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.

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

On 2/24/24 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.

Until you show a rebuttal for his argument, you are just lying.

Claim the question is wrong is NOT a rebutal, but just the circular
argument that you claim to be lies, so you are just admitting that your
whole argument for the past 20 years are LIES.

You are just proving your utter stupidity.

>
> // 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 ?

WRONG QUESTION, so LIE.

This is just Red Herring.

>
> Of the infinite set of implementations of H every one that
> transitions to H.qy derives an answer consistent with the
> behavior of H.

Which doesn't make it a "Halt Decider"

Proof by example is just a fallacy.

>
> 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 ?

Nope, LIE.

Ȟ is NOT a decider, so you are setting up an invalid problem.

You can't try to show a probem is invalid by using a machine that can't
possibly solve it.

>
> Of the infinite set of implementations of Ȟ none of them
> derives an answer consistent with the behavior of Ȟ.
>

So?

Why should it?

You are just proving yourself to be an IDIOT.

Why don't you work with the ACTUAL Halting Question?

Answer, because you can't and make your "arguement" seem at all reasonalbe.

So, you knock down your own strawman and then LIE that you are proving
something about the actual problem.

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.

The fact that NO H^ can be built that has an H that goest the correct
answer is NOT a proof that the problem is incorrect, just that the
Halting Question is uncomputable, which is a perfectly valid state for a
question.

Why do you think that the fact that no machine can compute a given
mathematical function makes the function "invalid"?

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?

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

On 2/24/2024 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

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

On 2/24/24 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.

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.

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.

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.

The Computation system is just powerful enough to do this.

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

> The Computation system is just powerful enough to do this.

What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
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?

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

On 2/24/24 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.

It is NOT H itself, but H is just a component of itself, so H^ isn't
contray to ITSELF.

Your problem is that you just rotely copy parts of the problem without
understanding them.

The specification on building H^ lead to a CONTRADICTION, not a
SELF-CONTRADICTION. The key point here is that assumption of existance
was on H, not H^, H^'s existance is conditional on the existance of H

You don't seem to understand the principle of Identity.

And either way, it shows that a Halt Decider can not exist, not that the
definition of what a Halt Decider is to do is contradictory, just
uncomputable.

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.
>
> 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.
>
> 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.
>
> 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.
>
> The Computation system is just powerful enough to do this.

ALL YOU HAVE IS CIRCULAR ARGUMENTS

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?

(a) Ȟ is wearing a green hat and doesn't like inputs with green hats.
(b) Ȟ ate too much spaghetti for lunch and is too full to compute halting.
(c) Ȟ is busy watching TV and doesn't want to be bothered.
(d) Input ⟨Ȟ⟩ ⟨Ȟ⟩ is self contradictory for TM Ȟ.

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

On 2/24/24 10:30 PM, 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

NOPE.

Proves you are STUPID.

And dont understand the meaning of a definiton.

>
>> 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
>

NOPE.

Proves you are STUPID.

And dont understand the meaning of a definiton.

>> 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

NOPE.

Proves you are STUPID.

And dont understand the meaning of a definiton.

>
>> 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

NOPE.

Proves you are STUPID.

And dont understand the meaning of a definiton.

>
>> The Computation system is just powerful enough to do this.
>
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> 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?
>
>

You are clearly too stupid to understand.

Maybe you should go to the highway and play with your ball, better at night.

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

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

On 2/24/24 10:30 PM, 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
>
>> The Computation system is just powerful enough to do this.
>
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
> 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?
>
>

Here is a question for you.

Don't use a "Circular" Arguemet.

Why is it light during the Day and not at night?

On 2/24/24 10:37 PM, 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.
>>
>> 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.
>>
>> 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.
>>
>> 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.
>>
>> The Computation system is just powerful enough to do this.
>
> ALL YOU HAVE IS CIRCULAR ARGUMENTS
>
> 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?
>
> (a) Ȟ is wearing a green hat and doesn't like inputs with green hats.
> (b) Ȟ ate too much spaghetti for lunch and is too full to compute halting.
> (c) Ȟ is busy watching TV and doesn't want to be bothered.
> (d) Input ⟨Ȟ⟩ ⟨Ȟ⟩ is self contradictory for TM Ȟ.
>

My answers are NOT "Circular"

They are just based on the definitions

If you don't accept the definition, you are just proving yourself stupid.

(and yes, ultimately definition become somewhat circular, as you get to
the fundamentals that don't have anything to define them on, this is the
limitation of langauge).

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

And are just lying you did.

As you have been told many times, but are too stupid to understand.

On 2/24/2024 9:37 PM, Richard Damon wrote:
> On 2/24/24 10:30 PM, 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
>
> NOPE.
>
> Proves you are STUPID.
>
> And dont understand the meaning of a definiton.
>
>
>>
>>> 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
>>
>
> NOPE.
>
> Proves you are STUPID.
>
> And dont understand the meaning of a definiton.
>
>>> 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
>
>
> NOPE.
>
> Proves you are STUPID.
>
> And dont understand the meaning of a definiton.
>
>>
>>> 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
>
>
> NOPE.
>
> Proves you are STUPID.
>
> And dont understand the meaning of a definiton.
>
>>
>>> The Computation system is just powerful enough to do this.
>>
>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>> 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?
>>
>>
>
> You are clearly too stupid to understand.

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?

*Since you cannot explain the details WHY Ȟ cannot*
*answer correctly your rebuttal of the reason that*
*I provided is proven to be vacuous*

That you continue to use ad hominem as your basis
makes you look much more ridiculously foolish than
you 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.

That you were able to reverse-engineer (my idea) above proves
that you are capable of much deeper understanding than most.

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

On 2/24/2024 9:43 PM, Richard Damon wrote:
> On 2/24/24 10:37 PM, 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.
>>>
>>> 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.
>>>
>>> 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.
>>>
>>> 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.
>>>
>>> The Computation system is just powerful enough to do this.
>>
>> ALL YOU HAVE IS CIRCULAR ARGUMENTS
>>
>> 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?
>>
>> (a) Ȟ is wearing a green hat and doesn't like inputs with green hats.
>> (b) Ȟ ate too much spaghetti for lunch and is too full to compute
>> halting.
>> (c) Ȟ is busy watching TV and doesn't want to be bothered.
>> (d) Input ⟨Ȟ⟩ ⟨Ȟ⟩ is self contradictory for TM Ȟ.
>>
>
> My answers are NOT "Circular"
>
> They are just based on the definitions
>
> If you don't accept the definition, you are just proving yourself stupid.
>
> (and yes, ultimately definition become somewhat circular, as you get to
> the fundamentals that don't have anything to define them on, this is the
> limitation of langauge).
>

These definitions have always been circular.
People that learn-by-rote simply learn them
by rote and thus never notice their circularity.

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

On 2/24/2024 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

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

On 2/24/24 10:47 PM, olcott wrote:
> On 2/24/2024 9:37 PM, Richard Damon wrote:
>> On 2/24/24 10:30 PM, 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
>>
>> NOPE.
>>
>> Proves you are STUPID.
>>
>> And dont understand the meaning of a definiton.
>>
>>
>>>
>>>> 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
>>>
>>
>> NOPE.
>>
>> Proves you are STUPID.
>>
>> And dont understand the meaning of a definiton.
>>
>>>> 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
>>
>>
>> NOPE.
>>
>> Proves you are STUPID.
>>
>> And dont understand the meaning of a definiton.
>>
>>>
>>>> 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
>>
>>
>> NOPE.
>>
>> Proves you are STUPID.
>>
>> And dont understand the meaning of a definiton.
>>
>>>
>>>> The Computation system is just powerful enough to do this.
>>>
>>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>>> What is it specifically about Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ that
>>> 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?
>>>
>>>
>>
>> You are clearly too stupid to understand.
>
> 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?

I have explained, and you don't understand,

This shows you are just too stupid to understand.

>
> *Since you cannot explain the details WHY Ȟ cannot*
> *answer correctly your rebuttal of the reason that*
> *I provided is proven to be vacuous*

That nothing exists that can compute a function is the DEFINITION of
non-computable.

There doesn't NEED to be a "reason" it doesn't exist.

You are just showing youself too stupid to understand the reasoning I
gave, so my best advice is you should just give up trying to understand
why something happens (or in this case, doesn't happen).

>
> That you continue to use ad hominem as your basis
> makes you look much more ridiculously foolish than
> you are.

You apparently don't know what ad homimem means.

I do NOT say you are wrong because you are stupid (or some other
attribue of your self, THAT would be ad mominem) I point out that you
are showing yourself stupid by repeating wrong statements, over and over.

If you look carefully at what I have said, it is always, you are wrong
because of ... stating reasons, and then and the fact that you continue
to make this errors makes you a stupid liar.

Perhaps the only time when I use stupid as a reason is here, where you
are asking why, after I have explained it, and I won't keep repeating
it, because it doesn't really matter, and you are clearly too stupid to
understand why things are the way they are.

We don't always get to know why things work out the way they do, we
might want to, but it isn't always given to us. We just need to learn
how things actually are.

Halting is NOT computable, That is a PROVEN fact. There are principles
we can learn that help explain why that is, but they don't change the
fact that it isn't computabe.

>
> 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.
>
> That you were able to reverse-engineer (my idea) above proves
> that you are capable of much deeper understanding than most.
>

It wasn't "Your Idea", it is the actual definition of how it always worked.

You are just to ignorant of the actual definitions to understand that.

You are just continuing to dig your hole deeper.

On 2/24/24 10:49 PM, olcott wrote:
> On 2/24/2024 9:43 PM, Richard Damon wrote:
>> On 2/24/24 10:37 PM, 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.
>>>>
>>>> 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.
>>>>
>>>> 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.
>>>>
>>>> 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.
>>>>
>>>> The Computation system is just powerful enough to do this.
>>>
>>> ALL YOU HAVE IS CIRCULAR ARGUMENTS
>>>
>>> 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?
>>>
>>> (a) Ȟ is wearing a green hat and doesn't like inputs with green hats.
>>> (b) Ȟ ate too much spaghetti for lunch and is too full to compute
>>> halting.
>>> (c) Ȟ is busy watching TV and doesn't want to be bothered.
>>> (d) Input ⟨Ȟ⟩ ⟨Ȟ⟩ is self contradictory for TM Ȟ.
>>>
>>
>> My answers are NOT "Circular"
>>
>> They are just based on the definitions
>>
>> If you don't accept the definition, you are just proving yourself stupid.
>>
>> (and yes, ultimately definition become somewhat circular, as you get
>> to the fundamentals that don't have anything to define them on, this
>> is the limitation of langauge).
>>
>
> These definitions have always been circular.
> People that learn-by-rote simply learn them
> by rote and thus never notice their circularity.
>

ALL definitions are in one sense cirular, as you reach a fundamental
basis that we run out of simpler words to describe it.

That is where we need to go to the actual first principles to understand
what things are.

You don't understand them, so everything seems confusing, but that is
YOUR fault for not seeking to understand the first principles before
working on problems in the field.

The basic definition of computable / non-computable is based on the
existance or non-existance of an specific descrite deterministic
algorithm that can produce the requried answer, for all possible inputs,
in finite time.

If such an algorithm exists, the problem is computable. If not, it is
non-computable.

Nothing "circular" about that, if you understand that basics of what an
algorithm is, and what it means for the required answer.

If you don't understand that, then you will get looped in circles, like
it seems you have.

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.

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.

>> 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?
>
> I have explained, and you don't understand,
>
> This shows you are just too stupid to understand.
>
>>
>> *Since you cannot explain the details WHY Ȟ cannot*
>> *answer correctly your rebuttal of the reason that*
>> *I provided is proven to be vacuous*
>
> That nothing exists that can compute a function is the DEFINITION of
> non-computable.
>

When you say that the reason why Ȟ cannot compute the halt status
of its own Turing Machine description because it is not computable
that is circular forming its conclusion on the basis of its premise.

> There doesn't NEED to be a "reason" it doesn't exist.
>

If it is not because Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ is self-contradictory
then there must be some other reason.

The only correct rebuttal is to find a better reason than the
one that I provided.

*Every truth requires a criterion measure*
*Every truth requires a criterion measure*
*Every truth requires a criterion measure*

Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ cannot compute the halt status of its
own Turing machine description because the halt status of
its own Turning machine description is not computable for Ȟ.

IS CIRCULAR --- IS CIRCULAR --- IS CIRCULAR
IS CIRCULAR --- IS CIRCULAR --- IS CIRCULAR
IS CIRCULAR --- IS CIRCULAR --- IS CIRCULAR

> If you look carefully at what I have said, it is always, you are wrong
> because of ... stating reasons,

Circular reasons are vacuous
Circular reasons are vacuous
Circular reasons are vacuous
Circular reasons are vacuous

and then and the fact that you continue
> to make this errors makes you a stupid liar.
>
> Perhaps the only time when I use stupid as a reason is here, where you
> are asking why, after I have explained it, and I won't keep repeating
> it, because it doesn't really matter, and you are clearly too stupid to
> understand why things are the way they are.
>
> We don't always get to know why things work out the way they do, we
> might want to, but it isn't always given to us. We just need to learn
> how things actually are.
>
> Halting is NOT computable, That is a PROVEN fact. There are principles
> we can learn that help explain why that is, but they don't change the
> fact that it isn't computabe.
>

Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ cannot compute the halt status of its
own Turing machine description because the input ⟨Ȟ⟩ ⟨Ȟ⟩ to
Ȟ is self-contradictory for Ȟ.

You are saying that Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ cannot compute the
halt status of its own Turing machine description because
it just can't do it. *That is the avoidance of the question*

There is no better reason why Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ cannot compute
the halt status of its own Turing machine description than the
one that I provided.

>>
>> 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.
>>
>> That you were able to reverse-engineer (my idea) above proves
>> that you are capable of much deeper understanding than most.
>>
>
> It wasn't "Your Idea", it is the actual definition of how it always worked.

(1) It was my idea years ago as proven by my postings in this forum.
No one else (that I am aware of) ever used it to refute Tarski.

(2) That it proves that Tarski is wrong by showing exactly how a
correct and consistent truth predicate can be defined within the
same formal system proves that it is not common knowledge.

(3) That pathological self-reference can be recognized within
the same formal system has been proven several ways.

This proves that Gödel didn't even understand that epistemological
antinomies must be rejected and thus not the basis for any proof.

....14 Every epistemological antinomy can likewise be used for a similar
undecidability proof...(Gödel 1931:43)

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

On 2/24/2024 10:12 PM, Richard Damon wrote:
> On 2/24/24 10:49 PM, olcott wrote:
>> On 2/24/2024 9:43 PM, Richard Damon wrote:
>>> On 2/24/24 10:37 PM, 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.
>>>>>
>>>>> 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.
>>>>>
>>>>> 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.
>>>>>
>>>>> 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.
>>>>>
>>>>> The Computation system is just powerful enough to do this.
>>>>
>>>> ALL YOU HAVE IS CIRCULAR ARGUMENTS
>>>>
>>>> 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?
>>>>
>>>> (a) Ȟ is wearing a green hat and doesn't like inputs with green hats.
>>>> (b) Ȟ ate too much spaghetti for lunch and is too full to compute
>>>> halting.
>>>> (c) Ȟ is busy watching TV and doesn't want to be bothered.
>>>> (d) Input ⟨Ȟ⟩ ⟨Ȟ⟩ is self contradictory for TM Ȟ.
>>>>
>>>
>>> My answers are NOT "Circular"
>>>
>>> They are just based on the definitions
>>>
>>> If you don't accept the definition, you are just proving yourself
>>> stupid.
>>>
>>> (and yes, ultimately definition become somewhat circular, as you get
>>> to the fundamentals that don't have anything to define them on, this
>>> is the limitation of langauge).
>>>
>>
>> These definitions have always been circular.
>> People that learn-by-rote simply learn them
>> by rote and thus never notice their circularity.
>>
>
> ALL definitions are in one sense cirular,

Not at all. It is all a knowledge true AKA acylic graph.

> as you reach a fundamental
> basis that we run out of simpler words to describe it.
>
> That is where we need to go to the actual first principles to understand
> what things are.
>

I independently invented {reasoning from first principles}
decades before I ever heard of it. I call my version categorically
exhaustive reasoning that makes gaps in reasoning easily detected.

When all reasoning is within the context of a complete set
of exhaustive categories then gaps in reasoning between
categories cannot possibly exist.

> You don't understand them, so everything seems confusing, but that is
> YOUR fault for not seeking to understand the first principles before
> working on problems in the field.
>

The ultimate first principle for all of my work is the
architectural foundation of analytic truth itself.

> The basic definition of computable / non-computable is based on the
> existance or non-existance of an specific descrite deterministic
> algorithm that can produce the requried answer, for all possible inputs,
> in finite time.

Not 100% exactly specific to Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩.
In this case the reason that Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ gets the wrong
answer is the Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ is self-contradictory for Ȟ.

If you can't find a better reason then that proves that my
reason is correct.

> If such an algorithm exists, the problem is computable. If not, it is
> non-computable.
>
> Nothing "circular" about that, if you understand that basics of what an
> algorithm is, and what it means for the required answer.
>

When you say that the reason that Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩
can't compute the right answer is that it is not computable
*THIS IS AS CIRCULAR AS CIRCULAR GETS*

> If you don't understand that, then you will get looped in circles, like
> it seems you have.

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

On 2/24/2024 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.

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.

--
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 12:09 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?

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

For a first point Ȟ is mot a "decider" (since it doesn't always halt),
so in one real sense, it doesn't "answer" a question, so why it can't
answer a question that we can't ask it becomes a category error.

>
>>> 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?
>>
>> I have explained, and you don't understand,
>>
>> This shows you are just too stupid to understand.
>>
>>>
>>> *Since you cannot explain the details WHY Ȟ cannot*
>>> *answer correctly your rebuttal of the reason that*
>>> *I provided is proven to be vacuous*
>>
>> That nothing exists that can compute a function is the DEFINITION of
>> non-computable.
>>
>
> When you say that the reason why Ȟ cannot compute the halt status
> of its own Turing Machine description because it is not computable
> that is circular forming its conclusion on the basis of its premise.

Nope. "Computable" and "Non-Computable" are base definition relative to
the existance or non-existance of a Turing Machine that computes that
function among the set of all possible Turing Machines.

There are Aleph-1 possible mappings that Turing Machines could be asked
to compute, and only Aleph-0 possible Turing Machines, so most mappings
are just not computable.

There need not be a "specific" why as to why one particular one isn't,
perhaps the better question to ask is why so many mappings we want to do
are computable, since the odds are against them.

The fact you think this is a circular definition shows you don't
understand the first principles of the field.

>
>> There doesn't NEED to be a "reason" it doesn't exist.
>>
>
> If it is not because Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ is self-contradictory
> then there must be some other reason.

>
> The only correct rebuttal is to find a better reason than the
> one that I provided.
>
> *Every truth requires a criterion measure*
> *Every truth requires a criterion measure*
> *Every truth requires a criterion measure*

Right, DOES the Turing Machine exist.

If so, Computable.

If not, non-computable.

>
> Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ cannot compute the halt status of its
> own Turing machine description because the halt status of
> its own Turning machine description is not computable for Ȟ.
>
> IS CIRCULAR --- IS CIRCULAR --- IS CIRCULAR
> IS CIRCULAR --- IS CIRCULAR --- IS CIRCULAR
> IS CIRCULAR --- IS CIRCULAR --- IS CIRCULAR
>
>> If you look carefully at what I have said, it is always, you are wrong
>> because of ... stating reasons,
>
> Circular reasons are vacuous
> Circular reasons are vacuous
> Circular reasons are vacuous
> Circular reasons are vacuous
>
> and then and the fact that you continue

And you prove yourself to be an idiot, by using unsupported logic
>> to make this errors makes you a stupid liar.

>>

>> Perhaps the only time when I use stupid as a reason is here, where you
>> are asking why, after I have explained it, and I won't keep repeating
>> it, because it doesn't really matter, and you are clearly too stupid
>> to understand why things are the way they are.
>>
>> We don't always get to know why things work out the way they do, we
>> might want to, but it isn't always given to us. We just need to learn
>> how things actually are.
>>
>> Halting is NOT computable, That is a PROVEN fact. There are principles
>> we can learn that help explain why that is, but they don't change the
>> fact that it isn't computabe.
>>
>
> Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ cannot compute the halt status of its
> own Turing machine description because the input ⟨Ȟ⟩ ⟨Ȟ⟩ to
> Ȟ is self-contradictory for Ȟ.
>
> You are saying that Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ cannot compute the
> halt status of its own Turing machine description because
> it just can't do it. *That is the avoidance of the question*
>
> There is no better reason why Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ cannot compute
> the halt status of its own Turing machine description than the
> one that I provided.
>
>>>
>>> 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.
>>>
>>> That you were able to reverse-engineer (my idea) above proves
>>> that you are capable of much deeper understanding than most.
>>>
>>
>> It wasn't "Your Idea", it is the actual definition of how it always
>> worked.
>
> (1) It was my idea years ago as proven by my postings in this forum.
> No one else (that I am aware of) ever used it to refute Tarski.
>
> (2) That it proves that Tarski is wrong by showing exactly how a
> correct and consistent truth predicate can be defined within the
> same formal system proves that it is not common knowledge.
>
> (3) That pathological self-reference can be recognized within
> the same formal system has been proven several ways.
>
> This proves that Gödel didn't even understand that epistemological
> antinomies must be rejected and thus not the basis for any proof.
>
> ...14 Every epistemological antinomy can likewise be used for a similar
> undecidability proof...(Gödel 1931:43)
>
>

Nope. Proven to be lies.

Answer me this, Why are you so stupid? Claiming you are not just proves
it because you have said so many incorrect things.

On 2/25/24 12:22 AM, olcott wrote:
> On 2/24/2024 10:12 PM, Richard Damon wrote:
>> On 2/24/24 10:49 PM, olcott wrote:
>>> On 2/24/2024 9:43 PM, Richard Damon wrote:
>>>> On 2/24/24 10:37 PM, 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.
>>>>>>
>>>>>> 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.
>>>>>>
>>>>>> 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.
>>>>>>
>>>>>> 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.
>>>>>>
>>>>>> The Computation system is just powerful enough to do this.
>>>>>
>>>>> ALL YOU HAVE IS CIRCULAR ARGUMENTS
>>>>>
>>>>> 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?
>>>>>
>>>>> (a) Ȟ is wearing a green hat and doesn't like inputs with green hats.
>>>>> (b) Ȟ ate too much spaghetti for lunch and is too full to compute
>>>>> halting.
>>>>> (c) Ȟ is busy watching TV and doesn't want to be bothered.
>>>>> (d) Input ⟨Ȟ⟩ ⟨Ȟ⟩ is self contradictory for TM Ȟ.
>>>>>
>>>>
>>>> My answers are NOT "Circular"
>>>>
>>>> They are just based on the definitions
>>>>
>>>> If you don't accept the definition, you are just proving yourself
>>>> stupid.
>>>>
>>>> (and yes, ultimately definition become somewhat circular, as you get
>>>> to the fundamentals that don't have anything to define them on, this
>>>> is the limitation of langauge).
>>>>
>>>
>>> These definitions have always been circular.
>>> People that learn-by-rote simply learn them
>>> by rote and thus never notice their circularity.
>>>
>>
>> ALL definitions are in one sense cirular,
>
> Not at all. It is all a knowledge true AKA acylic graph.

How do you define the root word?

>
>> as you reach a fundamental basis that we run out of simpler words to
>> describe it.
>>
>> That is where we need to go to the actual first principles to
>> understand what things are.
>>
>
> I independently invented {reasoning from first principles}
> decades before I ever heard of it. I call my version categorically
> exhaustive reasoning that makes gaps in reasoning easily detected.

But "invention" requires being actually first.

Your version seems totally defective as you don't see the gaps in your
own logic.

>
> When all reasoning is within the context of a complete set
> of exhaustive categories then gaps in reasoning between
> categories cannot possibly exist.

Then why do you have so many gaps in your logic?

>
>> You don't understand them, so everything seems confusing, but that is
>> YOUR fault for not seeking to understand the first principles before
>> working on problems in the field.
>>
>
> The ultimate first principle for all of my work is the
> architectural foundation of analytic truth itself.

And you have failed at that, because you don't seem to understand the
actual first principle of the field.

>
>> The basic definition of computable / non-computable is based on the
>> existance or non-existance of an specific descrite deterministic
>> algorithm that can produce the requried answer, for all possible
>> inputs, in finite time.
>
> Not 100% exactly specific to Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩.
> In this case the reason that Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ gets the wrong
> answer is the Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩ is self-contradictory for Ȟ.
>
> If you can't find a better reason then that proves that my
> reason is correct.

Not going to answer, as you are too stupid to understand.

Again, why is Pi == 3.14159...

>
>> If such an algorithm exists, the problem is computable. If not, it is
>> non-computable.
>>
>> Nothing "circular" about that, if you understand that basics of what
>> an algorithm is, and what it means for the required answer.
>>
>
> When you say that the reason that Ȟ applied to ⟨Ȟ⟩ ⟨Ȟ⟩
> can't compute the right answer is that it is not computable
> *THIS IS AS CIRCULAR AS CIRCULAR GETS*

Nope. It is the DEFINITION.

>
>> If you don't understand that, then you will get looped in circles,
>> like it seems you have.
>