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.
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.
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?
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
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?
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
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?
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.
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.
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
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)
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.
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
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.
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.
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.
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