On 2/25/24 12:35 AM, olcott wrote:
> On 2/24/2024 10:19 PM, Richard Damon wrote:
>> On 2/24/24 10:53 PM, olcott wrote:
>>> On 2/24/2024 9:43 PM, Richard Damon wrote:
>>>> On 2/24/24 10:39 PM, olcott wrote:
>>>>> On 2/24/2024 9:33 PM, Richard Damon wrote:
>>>>>> On 2/24/24 10:23 PM, olcott wrote:
>>>>>>> On 2/24/2024 9:15 PM, Richard Damon wrote:
>>>>>>>> On 2/24/24 9:34 PM, olcott wrote:
>>>>>>>>> I say that there are no implementations of Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩
>>>>>>>>> that return a value consistent with the behavior of Ȟ because
>>>>>>>>> the ⟨Ȟ⟩ ⟨Ȟ⟩ input to Ȟ is self-contradictory.
>>>>>>>>>
>>>>>>>>> It just occurred to me that rebuttals that cannot provide a sound
>>>>>>>>> (thus non-circular) alternative would be exposed as a lies.
>>>>>>>>>
>>>>>>>>
>>>>>>>> So,
>>>>>>>>
>>>>>>>> DEFINITION: A Halt Decider is a machine that gives the correct
>>>>>>>> answer about the halting behavior of the Computation Described
>>>>>>>> by its input.
>>>>>>>>
>>>>>>>> Thus
>>>>>>>>
>>>>>>>> H (M) (I) needs to go to Qy (and Halt) if M (I) halts and to go
>>>>>>>> to (and Halt) Qn if M (I) will not halt.
>>>>>>>>
>>>>>>>> We define H^ (M) to dupicate its input, then use a copy of H on
>>>>>>>> (M) (M) and then if that goes to Qy, H^ will loop forever, and
>>>>>>>> if it goes to Qn, then H^ Halts.
>>>>>>>>
>>>>>>>> We look at the computation H^ (H^), and ask H about it.
>>>>>>>>
>>>>>>>> You claim that H (H^) (H^) is correct in going to Qn, but when
>>>>>>>> we run that same H^ (H^) it's H goes to Qn and then H^ Halts, so
>>>>>>>> H^ (H^) Halts.
>>>>>>>>
>>>>>>>> You claim H was correct, but it went to Qn when the correct
>>>>>>>> answer is Qy
>>>>>>>>
>>>>>>>> How is that?
>>>>>>>>
>>>>>>>> Do you still hold that this incorrect answer is somehow correct?
>>>>>>>>
>>>>>>>> YOu need to use a SOUND argument to explain.
>>>>>>>>
>>>>>>>>
>>>>>>>> Note, H^ is NOT "Self-Contradictory" because it isn't defined to
>>>>>>>> contradict itself, but H.
>>>>>>>>
>>>>>>>
>>>>>>> I had a typo.
>>>>>>> Can you see that Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ is self-contradictory?
>>>>>>>
>>>>>>> Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qy ∞ // Ȟ applied to ⟨Ȟ⟩ halts
>>>>>>> Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qn   // Ȟ applied to ⟨Ȟ⟩ does not halt
>>>>>>>
>>>>>>
>>>>>> H^ is not "Self-Contradictory" because H^ is not defined to
>>>>>> "decide" anything to be contradictory to.
>>>>>>
>>>>>> H^ is H-Contradictory, and acheives this by using a copy of H
>>>>>> within itself.
>>>>>>
>>>>>
>>>>> There are no copies
>>>>>
>>>>
>>>> Then you haven't done it right
>>>
>>> I reformulated the problem so that Ȟ is merely the
>>> exact same H that has had a single change to make
>>> it self-contradictory. No copies simply Ȟ applied
>>> to its own machine description.
>>>
>>> Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qy ∞ // Ȟ applied to ⟨Ȟ⟩ halts
>>> Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qn   // Ȟ applied to ⟨Ȟ⟩ does not halt
>>>
>>>
>>
>> But the copy I mention wasn't the copy of the input, but the fact that
>> H^ has in it a copy of the original H.
>>
>> Note, you can't "change" a Turing machine to get "the same" machine
>> that is different.
>>
>> Turing Machines are "read only" and immutable. Once created that
>> machine will be that machine. If you try to change it, you are
>> changing a copy.
>>
>> Also, your problem definition is incorrect, as H^ (H^) (H^) as you
>> describe it is asking about H^ (H^) which will ask aobut H^ -
>>
>> so you don't get the same iteration of logic, as each evaluation
>> "consumes" one discription to be the program.
>>
>> That is why the proof program needs to duplicate its input.
>>
>> I havve explained this several times, but you seem to just not
>> understand the issue.
>>
>> H^ (H^) (H^) might Halt but H^ (H^) might not and so the above
>> computation might be correct to go to qn.
>>
>
> You are still dodging coming up with an actual rebuttal by
> showing a better reason why Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ gets the wrong
> answer.

Why is Pi = 3.14159... ?

>
> I say that X is because of Y and you say no I am wrong
> X is because of nothing. X cannot be because of nothing.
>
> Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ gets the wrong answer because of
> some reason not because of no reason.
>
> *You don't seem to understand that there must be a reason*
> *every truth requires a criterion measure*
>
> I admit its an impossible task, yet it is only impossible
> because the reason is that Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ really is
> self-contradictory.
>
>

So, you admit that Halt Deciding is impossibe.

Then why do you say you have solved it.

BECAUSE YOU ARE A PATHOLOGICAL LIAR.

On 25/02/24 03:34, olcott wrote:
> I say that there are no implementations of Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩
> that return a value consistent with the behavior of Ȟ because
> the ⟨Ȟ⟩ ⟨Ȟ⟩ input to Ȟ is self-contradictory.
>
> It just occurred to me that rebuttals that cannot provide a sound
> (thus non-circular) alternative would be exposed as a lies.
>
> // Linz Turing machine H --- H applied to ⟨H⟩
> H.q0 ⟨H⟩ ⟨H⟩ ⊢* H.qy    // H applied to ⟨H⟩ halts
> H.q0 ⟨H⟩ ⟨H⟩ ⊢* H.qn   // H applied to ⟨H⟩ does not halt
> Do you halt on your own Turing Machine description ?
>
> Of the infinite set of implementations of H every one that
> transitions to H.qy derives an answer consistent with the
> behavior of H.
>
> When we append an infinite loop to the H.qy state we derive Ȟ
>
> Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qy ∞ // Ȟ applied to ⟨Ȟ⟩ halts
> H.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qn     // Ȟ applied to ⟨Ȟ⟩ does not halt
> Do you halt on your own Turing Machine description ?
>
> Of the infinite set of implementations of Ȟ none of them
> derives an answer consistent with the behavior of Ȟ.
>
>

Which is exactly like saying: Of the infinite set of natural numbers
none of them satisfies the equation x=1-x.

On 25/02/24 04:30, olcott wrote:
> On 2/24/2024 9:27 PM, Richard Damon wrote:
>> On 2/24/24 10:19 PM, olcott wrote:
>>> On 2/24/2024 9:03 PM, Richard Damon wrote:
>>>> On 2/24/24 9:34 PM, olcott wrote:
>>>>> I say that there are no implementations of Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩
>>>>> that return a value consistent with the behavior of Ȟ because
>>>>> the ⟨Ȟ⟩ ⟨Ȟ⟩ input to Ȟ is self-contradictory.
>>>>
>>>> But that is POOP not Halting.
>>>>
>>>>>
>>>>> It just occurred to me that rebuttals that cannot provide a sound
>>>>> (thus non-circular) alternative would be exposed as a lies.
>>>>
>>>> Right, just as YOURS are.
>>>>
>>>> Alan Turing showed that it is impossible to create A Halt Decider
>>>> that correctly decides on the halting for all computations by giving
>>>> them a description of that computation.
>>>>
>>>> He does this by showing that for any possible decider that you might
>>>> try to create, that he could construct an input that it would get
>>>> wrong.
>>>
>>> *Rejected as circular*
>>>
>>> *That halting cannot be computed because halting is not*
>>> *computable and every other isomorphic answer is rejected*
>>> *in advance as circular*
>>>
>>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>>> causes none of the infinite set of implementations of Ȟ
>>> to derive an answer consistent with the behavior of Ȟ?
>>>
>>
>> Halting is not computabe, because for every possible computation,
>> there is at least one input that the decider will get wrong.
>>
>
> CIRCULAR
>
>> To prove that something is not computable, you need to show that for
>> every possible decider, there is at least one input that it gets wrong.
>>
>
> CIRCULAR
>
>> That is NOT a "circular" argument, that is the definition of a
>> categorical arguement.
>>
>> For ANY Halt Decider H, the input built from it by the template in the
>> Linz proof based on THAT Decider (which results in a SPECIFIC actual
>> machine), when given to THAT decider, it can be shown to get it wrong.
>>
>
> CIRCULAR
>
>> This happens, because it is possible to ALWAYS build an machine to
>> give to test the decider from the algorithm of the decider, and make
>> the test machine act contrary to the results of that machine.
>>
>
> CIRCULAR
>
None of these are circular. You sound like an idiot.

On 25/02/24 06:09, olcott wrote:
> On 2/24/2024 10:05 PM, Richard Damon wrote:
>>>>>>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>>>>>>> causes none of the infinite set of implementations of Ȟ
>>>>>>> to derive an answer consistent with the behavior of Ȟ?
>
>>>>>> For ANY Halt Decider H, the input built from it by the template in
>>>>>> the Linz proof based on THAT Decider (which results in a SPECIFIC
>>>>>> actual machine), when given to THAT decider, it can be shown to
>>>>>> get it wrong.
>
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that causes it
> to get the wrong answer? If it is not self-contradiction then it
> must be some other reason.

It's the code in Ȟ. For example, if implemented in x86, it's the result
of all the ADD, SUB, MOV, JNZ instructions, and all the other
instructions as well. Once you get to the end of the halt decider part,
the eax register (which holds function return values) will contain
either 1 or 0.

On 2/25/2024 8:53 AM, Mikko wrote:
> A problem is solved when it is proven that no solution exists.
> For the halting problem such proof is known.
>

The halting problem cannot be solved because no solution
exists is circular.

Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qy ∞ // Ȟ applied to ⟨Ȟ⟩ halts
H.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qn // Ȟ applied to ⟨Ȟ⟩ does not halt

What is it specifically about Ȟ that prevents every
implementation of itself from correctly reporting
on its own behavior?

*I say the reason is that Ȟ is defined to be self-contradictory*

Any disagreement requires an alternative reason. The reason
why cannot correctly report on its own behavior cannot be no
reason and it cannot be a circular reason.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 2/25/2024 8:58 AM, immibis wrote:
> On 25/02/24 03:34, olcott wrote:
>> I say that there are no implementations of Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩
>> that return a value consistent with the behavior of Ȟ because
>> the ⟨Ȟ⟩ ⟨Ȟ⟩ input to Ȟ is self-contradictory.
>>
>> It just occurred to me that rebuttals that cannot provide a sound
>> (thus non-circular) alternative would be exposed as a lies.
>>
>> // Linz Turing machine H --- H applied to ⟨H⟩
>> H.q0 ⟨H⟩ ⟨H⟩ ⊢* H.qy    // H applied to ⟨H⟩ halts
>> H.q0 ⟨H⟩ ⟨H⟩ ⊢* H.qn   // H applied to ⟨H⟩ does not halt
>> Do you halt on your own Turing Machine description ?
>>
>> Of the infinite set of implementations of H every one that
>> transitions to H.qy derives an answer consistent with the
>> behavior of H.
>>
>> When we append an infinite loop to the H.qy state we derive Ȟ
>>
>> Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qy ∞ // Ȟ applied to ⟨Ȟ⟩ halts
>> H.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qn     // Ȟ applied to ⟨Ȟ⟩ does not halt
>> Do you halt on your own Turing Machine description ?
>>
>> Of the infinite set of implementations of Ȟ none of them
>> derives an answer consistent with the behavior of Ȟ.
>>
>>
>
> Which is exactly like saying: Of the infinite set of natural numbers
> none of them satisfies the equation x=1-x.

Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qy ∞ // Ȟ applied to ⟨Ȟ⟩ halts
Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qn // Ȟ applied to ⟨Ȟ⟩ does not halt

What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
causes every implementation of Ȟ get the wrong answer.

I say that the reason is that Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ is
self-contradictory for Ȟ.

Any disagreement requires an alternative reason. The reason
why cannot correctly report on its own behavior cannot be no
reason and it cannot be a circular reason.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 2/25/2024 7:02 AM, Richard Damon wrote:
> On 2/25/24 12:09 AM, olcott wrote:
>> On 2/24/2024 10:05 PM, Richard Damon wrote:
>>>>>>>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>>>>>>>> causes none of the infinite set of implementations of Ȟ
>>>>>>>> to derive an answer consistent with the behavior of Ȟ?
>>
>>>>>>> For ANY Halt Decider H, the input built from it by the template
>>>>>>> in the Linz proof based on THAT Decider (which results in a
>>>>>>> SPECIFIC actual machine), when given to THAT decider, it can be
>>>>>>> shown to get it wrong.
>>
>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that causes it
>> to get the wrong answer? If it is not self-contradiction then it
>> must be some other reason.
>
> Why does there NEED to be a reason?
>

Analytic truth is an interlocking set of semantic
tautologies where expressions of language only acquire
a truth value on the basis of their relation top other
expressions of language.

> Why is Pi = 3.14159...? (other than that is its value)

It is stipulated that "Pi" refers to the value of the
ratio to the circumference to the diameter of a circle
and it is deduced that the decimal approximations of PI
require a certain sequence of mathematical steps.

Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qy ∞ // Ȟ applied to ⟨Ȟ⟩ halts
Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qn // Ȟ applied to ⟨Ȟ⟩ does not halt

What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
causes every implementation of Ȟ get the wrong answer.

It just can't do it is not a reason.

It just can't do it because Ȟ applied to ⟨Ȟ⟩ defines
a self-contradictory relation between itself and its
own Turing Machine Description <is> a reason.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 2/25/2024 7:02 AM, Richard Damon wrote:
> On 2/25/24 12:35 AM, olcott wrote:
>> On 2/24/2024 10:19 PM, Richard Damon wrote:
>>> On 2/24/24 10:53 PM, olcott wrote:
>>>> On 2/24/2024 9:43 PM, Richard Damon wrote:
>>>>> On 2/24/24 10:39 PM, olcott wrote:
>>>>>> On 2/24/2024 9:33 PM, Richard Damon wrote:
>>>>>>> On 2/24/24 10:23 PM, olcott wrote:
>>>>>>>> On 2/24/2024 9:15 PM, Richard Damon wrote:
>>>>>>>>> On 2/24/24 9:34 PM, olcott wrote:
>>>>>>>>>> I say that there are no implementations of Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩
>>>>>>>>>> that return a value consistent with the behavior of Ȟ because
>>>>>>>>>> the ⟨Ȟ⟩ ⟨Ȟ⟩ input to Ȟ is self-contradictory.
>>>>>>>>>>
>>>>>>>>>> It just occurred to me that rebuttals that cannot provide a sound
>>>>>>>>>> (thus non-circular) alternative would be exposed as a lies.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> So,
>>>>>>>>>
>>>>>>>>> DEFINITION: A Halt Decider is a machine that gives the correct
>>>>>>>>> answer about the halting behavior of the Computation Described
>>>>>>>>> by its input.
>>>>>>>>>
>>>>>>>>> Thus
>>>>>>>>>
>>>>>>>>> H (M) (I) needs to go to Qy (and Halt) if M (I) halts and to go
>>>>>>>>> to (and Halt) Qn if M (I) will not halt.
>>>>>>>>>
>>>>>>>>> We define H^ (M) to dupicate its input, then use a copy of H on
>>>>>>>>> (M) (M) and then if that goes to Qy, H^ will loop forever, and
>>>>>>>>> if it goes to Qn, then H^ Halts.
>>>>>>>>>
>>>>>>>>> We look at the computation H^ (H^), and ask H about it.
>>>>>>>>>
>>>>>>>>> You claim that H (H^) (H^) is correct in going to Qn, but when
>>>>>>>>> we run that same H^ (H^) it's H goes to Qn and then H^ Halts,
>>>>>>>>> so H^ (H^) Halts.
>>>>>>>>>
>>>>>>>>> You claim H was correct, but it went to Qn when the correct
>>>>>>>>> answer is Qy
>>>>>>>>>
>>>>>>>>> How is that?
>>>>>>>>>
>>>>>>>>> Do you still hold that this incorrect answer is somehow correct?
>>>>>>>>>
>>>>>>>>> YOu need to use a SOUND argument to explain.
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Note, H^ is NOT "Self-Contradictory" because it isn't defined
>>>>>>>>> to contradict itself, but H.
>>>>>>>>>
>>>>>>>>
>>>>>>>> I had a typo.
>>>>>>>> Can you see that Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ is self-contradictory?
>>>>>>>>
>>>>>>>> Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qy ∞ // Ȟ applied to ⟨Ȟ⟩ halts
>>>>>>>> Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qn   // Ȟ applied to ⟨Ȟ⟩ does not halt
>>>>>>>>
>>>>>>>
>>>>>>> H^ is not "Self-Contradictory" because H^ is not defined to
>>>>>>> "decide" anything to be contradictory to.
>>>>>>>
>>>>>>> H^ is H-Contradictory, and acheives this by using a copy of H
>>>>>>> within itself.
>>>>>>>
>>>>>>
>>>>>> There are no copies
>>>>>>
>>>>>
>>>>> Then you haven't done it right
>>>>
>>>> I reformulated the problem so that Ȟ is merely the
>>>> exact same H that has had a single change to make
>>>> it self-contradictory. No copies simply Ȟ applied
>>>> to its own machine description.
>>>>
>>>> Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qy ∞ // Ȟ applied to ⟨Ȟ⟩ halts
>>>> Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qn   // Ȟ applied to ⟨Ȟ⟩ does not halt
>>>>
>>>>
>>>
>>> But the copy I mention wasn't the copy of the input, but the fact
>>> that H^ has in it a copy of the original H.
>>>
>>> Note, you can't "change" a Turing machine to get "the same" machine
>>> that is different.
>>>
>>> Turing Machines are "read only" and immutable. Once created that
>>> machine will be that machine. If you try to change it, you are
>>> changing a copy.
>>>
>>> Also, your problem definition is incorrect, as H^ (H^) (H^) as you
>>> describe it is asking about H^ (H^) which will ask aobut H^ -
>>>
>>> so you don't get the same iteration of logic, as each evaluation
>>> "consumes" one discription to be the program.
>>>
>>> That is why the proof program needs to duplicate its input.
>>>
>>> I havve explained this several times, but you seem to just not
>>> understand the issue.
>>>
>>> H^ (H^) (H^) might Halt but H^ (H^) might not and so the above
>>> computation might be correct to go to qn.
>>>
>>
>> You are still dodging coming up with an actual rebuttal by
>> showing a better reason why Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ gets the wrong
>> answer.
>
> Why is Pi = 3.14159... ?
>
>>
>> I say that X is because of Y and you say no I am wrong
>> X is because of nothing. X cannot be because of nothing.
>>
>> Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ gets the wrong answer because of
>> some reason not because of no reason.
>>
>> *You don't seem to understand that there must be a reason*
>> *every truth requires a criterion measure*
>>
>> I admit its an impossible task, yet it is only impossible
>> because the reason is that Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ really is
>> self-contradictory.
>>
>>
>
> So, you admit that Halt Deciding is impossibe.
>
> Then why do you say you have solved it.
>
> BECAUSE YOU ARE A PATHOLOGICAL LIAR.

When it is understood that epistemological antinomies must be
rejected then the only class of inputs that prove halting is
undecidable cease to prove that halting is undecidable.

On 2/23/2024 9:22 PM, Richard Damon wrote:
> Yes, Epistemological antinomies, when given to a True Predicate, get
> "rejected" in a sense, the predicate returns FALSE.
>
> That doesn't mean the statement is false, just that it isn't true.
>
> It also doesn't mean the predicate doesn't answer.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

On 2/25/24 12:42 PM, olcott wrote:
> On 2/25/2024 7:02 AM, Richard Damon wrote:
>> On 2/25/24 12:09 AM, olcott wrote:
>>> On 2/24/2024 10:05 PM, Richard Damon wrote:
>>>>>>>>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>>>>>>>>> causes none of the infinite set of implementations of Ȟ
>>>>>>>>> to derive an answer consistent with the behavior of Ȟ?
>>>
>>>>>>>> For ANY Halt Decider H, the input built from it by the template
>>>>>>>> in the Linz proof based on THAT Decider (which results in a
>>>>>>>> SPECIFIC actual machine), when given to THAT decider, it can be
>>>>>>>> shown to get it wrong.
>>>
>>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that causes it
>>> to get the wrong answer? If it is not self-contradiction then it
>>> must be some other reason.
>>
>> Why does there NEED to be a reason?
>>
>
> Analytic truth is an interlocking set of semantic
> tautologies where expressions of language only acquire
> a truth value on the basis of their relation top other
> expressions of language.
>

But that doesn't mean there needs to be a reason that you can't make a
certain connection.

When you are dealing with unbounded domains, "Not Existing" might be
unprovable.

>> Why is Pi = 3.14159...? (other than that is its value)
>
> It is stipulated that "Pi" refers to the value of the
> ratio to the circumference to the diameter of a circle
> and it is deduced that the decimal approximations of PI
> require a certain sequence of mathematical steps.

But why is it THAT value?

We can compute it from the definition, but why is it THAT value.

>
> Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qy ∞ // Ȟ applied to ⟨Ȟ⟩ halts
> Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qn // Ȟ applied to ⟨Ȟ⟩ does not halt
>
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> causes every implementation of Ȟ get the wrong answer.
>
> It just can't do it is not a reason.

The reason is that this mapping (out of the aleph-1 possible mappings)
isn't one of the aleph-0 mapping that can be created with a computation.

Just like why Pi happens to be 3.14159...

>
> It just can't do it because Ȟ applied to ⟨Ȟ⟩ defines
> a self-contradictory relation between itself and its
> own Turing Machine Description <is> a reason.
>

Nope. BAD WORDS. It could be that Ȟ just isn't a "decider" and thus
can't be asked a question.

The fact that the piece you are quoting is built under the assumption
that an H that always decides correctly exists, and Ȟ acts contrary to
that H can be a reason (which isn't "Self-contradictory")

The Self-contradiction only appears when you try to create such an H,
which is what shows that such an H can't exist.

On 2/25/24 12:46 PM, olcott wrote:
> On 2/25/2024 7:02 AM, Richard Damon wrote:
>> On 2/25/24 12:35 AM, olcott wrote:
>>> On 2/24/2024 10:19 PM, Richard Damon wrote:
>>>> On 2/24/24 10:53 PM, olcott wrote:
>>>>> On 2/24/2024 9:43 PM, Richard Damon wrote:
>>>>>> On 2/24/24 10:39 PM, olcott wrote:
>>>>>>> On 2/24/2024 9:33 PM, Richard Damon wrote:
>>>>>>>> On 2/24/24 10:23 PM, olcott wrote:
>>>>>>>>> On 2/24/2024 9:15 PM, Richard Damon wrote:
>>>>>>>>>> On 2/24/24 9:34 PM, olcott wrote:
>>>>>>>>>>> I say that there are no implementations of Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩
>>>>>>>>>>> that return a value consistent with the behavior of Ȟ because
>>>>>>>>>>> the ⟨Ȟ⟩ ⟨Ȟ⟩ input to Ȟ is self-contradictory.
>>>>>>>>>>>
>>>>>>>>>>> It just occurred to me that rebuttals that cannot provide a
>>>>>>>>>>> sound
>>>>>>>>>>> (thus non-circular) alternative would be exposed as a lies.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> So,
>>>>>>>>>>
>>>>>>>>>> DEFINITION: A Halt Decider is a machine that gives the correct
>>>>>>>>>> answer about the halting behavior of the Computation Described
>>>>>>>>>> by its input.
>>>>>>>>>>
>>>>>>>>>> Thus
>>>>>>>>>>
>>>>>>>>>> H (M) (I) needs to go to Qy (and Halt) if M (I) halts and to
>>>>>>>>>> go to (and Halt) Qn if M (I) will not halt.
>>>>>>>>>>
>>>>>>>>>> We define H^ (M) to dupicate its input, then use a copy of H
>>>>>>>>>> on (M) (M) and then if that goes to Qy, H^ will loop forever,
>>>>>>>>>> and if it goes to Qn, then H^ Halts.
>>>>>>>>>>
>>>>>>>>>> We look at the computation H^ (H^), and ask H about it.
>>>>>>>>>>
>>>>>>>>>> You claim that H (H^) (H^) is correct in going to Qn, but when
>>>>>>>>>> we run that same H^ (H^) it's H goes to Qn and then H^ Halts,
>>>>>>>>>> so H^ (H^) Halts.
>>>>>>>>>>
>>>>>>>>>> You claim H was correct, but it went to Qn when the correct
>>>>>>>>>> answer is Qy
>>>>>>>>>>
>>>>>>>>>> How is that?
>>>>>>>>>>
>>>>>>>>>> Do you still hold that this incorrect answer is somehow correct?
>>>>>>>>>>
>>>>>>>>>> YOu need to use a SOUND argument to explain.
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Note, H^ is NOT "Self-Contradictory" because it isn't defined
>>>>>>>>>> to contradict itself, but H.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> I had a typo.
>>>>>>>>> Can you see that Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ is self-contradictory?
>>>>>>>>>
>>>>>>>>> Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qy ∞ // Ȟ applied to ⟨Ȟ⟩ halts
>>>>>>>>> Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qn   // Ȟ applied to ⟨Ȟ⟩ does not halt
>>>>>>>>>
>>>>>>>>
>>>>>>>> H^ is not "Self-Contradictory" because H^ is not defined to
>>>>>>>> "decide" anything to be contradictory to.
>>>>>>>>
>>>>>>>> H^ is H-Contradictory, and acheives this by using a copy of H
>>>>>>>> within itself.
>>>>>>>>
>>>>>>>
>>>>>>> There are no copies
>>>>>>>
>>>>>>
>>>>>> Then you haven't done it right
>>>>>
>>>>> I reformulated the problem so that Ȟ is merely the
>>>>> exact same H that has had a single change to make
>>>>> it self-contradictory. No copies simply Ȟ applied
>>>>> to its own machine description.
>>>>>
>>>>> Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qy ∞ // Ȟ applied to ⟨Ȟ⟩ halts
>>>>> Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qn   // Ȟ applied to ⟨Ȟ⟩ does not halt
>>>>>
>>>>>
>>>>
>>>> But the copy I mention wasn't the copy of the input, but the fact
>>>> that H^ has in it a copy of the original H.
>>>>
>>>> Note, you can't "change" a Turing machine to get "the same" machine
>>>> that is different.
>>>>
>>>> Turing Machines are "read only" and immutable. Once created that
>>>> machine will be that machine. If you try to change it, you are
>>>> changing a copy.
>>>>
>>>> Also, your problem definition is incorrect, as H^ (H^) (H^) as you
>>>> describe it is asking about H^ (H^) which will ask aobut H^ -
>>>>
>>>> so you don't get the same iteration of logic, as each evaluation
>>>> "consumes" one discription to be the program.
>>>>
>>>> That is why the proof program needs to duplicate its input.
>>>>
>>>> I havve explained this several times, but you seem to just not
>>>> understand the issue.
>>>>
>>>> H^ (H^) (H^) might Halt but H^ (H^) might not and so the above
>>>> computation might be correct to go to qn.
>>>>
>>>
>>> You are still dodging coming up with an actual rebuttal by
>>> showing a better reason why Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ gets the wrong
>>> answer.
>>
>> Why is Pi = 3.14159... ?
>>
>>>
>>> I say that X is because of Y and you say no I am wrong
>>> X is because of nothing. X cannot be because of nothing.
>>>
>>> Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ gets the wrong answer because of
>>> some reason not because of no reason.
>>>
>>> *You don't seem to understand that there must be a reason*
>>> *every truth requires a criterion measure*
>>>
>>> I admit its an impossible task, yet it is only impossible
>>> because the reason is that Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ really is
>>> self-contradictory.
>>>
>>>
>>
>> So, you admit that Halt Deciding is impossibe.
>>
>> Then why do you say you have solved it.
>>
>> BECAUSE YOU ARE A PATHOLOGICAL LIAR.
>
> When it is understood that epistemological antinomies must be
> rejected then the only class of inputs that prove halting is
> undecidable cease to prove that halting is undecidable.

Except that NO "Turing Machine" is an "Epistemological Antinomy" since
they are different types of things.

You are just proving your stupidity in making your claim, showing that
you don't understand the words you are using.

Note, you seem to conflate two different arguement.

Truth Predicates, are given statement, which CAN be epistemological
Antiomies", but can not be one themselves.

Halt Deciders are given description of Computations, which are the
description of a finite deterministic algorithm, and a finite set of
data, which can NOT be an Epistemological Antinomy.

Note, a different type of Turing Machine might define what its input
represents differently, and THAT Turing Machine might be able to be
given an Epistemological Antinomy as an input. One example of this would
be if we try to build a Turing Machine that computes the Truth Predicate.

The way you confuse these hust shows you really lack an understanding of
what these things actually are.

>
> On 2/23/2024 9:22 PM, Richard Damon wrote:
> > Yes, Epistemological antinomies, when given to a True Predicate, get
> > "rejected" in a sense, the predicate returns FALSE.
> >
> > That doesn't mean the statement is false, just that it isn't true.
> >
> > It also doesn't mean the predicate doesn't answer.
>
>

On 2/25/24 11:08 AM, olcott wrote:
> On 2/25/2024 8:53 AM, Mikko wrote:
>> A problem is solved when it is proven that no solution exists.
>> For the halting problem such proof is known.
>>
>
> The halting problem cannot be solved because no solution
> exists is circular.

Non-existance of something can be just an established fact.

There doesn't need to be a reason.

Can you give the reason you don't understand what you are talking about?

>
> Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qy ∞ // Ȟ applied to ⟨Ȟ⟩ halts
> H.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qn // Ȟ applied to ⟨Ȟ⟩ does not halt
>
> What is it specifically about Ȟ that prevents every
> implementation of itself from correctly reporting
> on its own behavior?
>
> *I say the reason is that Ȟ is defined to be self-contradictory*
>
> Any disagreement requires an alternative reason. The reason
> why cannot correctly report on its own behavior cannot be no
> reason and it cannot be a circular reason.
>
>

Nope.

That claim just proves you don't understand logic.

On 25/02/24 17:48, olcott wrote:
> On 2/25/2024 8:58 AM, immibis wrote:
>> On 25/02/24 03:34, olcott wrote:
>>> I say that there are no implementations of Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩
>>> that return a value consistent with the behavior of Ȟ because
>>> the ⟨Ȟ⟩ ⟨Ȟ⟩ input to Ȟ is self-contradictory.
>>>
>>> It just occurred to me that rebuttals that cannot provide a sound
>>> (thus non-circular) alternative would be exposed as a lies.
>>>
>>> // Linz Turing machine H --- H applied to ⟨H⟩
>>> H.q0 ⟨H⟩ ⟨H⟩ ⊢* H.qy    // H applied to ⟨H⟩ halts
>>> H.q0 ⟨H⟩ ⟨H⟩ ⊢* H.qn   // H applied to ⟨H⟩ does not halt
>>> Do you halt on your own Turing Machine description ?
>>>
>>> Of the infinite set of implementations of H every one that
>>> transitions to H.qy derives an answer consistent with the
>>> behavior of H.
>>>
>>> When we append an infinite loop to the H.qy state we derive Ȟ
>>>
>>> Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qy ∞ // Ȟ applied to ⟨Ȟ⟩ halts
>>> H.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qn     // Ȟ applied to ⟨Ȟ⟩ does not halt
>>> Do you halt on your own Turing Machine description ?
>>>
>>> Of the infinite set of implementations of Ȟ none of them
>>> derives an answer consistent with the behavior of Ȟ.
>>>
>>>
>>
>> Which is exactly like saying: Of the infinite set of natural numbers
>> none of them satisfies the equation x=1-x.
>
> Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qy ∞ // Ȟ applied to ⟨Ȟ⟩ halts
> Ȟ.q0 ⟨Ȟ⟩ ⟨Ȟ⟩ ⊢* Ȟ.qn // Ȟ applied to ⟨Ȟ⟩ does not halt

x=0 // 1-x=1
x=1 // 1-x=0

> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> causes every implementation of Ȟ get the wrong answer.

What is it specifically about 1-x that causes every value of x to get
the wrong answer.

> I say that the reason is that Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ is
> self-contradictory for  Ȟ.

I say that 1-x is self-contradictory for x.