On 10/27/2021 10:31 PM, André G. Isaak wrote:
> On 2021-10-27 20:18, olcott wrote:
>> On 10/27/2021 8:55 PM, Richard Damon wrote:
>>> On 10/27/21 9:39 PM, olcott wrote:
>>>> On 10/27/2021 6:46 PM, Richard Damon wrote:
>>>>>
>>>>> On 10/27/21 10:21 AM, olcott wrote:
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>> If the UTM simulation of the input to Ĥqx ⟨Ĥ⟩ ⟨Ĥ⟩ would never
>>>>>> reach its final state (whether or not this simulation is aborted)
>>>>>> then it is necessarily true that Ĥqx transitions to Ĥ.qn correctly.
>>>
>>> ^^^^^^^^^^^^^^^^^ NOTE WHAT YOU SAID HERE ^^^^^^^^^^^^^^^^^^^^^^^^
>>>
>>>>>
>>>>> Finally, you use the right words. We Know that UTM(<H^> <H^>) will
>>>>> act exactly like H^<H^>).
>>>>>
>>>>> We have that:
>>>>> q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
>>>>>
>>>>> Where the above initial q0 is that starting state of H^, so this is
>>>>> showing the behavior of H^ (<H^>) which we see ends up at H^qn
>>>>> which is defined as a terminal state for H^, so this PROVES that H^
>>>>> applied to <H^> is a Halting Computation.
>>>> That Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ halts is not the question.
>>>> The simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by Ĥq0 never halts is the answer.
>>>
>>> No, as you said earlier, it is what the input to H (or H^q0) would do
>>> if it was made the input to a UTM, NOT your 'non-utm' partial
>>> simulation H.
>>>
>>> The right answer to the Halting Problem is ALWAYS about the actual
>>> machine who has been converted into a representation to give to the
>>> decider, never about what happens in the decider.
>>
>> So then you are saying that when the halt decider does correctly
>> decide that its input never halts that it can still be wrong?
>>
>> That is irrational.
>
> You've got that completely ass-backwards. You're the one arguing that
> when your decider gets the wrong answer as defined by the problem it is
> somehow still deciding 'correctly'.
>

As long as the halt decider correctly decides that its input never halts

Then when we directly execute Ĥ applied to ⟨Ĥ⟩ and it:
(a) Halts.
(b) Destroys the whole universe.
(c) Converts all fascists to true Christianity.
(d) Circumnavigates the globe.
(e) Cures cancer.
(f) Raises the dead.

The halt decider's halt status decision about its input remains correct.

If we know that we have a black cat then saying that we don't have a
cat is irrational. Tautologies cannot possibly fail. If you believe
that they can fail then you must be irrational.

--
Copyright 2021 Pete Olcott

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

On 2021-10-27 22:14, olcott wrote:
> On 10/27/2021 10:31 PM, André G. Isaak wrote:
>> On 2021-10-27 20:18, olcott wrote:
>>> On 10/27/2021 8:55 PM, Richard Damon wrote:
>>>> On 10/27/21 9:39 PM, olcott wrote:
>>>>> On 10/27/2021 6:46 PM, Richard Damon wrote:
>>>>>>
>>>>>> On 10/27/21 10:21 AM, olcott wrote:
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>> If the UTM simulation of the input to Ĥqx ⟨Ĥ⟩ ⟨Ĥ⟩ would never
>>>>>>> reach its final state (whether or not this simulation is aborted)
>>>>>>> then it is necessarily true that Ĥqx transitions to Ĥ.qn correctly.
>>>>
>>>> ^^^^^^^^^^^^^^^^^ NOTE WHAT YOU SAID HERE ^^^^^^^^^^^^^^^^^^^^^^^^
>>>>
>>>>>>
>>>>>> Finally, you use the right words. We Know that UTM(<H^> <H^>) will
>>>>>> act exactly like H^<H^>).
>>>>>>
>>>>>> We have that:
>>>>>> q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
>>>>>>
>>>>>> Where the above initial q0 is that starting state of H^, so this
>>>>>> is showing the behavior of H^ (<H^>) which we see ends up at H^qn
>>>>>> which is defined as a terminal state for H^, so this PROVES that
>>>>>> H^ applied to <H^> is a Halting Computation.
>>>>> That Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ halts is not the question.
>>>>> The simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by Ĥq0 never halts is the answer.
>>>>
>>>> No, as you said earlier, it is what the input to H (or H^q0) would
>>>> do if it was made the input to a UTM, NOT your 'non-utm' partial
>>>> simulation H.
>>>>
>>>> The right answer to the Halting Problem is ALWAYS about the actual
>>>> machine who has been converted into a representation to give to the
>>>> decider, never about what happens in the decider.
>>>
>>> So then you are saying that when the halt decider does correctly
>>> decide that its input never halts that it can still be wrong?
>>>
>>> That is irrational.
>>
>> You've got that completely ass-backwards. You're the one arguing that
>> when your decider gets the wrong answer as defined by the problem it
>> is somehow still deciding 'correctly'.
>>
>
> As long as the halt decider correctly decides that its input never halts

But it doesn't decide this correctly. You keep *claiming* that it
decides correctly, but that's just your delusions talking.

> Then when we directly execute Ĥ applied to ⟨Ĥ⟩ and it:
> (a) Halts.
> (b) Destroys the whole universe.
> (c) Converts all fascists to true Christianity.
> (d) Circumnavigates the globe.
> (e) Cures cancer.
> (f) Raises the dead.
>
> The halt decider's halt status decision about its input remains correct.

No. If the halt decider's decision were correct, then we would expect
(g) when we directly execute Ĥ applied to ⟨Ĥ⟩ it runs forever, because
that's what the claim that Ĥ applied to ⟨Ĥ⟩ doesn't halt *means*, and
that's what the halt decider is supposed to be deciding.

> If we know that we have a black cat then saying that we don't have a > cat is irrational. Tautologies cannot possibly fail. If you believe
> that they can fail then you must be irrational.

But none of your claims consitute tautologies.

Besides, last I checked a 'black cat' was a WWII-era bomber. Or a type
of firecracker. Or a Marvel comic-book character. Or a brand of peanut
butter.

André

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

On 10/27/2021 11:32 PM, André G. Isaak wrote:
> On 2021-10-27 22:14, olcott wrote:
>> On 10/27/2021 10:31 PM, André G. Isaak wrote:
>>> On 2021-10-27 20:18, olcott wrote:
>>>> On 10/27/2021 8:55 PM, Richard Damon wrote:
>>>>> On 10/27/21 9:39 PM, olcott wrote:
>>>>>> On 10/27/2021 6:46 PM, Richard Damon wrote:
>>>>>>>
>>>>>>> On 10/27/21 10:21 AM, olcott wrote:
>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>> If the UTM simulation of the input to Ĥqx ⟨Ĥ⟩ ⟨Ĥ⟩ would never
>>>>>>>> reach its final state (whether or not this simulation is
>>>>>>>> aborted) then it is necessarily true that Ĥqx transitions to
>>>>>>>> Ĥ.qn correctly.
>>>>>
>>>>> ^^^^^^^^^^^^^^^^^ NOTE WHAT YOU SAID HERE ^^^^^^^^^^^^^^^^^^^^^^^^
>>>>>
>>>>>>>
>>>>>>> Finally, you use the right words. We Know that UTM(<H^> <H^>)
>>>>>>> will act exactly like H^<H^>).
>>>>>>>
>>>>>>> We have that:
>>>>>>> q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
>>>>>>>
>>>>>>> Where the above initial q0 is that starting state of H^, so this
>>>>>>> is showing the behavior of H^ (<H^>) which we see ends up at H^qn
>>>>>>> which is defined as a terminal state for H^, so this PROVES that
>>>>>>> H^ applied to <H^> is a Halting Computation.
>>>>>> That Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ halts is not the question.
>>>>>> The simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by Ĥq0 never halts is the answer.
>>>>>
>>>>> No, as you said earlier, it is what the input to H (or H^q0) would
>>>>> do if it was made the input to a UTM, NOT your 'non-utm' partial
>>>>> simulation H.
>>>>>
>>>>> The right answer to the Halting Problem is ALWAYS about the actual
>>>>> machine who has been converted into a representation to give to the
>>>>> decider, never about what happens in the decider.
>>>>
>>>> So then you are saying that when the halt decider does correctly
>>>> decide that its input never halts that it can still be wrong?
>>>>
>>>> That is irrational.
>>>
>>> You've got that completely ass-backwards. You're the one arguing that
>>> when your decider gets the wrong answer as defined by the problem it
>>> is somehow still deciding 'correctly'.
>>>
>>
>> As long as the halt decider correctly decides that its input never halts
>
> But it doesn't decide this correctly. You keep *claiming* that it
> decides correctly, but that's just your delusions talking.
>
>> Then when we directly execute Ĥ applied to ⟨Ĥ⟩ and it:
>> (a) Halts.
>> (b) Destroys the whole universe.
>> (c) Converts all fascists to true Christianity.
>> (d) Circumnavigates the globe.
>> (e) Cures cancer.
>> (f) Raises the dead.
>>
>> The halt decider's halt status decision about its input remains correct.
>
> No. If the halt decider's decision were correct, then we would expect
> (g) when we directly execute Ĥ applied to ⟨Ĥ⟩ it runs forever, because
> that's what the claim that Ĥ applied to ⟨Ĥ⟩ doesn't halt *means*, and
> that's what the halt decider is supposed to be deciding.
>
>> If we know that we have a black cat then saying that we don't have a >
>> cat is irrational. Tautologies cannot possibly fail. If you believe
>> that they can fail then you must be irrational.
>
> But none of your claims consitute tautologies.
>

When the halt decider correctly decides that the simulation of its input
never reaches the final state of this input (whether or not this
simulation is aborted) then the halt decider has decided the halt status
of this input correctly. If you don't understand that is a tautology
then you are not very bright.

> Besides, last I checked a 'black cat' was a WWII-era bomber. Or a type
> of firecracker. Or a Marvel comic-book character. Or a brand of peanut
> butter.
>
> André
>
>

--
Copyright 2021 Pete Olcott

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

On 10/28/21 12:55 AM, olcott wrote:
> On 10/27/2021 11:32 PM, André G. Isaak wrote:
>> On 2021-10-27 22:14, olcott wrote:
>>> On 10/27/2021 10:31 PM, André G. Isaak wrote:
>>>> On 2021-10-27 20:18, olcott wrote:
>>>>> On 10/27/2021 8:55 PM, Richard Damon wrote:
>>>>>> On 10/27/21 9:39 PM, olcott wrote:
>>>>>>> On 10/27/2021 6:46 PM, Richard Damon wrote:
>>>>>>>>
>>>>>>>> On 10/27/21 10:21 AM, olcott wrote:
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>> If the UTM simulation of the input to Ĥqx ⟨Ĥ⟩ ⟨Ĥ⟩ would never
>>>>>>>>> reach its final state (whether or not this simulation is
>>>>>>>>> aborted) then it is necessarily true that Ĥqx transitions to
>>>>>>>>> Ĥ.qn correctly.
>>>>>>

^^^^^^^^^^^^^^^^^ NOTE WHAT YOU SAID HERE ^^^^^^^^^^^^^^^^^^^^^^^^

>>>>>>
>>>>>>>>
>>>>>>>> Finally, you use the right words. We Know that UTM(<H^> <H^>)
>>>>>>>> will act exactly like H^<H^>).
>>>>>>>>
>>>>>>>> We have that:
>>>>>>>> q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
>>>>>>>>
>>>>>>>> Where the above initial q0 is that starting state of H^, so this
>>>>>>>> is showing the behavior of H^ (<H^>) which we see ends up at
>>>>>>>> H^qn which is defined as a terminal state for H^, so this PROVES
>>>>>>>> that H^ applied to <H^> is a Halting Computation.
>>>>>>> That Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ halts is not the question.
>>>>>>> The simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by Ĥq0 never halts is the answer.
>>>>>>
>>>>>> No, as you said earlier, it is what the input to H (or H^q0) would
>>>>>> do if it was made the input to a UTM, NOT your 'non-utm' partial
>>>>>> simulation H.
>>>>>>
>>>>>> The right answer to the Halting Problem is ALWAYS about the actual
>>>>>> machine who has been converted into a representation to give to
>>>>>> the decider, never about what happens in the decider.
>>>>>
>>>>> So then you are saying that when the halt decider does correctly
>>>>> decide that its input never halts that it can still be wrong?
>>>>>
>>>>> That is irrational.
>>>>
>>>> You've got that completely ass-backwards. You're the one arguing
>>>> that when your decider gets the wrong answer as defined by the
>>>> problem it is somehow still deciding 'correctly'.
>>>>
>>>
>>> As long as the halt decider correctly decides that its input never halts
>>
>> But it doesn't decide this correctly. You keep *claiming* that it
>> decides correctly, but that's just your delusions talking.
>>
>>> Then when we directly execute Ĥ applied to ⟨Ĥ⟩ and it:
>>> (a) Halts.
>>> (b) Destroys the whole universe.
>>> (c) Converts all fascists to true Christianity.
>>> (d) Circumnavigates the globe.
>>> (e) Cures cancer.
>>> (f) Raises the dead.
>>>
>>> The halt decider's halt status decision about its input remains correct.
>>
>> No. If the halt decider's decision were correct, then we would expect
>> (g) when we directly execute Ĥ applied to ⟨Ĥ⟩ it runs forever, because
>> that's what the claim that Ĥ applied to ⟨Ĥ⟩ doesn't halt *means*, and
>> that's what the halt decider is supposed to be deciding.
>>
>>> If we know that we have a black cat then saying that we don't have a
>>> > cat is irrational. Tautologies cannot possibly fail. If you believe
>>> that they can fail then you must be irrational.
>>
>> But none of your claims consitute tautologies.
>>
>
> When the halt decider correctly decides that the simulation of its input
> never reaches the final state of this input (whether or not this
> simulation is aborted) then the halt decider has decided the halt status
> of this input correctly. If you don't understand that is a tautology
> then you are not very bright.
>

The problem is you are using two different definitions of 'simulation'

The problem statement uses simulation by a real UTM (see your statement
marked above).

Your proof looks at the partial simulation done by H (or the copy of it
in H^). There REAL simulation by the UTM does reach a Halting State. The
partial simulation doesn't because it stops the simulation too soon
because the decider used unsound logic to 'show' that the simulation
won't halt (it treats it as a different machine built with a different
decider).

This is why H is wrong, its whole logic is about a different problem.
Maybe it is right about that other problem, but it isn't right about the
actual Halting Problem.

You are using properties of 'simulation' that depend on the simulation
being by an actual UTM that totally simulates its input, but then you
assume they also hold for your partial simulation also, which they don't.

It doesn't matter that it acts like one until it decides, to be one it
needs to NEVER stop.

THat is like claiming your immotal because you haven't died yet.

>> Besides, last I checked a 'black cat' was a WWII-era bomber. Or a type
>> of firecracker. Or a Marvel comic-book character. Or a brand of peanut
>> butter.
>>
>> André
>>
>>
>
>

On Thursday, 28 October 2021 at 00:34:50 UTC+1, richar...@gmail.com wrote:
> On 10/27/21 9:51 AM, olcott wrote:
> > On 10/27/2021 8:12 AM, Ben Bacarisse wrote:
> >> olcott <No...@NoWhere.com> writes:
> >>
> >>> On 10/26/2021 6:47 PM, Ben Bacarisse wrote:
> >>>> olcott <No...@NoWhere.com> writes:
> >>
> >>>> All you do is assert that H decides ⟨Ĥ⟩ ⟨Ĥ⟩ correctly.
> >>>>
> >>>>> Are you willing to admit that if it is shown that H correctly decides
> >>>>> the halt status of ⟨Ĥ⟩ ⟨Ĥ⟩ that Linz has been refuted or are you going
> >>>>> to dishonestly dodge this key point?
> >>>>
> >>>> I keep trying not to dodge it but you keep ignoring my answer. No
> >>>> proof
> >>>> is invalided by other truths. You might render all of mathematics
> >>>> inconsistent, but previously valid chains of reasoning remains valid..
> >>>>
> >>>> You made it plain that don't accept what a proof is when you stated
> >>>> that
> >>>> if A,B,C ⊦ X, then ~A,A,B,C ⊬ X. As far as I know you have never
> >>>> retracted this incorrect statement.
> >>>>
> >>>> As it happens, you can't show that Linz's H is correct about the input
> >>>> ⟨Ĥ⟩ ⟨Ĥ⟩, so the point is moot, but I have been trying to get you to
> >>>> note
> >>>> my answer to this hypothetical question for years, but you keep
> >>>> pretending not to notice.
> >>>
> >>> I ask you a YES / NO question and you dodge.
> >>
> >> How could you miss the answer yet again? Clearly it's no: I don't
> >> accept that if it is shown that H correctly decides the halt status of
> >> ⟨Ĥ⟩ ⟨Ĥ⟩ that Linz has been refuted.
> >
> > So you reject the idea of a semantic tautology.
> > If X says Y is true and I prove that Y is false then X could still be
> > correct. No wonder we are having trouble communicating, you are simply
> > irrational.
> >
> Perhaps in smaller words so you can understand.
>
> IF we have one proof that says X.
>
> And we then come up with another proof that says Not X.
>
> That second proof does NOT 'disprove' the first proof.
>
> What it does is point out that either there is a flaw in ONE of the
> proofs, or the logical system is invalid.
>
I remember one question from my Oxford entrance examination:

"The language of poetry is the language of the everyday".
"The language of the everyday is never the language of poetry".

Both statements are true, of course.

On 10/27/2021 6:34 PM, Richard Damon wrote:
> On 10/27/21 9:51 AM, olcott wrote:
>> On 10/27/2021 8:12 AM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 10/26/2021 6:47 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>>> All you do is assert that H decides ⟨Ĥ⟩ ⟨Ĥ⟩ correctly.
>>>>>
>>>>>> Are you willing to admit that if it is shown that H correctly decides
>>>>>> the halt status of ⟨Ĥ⟩ ⟨Ĥ⟩ that Linz has been refuted or are you
>>>>>> going
>>>>>> to dishonestly dodge this key point?
>>>>>
>>>>> I keep trying not to dodge it but you keep ignoring my answer.  No
>>>>> proof
>>>>> is invalided by other truths.  You might render all of mathematics
>>>>> inconsistent, but previously valid chains of reasoning remains valid.
>>>>>
>>>>> You made it plain that don't accept what a proof is when you stated
>>>>> that
>>>>> if A,B,C ⊦ X, then ~A,A,B,C ⊬ X.  As far as I know you have never
>>>>> retracted this incorrect statement.
>>>>>
>>>>> As it happens, you can't show that Linz's H is correct about the input
>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩, so the point is moot, but I have been trying to get you to
>>>>> note
>>>>> my answer to this hypothetical question for years, but you keep
>>>>> pretending not to notice.
>>>>
>>>> I ask you a YES / NO question and you dodge.
>>>
>>> How could you miss the answer yet again?  Clearly it's no: I don't
>>> accept that if it is shown that H correctly decides the halt status of
>>> ⟨Ĥ⟩ ⟨Ĥ⟩ that Linz has been refuted.
>>
>> So you reject the idea of a semantic tautology.
>> If X says Y is true and I prove that Y is false then X could still be
>> correct. No wonder we are having trouble communicating, you are simply
>> irrational.
>>
>
> Perhaps in smaller words so you can understand.
>
> IF we have one proof that says X.
>
> And we then come up with another proof that says Not X.
>
> That second proof does NOT 'disprove' the first proof.
>
These are the simple words that Ben cannot understand:
If Linz says that H cannot correctly decide the halt
status of ⟨Ĥ⟩ ⟨Ĥ⟩ and I show exactly how H correctly decides
the halt status of ⟨Ĥ⟩ ⟨Ĥ⟩ then I have proved Linz wrong.

https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
Since that time I did find the exact mistake that Linz made on the top
of page 320.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Linz incorrectly believes that Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ is deciding the halt status
of itself rather than the halt status of its input.

The thing that leads to his confusion is that on line 7 of the top of
page 320 he simply deletes most of the details of Ĥ: q0ŵ ⊢* Ĥqn, he
skips the point in Ĥ that actually determines the halt decider of its
input.

--
Copyright 2021 Pete Olcott

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

On Thursday, October 28, 2021 at 11:11:24 AM UTC-4, olcott wrote:
> On 10/27/2021 6:34 PM, Richard Damon wrote:
> > On 10/27/21 9:51 AM, olcott wrote:
> >> On 10/27/2021 8:12 AM, Ben Bacarisse wrote:
> >>> olcott <No...@NoWhere.com> writes:
> >>>
> >>>> On 10/26/2021 6:47 PM, Ben Bacarisse wrote:
> >>>>> olcott <No...@NoWhere.com> writes:
> >>>
> >>>>> All you do is assert that H decides ⟨Ĥ⟩ ⟨Ĥ⟩ correctly.
> >>>>>
> >>>>>> Are you willing to admit that if it is shown that H correctly decides
> >>>>>> the halt status of ⟨Ĥ⟩ ⟨Ĥ⟩ that Linz has been refuted or are you
> >>>>>> going
> >>>>>> to dishonestly dodge this key point?
> >>>>>
> >>>>> I keep trying not to dodge it but you keep ignoring my answer. No
> >>>>> proof
> >>>>> is invalided by other truths. You might render all of mathematics
> >>>>> inconsistent, but previously valid chains of reasoning remains valid.
> >>>>>
> >>>>> You made it plain that don't accept what a proof is when you stated
> >>>>> that
> >>>>> if A,B,C ⊦ X, then ~A,A,B,C ⊬ X. As far as I know you have never
> >>>>> retracted this incorrect statement.
> >>>>>
> >>>>> As it happens, you can't show that Linz's H is correct about the input
> >>>>> ⟨Ĥ⟩ ⟨Ĥ⟩, so the point is moot, but I have been trying to get you to
> >>>>> note
> >>>>> my answer to this hypothetical question for years, but you keep
> >>>>> pretending not to notice.
> >>>>
> >>>> I ask you a YES / NO question and you dodge.
> >>>
> >>> How could you miss the answer yet again? Clearly it's no: I don't
> >>> accept that if it is shown that H correctly decides the halt status of
> >>> ⟨Ĥ⟩ ⟨Ĥ⟩ that Linz has been refuted.
> >>
> >> So you reject the idea of a semantic tautology.
> >> If X says Y is true and I prove that Y is false then X could still be
> >> correct. No wonder we are having trouble communicating, you are simply
> >> irrational.
> >>
> >
> > Perhaps in smaller words so you can understand.
> >
> > IF we have one proof that says X.
> >
> > And we then come up with another proof that says Not X.
> >
> > That second proof does NOT 'disprove' the first proof.
> >
> These are the simple words that Ben cannot understand:
> If Linz says that H cannot correctly decide the halt
> status of ⟨Ĥ⟩ ⟨Ĥ⟩ and I show exactly how H correctly decides
> the halt status of ⟨Ĥ⟩ ⟨Ĥ⟩ then I have proved Linz wrong.
>
> https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
> Since that time I did find the exact mistake that Linz made on the top
> of page 320.
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> Linz incorrectly believes that Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ is deciding the halt status
> of itself rather than the halt status of its input.
>
> The thing that leads to his confusion is that on line 7 of the top of
> page 320 he simply deletes most of the details of Ĥ: q0ŵ ⊢* Ĥqn, he
> skips the point in Ĥ that actually determines the halt decider of its
> input.
> --
> Copyright 2021 Pete Olcott
>
> "Great spirits have always encountered violent opposition from mediocre
> minds." Einstein

+ If Linz says that
+ H cannot correctly
+ decide the halt

i am totally confused at this point in the parse.
daniel 2380 seven sigma claim again but
i have terrible troubles with negation.

On 10/28/2021 10:36 AM, Daniel Pehoushek wrote:
> On Thursday, October 28, 2021 at 11:11:24 AM UTC-4, olcott wrote:
>> On 10/27/2021 6:34 PM, Richard Damon wrote:
>>> On 10/27/21 9:51 AM, olcott wrote:
>>>> On 10/27/2021 8:12 AM, Ben Bacarisse wrote:
>>>>> olcott <No...@NoWhere.com> writes:
>>>>>
>>>>>> On 10/26/2021 6:47 PM, Ben Bacarisse wrote:
>>>>>>> olcott <No...@NoWhere.com> writes:
>>>>>
>>>>>>> All you do is assert that H decides ⟨Ĥ⟩ ⟨Ĥ⟩ correctly.
>>>>>>>
>>>>>>>> Are you willing to admit that if it is shown that H correctly decides
>>>>>>>> the halt status of ⟨Ĥ⟩ ⟨Ĥ⟩ that Linz has been refuted or are you
>>>>>>>> going
>>>>>>>> to dishonestly dodge this key point?
>>>>>>>
>>>>>>> I keep trying not to dodge it but you keep ignoring my answer. No
>>>>>>> proof
>>>>>>> is invalided by other truths. You might render all of mathematics
>>>>>>> inconsistent, but previously valid chains of reasoning remains valid.
>>>>>>>
>>>>>>> You made it plain that don't accept what a proof is when you stated
>>>>>>> that
>>>>>>> if A,B,C ⊦ X, then ~A,A,B,C ⊬ X. As far as I know you have never
>>>>>>> retracted this incorrect statement.
>>>>>>>
>>>>>>> As it happens, you can't show that Linz's H is correct about the input
>>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩, so the point is moot, but I have been trying to get you to
>>>>>>> note
>>>>>>> my answer to this hypothetical question for years, but you keep
>>>>>>> pretending not to notice.
>>>>>>
>>>>>> I ask you a YES / NO question and you dodge.
>>>>>
>>>>> How could you miss the answer yet again? Clearly it's no: I don't
>>>>> accept that if it is shown that H correctly decides the halt status of
>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩ that Linz has been refuted.
>>>>
>>>> So you reject the idea of a semantic tautology.
>>>> If X says Y is true and I prove that Y is false then X could still be
>>>> correct. No wonder we are having trouble communicating, you are simply
>>>> irrational.
>>>>
>>>
>>> Perhaps in smaller words so you can understand.
>>>
>>> IF we have one proof that says X.
>>>
>>> And we then come up with another proof that says Not X.
>>>
>>> That second proof does NOT 'disprove' the first proof.
>>>
>> These are the simple words that Ben cannot understand:
>> If Linz says that H cannot correctly decide the halt
>> status of ⟨Ĥ⟩ ⟨Ĥ⟩ and I show exactly how H correctly decides
>> the halt status of ⟨Ĥ⟩ ⟨Ĥ⟩ then I have proved Linz wrong.
>>
>> https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>> Since that time I did find the exact mistake that Linz made on the top
>> of page 320.
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>> Linz incorrectly believes that Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ is deciding the halt status
>> of itself rather than the halt status of its input.
>>
>> The thing that leads to his confusion is that on line 7 of the top of
>> page 320 he simply deletes most of the details of Ĥ: q0ŵ ⊢* Ĥqn, he
>> skips the point in Ĥ that actually determines the halt decider of its
>> input.
>> --
>> Copyright 2021 Pete Olcott
>>
>> "Great spirits have always encountered violent opposition from mediocre
>> minds." Einstein
>
> + If Linz says that
> + H cannot correctly
> + decide the halt
>
> i am totally confused at this point in the parse.
> daniel 2380 seven sigma claim again but
> i have terrible troubles with negation.
>

Linz says that Turing Machine H cannot possibly correctly decide the
actual halt status of Turing machine description ⟨Ĥ⟩ applied to ⟨Ĥ⟩.

As soon as I show all of the details of exactly how H correctly decides
the actual halt status of ⟨Ĥ⟩ ⟨Ĥ⟩ I have proved Linz wrong.

--
Copyright 2021 Pete Olcott

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

On 10/28/2021 8:52 AM, Malcolm McLean wrote:
> On Thursday, 28 October 2021 at 00:34:50 UTC+1, richar...@gmail.com wrote:
>> On 10/27/21 9:51 AM, olcott wrote:
>>> On 10/27/2021 8:12 AM, Ben Bacarisse wrote:
>>>> olcott <No...@NoWhere.com> writes:
>>>>
>>>>> On 10/26/2021 6:47 PM, Ben Bacarisse wrote:
>>>>>> olcott <No...@NoWhere.com> writes:
>>>>
>>>>>> All you do is assert that H decides ⟨Ĥ⟩ ⟨Ĥ⟩ correctly.
>>>>>>
>>>>>>> Are you willing to admit that if it is shown that H correctly decides
>>>>>>> the halt status of ⟨Ĥ⟩ ⟨Ĥ⟩ that Linz has been refuted or are you going
>>>>>>> to dishonestly dodge this key point?
>>>>>>
>>>>>> I keep trying not to dodge it but you keep ignoring my answer. No
>>>>>> proof
>>>>>> is invalided by other truths. You might render all of mathematics
>>>>>> inconsistent, but previously valid chains of reasoning remains valid.
>>>>>>
>>>>>> You made it plain that don't accept what a proof is when you stated
>>>>>> that
>>>>>> if A,B,C ⊦ X, then ~A,A,B,C ⊬ X. As far as I know you have never
>>>>>> retracted this incorrect statement.
>>>>>>
>>>>>> As it happens, you can't show that Linz's H is correct about the input
>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩, so the point is moot, but I have been trying to get you to
>>>>>> note
>>>>>> my answer to this hypothetical question for years, but you keep
>>>>>> pretending not to notice.
>>>>>
>>>>> I ask you a YES / NO question and you dodge.
>>>>
>>>> How could you miss the answer yet again? Clearly it's no: I don't
>>>> accept that if it is shown that H correctly decides the halt status of
>>>> ⟨Ĥ⟩ ⟨Ĥ⟩ that Linz has been refuted.
>>>
>>> So you reject the idea of a semantic tautology.
>>> If X says Y is true and I prove that Y is false then X could still be
>>> correct. No wonder we are having trouble communicating, you are simply
>>> irrational.
>>>
>> Perhaps in smaller words so you can understand.
>>
>> IF we have one proof that says X.
>>
>> And we then come up with another proof that says Not X.
>>
>> That second proof does NOT 'disprove' the first proof.
>>
>> What it does is point out that either there is a flaw in ONE of the
>> proofs, or the logical system is invalid.
>>
> I remember one question from my Oxford entrance examination:
>
> "The language of poetry is the language of the everyday".
> "The language of the everyday is never the language of poetry".
>
> Both statements are true, of course.
>

These are the simple words that Ben cannot understand:
If Linz says that H cannot correctly decide the halt
status of ⟨Ĥ⟩ ⟨Ĥ⟩ and I show exactly how H correctly decides
the halt status of ⟨Ĥ⟩ ⟨Ĥ⟩ then I have proved Linz wrong.

https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
Since that time I did find the exact mistake that Linz made on the top
of page 320.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Linz incorrectly believes that Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ is deciding the halt status
of itself rather than the halt status of its input.

The thing that leads to his confusion is that on line 7 of the top of
page 320 he simply deletes most of the details of Ĥ: q0ŵ ⊢* Ĥqn, he
skips the point in Ĥ that actually determines the halt decider of its input.

--
Copyright 2021 Pete Olcott

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

On 10/28/21 11:11 AM, olcott wrote:
> On 10/27/2021 6:34 PM, Richard Damon wrote:
>> On 10/27/21 9:51 AM, olcott wrote:
>>> On 10/27/2021 8:12 AM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 10/26/2021 6:47 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>>> All you do is assert that H decides ⟨Ĥ⟩ ⟨Ĥ⟩ correctly.
>>>>>>
>>>>>>> Are you willing to admit that if it is shown that H correctly
>>>>>>> decides
>>>>>>> the halt status of ⟨Ĥ⟩ ⟨Ĥ⟩ that Linz has been refuted or are you
>>>>>>> going
>>>>>>> to dishonestly dodge this key point?
>>>>>>
>>>>>> I keep trying not to dodge it but you keep ignoring my answer.  No
>>>>>> proof
>>>>>> is invalided by other truths.  You might render all of mathematics
>>>>>> inconsistent, but previously valid chains of reasoning remains valid.
>>>>>>
>>>>>> You made it plain that don't accept what a proof is when you
>>>>>> stated that
>>>>>> if A,B,C ⊦ X, then ~A,A,B,C ⊬ X.  As far as I know you have never
>>>>>> retracted this incorrect statement.
>>>>>>
>>>>>> As it happens, you can't show that Linz's H is correct about the
>>>>>> input
>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩, so the point is moot, but I have been trying to get you
>>>>>> to note
>>>>>> my answer to this hypothetical question for years, but you keep
>>>>>> pretending not to notice.
>>>>>
>>>>> I ask you a YES / NO question and you dodge.
>>>>
>>>> How could you miss the answer yet again?  Clearly it's no: I don't
>>>> accept that if it is shown that H correctly decides the halt status of
>>>> ⟨Ĥ⟩ ⟨Ĥ⟩ that Linz has been refuted.
>>>
>>> So you reject the idea of a semantic tautology.
>>> If X says Y is true and I prove that Y is false then X could still be
>>> correct. No wonder we are having trouble communicating, you are
>>> simply irrational.
>>>
>>
>> Perhaps in smaller words so you can understand.
>>
>> IF we have one proof that says X.
>>
>> And we then come up with another proof that says Not X.
>>
>> That second proof does NOT 'disprove' the first proof.
>>
> These are the simple words that Ben cannot understand:
> If Linz says that H cannot correctly decide the halt
> status of ⟨Ĥ⟩ ⟨Ĥ⟩ and I show exactly how H correctly decides
> the halt status of ⟨Ĥ⟩ ⟨Ĥ⟩ then I have proved Linz wrong.
>
> https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
> Since that time I did find the exact mistake that Linz made on the top
> of page 320.
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> Linz incorrectly believes that Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ is deciding the halt status
> of itself rather than the halt status of its input.
>
> The thing that leads to his confusion is that on line 7 of the top of
> page 320 he simply deletes most of the details of Ĥ: q0ŵ ⊢* Ĥqn, he
> skips the point in Ĥ that actually determines the halt decider of its
> input.
>

As you say, *IF* you coul show that H correctly decided the ACTUAL
haliting status of the computation represented by <H^> <H^> (you keep
maiking the mistake of saying the halting status of <H^><H^> which is a
non-sense sentance and 'inputs' don't themselves halt or not).

The problem is that <H^> <H^> represents H^(<H^>) which even you admit
will halt (because H aborts its simulation of a copy of it, but that
doesn't matter), so H saying it is non-halting is just wrong.

As you have recently commented, this can all be seen by putting this
input into a actual UTM, and we will see this exact same behavior (we
have to, as that is part of the definition of a UTM).

YOU seem to confuse this with the partial simulation done by H which
while H doen't simulate it to the non-halting state, that doesn't show
that the computation is non-halting. Thats like taking 100 steps on a
hike and stopping and saying this hike is impossible, that store down
the street must be an infinite distance away since you didn't reach it
in 100 steps.

As to that line 7, YOU are confused about it and obviously have lost
track of the logic.

H^ is built starting from H

H is the decider, not H^, but H^ will in its processing make a decsion.

Let us go step by step. H^ is going to be given the representation of
machine M, which Linz notes as wM

So we start at H^ wM

H^ then duplicates its input and then go to the beginning of its
embedded copy of the Halt decider H, so we get to H^q0 wM wM which is an
equivalent to H wM wM

H then does its work and end up in either its qn or qy (Linz adds xs or
ys to indicate that the contents of the tape doesn't matter).

H is supposed to go to qy if M wM will Halt and qn if M wM will never
halt, and since H^ has a copy of that same logic H^ will also end up in
H^qy if M wm will Halt or it will end up at H^qn if M wM will never
halt. (Note, this doens't mean that H didn't reach the halting state
before it aborted, it means that M wM will NEVER reach a halting state
no matter how long it is given to run).

We then note that we have changed the ending of the copy of H in H^ so
that qy is no longer a halting state, but becomes in infinite loop.

Thus we have that

H^ wM will not halt if M wM does Halt, and H^ wM will go to H^qn and
Halt if M wM does not Halt.

i.e. his term

q0 wM -> H^q0 wM wM -> H^ non-halting if M applied to wM halts and

q0 wM -> H^q0 wM wM -> H^ y1 qn y2 and Halts if M applied to wM does not
halt.

Now, we con convert this general description to a specific case by
making M be H^, and thus wM will be the representation he calls w^ (the
representation of H^)

Subsituting into the above we get:

q0 w^ -> H^q0 w^ w^ -> H^ non-halting if H^ applied to w^ is Halting and

q0 w^ -> H^q0 w^ w^ -> H^ y1 qn y2 (and Halts) if H^ applied to w^ is
non-halting.

Note, In general H^ doesn't decide about itself, but if w^ is given to
H^ the it just happens to need to do to it, so this isn't really
'self-reference' (because the input doesn't say 'H^') it just happens to
be it.

So, For H to be right, it needs to find an answer that works.

If H w^ w^ goes to qn, saying that H^ w^ is non-halting, then H^ w^ will
halt, and if H w^ w^ goes to qy, saying that H^ w^ will halt then H^ w^
will loop forever, so we see that there is no answer that H can give
that will be right.

The fact that H is a simulating Halt decider doesn't affect anything
here, because the logic doesn't depend at all about how H gets from its
q0 to qy or qn, and the right answer also doesn't depend on how H gets
to its answer.

H have just 4 possible ways it can behave:

H w^ w^ can go to qy, saying H^ w^ is halting, and H^ w^ will be
non-halting and thus H was wrong

H w^ w^ can go to qn, saying H^ w^ will not halt, and H^ w^ will then
Halt, and thus H was wrong.

H w^ w^ can just keep on computing forever and never get to qy or qn, at
which point it has failed to meet its requirement to be a decider.

or

H w^ w^ can halt in some other state, neither of qy or qn, and also fail
to meet its requirement to be a decider.

In all cases, H fails to be a correct decider, sometimes by being
incorrect, and sometime just failing at the definition of being a decider.

If you can point out a 5th option that H can do and give a correct
answer, please clearly state it.

Again, note that the right behavior of H is determined by the actual
behavor of H^ w^, not whether H simulates w^ w^ to a halting state,

An H that aborts its simulation does NOT prove non-halting, and if the
logic of H assumes that the copy of H w^ w^ in H^ will not abort its
simulation when the simulation H w^ w^ does, then it has used unsound logic.