Re: Clarification of Linz Ĥ Description [ untainted decision ]

[olcott]

On 10/15/2021 10:17 AM, Richard Damon wrote:
>
> On 10/15/21 10:47 AM, olcott wrote:
>> On 10/14/2021 11:29 PM, Richard Damon wrote:
>>> On 10/14/21 8:27 PM, olcott wrote:
>>>
>>>> Linz explicitly specified that H is any TM that gets the right answer.
>>>> He is incorrect that H does not exist.
>>>
>>> So, will you show an ACTUAL Turing Machine that gets the right answer?
>>>
>>> Not just something you claim must be sort of like a Turing Machine
>>> that givs something you hand wave to claim is the right answer.
>>>
>>> This is, of course, impossible, but you will try to make something
>>> that sort of seems to be the right answer, but you will hide the
>>> problems with a bunch of smoke.
>>>
>>>
>>> The key point that you seem to miss is that since H for you case only
>>> needs to get ONE machie right, H could just as easily ignore its
>>> input and give the 'right' answer.
>>>
>>> But Linz has shown that whatever answer H gives, it will be wrong by
>>> the nature of the construction of H^.
>>>
>>
>> Superficial analysis seems to indicate this, yet this is merely
>> superficial analysis. Only categorically exhaustive reasoning can test
>> every possibility that can possibly exist and do this in finite time.
>
> Right, and there are only 4 possible behaviors that H can have.
>
> It can halt in the state qy.
> It can halt in the state qn.
> It can halt in some other state and fail to meet its requirements.
> It can fail to halt in any state and fail to meet its requirements.
>
> What other possibility do you see for H to do.
>
> This IS a Categorically Exhaustive Reasoning.
H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to H.qy
because Ĥ.qx applied to ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to Ĥ.qn

https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation


--
Copyright 2021 Pete Olcott

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

[olcott]

On 10/19/2021 11:55 AM, Ben Bacarisse wrote:
> olcott <No...@NoWhere.com> writes:
>
>> On 10/18/2021 5:59 PM, Ben Bacarisse wrote:
>>> olcott <No...@NoWhere.com> writes:
>>>
>>>> On 10/17/2021 7:47 PM, Ben Bacarisse wrote:
>>>>> Quotes re-ordered for clarity...
>>>>> olcott <No...@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.
I am not vague about the point in the execution trace that matters.

As long as the simulating halt decider correctly decides that the
simulation of its input TM description never reaches its final state
whether or not this simulating halt decider aborts the simulation of
this input then this simulating halt decider does correctly decide that
this its input does not halt.

>> 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.

When it is known to be true that the simulated input to Ĥq0 Wm Wm never
reaches its final state whether or not the simulating halt decider
aborts its simulation of this input then we know that the state
transition to qn is correct.

>> 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.

As long as a simulating halt decider correctly decides that its input
never reaches its own final state then this simulating halt decider is
necessarily correct when it decides that this input never halts.

When we know that we have a black cat then we know that we have a cat.

> The only excuse for dropping what Linz says about these transition
> sequences is to deceive. Stop doing that.
>

Linz is either misleadingly vague or incorrect.
Whether or not a halt decider correctly decides the halt status of its
input only pertains to the halt status of this input and nothing else.

[olcott]

> 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.

Linz can only be referring to Ĥq0 Wm Wm the machine of the first
Wm being applied to the machine description of the second Wm.

A simulating halt decider H does correctly decide the halt status of its
input (P,P) if-and-only-if it correctly decides that its simulated input
P never reaches its final state whether or not it aborts the simulation
of this input.

If you can't understand this then you are hopelessly and irreconcilably
lost. Understanding this is a key mandatory prerequisite upon which all
subsequent understanding is based.

[olcott]

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:

>>> 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 that its halt decider at Ĥq0 has its required input pair.

> 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.

[olcott]

On 10/20/2021 11:46 AM, André G. Isaak wrote:

> 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

[olcott]

On 10/20/2021 4:40 PM, André G. Isaak wrote:
> On 2021-10-20 14:34, olcott wrote:
>> On 10/20/2021 3:06 PM, André G. Isaak wrote:
>
>>> But it is *not* guaranteed to halt, so it is not a decider at all.
>>> Not any kind of decider.
>>
>> If I make any thingamajig Whatchamacallit that correctly determines
>> the halt status of the "impossible" halting problem counter-example
>> input then this is very significant.
>
> But again, that's what your *H* is purported to do. It isn't what Ĥ is
> purported to do. I have been commenting on your insistence of calling Ĥ
> (or Ĥq0) a halt decider. It is not one. Linz doesn't describe it as one.
>
> If your going to make grandiose claims about having constructed a
> (partial) halt decider, at least restrict them to your H, not to Ĥ which
> isn't presented by Linz as a decider of anything. It's simply a TM which
> performs some relatively meaningless computation but which plays a role
> in his proof.
>
> André
>

Because H only correctly determines the halt status of ⟨Ĥ⟩ ⟨Ĥ⟩ on the
basis that Ĥq0 correctly detemines the halt status of ⟨Ĥ⟩ ⟨Ĥ⟩ it must be
understood that Ĥq0 correctly determines the halt status of ⟨Ĥ⟩ ⟨Ĥ⟩.

[olcott]

On 10/20/2021 8:37 PM, Ben Bacarisse wrote:
> olcott <No...@NoWhere.com> writes:
>

>>> This notation is hopeless in ASCII. Using the notation that has evolved
>>> in these threads (not my preferred one, but I don't want to complicate
>>> matters any more) Linz writes:
>>> Ĥ.q0 <M> ⊢* Ĥ.qx <M> <M> ⊢* y1 Ĥ.qn y2
>>> if M applied to <M> does not halt.
>>> And with these changes in notation you seem to be saying
>>>
>>>> Linz cannot be correctly referring to Ĥ applied to <M> because there

>>>> is no halt decider at q0.

>>> Yes, there is no halt decider at Ĥ.q0, but he is correctly referring to
>>> what Ĥ applied to <M> should do: Ĥ applied to <M> should transition to
>>> Ĥ.qn if (and only if) M (the TM encoded in the string input) applied to
>>> <M> (that exactly same string input) does not halt.
>>> Your next point helps explain how he comes to say this about Ĥ.
>>>
>>>> Linz can only be referring to Ĥ.qx <M> <M> the machine of the first
>>>> <M> being applied to the machine description of the second <M>.
>>>
>>> The annotation dose not specifically refer to what this sub-computation
>>> does. It simply gives the condition under which Ĥ will transition from
>>> Ĥ.q0 to Ĥ.qn.
>>
>> If a simulating halt decider H correctly determines that the
>> simulation of its input <M> applied to <M> never reaches its final

>> state whether or not this simulating halt decider aborts the
>> simulation of this input then this simulating halt decider does

>> correctly decide that this input never halts NO MATTER WHAT ELSE.
>
> This is not what Linz is saying.

Of course this is not what Linz is saying. Linz is still under the
misconception that Ĥ.q0 applied to <Ĥ> <Ĥ> would not correctly
transition to H.qn.

> If I can help with any further
> explanation, I'm happy to have a go but you need to stop repeating what
> you are saying and try to focus on what Linz is saying. Once you get
> what he is saying you will see that you can say anything you like about
> a subclass (those that "simulate") of the empty class of TMs (the halt
> deciders) that Linz is talking about. Atatements about no TMs are
> vacuously true (or irrelevant -- take your pick).
>
>> Anything and everything that is not input to this halt decider is 100%
>> totally irrelevant.
>
> Technically, there is no halt decider in the TM you are talking about,
> but that's a detail and you don't do details. Did you follow what I
> wrote about how the string input to the almost-decider at Ĥ.qx is
> entirely why Linz's annotation is correct? Here it is:
>
>>> But you are correct in that the annotation /derives/ from what we know
>>> about the modified copy of H embedded at Ĥ.qx. We know that
>>> H.q0 <M> s ⊢* y1 H.qn y2
>>> if M applied to s does not halt.
>>> and since the TM at Ĥ.qx is (in this regard at least) exactly like H we
>>> know that
>>> Ĥ.qx <M> <M> ⊢* y1 Ĥ.qn y2
>>> if M applied to <M> does not halt.
>>> Now, because
>>> Ĥ.q0 s ⊢* Ĥ.qx s s
>>> for /any/ string s, we know, specifically, that
>>> Ĥ.q0 <M> ⊢* Ĥ.qx <M> <M> ⊢* y1 Ĥ.qn y2
>>> if M applied to <M> does not halt.
>>> Can you now see where this correctly annotated line comes from? Can I
>>> explain any part of this some other way to make it clearer?
>
> I suspect you simply didn't read it since I am agreeing with you about
> the central role of the Ĥ.qx <M> <M> part in explaining why Linz is
> correct.
>

So you are agreeing that when Ĥ.qx simulates <Ĥ> applied to <Ĥ> and
correctly determines that the simulated <Ĥ> never reaches its final
state (whether or not Ĥ.qx stops simulating <Ĥ>) then Ĥ.qx correctly
aborts this simulation and transitions to qn ???

If we have to go over this same point thousands of times I will.
This one point proves that I have shown how to defeat the halting theorem.

Let's standardize on this notation so that new people will
be able to more easily see how it relates to the Linz text.
⟨Ĥ⟩ indicates the Turing machine description of Ĥ.



The machine at Ĥq0 transitions to Ĥqn if the simulation of ⟨Ĥ⟩ applied
to ⟨Ĥ⟩ would never reach its final state whether or not this simulation
is aborted.

// Adapted from bottom of page 319 (definition of Ĥ)
q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ∞
If the simulated input to Ĥq0 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ halts

q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
If the simulated input to Ĥq0 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt

https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf

All dialogue must be either agreeing with the above or pointing out
errors in the above. Changing the subject will generally be ignored
except that it will be pointed out as a dishonest dodge away from the
point.

[olcott]

On 10/21/2021 9:19 PM, Ben Bacarisse wrote:
> olcott <No...@NoWhere.com> writes:
>

>>> But until you understand what he is saying, you can't see why you are
>>> wrong.
>>
>> That Linz did not consider a simulating halt decider does not make me
>> wrong.
>
> Quite. You are wrong for the reasons I gave.
>
>> If any of this was wrong someone could point out an actual error:

>>
>> The machine at Ĥq0 transitions to Ĥqn if the simulation of ⟨Ĥ⟩ applied
>> to ⟨Ĥ⟩ would never reach its final state whether or not this
>> simulation is aborted.
>

> The error is that this is a statement about an sub-set of and empty set
> of Turing machines.
>

You are simply making a false assumption. You are beginning with the
premise that Linz is correct then concluding that Linz is correct on the
basis of this premise.

>> Adapted from bottom of page 319 (definition of Ĥ)

>> ⟨Ĥ⟩ indicates the Turing machine description of Ĥ.
>>

>> q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ∞
>> If the simulated input to Ĥq0 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ halts
>

> This is either the wrong annotation for Linz's Ĥ or a pointless
> re-wording of it.

>
>> q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
>> If the simulated input to Ĥq0 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt
>

> And again, this is the wrong annotation for Linz's Ĥ. Linz's Ĥ
> transitions to qn is (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
>

No you are wrong.
Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to Ĥqn when its input never halts.
You can't show that this is incorrect only because it is correct.

> You can't change the specification of what Ĥ does if you want to be
> taken seriously. You have to accept the logical consequences of his
> definitions or define your own class of machines.
>

The halt decider must only correctly decide whether or not its input
halts on its input. As long as this decision is correct then it is
impossible for anything else to show that the halt decider is incorrect.

The halt decider must only correctly decide whether or not its input
halts on its input. As long as this decision is correct then it is
impossible for anything else to show that the halt decider is incorrect.

The halt decider must only correctly decide whether or not its input
halts on its input. As long as this decision is correct then it is
impossible for anything else to show that the halt decider is incorrect.

The halt decider must only correctly decide whether or not its input
halts on its input. As long as this decision is correct then it is
impossible for anything else to show that the halt decider is incorrect.

The halt decider must only correctly decide whether or not its input
halts on its input. As long as this decision is correct then it is
impossible for anything else to show that the halt decider is incorrect.

[Jeff Barnett]

On 10/21/2021 8:33 PM, olcott wrote:

> The halt decider must only correctly decide whether or not its input
> halts on its input. As long as this decision is correct then it is
> impossible for anything else to show that the halt decider is incorrect.
>
> The halt decider must only correctly decide whether or not its input
> halts on its input. As long as this decision is correct then it is
> impossible for anything else to show that the halt decider is incorrect.
>
> The halt decider must only correctly decide whether or not its input
> halts on its input. As long as this decision is correct then it is
> impossible for anything else to show that the halt decider is incorrect.
>
> The halt decider must only correctly decide whether or not its input
> halts on its input. As long as this decision is correct then it is
> impossible for anything else to show that the halt decider is incorrect.
>
> The halt decider must only correctly decide whether or not its input
> halts on its input. As long as this decision is correct then it is
> impossible for anything else to show that the halt decider is incorrect.

<See my recent relevant to this one> N.B. Compulsive repetition such as
the above is strong evidence of fixation in the early, anal stages of
development. Even though it usually occurs in much younger humans, it is
possible (in your case) that either 1) you never passed out of this
stage or 2) you have regressed. Typically compulsions completely inhibit
the mind from processing other information of sensing the environment. I
think your regression has blocked your sanity in addition to your
ability to process simple adult tasks. It's no wonder your logic (not
simply as part of mathematics but as a part of ordinary life) is so
deeply flawed. You can't even engage in simple, pleasant dialogue
without tantrums and endless repetition.

I would suggest that you do some online searches to learn more about
your condition(s). Unfortunately I know better from your behavior here:
you would look something up, your eyes would glaze over since you can
not concentrate, then you would quote things you do not understand and
insist you had novel insights. All the while exhibiting all the signs of
a sick adult mired in the anal retentive stage with no self awareness
what so ever. Get help.
Jeff Barnett

[olcott]

On 10/25/2021 7:46 PM, Ben Bacarisse wrote:
> olcott <No...@NoWhere.com> writes:
>

>>>>> A proved theorem is not a "false assumption".
>>>>
>>>> Ah so you are back to your blasphemy. One can not rely on a proved
>>>> theorem as exactly equal to infallibility.
>>>
>>> You are being dishonest again. Setting aside the silly religious
>>> language, I am not asserting that any argument is infallible something
>>> that was obvious from what I wrote immediate after this.
>>> I am simply saying you have not shown any error in the argument. Until
>>> you show that, for example, 2+2 is not equal to 4, it is entirely
>>> reasonable to assert that it is. That is not blasphemy.
>>
>> I don't see how this is so difficult for you:
>> (a) If it is claimed that X cannot possibly be done
>> (b) I show X being correctly done
>> (c) then I have refuted X cannot possibly be done
>
> I don't think you honestly believe that the logic is difficult for me.

The alternative is that you are a liar and I prefer to give you the
benefit of the doubt.

> Neither do I think you have trouble knowing what part I am saying you
> have failed at (it's (b) by the way).

We have not gotten to B yet.
First we must have mutual agreement on the hypothetical.

Here is the long form of the hypothetical:

THIS STATEMENT IS NECESSARILY ALWAYS TRUE
Whenever simulating halt decider H correctly determines that input P
never reaches its final state (whether or not its simulation of P is
aborted) then H correctly decides that P never halts.

> But my objection to what you've been saying goes my deeper than a
> failure to show anything of value. You are being dishonest in that you
> claim to be talking about Linz's class of TMs, but you keep cutting key
> facts about them while trying to replace those facts with waffle of you
> your own invention.

THIS IS THE SHORT VERSION OF THE HYPOTHETICAL
If a halt decider correctly decides that its input never halts then the
halt decider did correct decide the halt status of its input.

You talk about me being dishonest when you continually refuse to agree
that when a halt decider is correct then it is not incorrect.