On 10/20/21 11:00 PM, olcott wrote:
> On 10/20/2021 9:56 PM, Richard Damon wrote:
>> On 10/20/21 10:22 PM, olcott wrote:
>>> On 10/20/2021 8:40 PM, Richard Damon wrote:

>>>> We don't need to talk hypothectical.
>>>>
>>> Yes we do. If the input to the halt decider is an infinite loop that
>>> would execute forever unless its simulation is aborted then the halt
>>> decider correctly aborts the simulation of this input and reports non
>>> halting.
>>
>> No, we don't. If the Halt decider ACTUALLY IS non-aborting (and not
>> just presumed to be when it isn't) then we have methods to actually
>> prove that it is an infinite loop.
>>
>
> Yes when the halt deciding basis that the simulation matches an infinite
> execution pattern then aborting is mandatory.

But to be right, so is getting the right answer.

That also means assuming that the copy of it inside the computation we
are analysing doesn't abort is an INCORRECT assumption.

When H sees a copy of itself, it can either get itself stuck in an
infinite loop (and fail) or it needs to stop that simulation at some
point and SOMEHOW figure out what answer that copy of H will return
without just simulating to get there.

That is the big problem. It needs the answer that it can't just simulate
to get to.

This shows that naive simulation is NOT the answer.

Yes, you can add patterns to detect many of the infinite executions, but
this will NEVER be good enough, as H^ will never match one of them, as
if you do put in a pattern that your simulation of H^(<H^>) matches,
then H will abort it, return non-halting, and H^ will halt, thus showing
that the pattern is NOT necessarily non-halting.

The claim that there has to be a pattern to use as the problem needs to
be solvable is NOT correct, as there doesn't need to be a way to solve
the problem, it is perfectly an acceptable condition that H just can't
give the right answer.

It turns out that ANY finite set of rules to use to decide on will
always be wrong and either incorrectly halt and think it is non-halting
when it is halting, or (if you add know to halt rules) incorrectly halt
and say you know this will halt in the future, and it will turn out to
be non-halting, or the decider just ends up running forever and never
halts at all.

This all PROVES (with a little bit more logic past this point) that in
Computation Theory, there DO exists some Computations that we can not
prove if they Halt or Not, but we also know that it is True that they
Either Halt or Not, so there are some Truths that are not Provable.

What this shows is that the logic of Mathematics is not compatible with
your 'precious' concept that all Truth is Provable. A logic system with
that definition, can not develop the fullness of Mathematics, but needs
to SIGNIFICANTLY limit its scope of what it can describe. Any attempt to
try to build to the fullness of Mathematics is doomed to collapse in a
pile of inconsistency or reach a barrier it just can't cross.

This was settled nearly a Century ago, and Mathematitions accept it now.

You are of course free to disagree, and work on the problem, and there
is a base of logical system that do restrict themselves to the concept
that all Truth must be provable. You WILL find that you reach the walls
described, and will NOT be able to get to anywhere near the fullness of
Mathematics, in fact, you can't even get the full properties of the
Natural Numbers, as Godel has proved.

Note, your arguement that Godel is wrong is incorrect, as by saying that
the statement isn't a Truth also is saying that other properties shown
to exist in the Natural Numbers can not be expressed in your system, and
thus your system fails to describe all of the properties of the Natural
Numbers.

Mathematitions, although they would love it if everything was provable,
more perfer to be able to talk about all the properties of the system
they are working with, so work under a different set of rules, and
accept that not everything that is True will be provable.

On 10/20/2021 6:29 PM, olcott wrote:
> On 10/20/2021 7:04 PM, Richard Damon wrote:

>> You have to take your choice, choose to restrict yourself to a
>> definition of truth that says that can't consider something to be true
>> if you can not prove it, and lose most of
>
> The conspiracy theories based on Nazi propaganda that could otherwise
> put an end to Democracy.

Bull. The only thing at risk is you advertising to the world that you
have shit for brains. You really must see your doctors - current plus a
psychiatrist - before your brain drips out in a smelly puddle.

Anyone who thinks Richard's statement deserve your idiotic retort would
need to be as nuts as you are. That just isn't likely at all. It's time,
actually way past time, for you to toilet train your mouth. You simply
are not connected to the conversations you start and want to dominate.

You throw (verbal) temper tantrums like a very young child. I presume
your bottom is as untrained as your mouth. First there's the child-like
attempt to impress your elders - anyone with a mental age north of 9;
then there's the humiliation of these friendlies trying to help you grow
up in a good way; following this stage comes the tantrums because you
can't distinguish friendly help to mature you from humiliation being
heaped upon an incapable child.

You've done this long enough to pass the stage/age where growth and
maturity should follow. It's immersion in the anal development stage for
the rest of your life. It's a pity that you will never shed the diapers
you wear on your bottom or as a skull cap.

You've now read what I think and I believe a vast majority of your
correspondents will have views of you much more similar to mine then
your self image. Now look above and reread the poop you tried to throw
at Richard. Notice: 1) it does not follow in any way from the prior
comments, 2) it was meant to be nasty, 3) it was meant to convince your
readers that you were insightful and on top of your game, 4) rather it
convinced everyone that you are a jerk with little reasoning power, and
5) it confirmed just about everyone's opinion of you as immature,
ignorant, and resistant to mental improvements.
--
Jeff Barnett

On 10/21/2021 12:23 AM, Jeff Barnett wrote:
> On 10/20/2021 6:29 PM, olcott wrote:
>> On 10/20/2021 7:04 PM, Richard Damon wrote:
>
>>> You have to take your choice, choose to restrict yourself to a
>>> definition of truth that says that can't consider something to be
>>> true if you can not prove it, and lose most of
>>
>> The conspiracy theories based on Nazi propaganda that could otherwise
>> put an end to Democracy.
>
> Bull. The only thing at risk is you advertising to the world that you
> have shit for brains. You really must see your doctors - current plus a
> psychiatrist - before your brain drips out in a smelly puddle.
>

One-third of Americans believe Biden won because of voter fraud: poll
https://thehill.com/homenews/campaign/559402-one-third-of-americans-believe-biden-won-because-of-voter-fraud-poll

Even though there is no evidence of such fraud.
There is evidence of Nazi propaganda techniques

https://en.wikipedia.org/wiki/Big_lie#Trump's_false_claim_of_a_stolen_election

Revelation 21:8 KJV
....all liars, shall have their part in the lake which burneth with fire
and brimstone: which is the second death.

--
Copyright 2021 Pete Olcott

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

On 10/21/2021 12:52 AM, olcott wrote:
> On 10/21/2021 12:23 AM, Jeff Barnett wrote:
>> On 10/20/2021 6:29 PM, olcott wrote:
>>> On 10/20/2021 7:04 PM, Richard Damon wrote:
>>
>>>> You have to take your choice, choose to restrict yourself to a
>>>> definition of truth that says that can't consider something to be
>>>> true if you can not prove it, and lose most of
>>>
>>> The conspiracy theories based on Nazi propaganda that could otherwise
>>> put an end to Democracy.
>>
>> Bull. The only thing at risk is you advertising to the world that you
>> have shit for brains. You really must see your doctors - current plus
>> a psychiatrist - before your brain drips out in a smelly puddle.
>>
>
> One-third of Americans believe Biden won because of voter fraud: poll
> https://thehill.com/homenews/campaign/559402-one-third-of-americans-believe-biden-won-because-of-voter-fraud-poll
>
>
> Even though there is no evidence of such fraud.
> There is evidence of Nazi propaganda techniques
>
> https://en.wikipedia.org/wiki/Big_lie#Trump's_false_claim_of_a_stolen_election
>
>
> Revelation 21:8 KJV
> ...all liars, shall have their part in the lake which burneth with fire
> and brimstone: which is the second death.

Given your illness, doesn't that quote from Revelations frighten you?
Perhaps you are expecting an exemption because of dissociative
personality. Yes? Note how you cut the vast majority of my message; yet,
your brain is so scrambled that you couldn't give a coherent response to
the part you elected to keep. Don't you see that you are in real trouble
- your brain isn't working right. You can't maintain a dialogue, you
can't keep on track, and just dissociated from reality when you must
face unpleasant facts about your self and condition.

Get help. Show the full text of the last few messages in this thread to
a mental help professional so they can see the full pattern of your
mental imbalance. Throw your self and all your resources into what ever
program (not the TM kind, dummy) that they suggest. Perhaps you can
regain your sanity in the remaining time allotted to you. As it is now,
you are just wasted space.
--
Jeff Barnett

olcott <NoOne@NoWhere.com> writes:

> On 10/20/2021 8:37 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 10/20/2021 7:00 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:

>>>>> 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.
>>>> 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.

But until you understand what he is saying, you can't see why you are
wrong. The result is that you will continue state things that will be
either vacuously true or irrelevant because they refer to an empty class
of TMs. Being correct about no Turing machines is just silly.

If you want to define your own class of TMs that decide something a bit
like halting, go ahead. Such a class may not be empty and your
statements about when the "hat" version will or will not transitions to
qn can be intelligently discussed (but /don't/ call that class H unless
you intend to deliberately confuse your readers).

If you continue to insist that your are using Ĥ to refer to the same
class of modified deciders that Linz does, you have to accept his
definitions and you have to understand the consequences of them.

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

No he is not. He is stating the conditions under which the transition
sequence you keep writing occurs. These follow from the definition of
Ĥ. You need to understand them if you don't want to keep wasting your
time.

>> 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 am wondering if you really read this. In many years of teaching this
material I never came across someone who didn't get it, but I suspect
you are deliberately trying not to.

>> 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 ???

Any statement about no Turing machines is vacuously true because I am
taking you at your word, that you use Ĥ to mean what Linz does: a
class of TMs that are trivial modifications of machines that do this:

H.q0 <M> s ⊢* x1 H.qy x2 if M applied to s halts, and
H.q0 <M> s ⊢* y1 H.qn y2 if M applied to s does not halt.

You don't like the consequences of that, but you can't just make those
consequences go away by stating more things about these non-existent
machines.

If, however, your often repeated waffle about simulation is intended to
define a different, non-empty, class of Turing machines, you need to
start there. Give it a name, J, say, and tell us what goes in the
.... parts of the specification:

J.q0 <M> s ⊢* x1 J.qy x2 if M ...
J.q0 <M> s ⊢* y1 J.qn y2 if M ...

> If we have to go over this same point thousands of times I will.

Sure. You can repeat what you say about no Turing machines as often as
you like, but if you repeat the line

Ĥ.q0 <M> ⊢* Ĥ.qx <M> <M> ⊢* y1 Ĥ.qn y2

I will keep adding the key condition under which this is defined to
occur:

if M applied to <M> does not halt.

> 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

Why change the name? You've been calling this state qx for years. At
least I think that's what you mean by Ĥq0. (If you mean the machine at
Ĥ.q0 then that's just called Ĥ.)

> 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 machine at Ĥ.qx transitions to Ĥ.qn if, and only if, Ĥ applied to
<Ĥ> does not halt:

Ĥ.qx <M> <M> ⊢* y1 Ĥ.qn y2
if M applied to <M> does not halt.

This follows from Linz's definitions and shows that no such TM can
exist. You can't make such a machine exist my saying more things about
it or about some special kinds of these non-existent machines.

Nothing you have said alters the fact that, for Linz-defined H, the Ĥ
version does this: Ĥ.qx <M> <M> ⊢* y1 Ĥ.qn y2 if (and only if) M applied
to <M> does not halt.

I know you don't get why this is the case, but unless you switch into
student mode and start reading what I say and asking intelligent
questions about the parts you don't think are right, I can't help.

> // 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
>
> All dialogue must be either agreeing with the above or pointing out
> errors in the above.

On 10/20/21 8:29 PM, olcott wrote:
> On 10/20/2021 7:04 PM, Richard Damon wrote:
>> On 10/20/21 9:29 AM, olcott wrote:
>>
>>> Then there is the third category where the reason that a math problem
>>> remains open is that there is a problem with the underpinnings of the
>>> philosophy of mathematics that no mathematician challenges because
>>> they are not philosophers thus simply accept these underpinnings as
>>> their given  basis. It doesn't matter that this makes math
>>> incoherent, they simply don't care about that.
>>>
>>> https://en.wikipedia.org/wiki/Ludwig_Wittgenstein was certainly no
>>> crank. He perfectly agrees that unless an analytical truth has been
>>> proven it simply does not count as true. True and unprovable is like
>>> a colorless black car, self-contradictory.
>>>
>>> In times like these where we have Nazi style propaganda that is very
>>> effectively undermining Democracy people really need to know that
>>> true requires provable. If there is no evidence of X then X DOES NOT
>>> COUNT AS TRUE.
>>>
>>> https://www.researchgate.net/publication/333907915_Proof_that_Wittgenstein_is_correct_about_Godel
>>>
>>>
>>
>> If this is your reasoning, then you haven't studied the history of
>> this problem, and are just repeating the mistakes of a century ago.
>>
>> If you want to claim that math is using the 'wrong' definition of
>> 'truth', then you need to go to the very beginning and start over with
>> a change in the definition.
>>
>> If you do, and then try to derive mathematics, you will find that you
>> can't prove anything close to the breadth of field of Mathematics and
>> keep your logic system consistent.
>>
>> You have to take your choice, choose to restrict yourself to a
>> definition of truth that says that can't consider something to be true
>> if you can not prove it, and lose most of
>
> The conspiracy theories based on Nazi propaganda that could otherwise
> put an end to Democracy.
>
>

STRAWMAN.

Belief and Truth are not the same thing.

Something can be True, and even if no one currently believes it, it is
still True.

Everyone can believe in something, but if it is not True, it isn't True.

At one point in the past, the common belief was that the earth was a
basically flat sheet. Just because everyone though it, didn't make it so.

Some philosophies make one of the Goals in life in to come to a better
alignment of our beliefs to what is actually True.

On Thursday, 21 October 2021 at 12:45:39 UTC+1, richar...@gmail.com wrote:
>
> At one point in the past, the common belief was that the earth was a
> basically flat sheet. Just because everyone though it, didn't make it so.
>
In the dim and distant past thatw as undoubtedly true. But as soon as you get into
the historical period, the Greeks knew that the Earth was round, and even had
a theoretically sound method for calculating its diameter (It involved taking
the angles of shadows in Greece and Egypt, and didn't give accurate results
because of measurement problems, but it was basically right).

On 10/21/2021 4:45 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 10/20/2021 8:37 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 10/20/2021 7:00 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>
>>>>>> 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.
>>>>> 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.
>
> 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.

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.

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

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

> The result is that you will continue state things that will be
> either vacuously true or irrelevant because they refer to an empty class
> of TMs. Being correct about no Turing machines is just silly.
>
> If you want to define your own class of TMs that decide something a bit
> like halting, go ahead. Such a class may not be empty and your
> statements about when the "hat" version will or will not transitions to
> qn can be intelligently discussed (but /don't/ call that class H unless
> you intend to deliberately confuse your readers).
>
> If you continue to insist that your are using Ĥ to refer to the same
> class of modified deciders that Linz does, you have to accept his
> definitions and you have to understand the consequences of them.
>
>> Linz is still under the misconception that Ĥ.q0 applied to <Ĥ> <Ĥ>
>> would not correctly transition to H.qn.
>
> No he is not. He is stating the conditions under which the transition
> sequence you keep writing occurs. These follow from the definition of
> Ĥ. You need to understand them if you don't want to keep wasting your
> time.
>

Because Linz ignores the idea of a simulating halt decider he cannot see
that ⟨Ĥ⟩ ⟨Ĥ⟩ presents infinitely nested simulation to ⟨Ĥ⟩ ⟨Ĥ⟩:

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

>>> 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 am wondering if you really read this.

I have standardized the notational conventions to this
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

> In many years of teaching this
> material I never came across someone who didn't get it, but I suspect
> you are deliberately trying not to.
>

Maybe if you were to try and find my mistake instead of perpetually
changing the subject you would begin to see what you are failing to
understand.

>>> 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 ???
>
> Any statement about no Turing machines is vacuously true because I am

As I already said at length in another thread the concept of vacuous
truth is simply incoherent.

This arises because Tarski convinced a bunch of gullible people that
Truth cannot be properly defined using his Undefinability Theorem:
https://liarparadox.org/Tarski_275_276.pdf

> taking you at your word, that you use Ĥ to mean what Linz does: a
> class of TMs that are trivial modifications of machines that do this:
>
> H.q0 <M> s ⊢* x1 H.qy x2 if M applied to s halts, and
> H.q0 <M> s ⊢* y1 H.qn y2 if M applied to s does not halt.
>

That you change the context away from simulating halting deciders is
dishonest.

I have standardized the notational conventions to this
q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ∞
If the simulated input to Ĥq0 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ halts

On 10/21/21 10:04 AM, olcott wrote:
> On 10/21/2021 4:45 AM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 10/20/2021 8:37 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 10/20/2021 7:00 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>
>>>>>>> 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.
>>>>>> 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.
>>
>> 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.
>
> 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.
>
> 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
>
> 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
>
>> The result is that you will continue state things that will be
>> either vacuously true or irrelevant because they refer to an empty class
>> of TMs.  Being correct about no Turing machines is just silly.
>>
>> If you want to define your own class of TMs that decide something a bit
>> like halting, go ahead.  Such a class may not be empty and your
>> statements about when the "hat" version will or will not transitions to
>> qn can be intelligently discussed (but /don't/ call that class H unless
>> you intend to deliberately confuse your readers).
>>
>> If you continue to insist that your are using Ĥ to refer to the same
>> class of modified deciders that Linz does, you have to accept his
>> definitions and you have to understand the consequences of them.
>>
>>> Linz is still under the misconception that Ĥ.q0 applied to <Ĥ> <Ĥ>
>>> would not correctly transition to H.qn.
>>
>> No he is not.  He is stating the conditions under which the transition
>> sequence you keep writing occurs.  These follow from the definition of
>> Ĥ.  You need to understand them if you don't want to keep wasting your
>> time.
>>
>
> Because Linz ignores the idea of a simulating halt decider he cannot see
> that ⟨Ĥ⟩ ⟨Ĥ⟩ presents infinitely nested simulation to ⟨Ĥ⟩ ⟨Ĥ⟩:

WHy do you say the Linz ignored it.

That the simulating Halt decider appears to get stuck in infinitely
nested simulations just shows that such a naive decider fails at its job.

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

But the machine whose representation it is simulating does (at least if
H is smart enough to abort the simulation and give the non-halting
answer as you claim). It is the actual machine, not the partial
simulation in H that matters. FAIL.

>>>> 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 am wondering if you really read this.
>
> I have standardized the notational conventions to this
> q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ∞
> If the simulated input to Ĥq0 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ halts

WRONG. That statement is POOP.

The Halting criteria (Which Linz gives) is if the Computation
represented by the input to the decider Halts.

H^(<H^>) DOES Halt (at least if H(<H^>,<H^> gives the non-halting answer
you claim) so H is wrong.

Only if H does something else, either not Halt and answer or answer
Halting is H^ non-Halting, and in those cases H is still wrong.

>
> q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
> If the simulated input to Ĥq0 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt
>
>> In many years of teaching this
>> material I never came across someone who didn't get it, but I suspect
>> you are deliberately trying not to.
>>
>
> Maybe if you were to try and find my mistake instead of perpetually
> changing the subject you would begin to see what you are failing to
> understand.
>

On 10/21/2021 6:15 PM, Richard Damon wrote:
> On 10/21/21 10:04 AM, olcott wrote:
>> On 10/21/2021 4:45 AM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 10/20/2021 8:37 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 10/20/2021 7:00 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>>>>>> 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.
>>>>>>> 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.
>>>
>>> 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.
>>
>> 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.
>>
>> 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
>>
>> 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
>>
>>> The result is that you will continue state things that will be
>>> either vacuously true or irrelevant because they refer to an empty class
>>> of TMs.  Being correct about no Turing machines is just silly.
>>>
>>> If you want to define your own class of TMs that decide something a bit
>>> like halting, go ahead.  Such a class may not be empty and your
>>> statements about when the "hat" version will or will not transitions to
>>> qn can be intelligently discussed (but /don't/ call that class H unless
>>> you intend to deliberately confuse your readers).
>>>
>>> If you continue to insist that your are using Ĥ to refer to the same
>>> class of modified deciders that Linz does, you have to accept his
>>> definitions and you have to understand the consequences of them.
>>>
>>>> Linz is still under the misconception that Ĥ.q0 applied to <Ĥ> <Ĥ>
>>>> would not correctly transition to H.qn.
>>>
>>> No he is not.  He is stating the conditions under which the transition
>>> sequence you keep writing occurs.  These follow from the definition of
>>> Ĥ.  You need to understand them if you don't want to keep wasting your
>>> time.
>>>
>>
>> Because Linz ignores the idea of a simulating halt decider he cannot
>> see that ⟨Ĥ⟩ ⟨Ĥ⟩ presents infinitely nested simulation to ⟨Ĥ⟩ ⟨Ĥ⟩:
>
> WHy do you say the Linz ignored it.
>
> That the simulating Halt decider appears to get stuck in infinitely
> nested simulations just shows that such a naive decider fails at its job.
>

A SELF-EVIDENT TRUTH is any expression of language that can be verified
as completely true entirely on the basis of the meaning of its words.**

SELF-EVIDENT TRUTH
When a simulating halt correctly decider detects that an input would
never reach its final state then this simulating halt decider is always
correct when it aborts the simulation of this input and reports not
halting.

**This also applies to formal language expressions that do not have any
words.

--
Copyright 2021 Pete Olcott

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

On 10/21/2021 6:30 PM, olcott wrote:
> On 10/21/2021 6:15 PM, Richard Damon wrote:
>> On 10/21/21 10:04 AM, olcott wrote:
>>> On 10/21/2021 4:45 AM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 10/20/2021 8:37 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 10/20/2021 7:00 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>>>>>> 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.
>>>>>>>> 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.
>>>>
>>>> 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.
>>>
>>> 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.
>>>
>>> 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
>>>
>>> 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
>>>
>>>> The result is that you will continue state things that will be
>>>> either vacuously true or irrelevant because they refer to an empty
>>>> class
>>>> of TMs.  Being correct about no Turing machines is just silly.
>>>>
>>>> If you want to define your own class of TMs that decide something a bit
>>>> like halting, go ahead.  Such a class may not be empty and your
>>>> statements about when the "hat" version will or will not transitions to
>>>> qn can be intelligently discussed (but /don't/ call that class H unless
>>>> you intend to deliberately confuse your readers).
>>>>
>>>> If you continue to insist that your are using Ĥ to refer to the same
>>>> class of modified deciders that Linz does, you have to accept his
>>>> definitions and you have to understand the consequences of them.
>>>>
>>>>> Linz is still under the misconception that Ĥ.q0 applied to <Ĥ> <Ĥ>
>>>>> would not correctly transition to H.qn.
>>>>
>>>> No he is not.  He is stating the conditions under which the transition
>>>> sequence you keep writing occurs.  These follow from the definition of
>>>> Ĥ.  You need to understand them if you don't want to keep wasting your
>>>> time.
>>>>
>>>
>>> Because Linz ignores the idea of a simulating halt decider he cannot
>>> see that ⟨Ĥ⟩ ⟨Ĥ⟩ presents infinitely nested simulation to ⟨Ĥ⟩ ⟨Ĥ⟩:
>>
>> WHy do you say the Linz ignored it.
>>
>> That the simulating Halt decider appears to get stuck in infinitely
>> nested simulations just shows that such a naive decider fails at its job.
>>
>
> A SELF-EVIDENT TRUTH is any expression of language that can be verified
> as completely true entirely on the basis of the meaning of its words.**
>
> SELF-EVIDENT TRUTH
> When a simulating halt correctly decider detects that an input would

When a simulating halt decider correctly detects that an input would

> never reach its final state then this simulating halt decider is always
> correct when it aborts the simulation of this input and reports not
> halting.
>
> **This also applies to formal language expressions that do not have any
> words.
>

--
Copyright 2021 Pete Olcott

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

olcott <NoOne@NoWhere.com> writes:

> On 10/21/2021 4:45 AM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 10/20/2021 8:37 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 10/20/2021 7:00 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>
>>>>>>> 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.
>>>>>> 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.
>> 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.

> 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.

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.

You can't do either and the keep the conversation going, so you have to
keep pretend that no one is pointing out an error.

> That you change the context away from simulating halting deciders is
> dishonest.

Linz's class of machines, H, includes any and all halting deciders,
"simulating" or otherwise, so I have not changed the subject. You have
narrowed it down to pointlessly talking about a subset of an already
empty set of Turing machines.

--
Ben.

On 10/21/2021 9:19 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 10/21/2021 4:45 AM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 10/20/2021 8:37 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 10/20/2021 7:00 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>>>>>> 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.
>>>>>>> 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.
>>> 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.

--
Copyright 2021 Pete Olcott

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

On 10/21/21 7:30 PM, olcott wrote:
> On 10/21/2021 6:15 PM, Richard Damon wrote:
>> On 10/21/21 10:04 AM, olcott wrote:
>>> On 10/21/2021 4:45 AM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 10/20/2021 8:37 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 10/20/2021 7:00 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>>>>>> 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.
>>>>>>>> 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.
>>>>
>>>> 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.
>>>
>>> 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.
>>>
>>> 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
>>>
>>> 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
>>>
>>>> The result is that you will continue state things that will be
>>>> either vacuously true or irrelevant because they refer to an empty
>>>> class
>>>> of TMs.  Being correct about no Turing machines is just silly.
>>>>
>>>> If you want to define your own class of TMs that decide something a bit
>>>> like halting, go ahead.  Such a class may not be empty and your
>>>> statements about when the "hat" version will or will not transitions to
>>>> qn can be intelligently discussed (but /don't/ call that class H unless
>>>> you intend to deliberately confuse your readers).
>>>>
>>>> If you continue to insist that your are using Ĥ to refer to the same
>>>> class of modified deciders that Linz does, you have to accept his
>>>> definitions and you have to understand the consequences of them.
>>>>
>>>>> Linz is still under the misconception that Ĥ.q0 applied to <Ĥ> <Ĥ>
>>>>> would not correctly transition to H.qn.
>>>>
>>>> No he is not.  He is stating the conditions under which the transition
>>>> sequence you keep writing occurs.  These follow from the definition of
>>>> Ĥ.  You need to understand them if you don't want to keep wasting your
>>>> time.
>>>>
>>>
>>> Because Linz ignores the idea of a simulating halt decider he cannot
>>> see that ⟨Ĥ⟩ ⟨Ĥ⟩ presents infinitely nested simulation to ⟨Ĥ⟩ ⟨Ĥ⟩:
>>
>> WHy do you say the Linz ignored it.
>>
>> That the simulating Halt decider appears to get stuck in infinitely
>> nested simulations just shows that such a naive decider fails at its job.
>>
>
> A SELF-EVIDENT TRUTH is any expression of language that can be verified
> as completely true entirely on the basis of the meaning of its words.**
>
> SELF-EVIDENT TRUTH
> When a simulating halt correctly decider detects that an input would
> never reach its final state then this simulating halt decider is always
> correct when it aborts the simulation of this input and reports not
> halting.
>
> **This also applies to formal language expressions that do not have any
> words.
>

And you can't jusg claim somethoing to be a 'Self-Evident Truth' unless
you can actually show it IS, and other agree to it.

If others (and not just one or two people) don't agree, then it is
obviously NOT self-evident.

You make LOTS of 'Claims' about stuff being self-evident when it isn't.

You have fundamental logical errors in your statement, so it really
isn't a self evident truth.

First, inputs don't do anything, they are just strings, which don't do
anything except be data for algorithms to do computations.

You apply this data to the wrong algorithm, the question of the halting
problem is what happens when this data is interpreted as the
representation of a machine and its input, and that machine is applied
to that input, and asks if this machine would halt or not.

An equivalent question would be what would happen if this input is given
to an actual UTM that simulates this input until it reaches a halting
state or continues running this simulation forever.

THAT is the question.

You are looking at what the partial incomplete simulation done by the
decider does. THAT IS WRONG.

A WRONG THING IS NEVER THE 'TRUTH', so can't be a 'self-evident truth'.

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

DV <xlt.pjw@gmail.com> wrote:
> On Wednesday, October 20, 2021 at 4:01:32 PM UTC-4, Alan Mackenzie wrote:
>> DV <xlt...@gmail.com> wrote:

>> >> That is also absurd, in the sense that the "right way" to conduct
>> >> one's self after having completed valuable mathematical work can
>> >> change with time, can be "warped" by circumstances, can be disguised
>> >> by mathematicians with different motives (including very legitimate
>> >> political ones, as I've made clear), and may not be well-known or even
>> >> easily quantified at any given time.

>> >> What if a mathematician has some sense of "the process" for assessing
>> >> how mathematical proofs are chosen to be looked at or not looked at,
>> >> despises this process, and acts in a way to change it?

>> >> Perhaps someone might reply that analysis of this process itself by
>> >> the establishment might lead to counter-measures, too.

>> >> Of course, I'm looking at "interest cues" regarding "does anyone want
>> >> to see the proof." That is the target of my satire, and I've made that
>> >> clear. If it is expected that I "be humbly respectful" to
>> >> mathematicians who literally defraud people like me every day, then I
>> >> have no interest in bowing to such criminals, however they present
>> >> themselves.
>> I think you'll find these "criminals" are humbly respectful towards each
>> other when they present their work. The number of mathematicians who
>> "literally defraud people" is likely very small, just as the number of
>> fraudsters in the wider population is also quite small.

> OK, I'll answer this.

> Aiding and abetting fraud is a crime where I'm from. Academic math
> people get benefit from their careers from participating in submissions
> to fraudulent academic journals as I have seen. I.e., I have seen
> academic journals be fraudulent.

I can only take that as hyperbole.

> Many academic math people are respectful no doubt. I have the right to
> criticize well-known people and establishments when they commit crimes,
> including disrespectfully, even if such accused (by me) criminals are
> polite about committing their crimes. I too am respectful to people
> who treat me with the respect I deserve. I am only a little bit
> sensitive.

"Crime" is a specific word with a specific meaning. It is something you
can be fined or imprisoned for. Other uses are somewhat metaphorical.

> Perhaps I should have said "literally aid and abet fraud." Benefiting
> from submitting to a fraudulent academic journal is criminal if you
> know about the fraud, I assert.

Perhaps you should not have broached the topic at all. You've made no
specific allegations, yet cast aspersions on all academic maths journals
and practitioners. That can't help improve your lot.

>> >> If you're suggesting that "if my cues changed," my proof would be
>> >> read, that is also absurd. I allege that academic math has no interest
>> >> in working with disabled people like me, ....

>> I think it's likely that disability status has no bearing on the
>> interest of academicians (is that a word?).

> I disagree with that ardently. How do you know anyway, how many
> disabled mathematicians have you met or interacted with?

None that I can think of. For that matter, none of my colleagues are
disabled, that I know about. Some of my friends are. A very
distinguished disabled mathematician died last year, so they do exist.

>> >> .... and that there is literally no reasonable way for me to get my
>> >> work looked at--there is very likely nothing I could say that would
>> >> get journals or anyone to read my proof and admit that it is
>> >> correct, at least not without tremendous effort.

>> Well, doing what you said above you wouldn't do would help quite a
>> bit.

>> >> It seems that no one has really challenged my claim about the
>> >> highly prejudiced nature of the mathematics establishment in a
>> >> meaningful way.

>> It isn't a productive argument to get into.

> Fine, you don't have to debate me on this.

>> >> Life is all about values and conditions. When idiots cannot measure
>> >> up, they are held accountable and denied access to math.

Most people, including "idiots", are exposed to and taught maths at
school.

>> >> In case you are wondering: This is not a JSH imitation thing. I
>> >> suspect JSH anticipated mathematicians like me, although I don't
>> >> claim to be the last such one.

>> >> The proof is absolutely correct, assuming I wrote the definitions
>> >> right.

>> Until it's checked by somebody else who's qualified and actively
>> looking for flaws, it's a bit premature to claim absolute correctness.

> Okay, you seem not to know completely what a mathematical proof is.

I think I do.

> Many undergraduate math majors who have graduated do not know this; I
> didn't know it when I graduated either. Further study was required for
> me, and I did it on my own.

> Let my clarify that: A mathematical proof, formally, is a sequence of
> well-formed formulas that, when the two inference rules of ZFC are
> properly applied, can be asserted to be "valid" in a precise sense and
> to terminate with the wf to be proved.

OK, but you've got to have some starting axioms, too.

> The most tremendously useful thing any mathematician can do to learn
> more about math proofs is: To learn logic and get a sense of what a
> correct proof is. Most proofs are written as "informal mathematical
> arguments" that any well-educated math person could easily see could be
> translated into a formal proof...a sequence of wfs, ending with the wf
> to be proved.

> The proof is absolutely correct, I've checked it repeatedly. The only
> possibility of incorrectness is if I misunderstood the problem
> description in the CMI write-up, which is the best exposition of the
> Hodge Conjecture.

I don't know what CMI means. You say your proof is a sequence of wffs
which end in the result. Do you actually understand these wffs, or is
your proof purely formal?

There are several possibilities of incorrectness. You may have made a
mistake. Even these "fraudulent" journals insist on having a second (at
least) person check work before accepting it, for this reason.

>> >> Will anyone ask to see it? If it looks right, will anyone read it?
>> >> How many of you are campaigning for Donald Trump?

>> That's the sort of remark which puts people off taking you seriously.

> I'm asserting that apathy about math and STEM hurts American
> liberalism, aiding DJT's campaign for presidency, like it or not.

That's a rather thin connection between maths and certain types of
politics.

> My assertion is made quite seriously, although I don't allege literal
> direct campaign activity.

>> >> I know you don't care about math or other people.

>> [ .... ]

> Maybe someone who posts on here does other than me, but I haven't seen it.

>> >> OK. Fine. I encourage anyone anywhere in the world to go ahead and
>> >> take the bait.

>> > WOW!! It looks like literally no one has BEGGED me to post my Hodge
>> > Conjecture proof! What a surprise!!

>> > Here, I'll help to facilitate this research breakthrough publication:

>> > "PLEASE, PLEASE, PLEASE...somebody ask me to publish my Hodge
>> > Conjecture proof! It is so important to so many people! You can
>> > pretend it is wrong! You can laugh to yourselves!! PLEASE!!!!! I BEG
>> > OF YOU!!!!!!!!!! USENET POSTERS AND READER, I beg you to PLEASE help
>> > important science, and ask for my Hodge Conjecture proof! Here are
>> > some possible ways to phrase it:

>> > - Please publish it.

>> > OR, if the p-word is something you're uncomfortable with...

>> > - Publish it.
>> > - Go do it.
>> > - I want to see it.
>> > - Fine, write it.
>> > - Do it.
>> > - Go.
>> > - Write it.

>> > GOSH, DOES ANYONE CARE AT ALL???"

>> > I'm awaiting replies....

>> Well, I'm not a mathematician, though I have a degree in maths. I don't
>> know what the Hodge Conjecture is. Perhaps it might help motivate people
>> if you briefly described what it is, and why it's important. Giving a
>> sober description of the conjecture would likely nudge people towards
>> believing you have a real proof. Some of your recent posts, I'm afraid,
>> are likely to nudge people in the opposite direction.

On Friday, October 22, 2021 at 12:56:58 PM UTC-4, Alan Mackenzie wrote:
> DV <xlt...@gmail.com> wrote:
> > On Wednesday, October 20, 2021 at 4:01:32 PM UTC-4, Alan Mackenzie wrote:
> >> DV <xlt...@gmail.com> wrote:
>
> >> >> That is also absurd, in the sense that the "right way" to conduct
> >> >> one's self after having completed valuable mathematical work can
> >> >> change with time, can be "warped" by circumstances, can be disguised
> >> >> by mathematicians with different motives (including very legitimate
> >> >> political ones, as I've made clear), and may not be well-known or even
> >> >> easily quantified at any given time.
>
> >> >> What if a mathematician has some sense of "the process" for assessing
> >> >> how mathematical proofs are chosen to be looked at or not looked at,
> >> >> despises this process, and acts in a way to change it?
>
> >> >> Perhaps someone might reply that analysis of this process itself by
> >> >> the establishment might lead to counter-measures, too.
>
> >> >> Of course, I'm looking at "interest cues" regarding "does anyone want
> >> >> to see the proof." That is the target of my satire, and I've made that
> >> >> clear. If it is expected that I "be humbly respectful" to
> >> >> mathematicians who literally defraud people like me every day, then I
> >> >> have no interest in bowing to such criminals, however they present
> >> >> themselves.
> >> I think you'll find these "criminals" are humbly respectful towards each
> >> other when they present their work. The number of mathematicians who
> >> "literally defraud people" is likely very small, just as the number of
> >> fraudsters in the wider population is also quite small.
>
> > OK, I'll answer this.
>
> > Aiding and abetting fraud is a crime where I'm from. Academic math
> > people get benefit from their careers from participating in submissions
> > to fraudulent academic journals as I have seen. I.e., I have seen
> > academic journals be fraudulent.
>
> I can only take that as hyperbole.
>

I don't mean it as hyperbole. I don't have a law degree and have not consulted a lawyer, but I think that inasmuch as "aid and abet" means "assist and encourage," my allegation makes perfect sense and could be supported by an interaction I had with the JACM based on a paper that contained correct proofs, was analyzed incorrectly, and, although my paper technically did not quite establish what I claimed it established due to a philosophical issue with ZFC, contained no errors. I encourage you not to repeat your unfounded claim that I "could not possibly know if it is correct" because I will not believe any claims that you are clairvoyant.

> > Many academic math people are respectful no doubt. I have the right to
> > criticize well-known people and establishments when they commit crimes,
> > including disrespectfully, even if such accused (by me) criminals are
> > polite about committing their crimes. I too am respectful to people
> > who treat me with the respect I deserve. I am only a little bit
> > sensitive.
>
> "Crime" is a specific word with a specific meaning. It is something you
> can be fined or imprisoned for. Other uses are somewhat metaphorical.
>

Yes, I meant it literally. It is possible that a court of law would disagree that it is criminal, but in the sense that my assertion that "fraud" is a crime--claiming to be a legitimate academic journal and then not analyzing academic articles truthfully--my claim is correct. Maybe the definitions of the word "aid" and "abet" are not fully known to me in the legal sense. As a citizen with a citizen's understanding of US law, of course I stand by my claim.

> > Perhaps I should have said "literally aid and abet fraud." Benefiting
> > from submitting to a fraudulent academic journal is criminal if you
> > know about the fraud, I assert.
>
> Perhaps you should not have broached the topic at all. You've made no
> specific allegations, yet cast aspersions on all academic maths journals
> and practitioners. That can't help improve your lot.
>

Of course I should have.

> >> >> If you're suggesting that "if my cues changed," my proof would be
> >> >> read, that is also absurd. I allege that academic math has no interest
> >> >> in working with disabled people like me, ....
>
> >> I think it's likely that disability status has no bearing on the
> >> interest of academicians (is that a word?).
>
> > I disagree with that ardently. How do you know anyway, how many
> > disabled mathematicians have you met or interacted with?
>
> None that I can think of. For that matter, none of my colleagues are
> disabled, that I know about. Some of my friends are. A very
> distinguished disabled mathematician died last year, so they do exist.
>

I'm pleased that you have disabled friends.

> >> >> .... and that there is literally no reasonable way for me to get my
> >> >> work looked at--there is very likely nothing I could say that would
> >> >> get journals or anyone to read my proof and admit that it is
> >> >> correct, at least not without tremendous effort.
>
> >> Well, doing what you said above you wouldn't do would help quite a
> >> bit.
>
> >> >> It seems that no one has really challenged my claim about the
> >> >> highly prejudiced nature of the mathematics establishment in a
> >> >> meaningful way.
>
> >> It isn't a productive argument to get into.
>
> > Fine, you don't have to debate me on this.
>
> >> >> Life is all about values and conditions. When idiots cannot measure
> >> >> up, they are held accountable and denied access to math.
>
> Most people, including "idiots", are exposed to and taught maths at
> school.
>

I meant novel math research.

> >> >> In case you are wondering: This is not a JSH imitation thing. I
> >> >> suspect JSH anticipated mathematicians like me, although I don't
> >> >> claim to be the last such one.
>
> >> >> The proof is absolutely correct, assuming I wrote the definitions
> >> >> right.
>
> >> Until it's checked by somebody else who's qualified and actively
> >> looking for flaws, it's a bit premature to claim absolute correctness.
>
> > Okay, you seem not to know completely what a mathematical proof is.
>
> I think I do.
>

Well, you acted like you really didn't and received an explanation from me....for free! Certainly, now you do.

> > Many undergraduate math majors who have graduated do not know this; I
> > didn't know it when I graduated either. Further study was required for
> > me, and I did it on my own.
>
> > Let my clarify that: A mathematical proof, formally, is a sequence of
> > well-formed formulas that, when the two inference rules of ZFC are
> > properly applied, can be asserted to be "valid" in a precise sense and
> > to terminate with the wf to be proved.
>
> OK, but you've got to have some starting axioms, too.
>

Yes.

> > The most tremendously useful thing any mathematician can do to learn
> > more about math proofs is: To learn logic and get a sense of what a
> > correct proof is. Most proofs are written as "informal mathematical
> > arguments" that any well-educated math person could easily see could be
> > translated into a formal proof...a sequence of wfs, ending with the wf
> > to be proved.
>
> > The proof is absolutely correct, I've checked it repeatedly. The only
> > possibility of incorrectness is if I misunderstood the problem
> > description in the CMI write-up, which is the best exposition of the
> > Hodge Conjecture.
>
> I don't know what CMI means. You say your proof is a sequence of wffs
> which end in the result. Do you actually understand these wffs, or is
> your proof purely formal?
>

"Clay Mathematics Institute," I don't like it any more. My proof is written informally, not as a list of ZFC wfs; the obvious metaphor is: Translating pseudocode for an algorithm into C++. I struggled with proofs in the past when dealing with math that was written in ways that represented effective exposition that I was uncomfortable with, because I did not have a deep understanding of mathematical logic. Working exercises in Chapters 1-2 of Mendelson changed that dramatically for me.

olcott <NoOne@NoWhere.com> writes:

> On 10/21/2021 9:19 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 10/21/2021 4:45 AM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 10/20/2021 8:37 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 10/20/2021 7:00 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>>>>>> 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.
>>>>>>>> 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.
>>>> 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.

A proved theorem is not a "false assumption". You have not invalidated
any proof of the halting theorem, much less all of them. You have not
even read Linz's real proof, only the simplistic one that you latched on
to all those years ago. Even there, I have taken you through the steps
(which you appear to agree with right up to the last, or second to last
step) and you declined to say what you think is invalid about it. You
just keep removing the key line that shows you are wrong.

>>> 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.

Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to Ĥqn if (only if) Ĥ applied to ⟨Ĥ⟩ does not
halt.

All I can do now is just correct the error. I've taken you through the
steps from the ones you have no objection to right up to this conclusion
and you simply reply by asserting contrary statements. I remain willing
to try to explain it, but you exhibit all the signs of someone
determined not to learn.

>> 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.

Childish.

Unless you tell me you were wrong when you said you are talking about
Linz's Ĥ, I will continue to assume that that is what you mean by the
symbol. Linz's Ĥ is derived from an H that decides all cases. From
that is follows using the simplest of logical steps that no such H (and
therefore Ĥ) can exist. A TM J that decides just one case involving Ĵ
would have a very different specification. I have invited you give it,
but of course you didn't.

If you want to stop talking about Linz's Ĥ, define what your "one case"
decider really does and we'll discuss it.

--
Ben.

On 10/22/2021 6:57 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 10/21/2021 9:19 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 10/21/2021 4:45 AM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 10/20/2021 8:37 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 10/20/2021 7:00 PM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>>>>>> 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.
>>>>>>>>> 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.
>>>>> 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.
>
> 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.

The problem with the proved theorem is that I have showed how it is
proved to be incorrect. Your whole "rebuttal" is simply assuming that I
must be wrong.

q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn

If the simulated input to Ĥq0 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never reaches its final
state (even if infinitely simulated) this makes the undecidable case
decidable as not halting.

You can't refute this because it is correct.

> You have not invalidated
> any proof of the halting theorem, much less all of them. You have not
> even read Linz's real proof, only the simplistic one that you latched on
> to all those years ago. Even there, I have taken you through the steps
> (which you appear to agree with right up to the last, or second to last
> step) and you declined to say what you think is invalid about it. You
> just keep removing the key line that shows you are wrong.
>

It seem like you are saying that a halt decider that does correctly
decide that its input never halts can still be wrong.
That would be a woefully foolish mistake on your part.

>>>> 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.
>
> Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to Ĥqn if (only if) Ĥ applied to ⟨Ĥ⟩ does not
> halt.
>
> All I can do now is just correct the error. I've taken you through the
> steps from the ones you have no objection to right up to this conclusion
> and you simply reply by asserting contrary statements. I remain willing
> to try to explain it, but you exhibit all the signs of someone
> determined not to learn.
>
>>> 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.
>
> Childish.
>
> Unless you tell me you were wrong when you said you are talking about
> Linz's Ĥ, I will continue to assume that that is what you mean by the
> symbol. Linz's Ĥ is derived from an H that decides all cases. From
> that is follows using the simplest of logical steps that no such H (and
> therefore Ĥ) can exist. A TM J that decides just one case involving Ĵ
> would have a very different specification. I have invited you give it,
> but of course you didn't.
>
> If you want to stop talking about Linz's Ĥ, define what your "one case"
> decider really does and we'll discuss it.
>

On 10/23/21 12:19 AM, olcott wrote:
> On 10/22/2021 6:57 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 10/21/2021 9:19 PM, Ben Bacarisse wrote:

>>> 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.
>>
>> 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.

So, if Proof can't be used to establish truth, I guess you just believe
that there isn't really anything that is true. That sore of makes sense
from you, if nothing can really ever be proved to be true, then there is
no problem in claiming things that are false to be true, so you couldn't
ever actually prove that either.

Just shows how you are a pathelogical liar.

>
> The problem with the proved theorem is that I have showed how it is
> proved to be incorrect. Your whole "rebuttal" is simply assuming that I
> must be wrong.
>

No, you haven't shown anything incorrect about the proof, you have just
shown a 'claimed' counter example.

> q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn

So H^(<H^>) is claimed to halt. THAT is what that line says,

H^.q0 <H^> goes to H^.qx/H.q0 <H^> <H^> (the claimed decider) which goes
to state qn where H^ Halts and H predicts that it won't ever halt (and
thus is wrong)
>
> If the simulated input to Ĥq0 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never reaches its final
> state (even if infinitely simulated) this makes the undecidable case
> decidable as not halting.
>

You have the wrong question. It is does the computation represented ever
halt or would this input if given to a real UTM halt, and in both cases
the answer is YES.

That fact that you can't make an H get there just proves that your
technique is insufficient to answer the question. H NEVER needed to be
able to simulate its input to a halting state for the computation to be
a Halting computation.

> You can't refute this because it is correct.

WRONG.

>
>
>>  You have not invalidated
>> any proof of the halting theorem, much less all of them.  You have not
>> even read Linz's real proof, only the simplistic one that you latched on
>> to all those years ago.  Even there, I have taken you through the steps
>> (which you appear to agree with right up to the last, or second to last
>> step) and you declined to say what you think is invalid about it.  You
>> just keep removing the key line that shows you are wrong.
>>
>
> It seem like you are saying that a halt decider that does correctly
> decide that its input never halts can still be wrong.
> That would be a woefully foolish mistake on your part.

IT is wrong if that input does halt, like H^(<H^>)

Just like your Poodle you entered into the cat show.

The Halting Question is NOT can the decider simulate the machine to a
Halting State, it is will the computation when run by itself reach the
Halting State.

Uness your decider IS a UTM, it doesn't have the property that its
simulation matches the behavior of the computation it is given. But, if
your decider actually IS a UTM, it can NEVER answer 'non-halting', as it
will continue to simulate the machine, and thus fails to be a proper
decider.

Which way are you wrong?

olcott <NoOne@NoWhere.com> writes:

> On 10/22/2021 6:57 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 10/21/2021 9:19 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 10/21/2021 4:45 AM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 10/20/2021 8:37 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> On 10/20/2021 7:00 PM, Ben Bacarisse wrote:
>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>>>>>> 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.
>>>>>>>>>> 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.
>>>>>> 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.
>>
>> 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.

> The problem with the proved theorem is that I have showed how it is
> proved to be incorrect.

Bad use of terms. To invalidate a theorem (which is what I think you
mean by "proved to be incorrect"), you must do more than look at one
proof. Why? Well you /have/ show Linz's proof to be invalid. You did
that years ago. So why are you continuing?

> Your whole "rebuttal" is simply assuming that I
> must be wrong.

No. I agreed, years ago, that you found an error in Linz's proof. The
halting theorem is unaffected since there are so many other correct
proof (indeed there is one in Linz that you have never mentioned because
I suspect you can't understand it) but you /have/ invalidated the proof
in Linz's that has so obsessed you.

> q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
>
> If the simulated input to Ĥq0 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never reaches its
> final state (even if infinitely simulated) this makes the undecidable
> case decidable as not halting.
>
> You can't refute this because it is correct.

What I assert is that this is not what Linz says about his Ĥ. I don't
care if what you also say correct or not. If it is correct about Linz's
Ĥ, it is correct about a non-existent Turing machine, and if it is not
correct about Linz's Ĥ, I have not reason to care about the TMs it is
correct about (and it means you were lying about using Ĥ only as Linz
does).

I have taken you through the steps by which is it obvious (to those that
know what the notation means) that

q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn

if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt. You can't point to the
step you think is wrong. All you do is, right at the end, state
something else. That's not how you do mathematics.

>> You have not invalidated any proof of the halting theorem, much less
>> all of them. You have not even read Linz's real proof, only the
>> simplistic one that you latched on to all those years ago. Even
>> there, I have taken you through the steps (which you appear to agree
>> with right up to the last, or second to last step) and you declined
>> to say what you think is invalid about it. You just keep removing
>> the key line that shows you are wrong.
>
> It seem like you are saying that a halt decider that does correctly
> decide that its input never halts can still be wrong. That would be a
> woefully foolish mistake on your part.

Your track record in knowing what I am saying is very poor. The way to
deal with this is to ask intelligent questions so that I can explain in
more detail. I am open to such questions if you have any.

--
Ben.

On 10/24/2021 6:30 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 10/22/2021 6:57 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 10/21/2021 9:19 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 10/21/2021 4:45 AM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 10/20/2021 8:37 PM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> On 10/20/2021 7:00 PM, Ben Bacarisse wrote:
>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>>>>>> 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.
>>>>>>>>>>> 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.
>>>>>>> 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.
>>>
>>> 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

--
Copyright 2021 Pete Olcott

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

On 10/25/21 10:46 AM, olcott wrote:
> On 10/24/2021 6:30 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 10/22/2021 6:57 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 10/21/2021 9:19 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 10/21/2021 4:45 AM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> On 10/20/2021 8:37 PM, Ben Bacarisse wrote:
>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>
>>>>>>>>>>> On 10/20/2021 7:00 PM, Ben Bacarisse wrote:
>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>>>>>> 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.
>>>>>>>>>>>> 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.
>>>>>>>> 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.
>>>>
>>>> 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
>

Except that you haven't actually shown (b)

H(<H^>,<H^>) says that H^(<H^>) will be non-halting but H^(<H^>) is halting.

You have just shown that the partial simulation of H doesn't see this
halting state because it aborts its simulation too soon. In fact you
prove that no H can simulate long enough to see that halting state, but
Linz established that already, as if H could simulate to that point, it
could accurated answer Halting and be right.

olcott <NoOne@NoWhere.com> writes:

> On 10/24/2021 6:30 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 10/22/2021 6:57 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 10/21/2021 9:19 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 10/21/2021 4:45 AM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> On 10/20/2021 8:37 PM, Ben Bacarisse wrote:
>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>
>>>>>>>>>>> On 10/20/2021 7:00 PM, Ben Bacarisse wrote:
>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>>>>>> 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.
>>>>>>>>>>>> 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.
>>>>>>>> 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.
>>>>
>>>> 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.
Neither do I think you have trouble knowing what part I am saying you
have failed at (it's (b) by the way).

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.

--
Ben.

On 10/25/2021 7:46 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 10/24/2021 6:30 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 10/22/2021 6:57 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 10/21/2021 9:19 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 10/21/2021 4:45 AM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> On 10/20/2021 8:37 PM, Ben Bacarisse wrote:
>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>>
>>>>>>>>>>>> On 10/20/2021 7:00 PM, Ben Bacarisse wrote:
>>>>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>>>>>> 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.
>>>>>>>>>>>>> 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.
>>>>>>>>> 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.
>>>>>
>>>>> 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.

--
Copyright 2021 Pete Olcott

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