0. Nor anywhere else in Ĥ.

It "no" path is the H halt decider is at Ĥq0 of Ĥ.

>
>>> Since it always passes through state Ĥq0, you can talk about the path
>>> beginning at Ĥq0 if you want, but that has no bearing on the point
>>> being made, which is that final state qn is only reached in cases
>>> where M applied to Wm does not halt.
>>>
>>> André
>>>
>>
>> As long as the simulating halt decider at Ĥq0 correctly decides that
>> its simulated input Wm applied to Wm never reaches the final state of
>> this simulated input (whether or not the simulating halt decider
>> aborts its simulation of this input) then this simulating halt decider
>> at Ĥq0 is correct when it transitions to qn.
>
> But that's not the criterion given by Linz. He specifies that qn is
> reached if M applied to Wm does not halt.

q0 Wm ⊢* Ĥq0 Wm Wm ⊢* Ĥ y1 qn y2 // The "no" path of TM Ĥ

He either means the M represented by the first Wm after Ĥq0 or he is
simply incorrect. As long as a halt decider correctly decides the halt
status of its input then this halt decider is necessary correct.

>
> If you have to change that last bit, you're not talking about Linz's Ĥ.
>
> André
>

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein