∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
There is some input TMD to every H such that
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
When we disallow decider/input pairs that are incorrect
questions where both YES and NO are the wrong answer
(the same way the ZFC disallowed self-referential sets) then
pathological inputs are not allowed to come into existence.
Does the barber that shaves everyone that does not shave
themselves shave himself? is rejected as an incorrect question. https://en.wikipedia.org/wiki/Barber_paradox#
∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
There is some input TMD to every H such that
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
When we disallow decider/input pairs that are incorrect
questions where both YES and NO are the wrong answer
Does the barber that shaves everyone that does not shave
themselves shave himself? is rejected as an incorrect question.
∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
There is some input TMD to every H such that
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
When we disallow decider/input pairs that are incorrect
questions where both YES and NO are the wrong answer
(the same way the ZFC disallowed self-referential sets) then
pathological inputs are not allowed to come into existence.
Does the barber that shaves everyone that does not shave
themselves shave himself? is rejected as an incorrect question. https://en.wikipedia.org/wiki/Barber_paradox#
On 3/12/2024 1:41 PM, Richard Damon wrote:
On 3/12/24 11:12 AM, olcott wrote:
∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
There is some input TMD to every H such that
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
When we disallow decider/input pairs that are incorrect
questions where both YES and NO are the wrong answer
(the same way the ZFC disallowed self-referential sets) then
pathological inputs are not allowed to come into existence.
But each of the questions in your set, that for each H^ built on each
SPECIFIC H, there is a correct answer, just not the one that H gives,
so the ALL the questions are correct.
*Halts(TMD) means true if TMD actually halts and false otherwise*
∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
(H TMD ⊢* H.qy != Halts(TMD)) ∧ (H TMD ⊢* H.qn != Halts(TMD))
The questions in my set are only the H/TMD pairs specified above.
Every decider/input pair (referenced in the above set) has a
corresponding decider/input pair that only differs by the return
value of its decider.
That both of these decider/input pairs get the wrong answer proves
that their question was incorrect because the opposite answer to the
same question is also proven to be incorrect.
On 3/12/2024 4:23 PM, Richard Damon wrote:
On 3/12/24 1:11 PM, olcott wrote:
On 3/12/2024 2:40 PM, Richard Damon wrote:
On 3/12/24 12:02 PM, olcott wrote:
On 3/12/2024 1:31 PM, immibis wrote:No, because a given H will only go to one of the answers. THAT will
On 12/03/24 19:12, olcott wrote:
∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
There is some input TMD to every H such that
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
And it can be a different TMD to each H.
When we disallow decider/input pairs that are incorrect
questions where both YES and NO are the wrong answer
Once we understand that either YES or NO is the right answer, the
whole rebuttal is tossed out as invalid and incorrect.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ >>>>
be wrong, and the other one right.
∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Not exactly. A pair of otherwise identical machines that
(that are contained within the above specified set)
only differ by return value will both be wrong on the
same pathological input.
You mean a pair of DIFFERENT machines. Any difference is different.
Every decider/input pair (referenced in the above set) has a
corresponding decider/input pair that only differs by the return
value of its decider.
That both of these decider/input pairs get the wrong answer proves
that their question was incorrect because the opposite answer to the
same question is also proven to be incorrect.
On 3/12/2024 5:30 PM, Richard Damon wrote:
On 3/12/24 2:34 PM, olcott wrote:∀ H ∈ Turing_Machines_Returning_Boolean
On 3/12/2024 4:23 PM, Richard Damon wrote:
On 3/12/24 1:11 PM, olcott wrote:
Not exactly. A pair of otherwise identical machines that
(that are contained within the above specified set)
only differ by return value will both be wrong on the
same pathological input.
You mean a pair of DIFFERENT machines. Any difference is different.
Every decider/input pair (referenced in the above set) has a
corresponding decider/input pair that only differs by the return
value of its decider.
Nope.
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Every H/TMD pair (referenced in the above set) has a
corresponding H/TMD pair that only differs by the return
value of its Boolean_TM.
That both of these H/TMD pairs get the wrong answer proves that
their question was incorrect because the opposite answer to the
same question is also proven to be incorrect.
On 3/12/2024 5:30 PM, Richard Damon wrote:
On 3/12/24 2:34 PM, olcott wrote:∀ H ∈ Turing_Machines_Returning_Boolean
On 3/12/2024 4:23 PM, Richard Damon wrote:
On 3/12/24 1:11 PM, olcott wrote:
On 3/12/2024 2:40 PM, Richard Damon wrote:
On 3/12/24 12:02 PM, olcott wrote:
On 3/12/2024 1:31 PM, immibis wrote:No, because a given H will only go to one of the answers. THAT
On 12/03/24 19:12, olcott wrote:
∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
There is some input TMD to every H such that
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
And it can be a different TMD to each H.
When we disallow decider/input pairs that are incorrect
questions where both YES and NO are the wrong answer
Once we understand that either YES or NO is the right answer,
the whole rebuttal is tossed out as invalid and incorrect.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not
halt
BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ >>>>>>
will be wrong, and the other one right.
∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Not exactly. A pair of otherwise identical machines that
(that are contained within the above specified set)
only differ by return value will both be wrong on the
same pathological input.
You mean a pair of DIFFERENT machines. Any difference is different.
Every decider/input pair (referenced in the above set) has a
corresponding decider/input pair that only differs by the return
value of its decider.
Nope.
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Every H/TMD pair (referenced in the above set) has a
corresponding H/TMD pair that only differs by the return
value of its Boolean_TM.
That both of these H/TMD pairs get the wrong answer proves that
their question was incorrect because the opposite answer to the
same question is also proven to be incorrect.
On 3/12/2024 6:05 PM, immibis wrote:
On 12/03/24 23:53, olcott wrote:
On 3/12/2024 5:30 PM, Richard Damon wrote:
On 3/12/24 2:34 PM, olcott wrote:∀ H ∈ Turing_Machines_Returning_Boolean
On 3/12/2024 4:23 PM, Richard Damon wrote:
On 3/12/24 1:11 PM, olcott wrote:Every decider/input pair (referenced in the above set) has a
Not exactly. A pair of otherwise identical machines that
(that are contained within the above specified set)
only differ by return value will both be wrong on the
same pathological input.
You mean a pair of DIFFERENT machines. Any difference is different. >>>>>
corresponding decider/input pair that only differs by the return
value of its decider.
Nope.
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Every H/TMD pair (referenced in the above set) has a
corresponding H/TMD pair that only differs by the return
value of its Boolean_TM.
That both of these H/TMD pairs get the wrong answer proves that
their question was incorrect because the opposite answer to the
same question is also proven to be incorrect.
Nobody knows what the fuck you are talking about. You have to actually
explain it. The same machine always gives the same return value on the
same input.
It has taken me twenty years to translate my intuitions into
words that can possibly understood.
A pair of Turing Machines that return Boolean that are identical
besides their return value that cannot decide some property of
the same input are being asked the same YES/NO question having
no correct YES/NO answer.
On 3/12/2024 6:38 PM, immibis wrote:
On 13/03/24 00:24, olcott wrote:
On 3/12/2024 6:05 PM, immibis wrote:
On 12/03/24 23:53, olcott wrote:
On 3/12/2024 5:30 PM, Richard Damon wrote:
On 3/12/24 2:34 PM, olcott wrote:∀ H ∈ Turing_Machines_Returning_Boolean
On 3/12/2024 4:23 PM, Richard Damon wrote:
On 3/12/24 1:11 PM, olcott wrote:Every decider/input pair (referenced in the above set) has a
Not exactly. A pair of otherwise identical machines that
(that are contained within the above specified set)
only differ by return value will both be wrong on the
same pathological input.
You mean a pair of DIFFERENT machines. Any difference is different. >>>>>>>
corresponding decider/input pair that only differs by the return >>>>>>> value of its decider.
Nope.
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Every H/TMD pair (referenced in the above set) has a
corresponding H/TMD pair that only differs by the return
value of its Boolean_TM.
That both of these H/TMD pairs get the wrong answer proves that
their question was incorrect because the opposite answer to the
same question is also proven to be incorrect.
Nobody knows what the fuck you are talking about. You have to
actually explain it. The same machine always gives the same return
value on the same input.
It has taken me twenty years to translate my intuitions into
words that can possibly understood.
You failed.
A pair of Turing Machines that return Boolean that are identical
besides their return value that cannot decide some property of
the same input are being asked the same YES/NO question having
no correct YES/NO answer.
https://en.wikipedia.org/wiki/Turing_machine#Formal_definition
A Turing machine is ⟨Q, Γ, b, Σ, δ, q0, F⟩
Show me two ⟨Q, Γ, b, Σ, δ, q0, F⟩ that are identical besides their >> return value.
You can't because you are talking nonsense. they don't exist.
Turing machine descriptions that are identical finite strings
except for the the 1/0 that they write the their exact same
tape relative location.
On 3/12/2024 7:10 PM, immibis wrote:
On 13/03/24 00:56, olcott wrote:Exactly one element of Q differs by writing a 1 instead of a 0.
On 3/12/2024 6:38 PM, immibis wrote:
On 13/03/24 00:24, olcott wrote:
On 3/12/2024 6:05 PM, immibis wrote:
On 12/03/24 23:53, olcott wrote:
On 3/12/2024 5:30 PM, Richard Damon wrote:
On 3/12/24 2:34 PM, olcott wrote:∀ H ∈ Turing_Machines_Returning_Boolean
On 3/12/2024 4:23 PM, Richard Damon wrote:
On 3/12/24 1:11 PM, olcott wrote:
Not exactly. A pair of otherwise identical machines that >>>>>>>>>>> (that are contained within the above specified set)
only differ by return value will both be wrong on the
same pathological input.
You mean a pair of DIFFERENT machines. Any difference is
different.
Every decider/input pair (referenced in the above set) has a >>>>>>>>> corresponding decider/input pair that only differs by the return >>>>>>>>> value of its decider.
Nope.
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Every H/TMD pair (referenced in the above set) has a
corresponding H/TMD pair that only differs by the return
value of its Boolean_TM.
That both of these H/TMD pairs get the wrong answer proves that
their question was incorrect because the opposite answer to the
same question is also proven to be incorrect.
Nobody knows what the fuck you are talking about. You have to
actually explain it. The same machine always gives the same return >>>>>> value on the same input.
It has taken me twenty years to translate my intuitions into
words that can possibly understood.
You failed.
A pair of Turing Machines that return Boolean that are identical
besides their return value that cannot decide some property of
the same input are being asked the same YES/NO question having
no correct YES/NO answer.
https://en.wikipedia.org/wiki/Turing_machine#Formal_definition
A Turing machine is ⟨Q, Γ, b, Σ, δ, q0, F⟩
Show me two ⟨Q, Γ, b, Σ, δ, q0, F⟩ that are identical besides their >>>> return value.
You can't because you are talking nonsense. they don't exist.
Turing machine descriptions that are identical finite strings
except for the the 1/0 that they write the their exact same
tape relative location.
So which part of ⟨Q, Γ, b, Σ, δ, q0, F⟩ is different?
On 3/12/2024 8:05 PM, immibis wrote:
On 13/03/24 01:18, olcott wrote:They are identical except for their return value that is specified
On 3/12/2024 7:10 PM, immibis wrote:
So which part of ⟨Q, Γ, b, Σ, δ, q0, F⟩ is different?Exactly one element of Q differs by writing a 1 instead of a 0.
That's part of δ but this mistake doesn't matter.
It wasn't clear whether you were talking about a Turing machine that
was somehow identical but gave a different return value, or one that
was not identical. Now you have explained it is not identical.
in a single state that is different.
*This means that they implement the exact same algorithm*
On 3/12/2024 8:05 PM, immibis wrote:
On 13/03/24 01:18, olcott wrote:They are identical except for their return value that is specified
On 3/12/2024 7:10 PM, immibis wrote:
On 13/03/24 00:56, olcott wrote:Exactly one element of Q differs by writing a 1 instead of a 0.
On 3/12/2024 6:38 PM, immibis wrote:
On 13/03/24 00:24, olcott wrote:
On 3/12/2024 6:05 PM, immibis wrote:
On 12/03/24 23:53, olcott wrote:
On 3/12/2024 5:30 PM, Richard Damon wrote:
On 3/12/24 2:34 PM, olcott wrote:∀ H ∈ Turing_Machines_Returning_Boolean
On 3/12/2024 4:23 PM, Richard Damon wrote:
On 3/12/24 1:11 PM, olcott wrote:
Not exactly. A pair of otherwise identical machines that >>>>>>>>>>>>> (that are contained within the above specified set)
only differ by return value will both be wrong on the >>>>>>>>>>>>> same pathological input.
You mean a pair of DIFFERENT machines. Any difference is >>>>>>>>>>>> different.
Every decider/input pair (referenced in the above set) has a >>>>>>>>>>> corresponding decider/input pair that only differs by the return >>>>>>>>>>> value of its decider.
Nope.
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Every H/TMD pair (referenced in the above set) has a
corresponding H/TMD pair that only differs by the return
value of its Boolean_TM.
That both of these H/TMD pairs get the wrong answer proves that >>>>>>>>> their question was incorrect because the opposite answer to the >>>>>>>>> same question is also proven to be incorrect.
Nobody knows what the fuck you are talking about. You have to
actually explain it. The same machine always gives the same
return value on the same input.
It has taken me twenty years to translate my intuitions into
words that can possibly understood.
You failed.
A pair of Turing Machines that return Boolean that are identical >>>>>>> besides their return value that cannot decide some property of
the same input are being asked the same YES/NO question having
no correct YES/NO answer.
https://en.wikipedia.org/wiki/Turing_machine#Formal_definition
A Turing machine is ⟨Q, Γ, b, Σ, δ, q0, F⟩
Show me two ⟨Q, Γ, b, Σ, δ, q0, F⟩ that are identical besides >>>>>> their return value.
You can't because you are talking nonsense. they don't exist.
Turing machine descriptions that are identical finite strings
except for the the 1/0 that they write the their exact same
tape relative location.
So which part of ⟨Q, Γ, b, Σ, δ, q0, F⟩ is different?
That's part of δ but this mistake doesn't matter.
It wasn't clear whether you were talking about a Turing machine that
was somehow identical but gave a different return value, or one that
was not identical. Now you have explained it is not identical.
in a single state that is different.
*This means that they implement the exact same algorithm*
A protocol can be defined so that Turing machine descriptions always implement their return value in their state with the largest Natural
Number value. This allows other Turing machines to determine identical algorithms except for return value.
On 3/12/2024 6:11 PM, Richard Damon wrote:
On 3/12/24 3:53 PM, olcott wrote:
On 3/12/2024 5:30 PM, Richard Damon wrote:
On 3/12/24 2:34 PM, olcott wrote:∀ H ∈ Turing_Machines_Returning_Boolean
On 3/12/2024 4:23 PM, Richard Damon wrote:
On 3/12/24 1:11 PM, olcott wrote:Every decider/input pair (referenced in the above set) has a
On 3/12/2024 2:40 PM, Richard Damon wrote:
On 3/12/24 12:02 PM, olcott wrote:
On 3/12/2024 1:31 PM, immibis wrote:No, because a given H will only go to one of the answers. THAT >>>>>>>> will be wrong, and the other one right.
On 12/03/24 19:12, olcott wrote:
∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
There is some input TMD to every H such that
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
And it can be a different TMD to each H.
When we disallow decider/input pairs that are incorrect
questions where both YES and NO are the wrong answer
Once we understand that either YES or NO is the right answer, >>>>>>>>>> the whole rebuttal is tossed out as invalid and incorrect. >>>>>>>>>>
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does
not halt
BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ >>>>>>>>
∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Not exactly. A pair of otherwise identical machines that
(that are contained within the above specified set)
only differ by return value will both be wrong on the
same pathological input.
You mean a pair of DIFFERENT machines. Any difference is different. >>>>>
corresponding decider/input pair that only differs by the return
value of its decider.
Nope.
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Every H/TMD pair (referenced in the above set) has a
corresponding H/TMD pair that only differs by the return
value of its Boolean_TM.
That isn't in the set above.
Nope, since both aren't in the set selected.
That both of these H/TMD pairs get the wrong answer proves that
their question was incorrect because the opposite answer to the
same question is also proven to be incorrect.
When they are deciders that must get the correct answer both
of them are not in the set.
When they are Turing_Machines_Returning_Boolean the this
set inherently includes identical pairs that only differ
by return value.
You just don't understand what that statement is saying.No the problem is that you are not paying attention.
I've expalined it, but it seems over you head.
For Every H, we show we can find at least one input (chosen just forWhen we use machine templates then we can see instances of
that machine) that it will get wrong.
the same machine that only differs by return value where both
get the wrong answer on the same input. By same input I mean
the same finite string of numerical values.
On 3/12/2024 9:17 PM, immibis wrote:
On 13/03/24 02:47, olcott wrote:
On 3/12/2024 8:05 PM, immibis wrote:
On 13/03/24 01:18, olcott wrote:They are identical except for their return value that is specified
On 3/12/2024 7:10 PM, immibis wrote:
So which part of ⟨Q, Γ, b, Σ, δ, q0, F⟩ is different?Exactly one element of Q differs by writing a 1 instead of a 0.
That's part of δ but this mistake doesn't matter.
It wasn't clear whether you were talking about a Turing machine that
was somehow identical but gave a different return value, or one that
was not identical. Now you have explained it is not identical.
in a single state that is different.
*This means that they implement the exact same algorithm*
OK. Well, one of them gets the right answer and one of them gets the
wrong answer. What is the confusion?
The Linz Ĥ.H machine gets the wrong answer on its own
machine description no matter how its Linz H is defined.
This means that it gets the wrong answer on YES and the
wrong answer on NO.
On 3/12/2024 6:38 PM, immibis wrote:
On 13/03/24 00:24, olcott wrote:
On 3/12/2024 6:05 PM, immibis wrote:
On 12/03/24 23:53, olcott wrote:
On 3/12/2024 5:30 PM, Richard Damon wrote:
On 3/12/24 2:34 PM, olcott wrote:∀ H ∈ Turing_Machines_Returning_Boolean
On 3/12/2024 4:23 PM, Richard Damon wrote:
On 3/12/24 1:11 PM, olcott wrote:Every decider/input pair (referenced in the above set) has a
Not exactly. A pair of otherwise identical machines that
(that are contained within the above specified set)
only differ by return value will both be wrong on the
same pathological input.
You mean a pair of DIFFERENT machines. Any difference is different. >>>>>>>
corresponding decider/input pair that only differs by the return >>>>>>> value of its decider.
Nope.
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Every H/TMD pair (referenced in the above set) has a
corresponding H/TMD pair that only differs by the return
value of its Boolean_TM.
That both of these H/TMD pairs get the wrong answer proves that
their question was incorrect because the opposite answer to the
same question is also proven to be incorrect.
Nobody knows what the fuck you are talking about. You have to
actually explain it. The same machine always gives the same return
value on the same input.
It has taken me twenty years to translate my intuitions into
words that can possibly understood.
You failed.
A pair of Turing Machines that return Boolean that are identical
besides their return value that cannot decide some property of
the same input are being asked the same YES/NO question having
no correct YES/NO answer.
https://en.wikipedia.org/wiki/Turing_machine#Formal_definition
A Turing machine is ⟨Q, Γ, b, Σ, δ, q0, F⟩
Show me two ⟨Q, Γ, b, Σ, δ, q0, F⟩ that are identical besides their >> return value.
You can't because you are talking nonsense. they don't exist.
Turing machine descriptions that are identical finite strings
except for the the 1/0 that they write the their exact same
tape relative location.
On 3/12/2024 9:17 PM, immibis wrote:
On 13/03/24 02:47, olcott wrote:
On 3/12/2024 8:05 PM, immibis wrote:
On 13/03/24 01:18, olcott wrote:They are identical except for their return value that is specified
On 3/12/2024 7:10 PM, immibis wrote:
So which part of ⟨Q, Γ, b, Σ, δ, q0, F⟩ is different?Exactly one element of Q differs by writing a 1 instead of a 0.
That's part of δ but this mistake doesn't matter.
It wasn't clear whether you were talking about a Turing machine that
was somehow identical but gave a different return value, or one that
was not identical. Now you have explained it is not identical.
in a single state that is different.
*This means that they implement the exact same algorithm*
OK. Well, one of them gets the right answer and one of them gets the
wrong answer. What is the confusion?
The Linz Ĥ.H machine gets the wrong answer on its own
machine description no matter how its Linz H is defined.
This means that it gets the wrong answer on YES and the
wrong answer on NO.
On 3/12/2024 10:04 PM, immibis wrote:
On 13/03/24 03:41, olcott wrote:None-the-less they only get the wrong answer because of the
On 3/12/2024 9:17 PM, immibis wrote:
On 13/03/24 02:47, olcott wrote:
On 3/12/2024 8:05 PM, immibis wrote:
On 13/03/24 01:18, olcott wrote:They are identical except for their return value that is specified
On 3/12/2024 7:10 PM, immibis wrote:
So which part of ⟨Q, Γ, b, Σ, δ, q0, F⟩ is different?Exactly one element of Q differs by writing a 1 instead of a 0.
That's part of δ but this mistake doesn't matter.
It wasn't clear whether you were talking about a Turing machine
that was somehow identical but gave a different return value, or
one that was not identical. Now you have explained it is not
identical.
in a single state that is different.
*This means that they implement the exact same algorithm*
OK. Well, one of them gets the right answer and one of them gets the
wrong answer. What is the confusion?
The Linz Ĥ.H machine gets the wrong answer on its own
machine description no matter how its Linz H is defined.
This means that it gets the wrong answer on YES and the
wrong answer on NO.
This means that two different machines get two different wrong answers
on two different inputs. The fact that 1+1=4 is wrong and 2+2=2 is
wrong does not mean that 2+2=4 is wrong or 1+1=2 is wrong.
epistemological antinomy AKA incorrect question.
On 3/12/2024 10:49 PM, Richard Damon wrote:
On 3/12/24 7:41 PM, olcott wrote:
On 3/12/2024 9:17 PM, immibis wrote:
On 13/03/24 02:47, olcott wrote:
On 3/12/2024 8:05 PM, immibis wrote:
On 13/03/24 01:18, olcott wrote:They are identical except for their return value that is specified
On 3/12/2024 7:10 PM, immibis wrote:
So which part of ⟨Q, Γ, b, Σ, δ, q0, F⟩ is different?Exactly one element of Q differs by writing a 1 instead of a 0.
That's part of δ but this mistake doesn't matter.
It wasn't clear whether you were talking about a Turing machine
that was somehow identical but gave a different return value, or
one that was not identical. Now you have explained it is not
identical.
in a single state that is different.
*This means that they implement the exact same algorithm*
OK. Well, one of them gets the right answer and one of them gets the
wrong answer. What is the confusion?
The Linz Ĥ.H machine gets the wrong answer on its own
machine description no matter how its Linz H is defined.
This means that it gets the wrong answer on YES and the
wrong answer on NO.
Not quite. It always gets the wrong answer, but only one of them for
each quesiton.
They all gets the wrong answer on a whole class of questions because epistemological antinomies are not rejected as semantically invalid input.
For EACH SEPARATE definition of H, and thus H^, we have a different
question.
They are all the same epistemological antinomy category of question.
Note, the machine H^ isn't DEFINED to just get H^ as an input.
H^ is defined to get as an input, the description of ANY Turing
Machine, and to ask H what that machine applied to its description
will do, and then it does the opposite.
Thus, for every different H we go to test, we get a DIFFERENT H^
machine. and when we look at the question to H (or H^.H) about the
description (H^) (H^),
If H (H^) (H^) goes to qn, then H^ (H^) goes to qn too and halts, so
the correct answer would have been to go to qy.
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy too, and loops, so
the correct answer would have been to go to qn.
So, each case HAS a correct answer, just not the one that H (or H^.H)
goes to,
So yes, which ever one it goes to (and a given machine will only go to
one with this input) will be wrong, but the other one would have been
right, and an H* machine that answer the opposite of H would have been
correct.
On 3/12/2024 9:21 PM, Richard Damon wrote:
On 3/12/24 6:47 PM, olcott wrote:
On 3/12/2024 8:05 PM, immibis wrote:
On 13/03/24 01:18, olcott wrote:They are identical except for their return value that is specified
On 3/12/2024 7:10 PM, immibis wrote:
On 13/03/24 00:56, olcott wrote:Exactly one element of Q differs by writing a 1 instead of a 0.
On 3/12/2024 6:38 PM, immibis wrote:
On 13/03/24 00:24, olcott wrote:
On 3/12/2024 6:05 PM, immibis wrote:
On 12/03/24 23:53, olcott wrote:
On 3/12/2024 5:30 PM, Richard Damon wrote:
On 3/12/24 2:34 PM, olcott wrote:∀ H ∈ Turing_Machines_Returning_Boolean
On 3/12/2024 4:23 PM, Richard Damon wrote:
On 3/12/24 1:11 PM, olcott wrote:
Not exactly. A pair of otherwise identical machines that >>>>>>>>>>>>>>> (that are contained within the above specified set) >>>>>>>>>>>>>>> only differ by return value will both be wrong on the >>>>>>>>>>>>>>> same pathological input.
You mean a pair of DIFFERENT machines. Any difference is >>>>>>>>>>>>>> different.
Every decider/input pair (referenced in the above set) has a >>>>>>>>>>>>> corresponding decider/input pair that only differs by the >>>>>>>>>>>>> return
value of its decider.
Nope.
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Every H/TMD pair (referenced in the above set) has a
corresponding H/TMD pair that only differs by the return >>>>>>>>>>> value of its Boolean_TM.
That both of these H/TMD pairs get the wrong answer proves that >>>>>>>>>>> their question was incorrect because the opposite answer to the >>>>>>>>>>> same question is also proven to be incorrect.
Nobody knows what the fuck you are talking about. You have to >>>>>>>>>> actually explain it. The same machine always gives the same >>>>>>>>>> return value on the same input.
It has taken me twenty years to translate my intuitions into >>>>>>>>> words that can possibly understood.
You failed.
A pair of Turing Machines that return Boolean that are identical >>>>>>>>> besides their return value that cannot decide some property of >>>>>>>>> the same input are being asked the same YES/NO question having >>>>>>>>> no correct YES/NO answer.
https://en.wikipedia.org/wiki/Turing_machine#Formal_definition >>>>>>>> A Turing machine is ⟨Q, Γ, b, Σ, δ, q0, F⟩
Show me two ⟨Q, Γ, b, Σ, δ, q0, F⟩ that are identical besides >>>>>>>> their return value.
You can't because you are talking nonsense. they don't exist.
Turing machine descriptions that are identical finite strings
except for the the 1/0 that they write the their exact same
tape relative location.
So which part of ⟨Q, Γ, b, Σ, δ, q0, F⟩ is different?
That's part of δ but this mistake doesn't matter.
It wasn't clear whether you were talking about a Turing machine that
was somehow identical but gave a different return value, or one that
was not identical. Now you have explained it is not identical.
in a single state that is different.
*This means that they implement the exact same algorithm*
Nope, because the algorithm include the final transition to the output.
The decision criteria is identical
A protocol can be defined so that Turing machine descriptions always
implement their return value in their state with the largest Natural
Number value. This allows other Turing machines to determine identical
algorithms except for return value.
State doesn't have defined "numbers".
It must have something like this.
On 3/12/2024 10:33 PM, Richard Damon wrote:
On 3/12/24 4:56 PM, olcott wrote:
On 3/12/2024 6:38 PM, immibis wrote:
On 13/03/24 00:24, olcott wrote:
On 3/12/2024 6:05 PM, immibis wrote:
On 12/03/24 23:53, olcott wrote:
On 3/12/2024 5:30 PM, Richard Damon wrote:
On 3/12/24 2:34 PM, olcott wrote:∀ H ∈ Turing_Machines_Returning_Boolean
On 3/12/2024 4:23 PM, Richard Damon wrote:
On 3/12/24 1:11 PM, olcott wrote:
Not exactly. A pair of otherwise identical machines that >>>>>>>>>>> (that are contained within the above specified set)
only differ by return value will both be wrong on the
same pathological input.
You mean a pair of DIFFERENT machines. Any difference is
different.
Every decider/input pair (referenced in the above set) has a >>>>>>>>> corresponding decider/input pair that only differs by the return >>>>>>>>> value of its decider.
Nope.
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Every H/TMD pair (referenced in the above set) has a
corresponding H/TMD pair that only differs by the return
value of its Boolean_TM.
That both of these H/TMD pairs get the wrong answer proves that
their question was incorrect because the opposite answer to the
same question is also proven to be incorrect.
Nobody knows what the fuck you are talking about. You have to
actually explain it. The same machine always gives the same return >>>>>> value on the same input.
It has taken me twenty years to translate my intuitions into
words that can possibly understood.
You failed.
A pair of Turing Machines that return Boolean that are identical
besides their return value that cannot decide some property of
the same input are being asked the same YES/NO question having
no correct YES/NO answer.
https://en.wikipedia.org/wiki/Turing_machine#Formal_definition
A Turing machine is ⟨Q, Γ, b, Σ, δ, q0, F⟩
Show me two ⟨Q, Γ, b, Σ, δ, q0, F⟩ that are identical besides their >>>> return value.
You can't because you are talking nonsense. they don't exist.
Turing machine descriptions that are identical finite strings
except for the the 1/0 that they write the their exact same
tape relative location.
So they aren't identical.
"Identical except ..." means DIFFERENT.
So you LIE
Not at all. I did not know these details until immbis
pressed me for them. I did know these details until
I encoded them as quintuples. http://www2.lns.mit.edu/~dsw/turing/doc/tm_manual.txt
On 03/12/2024 07:52 PM, olcott wrote:
On 3/12/2024 9:28 PM, Richard Damon wrote:
On 3/12/24 4:31 PM, olcott wrote:
On 3/12/2024 6:11 PM, Richard Damon wrote:
On 3/12/24 3:53 PM, olcott wrote:
On 3/12/2024 5:30 PM, Richard Damon wrote:
On 3/12/24 2:34 PM, olcott wrote:∀ H ∈ Turing_Machines_Returning_Boolean
On 3/12/2024 4:23 PM, Richard Damon wrote:
On 3/12/24 1:11 PM, olcott wrote:
On 3/12/2024 2:40 PM, Richard Damon wrote:
On 3/12/24 12:02 PM, olcott wrote:
On 3/12/2024 1:31 PM, immibis wrote:
On 12/03/24 19:12, olcott wrote:
∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD) >>>>>>>>>>>>>>
There is some input TMD to every H such that
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
And it can be a different TMD to each H.
When we disallow decider/input pairs that are incorrect >>>>>>>>>>>>>> questions where both YES and NO are the wrong answer >>>>>>>>>>>>>Once we understand that either YES or NO is the right >>>>>>>>>>>>> answer, the whole rebuttal is tossed out as invalid and >>>>>>>>>>>>> incorrect.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does
not halt
BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
No, because a given H will only go to one of the answers. THAT >>>>>>>>>>> will be wrong, and the other one right.
∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Not exactly. A pair of otherwise identical machines that
(that are contained within the above specified set)
only differ by return value will both be wrong on the
same pathological input.
You mean a pair of DIFFERENT machines. Any difference is
different.
Every decider/input pair (referenced in the above set) has a
corresponding decider/input pair that only differs by the return >>>>>>>> value of its decider.
Nope.
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Every H/TMD pair (referenced in the above set) has a
corresponding H/TMD pair that only differs by the return
value of its Boolean_TM.
That isn't in the set above.
Nope, since both aren't in the set selected.
That both of these H/TMD pairs get the wrong answer proves that
their question was incorrect because the opposite answer to the
same question is also proven to be incorrect.
When they are deciders that must get the correct answer both
of them are not in the set.
*IF* they are correct decider.
WHen we select from all Turing Machine Deciders, there is no
requirement that any of them get any particular answer right.
So, ALL deciders are in the set that we cycle through and apply the
following logic to ALL of them.
Each is them paired with an input that it will get wrong, and the
existance of the input was what as just proven, the ^ template
When they are Turing_Machines_Returning_Boolean the this
set inherently includes identical pairs that only differ
by return value.
But in the step of select and input that they will get wrong, they
will be givne DIFFERENT inputs.
You just don't understand what that statement is saying.No the problem is that you are not paying attention.
I've expalined it, but it seems over you head.
No, you keep on making STUPID mistakes, like thinking that select a
input that the machine will get wrong needs to be the same for two
differnt machines.
For Every H, we show we can find at least one input (chosen just for >>>>> that machine) that it will get wrong.When we use machine templates then we can see instances of
the same machine that only differs by return value where both
get the wrong answer on the same input. By same input I mean
the same finite string of numerical values.
But if they returned differnt values, they will have different
descriptions.
Otherwise, how could a UTM get the right answer, since it only gets
the description.
We can get around all of this stuff by simply using this criteria:
Date 10/13/2022 11:29:23 AM
*MIT Professor Michael Sipser agreed this verbatim paragraph is correct*
(He has neither reviewed nor agreed to anything else in this paper)
(a) If simulating halt decider H correctly simulates its input D until H
correctly determines that its simulated D would never stop running
unless aborted then
(b) H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
*When we apply this criteria* (elaborated above)
Will you halt if you never abort your simulation?
*Then the halting problem is conquered*
When two different machines implementing this criteria
get different results from identical inputs then we
know that Pathological Self-Reference has been detected.
We don't even need to know that for:
*denial-of-service-attack detection*
*NO always means reject as unsafe*
But, Halting theorem never said "there's an input that halts
all machines", it just says "for any machine, there's an input
that halts it".
Where "halt the machine" means "put it in an infinite loop".
So, rather, Halting theorem never said "there's an input that
exhausts all machines", it just says, "for any machine, there's
an input that exhausts it".
I still don't see how that would be with infinite tapes though,
without a means of checking all the way right the tape in one
step, i.e. that it's 1's or 0's or any pattern at all, any
input that unbounded with respect to the machine basically
exhausts it or where the machine would run in the unbounded.
Of course any finite tape, has a static analysis that is
not infinite, that decides whether or not it halts
(or, loops, or grows, the state space of the decider).
Static analysis has to either enumerate or _infer_ the
state-space, where equal values in what's determined
the idempotent can detect loops, while inequalities
or proven divergence, ..., can detect unbounded growth.
Now, proving convergence or divergence is its own kind
of thing. For example, there are series that converge
very slowly, and series that diverge very slowly. These
are subtly intractable to analysis.
Then the usual idea of progressions that rapidly grow
yet return to what's detectable, are pathological to
analysis, only in resources not correctness or completion,
vis-a-vis the subtle intractability of the convergence or
divergence what either halts, or loops, or grows unboundedly.
Separately "not-halts" into "loops or grows unboundedly",
has that really for most matters of interest of understanding
the state-space of programs, is "halts or enters loops"
and not "grows unboundedly".
I.e. if the Halting problem is basically subject the
subtle intractability of slow convergence, that otherwise
it can just infer divergence and decide, practically
it's sort of more relevant what the machine would be
doing on the input on the tape, then with respect to
beyond the Turing theory, of the state of the read-head,
what happens when somebody modifies the tape, or events,
the write-head.
Anyways though for bounded inputs, besides slow divergence,
it's to be made clear that _most all_ and _almost all_
programs _are_ decided their behavior by static analysis.
Though, "most all" and "almost all" might be a bit strong,
but pretty much all that don't involve "the subtle intractability
of slow divergence".
Giving the idea that an existence result
is in any way the expected result here
seems sort of the root of this dilem-na.
(Though that the real numbers in ZFC have a well-ordering
and if they had a normal ordering that was a well-ordering,
that would be a thing, because ZFC has a well-ordering of
[0,1], but can't give one.)
On 3/12/2024 9:28 PM, Richard Damon wrote:
On 3/12/24 4:31 PM, olcott wrote:
On 3/12/2024 6:11 PM, Richard Damon wrote:
On 3/12/24 3:53 PM, olcott wrote:
On 3/12/2024 5:30 PM, Richard Damon wrote:
On 3/12/24 2:34 PM, olcott wrote:∀ H ∈ Turing_Machines_Returning_Boolean
On 3/12/2024 4:23 PM, Richard Damon wrote:
On 3/12/24 1:11 PM, olcott wrote:Every decider/input pair (referenced in the above set) has a
On 3/12/2024 2:40 PM, Richard Damon wrote:
On 3/12/24 12:02 PM, olcott wrote:
On 3/12/2024 1:31 PM, immibis wrote:
On 12/03/24 19:12, olcott wrote:
∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
There is some input TMD to every H such that
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
And it can be a different TMD to each H.
When we disallow decider/input pairs that are incorrect >>>>>>>>>>>>> questions where both YES and NO are the wrong answer
Once we understand that either YES or NO is the right
answer, the whole rebuttal is tossed out as invalid and >>>>>>>>>>>> incorrect.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does
not halt
BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
No, because a given H will only go to one of the answers. THAT >>>>>>>>>> will be wrong, and the other one right.
∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Not exactly. A pair of otherwise identical machines that
(that are contained within the above specified set)
only differ by return value will both be wrong on the
same pathological input.
You mean a pair of DIFFERENT machines. Any difference is different. >>>>>>>
corresponding decider/input pair that only differs by the return >>>>>>> value of its decider.
Nope.
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Every H/TMD pair (referenced in the above set) has a
corresponding H/TMD pair that only differs by the return
value of its Boolean_TM.
That isn't in the set above.
Nope, since both aren't in the set selected.
That both of these H/TMD pairs get the wrong answer proves that
their question was incorrect because the opposite answer to the
same question is also proven to be incorrect.
When they are deciders that must get the correct answer both
of them are not in the set.
*IF* they are correct decider.
WHen we select from all Turing Machine Deciders, there is no
requirement that any of them get any particular answer right.
So, ALL deciders are in the set that we cycle through and apply the
following logic to ALL of them.
Each is them paired with an input that it will get wrong, and the
existance of the input was what as just proven, the ^ template
When they are Turing_Machines_Returning_Boolean the this
set inherently includes identical pairs that only differ
by return value.
But in the step of select and input that they will get wrong, they
will be givne DIFFERENT inputs.
You just don't understand what that statement is saying.No the problem is that you are not paying attention.
I've expalined it, but it seems over you head.
No, you keep on making STUPID mistakes, like thinking that select a
input that the machine will get wrong needs to be the same for two
differnt machines.
For Every H, we show we can find at least one input (chosen just forWhen we use machine templates then we can see instances of
that machine) that it will get wrong.
the same machine that only differs by return value where both
get the wrong answer on the same input. By same input I mean
the same finite string of numerical values.
But if they returned differnt values, they will have different
descriptions.
Otherwise, how could a UTM get the right answer, since it only gets
the description.
We can get around all of this stuff by simply using this criteria:
Date 10/13/2022 11:29:23 AM
*MIT Professor Michael Sipser agreed this verbatim paragraph is correct*
(He has neither reviewed nor agreed to anything else in this paper)
(a) If simulating halt decider H correctly simulates its input D until H correctly determines that its simulated D would never stop running
unless aborted then
(b) H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
*When we apply this criteria* (elaborated above)
Will you halt if you never abort your simulation?
*Then the halting problem is conquered*
When two different machines implementing this criteria
get different results from identical inputs then we
know that Pathological Self-Reference has been detected.
We don't even need to know that for:
*denial-of-service-attack detection*
*NO always means reject as unsafe*
On 3/12/2024 11:16 PM, Richard Damon wrote:
On 3/12/24 8:41 PM, olcott wrote:
On 3/12/2024 10:33 PM, Richard Damon wrote:
On 3/12/24 4:56 PM, olcott wrote:
On 3/12/2024 6:38 PM, immibis wrote:
On 13/03/24 00:24, olcott wrote:
On 3/12/2024 6:05 PM, immibis wrote:
On 12/03/24 23:53, olcott wrote:
On 3/12/2024 5:30 PM, Richard Damon wrote:
On 3/12/24 2:34 PM, olcott wrote:∀ H ∈ Turing_Machines_Returning_Boolean
On 3/12/2024 4:23 PM, Richard Damon wrote:
On 3/12/24 1:11 PM, olcott wrote:
Not exactly. A pair of otherwise identical machines that >>>>>>>>>>>>> (that are contained within the above specified set)
only differ by return value will both be wrong on the >>>>>>>>>>>>> same pathological input.
You mean a pair of DIFFERENT machines. Any difference is >>>>>>>>>>>> different.
Every decider/input pair (referenced in the above set) has a >>>>>>>>>>> corresponding decider/input pair that only differs by the return >>>>>>>>>>> value of its decider.
Nope.
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Every H/TMD pair (referenced in the above set) has a
corresponding H/TMD pair that only differs by the return
value of its Boolean_TM.
That both of these H/TMD pairs get the wrong answer proves that >>>>>>>>> their question was incorrect because the opposite answer to the >>>>>>>>> same question is also proven to be incorrect.
Nobody knows what the fuck you are talking about. You have to
actually explain it. The same machine always gives the same
return value on the same input.
It has taken me twenty years to translate my intuitions into
words that can possibly understood.
You failed.
A pair of Turing Machines that return Boolean that are identical >>>>>>> besides their return value that cannot decide some property of
the same input are being asked the same YES/NO question having
no correct YES/NO answer.
https://en.wikipedia.org/wiki/Turing_machine#Formal_definition
A Turing machine is ⟨Q, Γ, b, Σ, δ, q0, F⟩
Show me two ⟨Q, Γ, b, Σ, δ, q0, F⟩ that are identical besides >>>>>> their return value.
You can't because you are talking nonsense. they don't exist.
Turing machine descriptions that are identical finite strings
except for the the 1/0 that they write the their exact same
tape relative location.
So they aren't identical.
"Identical except ..." means DIFFERENT.
So you LIE
Not at all. I did not know these details until immbis
pressed me for them. I did know these details until
I encoded them as quintuples.
http://www2.lns.mit.edu/~dsw/turing/doc/tm_manual.txt
So, you still have been trying to say that the two machines that do
the opposite have the same identical algorithm.
Note to write a 1 or a 0 requires different quintuples.
The have the exact same steps except their final write to tape step.
On 3/12/2024 11:14 PM, Richard Damon wrote:
On 3/12/24 7:44 PM, olcott wrote:The criteria is identical the answer is reversed.
On 3/12/2024 9:21 PM, Richard Damon wrote:
On 3/12/24 6:47 PM, olcott wrote:The decision criteria is identical
On 3/12/2024 8:05 PM, immibis wrote:
On 13/03/24 01:18, olcott wrote:They are identical except for their return value that is specified
On 3/12/2024 7:10 PM, immibis wrote:
On 13/03/24 00:56, olcott wrote:Exactly one element of Q differs by writing a 1 instead of a 0.
On 3/12/2024 6:38 PM, immibis wrote:
On 13/03/24 00:24, olcott wrote:Turing machine descriptions that are identical finite strings >>>>>>>>> except for the the 1/0 that they write the their exact same
On 3/12/2024 6:05 PM, immibis wrote:
On 12/03/24 23:53, olcott wrote:
On 3/12/2024 5:30 PM, Richard Damon wrote:
On 3/12/24 2:34 PM, olcott wrote:∀ H ∈ Turing_Machines_Returning_Boolean
On 3/12/2024 4:23 PM, Richard Damon wrote:
On 3/12/24 1:11 PM, olcott wrote:
Not exactly. A pair of otherwise identical machines that >>>>>>>>>>>>>>>>> (that are contained within the above specified set) >>>>>>>>>>>>>>>>> only differ by return value will both be wrong on the >>>>>>>>>>>>>>>>> same pathological input.
You mean a pair of DIFFERENT machines. Any difference is >>>>>>>>>>>>>>>> different.
Every decider/input pair (referenced in the above set) has a >>>>>>>>>>>>>>> corresponding decider/input pair that only differs by the >>>>>>>>>>>>>>> return
value of its decider.
Nope.
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Every H/TMD pair (referenced in the above set) has a >>>>>>>>>>>>> corresponding H/TMD pair that only differs by the return >>>>>>>>>>>>> value of its Boolean_TM.
That both of these H/TMD pairs get the wrong answer proves >>>>>>>>>>>>> that
their question was incorrect because the opposite answer to >>>>>>>>>>>>> the
same question is also proven to be incorrect.
Nobody knows what the fuck you are talking about. You have >>>>>>>>>>>> to actually explain it. The same machine always gives the >>>>>>>>>>>> same return value on the same input.
It has taken me twenty years to translate my intuitions into >>>>>>>>>>> words that can possibly understood.
You failed.
A pair of Turing Machines that return Boolean that are identical >>>>>>>>>>> besides their return value that cannot decide some property of >>>>>>>>>>> the same input are being asked the same YES/NO question having >>>>>>>>>>> no correct YES/NO answer.
https://en.wikipedia.org/wiki/Turing_machine#Formal_definition >>>>>>>>>> A Turing machine is ⟨Q, Γ, b, Σ, δ, q0, F⟩
Show me two ⟨Q, Γ, b, Σ, δ, q0, F⟩ that are identical besides >>>>>>>>>> their return value.
You can't because you are talking nonsense. they don't exist. >>>>>>>>>
tape relative location.
So which part of ⟨Q, Γ, b, Σ, δ, q0, F⟩ is different?
That's part of δ but this mistake doesn't matter.
It wasn't clear whether you were talking about a Turing machine
that was somehow identical but gave a different return value, or
one that was not identical. Now you have explained it is not
identical.
in a single state that is different.
*This means that they implement the exact same algorithm*
Nope, because the algorithm include the final transition to the output. >>>
No, the decision criteria is reversed. The generation of the output is
part of the algortihm
You are so dumb you don't understand that.There are two way of looking at this.
The criteria has the exact same steps and the answer is reversed.
A protocol can be defined so that Turing machine descriptions always >>>>> implement their return value in their state with the largest Natural >>>>> Number value. This allows other Turing machines to determine identical >>>>> algorithms except for return value.
State doesn't have defined "numbers".
It must have something like this.
Nope. States are SYMBOLS / NAMES. They can be encoded into numbers,
but are not themselves numbers.
Not when compiled into machine code.
On 3/12/2024 11:25 PM, Richard Damon wrote:
On 3/12/24 7:52 PM, olcott wrote:
On 3/12/2024 9:28 PM, Richard Damon wrote:
On 3/12/24 4:31 PM, olcott wrote:
On 3/12/2024 6:11 PM, Richard Damon wrote:
On 3/12/24 3:53 PM, olcott wrote:
On 3/12/2024 5:30 PM, Richard Damon wrote:
On 3/12/24 2:34 PM, olcott wrote:∀ H ∈ Turing_Machines_Returning_Boolean
On 3/12/2024 4:23 PM, Richard Damon wrote:
On 3/12/24 1:11 PM, olcott wrote:
On 3/12/2024 2:40 PM, Richard Damon wrote:
On 3/12/24 12:02 PM, olcott wrote:
On 3/12/2024 1:31 PM, immibis wrote:
On 12/03/24 19:12, olcott wrote:
∀ H ∈ Turing_Machine_DecidersAnd it can be a different TMD to each H.
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD) >>>>>>>>>>>>>>>
There is some input TMD to every H such that
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD) >>>>>>>>>>>>>>
When we disallow decider/input pairs that are incorrect >>>>>>>>>>>>>>> questions where both YES and NO are the wrong answer >>>>>>>>>>>>>>Once we understand that either YES or NO is the right >>>>>>>>>>>>>> answer, the whole rebuttal is tossed out as invalid and >>>>>>>>>>>>>> incorrect.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩
does not halt
BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
No, because a given H will only go to one of the answers. >>>>>>>>>>>> THAT will be wrong, and the other one right.
∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Not exactly. A pair of otherwise identical machines that >>>>>>>>>>> (that are contained within the above specified set)
only differ by return value will both be wrong on the
same pathological input.
You mean a pair of DIFFERENT machines. Any difference is
different.
Every decider/input pair (referenced in the above set) has a >>>>>>>>> corresponding decider/input pair that only differs by the return >>>>>>>>> value of its decider.
Nope.
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
Every H/TMD pair (referenced in the above set) has a
corresponding H/TMD pair that only differs by the return
value of its Boolean_TM.
That isn't in the set above.
Nope, since both aren't in the set selected.
That both of these H/TMD pairs get the wrong answer proves that
their question was incorrect because the opposite answer to the
same question is also proven to be incorrect.
When they are deciders that must get the correct answer both
of them are not in the set.
*IF* they are correct decider.
WHen we select from all Turing Machine Deciders, there is no
requirement that any of them get any particular answer right.
So, ALL deciders are in the set that we cycle through and apply the
following logic to ALL of them.
Each is them paired with an input that it will get wrong, and the
existance of the input was what as just proven, the ^ template
When they are Turing_Machines_Returning_Boolean the this
set inherently includes identical pairs that only differ
by return value.
But in the step of select and input that they will get wrong, they
will be givne DIFFERENT inputs.
You just don't understand what that statement is saying.No the problem is that you are not paying attention.
I've expalined it, but it seems over you head.
No, you keep on making STUPID mistakes, like thinking that select a
input that the machine will get wrong needs to be the same for two
differnt machines.
For Every H, we show we can find at least one input (chosen justWhen we use machine templates then we can see instances of
for that machine) that it will get wrong.
the same machine that only differs by return value where both
get the wrong answer on the same input. By same input I mean
the same finite string of numerical values.
But if they returned differnt values, they will have different
descriptions.
Otherwise, how could a UTM get the right answer, since it only gets
the description.
We can get around all of this stuff by simply using this criteria:
Date 10/13/2022 11:29:23 AM
*MIT Professor Michael Sipser agreed this verbatim paragraph is correct* >>> (He has neither reviewed nor agreed to anything else in this paper)
(a) If simulating halt decider H correctly simulates its input D
until H correctly determines that its simulated D would never stop
running unless aborted then
(b) H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
*When we apply this criteria* (elaborated above)
Will you halt if you never abort your simulation?
*Then the halting problem is conquered*
When two different machines implementing this criteria
get different results from identical inputs then we
know that Pathological Self-Reference has been detected.
We don't even need to know that for:
*denial-of-service-attack detection*
*NO always means reject as unsafe*
Which means, when you understand what the words ACTUALLY MEAN, is that
H is allowed to abort is simulation, if it can CORRECTLY DETERMINE,
that the ACTUAL CORRECT SIMULATION (ie with a different machine that
never stops simulating, ie a UTM) would never stop.
The key is whether or not it can be thwarted by any of
the conventional halting problem proofs.
*This is stipulated to be the criteria*
Will you halt if you never abort your simulation?
Can it be thwarted by any of the conventional HP proofs?
This giving to the UTM doesn't change the machine represented on the
tape, fot H^ it is still using the copy of H, as you finally decided
to make it, so if that aborts and returns qn, then H^'s submachine
copy of H at H^.H will ALSO abort its simulation and return to qn, and
H^ will then Halt, making H wrong, because it didn't CORRECTLY
DETERMINE that the UTM simulating this input would never halt.
So, your definition of DOS detection is allowing false positives for
non-halting detection, which is fine if you don't mind some wrong
answers, but doesn't make it a HALT DECIDER.
You are just ADMITTING that you have been LYING about this, (LYING
since you claim it applies to the Linz proof, which is about Halt
Deciders, not DOS detectors).
On 3/12/2024 10:49 PM, Richard Damon wrote:
Not quite. It always gets the wrong answer, but only one of them for
each quesiton.
They all gets the wrong answer on a whole class of questions
because epistemological antinomies are not rejected as semantically invalid input.
For EACH SEPARATE definition of H, and thus H^, we have a different
question.
They are all the same epistemological antinomy category of question.
There is no problem with making a H^ from an H, it is built from
totally legal steps.
On 3/12/2024 5:20 AM, Mikko wrote:
On 2024-03-11 16:02:28 +0000, olcott said:
On 3/11/2024 10:12 AM, Mikko wrote:
On 2024-03-11 14:54:34 +0000, olcott said:
On 3/11/2024 5:45 AM, Mikko wrote:
On 2024-03-11 04:38:40 +0000, olcott said:
On 3/10/2024 11:10 PM, Richard Damon wrote:An incorrect answer does not mean that the question is incorrect.
On 3/10/24 8:33 PM, olcott wrote:
On 3/10/2024 9:13 PM, Richard Damon wrote:
On 3/10/24 7:05 PM, olcott wrote:
On 3/10/2024 8:52 PM, Richard Damon wrote:
On 3/10/24 6:00 PM, olcott wrote:
On 3/10/2024 2:23 PM, Richard Damon wrote:
On 3/10/24 11:23 AM, olcott wrote:
On 3/10/2024 12:55 PM, Richard Damon wrote:
On 3/10/24 10:17 AM, olcott wrote:If no source can be cited that shows a simulating halt decider can
On 3/10/2024 12:08 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>> On 3/10/24 9:52 AM, olcott wrote:Since, BY THE DEFINITIONS of what H MUST do to be correct, and what H^
On 3/10/2024 10:50 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/10/24 7:28 AM, olcott wrote:NOPE.
On 3/10/2024 12:16 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/9/24 9:49 PM, olcott wrote:
On 3/9/2024 11:36 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/9/24 9:14 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/9/2024 10:55 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/9/24 8:30 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/9/2024 7:40 PM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 10/03/24 02:37, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/9/2024 7:32 PM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 10/03/24 02:29, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/9/2024 7:24 PM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 10/03/24 01:30, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/9/2024 6:24 PM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 10/03/24 01:22, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/9/2024 5:57 PM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 10/03/24 00:26, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/9/2024 5:10 PM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 9/03/24 23:22, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/9/2024 3:50 PM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 9/03/24 22:34, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> What criteria would you use so that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ knows what
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ haltsIt is objectively true that Ĥ.H can get stuck in recursiveSimulating halt deciders must make sure that they themselveswrong answer to provide? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same objective criteria thatH ⟨Ĥ⟩ uses. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
do not get stuck in infinite execution. This means that they
must abort every simulation that cannot possibly otherwise halt.
This requires Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ to abort its simulation and does not
require H ⟨Ĥ⟩ ⟨Ĥ⟩ to abort its simulation when Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ aborts
its simulation. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ does simulate itself in recursive simulation H ⟨Ĥ⟩ ⟨Ĥ⟩
does not simulate itself in recursive simulation.
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same objective criteria that
H ⟨Ĥ⟩ uses. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*Only because Ĥ.H is embedded within Ĥ and H is not*
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly get stuck in recursive simulation and
H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly get stuck in recursive simulation.
You dishonestly ignored that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact
same OBJECTIVE criteria that H ⟨Ĥ⟩ uses.
The above is true no matter what criteria that is used
as long as H is a simulating halt decider. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Objective criteria cannot vary based on who the subject is. They are
objective. The answer to different people is the same answer if the
criteria are objective. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
simulation because Ĥ copies its input thus never runs
out of params. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
It is objectively true that Ĥ cannot possibly get stuck
in recursive because H does not copy its input thus runs
out of params. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Wrong. Dead wrong. Stupidly wrong. So wrong that a dead monkey could do
better. Write the Olcott machine (not x86utm) code for Ĥ and I would
show you.
*In other words you are denying these verified facts*
*In other words you are denying these verified facts*
*In other words you are denying these verified facts*
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
That's not a verified fact, that's just something you want to be true.
∞ means infinite loop. Infinite loop doesn't halt. You see how stupid
it is, to say that an infinite loop halts? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Execution trace of Ĥ applied to ⟨Ĥ⟩ >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> (a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
(b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process
Execution trace of H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ BECAUSE IT IS PRECISELY
IDENTICAL TO STEPS B AND C: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> > (b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at Ĥ's own simulated ⟨Ĥ.q0⟩ to repeat the process
*Yes and the key step of copying its input is left out so*
*H ⟨Ĥ⟩ ⟨Ĥ⟩ runs out of params and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ never runs out of params*
that isn't how any of this works. Do you even know what words mean?
(b) and (c) are not the same as (1) and (2) >>>>>>>>>>>>>>>>>>>>>>>>>>> Execution trace of H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ >>>>>>>>>>>>>>>>>>>>>>>>>>> (1) H applied ⟨Ĥ⟩ ⟨Ĥ⟩ simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(2) which begins at simulated ⟨Ĥ.q0⟩ >>>>>>>>>>>>>>>>>>>>>>>>>>> (a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
(b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process
This means that Turing machine H ⟨Ĥ⟩ ⟨Ĥ⟩ can see one more execution
trace of Ĥ ⟨Ĥ⟩ than its simulated Turing machine Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can see.
Nope, your just being stuupid, perhaps intentionally.
(c) just moves around to its simulation of a >>>>>>>>>>>>>>>>>>>>>>>>>>
(a) H^.q0 (H^)
H^ then makes a copy of its inp >>>>>>>>>>>>>>>>>>>>>>>>>>
(b) H^.H (H^) (H^) == (1) H (H^) (H^) >>>>>>>>>>>>>>>>>>>>>>>>>> The algorithm of H begins a simulation of its input, watching the
behaior of H^ (H^)
(c) = (2)
Which begins at the simulation of H^.q0 (H^) >>>>>>>>>>>>>>>>>>>>>>>>>>
(d = sim a) = (sim a)
Ths Simulated H^.q0 (H^) makes a copy of its input >>>>>>>>>>>>>>>>>>>>>>>>>>
(e = sim b) = (sim b)
The Simulated H^.H (H^) (H^) has is H begin the simulation of its input ...
and so on.
Both machine see EXACTLY the same level of details. >>>>>>>>>>>>>>>>>>>>>>>>>>
Yes, the top level H is farther along at any given time then its
simulated machine, and that is H's problem, it has to act before it
sees how its simulation will respond to its copy of its actions.
Thus, if it stops, it needs to make its decision "blind" and not with
an idea of how the machine it is simulating will perform.
If it doesn't stop, the level of recursion just keeps growing and no
answer ever comes out.
The earliest point that H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly see to abort
its simulation is immediately before Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ would begin
its simulation. Right before its cycle repeats the first time.
So?
If it DOES abort there, then so will H^.H when it gets to that point in
its simulation, which will be AFTER The point that H has stopped
simulating it, so H doesn't know what H^ will do. >>>>>>>>>>>>>>>>>>>>>>>>
Thus, if H DOES abort there, we presume from your previous answer it
will think the input will not halt and answer qn. >>>>>>>>>>>>>>>>>>>>>>>>
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
H ⟨Ĥ⟩ ⟨Ĥ⟩ aborts right after Ĥ.Hq0 before it simulates ⟨Ĥ⟩ ⟨Ĥ⟩.
And if it does, as I said below, so will H^.H when it is run.
Yes.
And thus, H^.H will give the same answer as H, >>>>>>>>>>>>>>>>>>>> so H^ will act contrary to what H says, >>>>>>>>>>>>>>>>>>>> so H will give the wrong answer.
Unlike anything else that anyone else has ever done both H ⟨Ĥ⟩ ⟨Ĥ⟩
and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly determine that they must abort their own
simulation to prevent their own infinite execution. >>>>>>>>>>>>>>>>>>
If no source can be cited then the Olcott thesis >>>>>>>>>>>>>>>>> "that no one did this before" remains unrefuted. >>>>>>>>>>>>>>>>
WILL do by its design, as shown in the Linz Proof. >>>>>>>>>>>>>>>
correctly determine that it must abort its simulation of the Halting
Problem's pathological input to prevent its own non-termination, then
innovation remains attributable to me.
<snip>
Of course it can abort its simulation.
It just needs some way to get the right answer.
*I have always been using this long before I read about it* >>>>>>>>>>>>> blind variation and selective retention (BVSR)...
Two common phenomena characterize BVSR thinking: superfluity and >>>>>>>>>>>>> backtracking. Superfluity means that the creator generates a variety of
ideas, one or more of which turn out to be useless.
But if you have mo idea how things actually works, this seems to just
generate random noise.
Nope, I remember talk of that when I was in college, and they showed
Backtracking signifies that the creator must often return to an earlier
approach after blindly going off in the wrong direction. >>>>>>>>>>>>> https://www.scientificamerican.com/article/the-science-of-genius/ >>>>>>>>>>>>>
*I am aware of no one else that had the idea to apply a simulating*
*termination analyzer to the halting problem counter-example input*
Professor Hehner had a seed of this idea before I did. >>>>>>>>>>>>
why it can't work.
From a programmer's point of view, if we apply an interpreter to a
program text that includes a call to that same interpreter with that
same text as argument, then we have an infinite loop. A halting program
has some of the same character as an interpreter: it applies to texts
through abstract interpretation. Unsurprisingly, if we apply a halting
program to a program text that includes a call to that same halting
program with that same text as argument, then we have an infinite loop.
https://www.cs.toronto.edu/~hehner/PHP.pdf
You THINK so, but if the interpreter is a CONDITIONAL interpreter, that
doesn't hold.
You seem to miss that fact.
*Turing Machine and Olcott machine implementations seem to be dead*
*This the (possibly augmented) RASP machine equivalent of x86* >>>>>>>>>>>>> Every machine must be able to get its own machine address. >>>>>>>>>>>>>
And the reason it is a dead end is they make it too hard for you to cheat.
You need to hide that your H is trying to get in some extra information
to hide that the embedded version of H doesn't give the same answer,
which just shows that your H^ is built wrong.
My C code proves these two have different behavior:
(a) H1(D,D) + H1_machine_address
(b) H(D,D) + H_machine_address
H1(D,D) does correctly determine the halt status of D(D) because >>>>>>>>>>> H(D,D) does NOT correctly determine the halt status of D(D). >>>>>>>>>>>
I say:
H1(D,D) is isomorphic to H ⟨Ĥ⟩ ⟨Ĥ⟩
H(D,D) is isomorphic to Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
immibis disagrees.
Correct reasoning will show who is correct.
Yes, and since H1 is a different computation than H, it getting the >>>>>>>>>> right answer doesn't keep H from being broken.
We can then make a D1 to break H1.
I think that immibis already said that and I did not notice
the significance of it at the time.
Of course.
Then we are back to undecidability being incorrectly construed >>>>>>>>> as an actual limit to computation.
Strange definition of not an actual limit if not being able to do >>>>>>>> something isn't a limit.
Professor's Hehner and Stoddart have only construed this as
applying to the Halting Problem's pathological input.
We three perfectly agree on this as it pertains to the
Halting Problem.
That two full PhD professors of computer science and I all
agree on this shows that I am not a crackpot/crank on this.
I think that all of the other options may now be exhausted.
I am very happy that I quit tolerating the [change the subject] >>>>>>>>> form of rebuttal that wasted 15 years with Ben Bacarisse.
*The focus now must be on finding the best words that prove* >>>>>>>>> *this original position of mine (thus the concurring positions* >>>>>>>>> *of professors Hehner and Stoddart) is correct*
Go knock yourself out on that.
Alan Turing's Halting Problem is incorrectly formed (PART-TWO) sci.logic
On 6/20/2004 11:31 AM, Peter Olcott wrote:
PREMISES:
(1) The Halting Problem was specified in such a way that a solution
was defined to be impossible.
Nope.
The PROBLEM is the question if a machine can compute the Halting Function.
The answer to that, has turned out to be NO.
When the problem was first being posed, it was hoped the answer woudl >>>>>>>> be yes, so it couldn't have bee made specifically to make it impossible.
The Halting QUESTION, has an answer for every input that it the >>>>>>>> description of an actual algorithm applied to an actual data input. >>>>>>>>
Note, Not a "Template" that gets appled to the decider, that IS an >>>>>>>> invalid question, and impossible to build a description of a Turing >>>>>>>> Machine to ask that.
Thus, when you admitted that your input wasn't actually a description >>>>>>>> of a program, but just a template, you were admitting that you were >>>>>>>> lying about working on the Halting Problem, as your input isn't of the >>>>>>>> right type.
Yes, asking about a template IS an invalid question.
;
(2) The set of questions that are defined to not have any possible
correct answer(s) forms a proper subset of all possible questions.
…
And, when you are asking the actual Halting Question, about a specific >>>>>>>> machine and input, like a SPECIFIC H^, built to foil a SPECIIFIC H, >>>>>>>> then that input has a specific and definate behavior and there is a >>>>>>>> specific and definate answer (That depends on the H that you chose to >>>>>>>> build it on, but not the decider you are asking the question to). >>>>>>>>
CONCLUSION:
Therefore the Halting Problem is an ill-formed question. >>>>>>>>> >
Nope, as explained above. You are just showing that you never
understood the actual question or what any of the theory actually >>>>>>>> means, and have just wasted the last decades of your life on a stupid >>>>>>>> misconception of your own.
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Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
That YES and NO are the wrong answer for each implementation of
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ specified by the second ⊢* state transition proves that the
questions asked of these machine/inputs pairs are incorrect questions. >>>>>>
When every element of the infinite set of every possible
implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer then
there is something wrong with the question.
According to the definition by Linz there is only one Ĥ for each H.
Anyway, a wrong answer, even if given in large quentities, does
not make the question worng.
None-the-less the infinite set of every implementation of
H/Ĥ.H cannot possibly get an answer that is consistent with
the behavior of of Ĥ ⟨Ĥ⟩.
The infinite set is not expected to answer. And the set is
infinite only if you include defective imiplementations.
Anyway, no memeber of the set of implementations can get
an answer that is consistent with the corresponding counter
example.
∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
There is some input TMD to every H such that
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
When we disallow decider/input pairs that are incorrect
questions where both YES and NO are the wrong answer
(the same way the ZFC disallowed self-referential sets) then
pathological inputs are not allowed to come into existence.
Does the barber that shaves everyone that does not shave
themselves shave himself? is rejected as an incorrect question. https://en.wikipedia.org/wiki/Barber_paradox#
On 3/13/2024 12:04 PM, Mikko wrote:
On 2024-03-12 18:00:11 +0000, Richard Damon said:
There is no problem with making a H^ from an H, it is built from
totally legal steps.
Another way to say the same is that if H^ cannot be built
the H does not exist.
The proof of the halting problem assumes a universal halt test
exists and then provides S as an example of a program that the
test cannot handle. But S is not a program at all. It is not
even a conceptual object, and this is due to inconsistencies
in the specification of the halting function. (Stoddart: 2017)
On 3/13/2024 12:02 PM, Mikko wrote:
On 2024-03-12 15:52:10 +0000, olcott said:
The self-contradictory instances are all incorrect.∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
There is some input TMD to every H such that
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
When we disallow decider/input pairs that are incorrect
questions where both YES and NO are the wrong answer
(the same way the ZFC disallowed self-referential sets) then
pathological inputs are not allowed to come into existence.
No question of the type "Does T(I) halt?" is incorrect
in that sense.
The answer of YES affirms a false presupposition thus is incorrect.Does the barber that shaves everyone that does not shave
themselves shave himself? is rejected as an incorrect question.
https://en.wikipedia.org/wiki/Barber_paradox#
The answer "Yes" is correct because no such barber fails to
shave himself.
The answer "No" is correct because no such barber shaves himself.The answer of NO affirms a false presupposition thus is incorrect.
Because the qeustion can be correctly answered "Yes" or "No"
the question is not incorrect.
On 3/13/2024 11:44 AM, immibis wrote:
On 13/03/24 04:55, olcott wrote:
On 3/12/2024 10:49 PM, Richard Damon wrote:
Not quite. It always gets the wrong answer, but only one of them for
each quesiton.
They all gets the wrong answer on a whole class of questions
Wrong. You said. yourself. that H1 gets the right answer for D.
Since it is a logical impossibility to determine the truth
value of a self-contradictory expression the requirement
for H to do this is bogus.
because epistemological antinomies are not rejected as semantically
invalid input.
For EACH SEPARATE definition of H, and thus H^, we have a different
question.
They are all the same epistemological antinomy category of question.
1+1 and 2+2 are in the same category of question. This does not mean
that when Billy says that 1+1=4 and Bobby says that 2+2=2 that these
questions have no correct answers.
On 3/13/2024 12:04 PM, Mikko wrote:
On 2024-03-12 18:00:11 +0000, Richard Damon said:
There is no problem with making a H^ from an H, it is built from
totally legal steps.
Another way to say the same is that if H^ cannot be built
the H does not exist.
The proof of the halting problem assumes a universal halt test
exists and then provides S as an example of a program that the
test cannot handle. But S is not a program at all. It is not
even a conceptual object, and this is due to inconsistencies
in the specification of the halting function. (Stoddart: 2017)
On 3/13/2024 2:15 PM, immibis wrote:
On 13/03/24 18:11, olcott wrote:
On 3/13/2024 12:04 PM, Mikko wrote:
On 2024-03-12 18:00:11 +0000, Richard Damon said:
There is no problem with making a H^ from an H, it is built from
totally legal steps.
Another way to say the same is that if H^ cannot be built
the H does not exist.
The proof of the halting problem assumes a universal halt test
exists and then provides S as an example of a program that the
test cannot handle. But S is not a program at all. It is not
even a conceptual object, and this is due to inconsistencies
in the specification of the halting function. (Stoddart: 2017)
This is obviously bullshit. Why do you lie?
That was cited as a statement from Bill Stoddart and a direct quote
from this paper.
Bill Stoddart. The Halting Paradox
20 December 2017
https://arxiv.org/abs/1906.05340
arXiv:1906.05340 [cs.LO]
That you call me a liar about this meets the required definition of
reckless disregard for the truth required for libel and slander. https://dictionary.findlaw.com/definition/reckless-disregard-of-the-truth.html
On 3/13/2024 2:15 PM, immibis wrote:
On 13/03/24 18:11, olcott wrote:
On 3/13/2024 12:04 PM, Mikko wrote:
On 2024-03-12 18:00:11 +0000, Richard Damon said:
There is no problem with making a H^ from an H, it is built from
totally legal steps.
Another way to say the same is that if H^ cannot be built
the H does not exist.
The proof of the halting problem assumes a universal halt test
exists and then provides S as an example of a program that the
test cannot handle. But S is not a program at all. It is not
even a conceptual object, and this is due to inconsistencies
in the specification of the halting function. (Stoddart: 2017)
This is obviously bullshit. Why do you lie?
That was cited as a statement from Bill Stoddart and a direct quote
from this paper.
Bill Stoddart. The Halting Paradox
20 December 2017
https://arxiv.org/abs/1906.05340
arXiv:1906.05340 [cs.LO]
That you call me a liar about this meets the required definition of
reckless disregard for the truth required for libel and slander. https://dictionary.findlaw.com/definition/reckless-disregard-of-the-truth.html
On 3/13/2024 12:52 PM, Richard Damon wrote:
On 3/13/24 10:08 AM, olcott wrote:
On 3/13/2024 11:44 AM, immibis wrote:
On 13/03/24 04:55, olcott wrote:
On 3/12/2024 10:49 PM, Richard Damon wrote:
Not quite. It always gets the wrong answer, but only one of them
for each quesiton.
They all gets the wrong answer on a whole class of questions
Wrong. You said. yourself. that H1 gets the right answer for D.
Since it is a logical impossibility to determine the truth
value of a self-contradictory expression the requirement
for H to do this is bogus.
Shows you are just a LIAR, as there IS a truth value to the expression
that is the requirment for ANY SPECIFIC H.
*Lying about me being a liar may possibly cost your soul*
*Lying about me being a liar may possibly cost your soul*
*Lying about me being a liar may possibly cost your soul*
There is no mapping from H(D,D) to Halts(D,D) that exists.
This proves that H(D,D) is being asked an incorrect question.
On 3/13/2024 4:32 PM, Richard Damon wrote:
On 3/13/24 1:59 PM, olcott wrote:Revelations 21:8 NRSV
On 3/13/2024 3:40 PM, immibis wrote:
On 13/03/24 20:24, olcott wrote:It is not even me speaking so it can't be me lying.
On 3/13/2024 2:15 PM, immibis wrote:
On 13/03/24 18:11, olcott wrote:
On 3/13/2024 12:04 PM, Mikko wrote:
On 2024-03-12 18:00:11 +0000, Richard Damon said:
There is no problem with making a H^ from an H, it is built
from totally legal steps.
Another way to say the same is that if H^ cannot be built
the H does not exist.
The proof of the halting problem assumes a universal halt test
exists and then provides S as an example of a program that the
test cannot handle. But S is not a program at all. It is not
even a conceptual object, and this is due to inconsistencies
in the specification of the halting function. (Stoddart: 2017)
This is obviously bullshit. Why do you lie?
That was cited as a statement from Bill Stoddart and a direct quote
from this paper.
Bill Stoddart. The Halting Paradox
20 December 2017
https://arxiv.org/abs/1906.05340
arXiv:1906.05340 [cs.LO]
And it's obviously bullshit. Why do you lie?
Repeating another lies is a form of lie.
...all liars, their place will be in the lake that burns
with fire and sulphur, which is the second death.’
Yet again Trollish behaviorThat you call me a liar about this meets the required definition of
reckless disregard for the truth required for libel and slander.
https://dictionary.findlaw.com/definition/reckless-disregard-of-the-truth.html
Does that mean you libelled and slandered the halting problem?
On 3/13/2024 4:32 PM, Richard Damon wrote:
On 3/13/24 12:24 PM, olcott wrote:I believe what I say or I wouldn't say it.
On 3/13/2024 2:15 PM, immibis wrote:
On 13/03/24 18:11, olcott wrote:
On 3/13/2024 12:04 PM, Mikko wrote:
On 2024-03-12 18:00:11 +0000, Richard Damon said:
There is no problem with making a H^ from an H, it is built from >>>>>>> totally legal steps.
Another way to say the same is that if H^ cannot be built
the H does not exist.
The proof of the halting problem assumes a universal halt test
exists and then provides S as an example of a program that the
test cannot handle. But S is not a program at all. It is not
even a conceptual object, and this is due to inconsistencies
in the specification of the halting function. (Stoddart: 2017)
This is obviously bullshit. Why do you lie?
That was cited as a statement from Bill Stoddart and a direct quote
from this paper.
Bill Stoddart. The Halting Paradox
20 December 2017
https://arxiv.org/abs/1906.05340
arXiv:1906.05340 [cs.LO]
That you call me a liar about this meets the required definition of
reckless disregard for the truth required for libel and slander.
https://dictionary.findlaw.com/definition/reckless-disregard-of-the-truth.html
And repeating lies and error just shows that your lie and believe the
errors.
Revelations 21:8 NRSV
...all liars, their place will be in the lake that burns
with fire and sulphur, which is the second death.’
Since it has been pointed out that he was WRONG in his statement,He called me a liar because he did not bother to even notice
there is no "reckless disreguard for the truth"
that I was quoting someone else's words this is definitely
a "reckless disregard for the truth".
YOU are the one that shows a "reckless disregard for the truth",
(because you refuse to learn the truth) and thus YOUR statements could
make you subject to Libel and Slander.
I did not commit libel against the halting problem as immbis suggested.
On 3/13/2024 3:40 PM, immibis wrote:
Does that mean you libelled and slandered the halting problem?
On 3/13/2024 4:39 PM, Richard Damon wrote:
On 3/13/24 1:52 PM, olcott wrote:I invented it so I get to stipulate its meaning. https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
On 3/13/2024 12:52 PM, Richard Damon wrote:
On 3/13/24 10:08 AM, olcott wrote:
On 3/13/2024 11:44 AM, immibis wrote:
On 13/03/24 04:55, olcott wrote:
On 3/12/2024 10:49 PM, Richard Damon wrote:
Not quite. It always gets the wrong answer, but only one of them >>>>>>>> for each quesiton.
They all gets the wrong answer on a whole class of questions
Wrong. You said. yourself. that H1 gets the right answer for D.
Since it is a logical impossibility to determine the truth
value of a self-contradictory expression the requirement
for H to do this is bogus.
Shows you are just a LIAR, as there IS a truth value to the
expression that is the requirment for ANY SPECIFIC H.
*Lying about me being a liar may possibly cost your soul*
*Lying about me being a liar may possibly cost your soul*
*Lying about me being a liar may possibly cost your soul*
There is no mapping from H(D,D) to Halts(D,D) that exists.
This proves that H(D,D) is being asked an incorrect question.
Why, because it is NOT a LIE.
You don't even know the definiton of an incorrect question.
The fact that there DOES exist a mapping Halt(M,d) that maps allThat part is true.
Turing Machines and there input to a result of Halting / Non-Halting
for EVERY member of that input set, means tha Halts is a valid mapping
to ask a decider to try to decider.
Likewise when you ask a man that has never been married:
Have you stopped beating tour wife?
There are some men that have stopped beating their wife.
The fact that no machine that exist that computes that mapping, meansLikewise the square-root of an actual banana is also uncomputable
the mapping is uncomputable.
for the same logical impossibility reason.
Perhaps you should actually learn the technical meaning of the termsRevelations 21:8 NRSV
before you state LIES about them.
...all liars, their place will be in the lake that burns
with fire and sulphur, which is the second death.’
I am pointing out and correcting the incoherence of the
meaning of these terms.
Yes, I guess it would be an incorrect question to ask what a correct
halt decider computes for a given input, the best I can interprete
your statement, since no H that does that can exist, but that isn't
what the question asks.
You are just stuck in the LIES you have created with your strawmen and
ignorance of the subject you are talking about.
*You know that I am not saying anything that I do not believe*
Revelations 21:8 NRSV
...all liars, their place will be in the lake that burns
with fire and sulphur, which is the second death.’
On 3/13/2024 5:43 PM, Richard Damon wrote:
On 3/13/24 2:54 PM, olcott wrote:
That part is true.
Likewise when you ask a man that has never been married:
Have you stopped beating tour wife?
There are some men that have stopped beating their wife.
Right, because that question include a presumption of something not
actually present.
Although there is a mapping from some men to YES/NO
there is no mapping from never unmarried men to YES/NO
thus the question is incorrect for all unmarried men.
Although there is a mapping from some TM/input pairs to YES/NO
there is no mapping from H/D to YES/NO
thus the question is incorrect for H/D
On 3/13/2024 12:04 PM, Mikko wrote:
On 2024-03-12 18:00:11 +0000, Richard Damon said:
There is no problem with making a H^ from an H, it is built from
totally legal steps.
Another way to say the same is that if H^ cannot be built
the H does not exist.
The proof of the halting problem assumes a universal halt test
exists and then provides S as an example of a program that the
test cannot handle. But S is not a program at all. It is not
even a conceptual object, and this is due to inconsistencies
in the specification of the halting function. (Stoddart: 2017)
On 13/03/24 18:11, olcott wrote:
On 3/13/2024 12:04 PM, Mikko wrote:
On 2024-03-12 18:00:11 +0000, Richard Damon said:
There is no problem with making a H^ from an H, it is built from
totally legal steps.
Another way to say the same is that if H^ cannot be built
the H does not exist.
The proof of the halting problem assumes a universal halt test
exists and then provides S as an example of a program that the
test cannot handle. But S is not a program at all. It is not
even a conceptual object, and this is due to inconsistencies
in the specification of the halting function. (Stoddart: 2017)
This is obviously bullshit. Why do you lie?
On 3/13/2024 12:02 PM, Mikko wrote:
On 2024-03-12 15:52:10 +0000, olcott said:The self-contradictory instances are all incorrect.
On 3/12/2024 5:20 AM, Mikko wrote:
On 2024-03-11 16:02:28 +0000, olcott said:
On 3/11/2024 10:12 AM, Mikko wrote:
On 2024-03-11 14:54:34 +0000, olcott said:
On 3/11/2024 5:45 AM, Mikko wrote:
On 2024-03-11 04:38:40 +0000, olcott said:
On 3/10/2024 11:10 PM, Richard Damon wrote:
On 3/10/24 8:33 PM, olcott wrote:
On 3/10/2024 9:13 PM, Richard Damon wrote:
On 3/10/24 7:05 PM, olcott wrote:
On 3/10/2024 8:52 PM, Richard Damon wrote:
On 3/10/24 6:00 PM, olcott wrote:
On 3/10/2024 2:23 PM, Richard Damon wrote:But if you have mo idea how things actually works, this seems to just
On 3/10/24 11:23 AM, olcott wrote:
On 3/10/2024 12:55 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>> On 3/10/24 10:17 AM, olcott wrote:
If no source can be cited that shows a simulating halt decider canOn 3/10/2024 12:08 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/10/24 9:52 AM, olcott wrote:Since, BY THE DEFINITIONS of what H MUST do to be correct, and what H^
On 3/10/2024 10:50 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/10/24 7:28 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/10/2024 12:16 AM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/9/24 9:49 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/9/2024 11:36 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/9/24 9:14 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/9/2024 10:55 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/9/24 8:30 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/9/2024 7:40 PM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 10/03/24 02:37, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/9/2024 7:32 PM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 10/03/24 02:29, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/9/2024 7:24 PM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 10/03/24 01:30, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/9/2024 6:24 PM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 10/03/24 01:22, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/9/2024 5:57 PM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 10/03/24 00:26, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/9/2024 5:10 PM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 9/03/24 23:22, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/9/2024 3:50 PM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 9/03/24 22:34, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> What criteria would you use so that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ knows whatNOPE.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ haltsIt is objectively true that Ĥ.H can get stuck in recursive*Only because Ĥ.H is embedded within Ĥ and H is not*H ⟨Ĥ⟩ uses. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>do not get stuck in infinite execution. This means that theywrong answer to provide? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same objective criteria thatH ⟨Ĥ⟩ uses. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Simulating halt deciders must make sure that they themselves
must abort every simulation that cannot possibly otherwise halt.
require H ⟨Ĥ⟩ ⟨Ĥ⟩ to abort its simulation when Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ abortsThis requires Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ to abort its simulation and does not
its simulation. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ does simulate itself in recursive simulation H ⟨Ĥ⟩ ⟨Ĥ⟩
does not simulate itself in recursive simulation.
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same objective criteria that
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly get stuck in recursive simulation and
H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly get stuck in recursive simulation.
You dishonestly ignored that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact
same OBJECTIVE criteria that H ⟨Ĥ⟩ uses.
The above is true no matter what criteria that is used
as long as H is a simulating halt decider.
Objective criteria cannot vary based on who the subject is. They are
objective. The answer to different people is the same answer if the
criteria are objective. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
simulation because Ĥ copies its input thus never runs
out of params. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
It is objectively true that Ĥ cannot possibly get stuck
in recursive because H does not copy its input thus runs
out of params. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Wrong. Dead wrong. Stupidly wrong. So wrong that a dead monkey could do
better. Write the Olcott machine (not x86utm) code for Ĥ and I would
show you.
*In other words you are denying these verified facts*
*In other words you are denying these verified facts*
*In other words you are denying these verified facts*
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
That's not a verified fact, that's just something you want to be true.
∞ means infinite loop. Infinite loop doesn't halt. You see how stupid
it is, to say that an infinite loop halts? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Execution trace of Ĥ applied to ⟨Ĥ⟩ >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> (a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
(b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process
Execution trace of H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ BECAUSE IT IS PRECISELY
IDENTICAL TO STEPS B AND C: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> > (b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at Ĥ's own simulated ⟨Ĥ.q0⟩ to repeat the process
*Yes and the key step of copying its input is left out so*
*H ⟨Ĥ⟩ ⟨Ĥ⟩ runs out of params and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ never runs out of params*
that isn't how any of this works. Do you even know what words mean?
(b) and (c) are not the same as (1) and (2) >>>>>>>>>>>>>>>>>>>>>>>>>>>>> Execution trace of H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ >>>>>>>>>>>>>>>>>>>>>>>>>>>>> (1) H applied ⟨Ĥ⟩ ⟨Ĥ⟩ simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(2) which begins at simulated ⟨Ĥ.q0⟩ >>>>>>>>>>>>>>>>>>>>>>>>>>>>> (a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
(b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process
This means that Turing machine H ⟨Ĥ⟩ ⟨Ĥ⟩ can see one more execution
trace of Ĥ ⟨Ĥ⟩ than its simulated Turing machine Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can see.
Nope, your just being stuupid, perhaps intentionally.
(c) just moves around to its simulation of a >>>>>>>>>>>>>>>>>>>>>>>>>>>>
(a) H^.q0 (H^)
H^ then makes a copy of its inp >>>>>>>>>>>>>>>>>>>>>>>>>>>>
(b) H^.H (H^) (H^) == (1) H (H^) (H^) >>>>>>>>>>>>>>>>>>>>>>>>>>>> The algorithm of H begins a simulation of its input, watching the
behaior of H^ (H^)
(c) = (2)
Which begins at the simulation of H^.q0 (H^) >>>>>>>>>>>>>>>>>>>>>>>>>>>>
(d = sim a) = (sim a)
Ths Simulated H^.q0 (H^) makes a copy of its input >>>>>>>>>>>>>>>>>>>>>>>>>>>>
(e = sim b) = (sim b)
The Simulated H^.H (H^) (H^) has is H begin the simulation of its input ...
and so on.
Both machine see EXACTLY the same level of details.
Yes, the top level H is farther along at any given time then its
simulated machine, and that is H's problem, it has to act before it
sees how its simulation will respond to its copy of its actions.
Thus, if it stops, it needs to make its decision "blind" and not with
an idea of how the machine it is simulating will perform.
If it doesn't stop, the level of recursion just keeps growing and no
answer ever comes out.
The earliest point that H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly see to abort
its simulation is immediately before Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ would begin
its simulation. Right before its cycle repeats the first time.
So?
If it DOES abort there, then so will H^.H when it gets to that point in
its simulation, which will be AFTER The point that H has stopped
simulating it, so H doesn't know what H^ will do. >>>>>>>>>>>>>>>>>>>>>>>>>>
Thus, if H DOES abort there, we presume from your previous answer it
will think the input will not halt and answer qn. >>>>>>>>>>>>>>>>>>>>>>>>>>
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
H ⟨Ĥ⟩ ⟨Ĥ⟩ aborts right after Ĥ.Hq0 before it simulates ⟨Ĥ⟩ ⟨Ĥ⟩.
And if it does, as I said below, so will H^.H when it is run.
Yes.
And thus, H^.H will give the same answer as H, >>>>>>>>>>>>>>>>>>>>>> so H^ will act contrary to what H says, >>>>>>>>>>>>>>>>>>>>>> so H will give the wrong answer.
Unlike anything else that anyone else has ever done both H ⟨Ĥ⟩ ⟨Ĥ⟩
and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly determine that they must abort their own
simulation to prevent their own infinite execution. >>>>>>>>>>>>>>>>>>>>
If no source can be cited then the Olcott thesis >>>>>>>>>>>>>>>>>>> "that no one did this before" remains unrefuted. >>>>>>>>>>>>>>>>>>
WILL do by its design, as shown in the Linz Proof. >>>>>>>>>>>>>>>>>
correctly determine that it must abort its simulation of the Halting
Problem's pathological input to prevent its own non-termination, then
innovation remains attributable to me.
<snip>
Of course it can abort its simulation.
It just needs some way to get the right answer. >>>>>>>>>>>>>>>>
*I have always been using this long before I read about it* >>>>>>>>>>>>>>> blind variation and selective retention (BVSR)... >>>>>>>>>>>>>>> Two common phenomena characterize BVSR thinking: superfluity and
backtracking. Superfluity means that the creator generates a variety of
ideas, one or more of which turn out to be useless. >>>>>>>>>>>>>>
generate random noise.
Nope, I remember talk of that when I was in college, and they showed
Backtracking signifies that the creator must often return to an earlier
approach after blindly going off in the wrong direction. >>>>>>>>>>>>>>> https://www.scientificamerican.com/article/the-science-of-genius/
*I am aware of no one else that had the idea to apply a simulating*
*termination analyzer to the halting problem counter-example input*
Professor Hehner had a seed of this idea before I did. >>>>>>>>>>>>>>
why it can't work.
From a programmer's point of view, if we apply an interpreter to a
program text that includes a call to that same interpreter with that
same text as argument, then we have an infinite loop. A halting program
has some of the same character as an interpreter: it applies to texts
through abstract interpretation. Unsurprisingly, if we apply a halting
program to a program text that includes a call to that same halting
program with that same text as argument, then we have an infinite loop.
https://www.cs.toronto.edu/~hehner/PHP.pdf
You THINK so, but if the interpreter is a CONDITIONAL interpreter, that
doesn't hold.
You seem to miss that fact.
*Turing Machine and Olcott machine implementations seem to be dead*
*This the (possibly augmented) RASP machine equivalent of x86* >>>>>>>>>>>>>>> Every machine must be able to get its own machine address. >>>>>>>>>>>>>>>
And the reason it is a dead end is they make it too hard for you to cheat.
You need to hide that your H is trying to get in some extra information
to hide that the embedded version of H doesn't give the same answer,
which just shows that your H^ is built wrong.
My C code proves these two have different behavior:
(a) H1(D,D) + H1_machine_address
(b) H(D,D) + H_machine_address
H1(D,D) does correctly determine the halt status of D(D) because >>>>>>>>>>>>> H(D,D) does NOT correctly determine the halt status of D(D). >>>>>>>>>>>>>
I say:
H1(D,D) is isomorphic to H ⟨Ĥ⟩ ⟨Ĥ⟩
H(D,D) is isomorphic to Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
immibis disagrees.
Correct reasoning will show who is correct.
Yes, and since H1 is a different computation than H, it getting the
right answer doesn't keep H from being broken.
We can then make a D1 to break H1.
I think that immibis already said that and I did not notice >>>>>>>>>>> the significance of it at the time.
Of course.
Then we are back to undecidability being incorrectly construed >>>>>>>>>>> as an actual limit to computation.
Strange definition of not an actual limit if not being able to do >>>>>>>>>> something isn't a limit.
Professor's Hehner and Stoddart have only construed this as >>>>>>>>>>> applying to the Halting Problem's pathological input.
We three perfectly agree on this as it pertains to the
Halting Problem.
That two full PhD professors of computer science and I all >>>>>>>>>>> agree on this shows that I am not a crackpot/crank on this. >>>>>>>>>>> I think that all of the other options may now be exhausted. >>>>>>>>>>>
I am very happy that I quit tolerating the [change the subject] >>>>>>>>>>> form of rebuttal that wasted 15 years with Ben Bacarisse. >>>>>>>>>>>
*The focus now must be on finding the best words that prove* >>>>>>>>>>> *this original position of mine (thus the concurring positions* >>>>>>>>>>> *of professors Hehner and Stoddart) is correct*
Go knock yourself out on that.
Alan Turing's Halting Problem is incorrectly formed (PART-TWO) sci.logic
On 6/20/2004 11:31 AM, Peter Olcott wrote:
PREMISES:
(1) The Halting Problem was specified in such a way that a solution
was defined to be impossible.
Nope.
The PROBLEM is the question if a machine can compute the Halting Function.
The answer to that, has turned out to be NO.
When the problem was first being posed, it was hoped the answer woudl
be yes, so it couldn't have bee made specifically to make it impossible.
The Halting QUESTION, has an answer for every input that it the >>>>>>>>>> description of an actual algorithm applied to an actual data input. >>>>>>>>>>
Note, Not a "Template" that gets appled to the decider, that IS an >>>>>>>>>> invalid question, and impossible to build a description of a Turing >>>>>>>>>> Machine to ask that.
Thus, when you admitted that your input wasn't actually a description
of a program, but just a template, you were admitting that you were >>>>>>>>>> lying about working on the Halting Problem, as your input isn't of the
right type.
Yes, asking about a template IS an invalid question.
;
(2) The set of questions that are defined to not have any possible
correct answer(s) forms a proper subset of all possible questions.
…
And, when you are asking the actual Halting Question, about a specific
machine and input, like a SPECIFIC H^, built to foil a SPECIIFIC H, >>>>>>>>>> then that input has a specific and definate behavior and there is a >>>>>>>>>> specific and definate answer (That depends on the H that you chose to
build it on, but not the decider you are asking the question to). >>>>>>>>>>
CONCLUSION:
Therefore the Halting Problem is an ill-formed question. >>>>>>>>>>> >
Nope, as explained above. You are just showing that you never >>>>>>>>>> understood the actual question or what any of the theory actually >>>>>>>>>> means, and have just wasted the last decades of your life on a stupid
misconception of your own.
USENET Message-ID:
<kZiBc.103407$Gx4.18142@bgtnsc04-news.ops.worldnet.att.net> >>>>>>>>>>>
*Direct Link to original message*
http://al.howardknight.net/?STYPE=msgid&MSGI=%3CkZiBc.103407%24Gx4.18142%40bgtnsc04-news.ops.worldnet.att.net%3E+
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
That YES and NO are the wrong answer for each implementation of >>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ specified by the second ⊢* state transition proves that the
questions asked of these machine/inputs pairs are incorrect questions.
An incorrect answer does not mean that the question is incorrect. >>>>>>>>
When every element of the infinite set of every possible
implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer then >>>>>>> there is something wrong with the question.
According to the definition by Linz there is only one Ĥ for each H. >>>>>> Anyway, a wrong answer, even if given in large quentities, does
not make the question worng.
None-the-less the infinite set of every implementation of
H/Ĥ.H cannot possibly get an answer that is consistent with
the behavior of of Ĥ ⟨Ĥ⟩.
The infinite set is not expected to answer. And the set is
infinite only if you include defective imiplementations.
Anyway, no memeber of the set of implementations can get
an answer that is consistent with the corresponding counter
example.
∀ H ∈ Turing_Machine_Deciders
∃ TMD ∈ Turing_Machine_Descriptions |
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
There is some input TMD to every H such that
Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
When we disallow decider/input pairs that are incorrect
questions where both YES and NO are the wrong answer
(the same way the ZFC disallowed self-referential sets) then
pathological inputs are not allowed to come into existence.
No question of the type "Does T(I) halt?" is incorrect
in that sense.
Does the barber that shaves everyone that does not shave
themselves shave himself? is rejected as an incorrect question.
https://en.wikipedia.org/wiki/Barber_paradox#
The answer "Yes" is correct because no such barber fails to
shave himself.
The answer of YES affirms a false presupposition thus is incorrect.
The answer "No" is correct because no such barber shaves himself.The answer of NO affirms a false presupposition thus is incorrect.
On 3/13/2024 5:43 PM, Richard Damon wrote:
On 3/13/24 2:54 PM, olcott wrote:
On 3/13/2024 4:39 PM, Richard Damon wrote:
On 3/13/24 1:52 PM, olcott wrote:I invented it so I get to stipulate its meaning.
On 3/13/2024 12:52 PM, Richard Damon wrote:
On 3/13/24 10:08 AM, olcott wrote:
On 3/13/2024 11:44 AM, immibis wrote:
On 13/03/24 04:55, olcott wrote:
On 3/12/2024 10:49 PM, Richard Damon wrote:
Not quite. It always gets the wrong answer, but only one of >>>>>>>>>> them for each quesiton.
They all gets the wrong answer on a whole class of questions
Wrong. You said. yourself. that H1 gets the right answer for D. >>>>>>>>
Since it is a logical impossibility to determine the truth
value of a self-contradictory expression the requirement
for H to do this is bogus.
Shows you are just a LIAR, as there IS a truth value to the
expression that is the requirment for ANY SPECIFIC H.
*Lying about me being a liar may possibly cost your soul*
*Lying about me being a liar may possibly cost your soul*
*Lying about me being a liar may possibly cost your soul*
There is no mapping from H(D,D) to Halts(D,D) that exists.
This proves that H(D,D) is being asked an incorrect question.
Why, because it is NOT a LIE.
You don't even know the definiton of an incorrect question.
https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
Nope, common technical term.
Cite a source.
The fact that there DOES exist a mapping Halt(M,d) that maps allThat part is true.
Turing Machines and there input to a result of Halting / Non-Halting
for EVERY member of that input set, means tha Halts is a valid
mapping to ask a decider to try to decider.
Likewise when you ask a man that has never been married:
Have you stopped beating tour wife?
There are some men that have stopped beating their wife.
Right, because that question include a presumption of something not
actually present.
Although there is a mapping from some men to YES/NO
there is no mapping from never unmarried men to YES/NO
thus the question is incorrect for all unmarried men.
Although there is a mapping from some TM/input pairs to YES/NO
there is no mapping from H/D to YES/NO
thus the question is incorrect for H/D
On 3/14/2024 12:33 PM, Richard Damon wrote:
On 3/13/24 4:04 PM, olcott wrote:
On 3/13/2024 5:43 PM, Richard Damon wrote:
On 3/13/24 2:54 PM, olcott wrote:
On 3/13/2024 4:39 PM, Richard Damon wrote:
On 3/13/24 1:52 PM, olcott wrote:I invented it so I get to stipulate its meaning.
On 3/13/2024 12:52 PM, Richard Damon wrote:
On 3/13/24 10:08 AM, olcott wrote:
On 3/13/2024 11:44 AM, immibis wrote:
On 13/03/24 04:55, olcott wrote:
On 3/12/2024 10:49 PM, Richard Damon wrote:Wrong. You said. yourself. that H1 gets the right answer for D. >>>>>>>>>>
Not quite. It always gets the wrong answer, but only one of >>>>>>>>>>>> them for each quesiton.
They all gets the wrong answer on a whole class of questions >>>>>>>>>>
Since it is a logical impossibility to determine the truth
value of a self-contradictory expression the requirement
for H to do this is bogus.
Shows you are just a LIAR, as there IS a truth value to the
expression that is the requirment for ANY SPECIFIC H.
*Lying about me being a liar may possibly cost your soul*
*Lying about me being a liar may possibly cost your soul*
*Lying about me being a liar may possibly cost your soul*
There is no mapping from H(D,D) to Halts(D,D) that exists.
This proves that H(D,D) is being asked an incorrect question.
Why, because it is NOT a LIE.
You don't even know the definiton of an incorrect question.
https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
Nope, common technical term.
Cite a source.
The fact that there DOES exist a mapping Halt(M,d) that maps allThat part is true.
Turing Machines and there input to a result of Halting /
Non-Halting for EVERY member of that input set, means tha Halts is >>>>>> a valid mapping to ask a decider to try to decider.
Likewise when you ask a man that has never been married:
Have you stopped beating tour wife?
There are some men that have stopped beating their wife.
Right, because that question include a presumption of something not
actually present.
Although there is a mapping from some men to YES/NO
there is no mapping from never unmarried men to YES/NO
thus the question is incorrect for all unmarried men.
Although there is a mapping from some TM/input pairs to YES/NO
there is no mapping from H/D to YES/NO
thus the question is incorrect for H/D
Except that the mapping requested is about the INPUTS to H, not H itsef.
In order to see that it is an incorrect question we must examine
the question in detail. Making sure to always ignore this key detail
<is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
THERE IS a mapping from ALL inputs to H(<M>,d) to a corrct Yes/No
answer as determined by Halts(M,d), or in this specific case the
inputs to H(D,D) to Halts (D,D).
If H(D,D) returns 0, then the Mapping of the input D,D to Halts(D,D)
is to Yes.
If H(D,D) returns 1, then the Mapping of the input D,D to Halts(D,D)
is No.
So, the mapping asked for in the ACTUAL question exist.
That H(D,D) doesn't give the right answer just shows that H is wrong
for the input.
The fact that there exists a D for every H such that it will get the
answer wrong proves the Mapping described in the Halting Problem is
uncomputable, not "invalid" (the fact that it actually exists, makes
it valid).
On 3/14/2024 3:20 PM, Richard Damon wrote:
On 3/14/24 12:32 PM, olcott wrote:
On 3/14/2024 12:33 PM, Richard Damon wrote:
On 3/13/24 4:04 PM, olcott wrote:
On 3/13/2024 5:43 PM, Richard Damon wrote:
On 3/13/24 2:54 PM, olcott wrote:
On 3/13/2024 4:39 PM, Richard Damon wrote:Nope, common technical term.
On 3/13/24 1:52 PM, olcott wrote:I invented it so I get to stipulate its meaning.
On 3/13/2024 12:52 PM, Richard Damon wrote:
On 3/13/24 10:08 AM, olcott wrote:
On 3/13/2024 11:44 AM, immibis wrote:
On 13/03/24 04:55, olcott wrote:
On 3/12/2024 10:49 PM, Richard Damon wrote:Wrong. You said. yourself. that H1 gets the right answer for D. >>>>>>>>>>>>
Not quite. It always gets the wrong answer, but only one >>>>>>>>>>>>>> of them for each quesiton.
They all gets the wrong answer on a whole class of questions >>>>>>>>>>>>
Since it is a logical impossibility to determine the truth >>>>>>>>>>> value of a self-contradictory expression the requirement >>>>>>>>>>> for H to do this is bogus.
Shows you are just a LIAR, as there IS a truth value to the >>>>>>>>>> expression that is the requirment for ANY SPECIFIC H.
*Lying about me being a liar may possibly cost your soul*
*Lying about me being a liar may possibly cost your soul*
*Lying about me being a liar may possibly cost your soul*
There is no mapping from H(D,D) to Halts(D,D) that exists.
This proves that H(D,D) is being asked an incorrect question. >>>>>>>>>
Why, because it is NOT a LIE.
You don't even know the definiton of an incorrect question.
https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ >>>>>>
Cite a source.
The fact that there DOES exist a mapping Halt(M,d) that maps all >>>>>>>> Turing Machines and there input to a result of Halting /That part is true.
Non-Halting for EVERY member of that input set, means tha Halts >>>>>>>> is a valid mapping to ask a decider to try to decider.
Likewise when you ask a man that has never been married:
Have you stopped beating tour wife?
There are some men that have stopped beating their wife.
Right, because that question include a presumption of something
not actually present.
Although there is a mapping from some men to YES/NO
there is no mapping from never unmarried men to YES/NO
thus the question is incorrect for all unmarried men.
Although there is a mapping from some TM/input pairs to YES/NO
there is no mapping from H/D to YES/NO
thus the question is incorrect for H/D
Except that the mapping requested is about the INPUTS to H, not H
itsef.
In order to see that it is an incorrect question we must examine
the question in detail. Making sure to always ignore this key detail
<is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
Which isn;t the question at all, so you are just shown to be a stupid
liar.
The QUESTION is:
Does the machine and input described by this input, Halt when run?
The question posed to Ĥ.H has no correct answer, thus not the
same question at all.
How long are you going to keep trying to get away with this
strawman deception?
On 3/14/2024 3:54 PM, Richard Damon wrote:
On 3/14/24 1:26 PM, olcott wrote:
On 3/14/2024 3:20 PM, Richard Damon wrote:
On 3/14/24 12:32 PM, olcott wrote:
On 3/14/2024 12:33 PM, Richard Damon wrote:
On 3/13/24 4:04 PM, olcott wrote:
On 3/13/2024 5:43 PM, Richard Damon wrote:
On 3/13/24 2:54 PM, olcott wrote:
On 3/13/2024 4:39 PM, Richard Damon wrote:Nope, common technical term.
On 3/13/24 1:52 PM, olcott wrote:I invented it so I get to stipulate its meaning.
On 3/13/2024 12:52 PM, Richard Damon wrote:
On 3/13/24 10:08 AM, olcott wrote:
On 3/13/2024 11:44 AM, immibis wrote:
On 13/03/24 04:55, olcott wrote:
On 3/12/2024 10:49 PM, Richard Damon wrote:Wrong. You said. yourself. that H1 gets the right answer >>>>>>>>>>>>>> for D.
Not quite. It always gets the wrong answer, but only one >>>>>>>>>>>>>>>> of them for each quesiton.
They all gets the wrong answer on a whole class of questions >>>>>>>>>>>>>>
Since it is a logical impossibility to determine the truth >>>>>>>>>>>>> value of a self-contradictory expression the requirement >>>>>>>>>>>>> for H to do this is bogus.
Shows you are just a LIAR, as there IS a truth value to the >>>>>>>>>>>> expression that is the requirment for ANY SPECIFIC H.
*Lying about me being a liar may possibly cost your soul* >>>>>>>>>>> *Lying about me being a liar may possibly cost your soul* >>>>>>>>>>> *Lying about me being a liar may possibly cost your soul* >>>>>>>>>>>
There is no mapping from H(D,D) to Halts(D,D) that exists. >>>>>>>>>>> This proves that H(D,D) is being asked an incorrect question. >>>>>>>>>>>
Why, because it is NOT a LIE.
You don't even know the definiton of an incorrect question. >>>>>>>>>>
https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ >>>>>>>>
Cite a source.
The fact that there DOES exist a mapping Halt(M,d) that maps >>>>>>>>>> all Turing Machines and there input to a result of Halting / >>>>>>>>>> Non-Halting for EVERY member of that input set, means thaThat part is true.
Halts is a valid mapping to ask a decider to try to decider. >>>>>>>>>>
Likewise when you ask a man that has never been married:
Have you stopped beating tour wife?
There are some men that have stopped beating their wife.
Right, because that question include a presumption of something >>>>>>>> not actually present.
Although there is a mapping from some men to YES/NO
there is no mapping from never unmarried men to YES/NO
thus the question is incorrect for all unmarried men.
Although there is a mapping from some TM/input pairs to YES/NO
there is no mapping from H/D to YES/NO
thus the question is incorrect for H/D
Except that the mapping requested is about the INPUTS to H, not H
itsef.
In order to see that it is an incorrect question we must examine
the question in detail. Making sure to always ignore this key detail >>>>> <is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
Which isn;t the question at all, so you are just shown to be a
stupid liar.
The QUESTION is:
Does the machine and input described by this input, Halt when run?
The question posed to Ĥ.H has no correct answer, thus not the
same question at all.
But it DOES.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct and make sure that
you explain why this element is correct and don't try to switch
to any other element outside of the above specified set.
On 3/14/2024 4:55 PM, Richard Damon wrote:
On 3/14/24 1:59 PM, olcott wrote:
On 3/14/2024 3:54 PM, Richard Damon wrote:
On 3/14/24 1:26 PM, olcott wrote:
On 3/14/2024 3:20 PM, Richard Damon wrote:
On 3/14/24 12:32 PM, olcott wrote:
On 3/14/2024 12:33 PM, Richard Damon wrote:
On 3/13/24 4:04 PM, olcott wrote:
On 3/13/2024 5:43 PM, Richard Damon wrote:
On 3/13/24 2:54 PM, olcott wrote:
On 3/13/2024 4:39 PM, Richard Damon wrote:Nope, common technical term.
On 3/13/24 1:52 PM, olcott wrote:I invented it so I get to stipulate its meaning.
On 3/13/2024 12:52 PM, Richard Damon wrote:
On 3/13/24 10:08 AM, olcott wrote:
On 3/13/2024 11:44 AM, immibis wrote:
On 13/03/24 04:55, olcott wrote:
On 3/12/2024 10:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>
Not quite. It always gets the wrong answer, but only >>>>>>>>>>>>>>>>>> one of them for each quesiton.
They all gets the wrong answer on a whole class of >>>>>>>>>>>>>>>>> questions
Wrong. You said. yourself. that H1 gets the right answer >>>>>>>>>>>>>>>> for D.
Since it is a logical impossibility to determine the truth >>>>>>>>>>>>>>> value of a self-contradictory expression the requirement >>>>>>>>>>>>>>> for H to do this is bogus.
Shows you are just a LIAR, as there IS a truth value to >>>>>>>>>>>>>> the expression that is the requirment for ANY SPECIFIC H. >>>>>>>>>>>>>>
*Lying about me being a liar may possibly cost your soul* >>>>>>>>>>>>> *Lying about me being a liar may possibly cost your soul* >>>>>>>>>>>>> *Lying about me being a liar may possibly cost your soul* >>>>>>>>>>>>>
There is no mapping from H(D,D) to Halts(D,D) that exists. >>>>>>>>>>>>> This proves that H(D,D) is being asked an incorrect question. >>>>>>>>>>>>>
Why, because it is NOT a LIE.
You don't even know the definiton of an incorrect question. >>>>>>>>>>>>
https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ >>>>>>>>>>
Cite a source.
The fact that there DOES exist a mapping Halt(M,d) that maps >>>>>>>>>>>> all Turing Machines and there input to a result of Halting / >>>>>>>>>>>> Non-Halting for EVERY member of that input set, means tha >>>>>>>>>>>> Halts is a valid mapping to ask a decider to try to decider. >>>>>>>>>>>>That part is true.
Likewise when you ask a man that has never been married: >>>>>>>>>>> Have you stopped beating tour wife?
There are some men that have stopped beating their wife.
Right, because that question include a presumption of
something not actually present.
Although there is a mapping from some men to YES/NO
there is no mapping from never unmarried men to YES/NO
thus the question is incorrect for all unmarried men.
Although there is a mapping from some TM/input pairs to YES/NO >>>>>>>>> there is no mapping from H/D to YES/NO
thus the question is incorrect for H/D
Except that the mapping requested is about the INPUTS to H, not >>>>>>>> H itsef.
In order to see that it is an incorrect question we must examine >>>>>>> the question in detail. Making sure to always ignore this key detail >>>>>>> <is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not
halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
Which isn;t the question at all, so you are just shown to be a
stupid liar.
The QUESTION is:
Does the machine and input described by this input, Halt when run?
The question posed to Ĥ.H has no correct answer, thus not the
same question at all.
But it DOES.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct and make sure that
you explain why this element is correct and don't try to switch
to any other element outside of the above specified set.
I didn't say there was.
Then you understand that each question posed to each Ĥ.H in the
above set has no correct answer only because each of these answers
are contradicted by the machine that H is contained within.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct and make sure that
you explain why this element is correct and don't try to switch
to any other element outside of the above specified set.
The question posed to Ĥ.H has no correct answer, thus not the
same question at all.
On 3/14/2024 5:37 PM, Richard Damon wrote:
No, YOU don't understand that the IS a correct answer, just not the
one that H (or H^.H ) happens to give.
Then show me which contradicted answer is correct.
On 3/14/2024 5:37 PM, Richard Damon wrote:
On 3/14/24 3:04 PM, olcott wrote:
On 3/14/2024 4:55 PM, Richard Damon wrote:
On 3/14/24 1:59 PM, olcott wrote:
On 3/14/2024 3:54 PM, Richard Damon wrote:
On 3/14/24 1:26 PM, olcott wrote:
On 3/14/2024 3:20 PM, Richard Damon wrote:
On 3/14/24 12:32 PM, olcott wrote:The question posed to Ĥ.H has no correct answer, thus not the
On 3/14/2024 12:33 PM, Richard Damon wrote:Which isn;t the question at all, so you are just shown to be a >>>>>>>> stupid liar.
On 3/13/24 4:04 PM, olcott wrote:
On 3/13/2024 5:43 PM, Richard Damon wrote:
On 3/13/24 2:54 PM, olcott wrote:
On 3/13/2024 4:39 PM, Richard Damon wrote:Nope, common technical term.
On 3/13/24 1:52 PM, olcott wrote:I invented it so I get to stipulate its meaning.
On 3/13/2024 12:52 PM, Richard Damon wrote:
On 3/13/24 10:08 AM, olcott wrote:
On 3/13/2024 11:44 AM, immibis wrote:
On 13/03/24 04:55, olcott wrote:
On 3/12/2024 10:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>
Not quite. It always gets the wrong answer, but only >>>>>>>>>>>>>>>>>>>> one of them for each quesiton.
They all gets the wrong answer on a whole class of >>>>>>>>>>>>>>>>>>> questions
Wrong. You said. yourself. that H1 gets the right >>>>>>>>>>>>>>>>>> answer for D.
Since it is a logical impossibility to determine the truth >>>>>>>>>>>>>>>>> value of a self-contradictory expression the requirement >>>>>>>>>>>>>>>>> for H to do this is bogus.
Shows you are just a LIAR, as there IS a truth value to >>>>>>>>>>>>>>>> the expression that is the requirment for ANY SPECIFIC H. >>>>>>>>>>>>>>>>
*Lying about me being a liar may possibly cost your soul* >>>>>>>>>>>>>>> *Lying about me being a liar may possibly cost your soul* >>>>>>>>>>>>>>> *Lying about me being a liar may possibly cost your soul* >>>>>>>>>>>>>>>
There is no mapping from H(D,D) to Halts(D,D) that exists. >>>>>>>>>>>>>>> This proves that H(D,D) is being asked an incorrect >>>>>>>>>>>>>>> question.
Why, because it is NOT a LIE.
You don't even know the definiton of an incorrect question. >>>>>>>>>>>>>>
https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ >>>>>>>>>>>>
Cite a source.
Right, because that question include a presumption of
The fact that there DOES exist a mapping Halt(M,d) that >>>>>>>>>>>>>> maps all Turing Machines and there input to a result of >>>>>>>>>>>>>> Halting / Non-Halting for EVERY member of that input set, >>>>>>>>>>>>>> means tha Halts is a valid mapping to ask a decider to try >>>>>>>>>>>>>> to decider.That part is true.
Likewise when you ask a man that has never been married: >>>>>>>>>>>>> Have you stopped beating tour wife?
There are some men that have stopped beating their wife. >>>>>>>>>>>>
something not actually present.
Although there is a mapping from some men to YES/NO
there is no mapping from never unmarried men to YES/NO
thus the question is incorrect for all unmarried men.
Although there is a mapping from some TM/input pairs to YES/NO >>>>>>>>>>> there is no mapping from H/D to YES/NO
thus the question is incorrect for H/D
Except that the mapping requested is about the INPUTS to H, >>>>>>>>>> not H itsef.
In order to see that it is an incorrect question we must examine >>>>>>>>> the question in detail. Making sure to always ignore this key >>>>>>>>> detail
<is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does
not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) >>>>>>>>
The QUESTION is:
Does the machine and input described by this input, Halt when run? >>>>>>>
same question at all.
But it DOES.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct and make sure that
you explain why this element is correct and don't try to switch
to any other element outside of the above specified set.
I didn't say there was.
Then you understand that each question posed to each Ĥ.H in the
above set has no correct answer only because each of these answers
are contradicted by the machine that H is contained within.
No, YOU don't understand that the IS a correct answer, just not the
one that H (or H^.H ) happens to give.
Then show me which contradicted answer is correct.
On 3/14/2024 7:23 PM, immibis wrote:
On 14/03/24 21:59, olcott wrote:In any language that you are familiar with.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct and make sure that
you explain why this element is correct and don't try to switch
to any other element outside of the above specified set.
Syntactically invalid question.
On 3/14/2024 7:22 PM, immibis wrote:
On 14/03/24 21:26, olcott wrote:
The question posed to Ĥ.H has no correct answer, thus not the
same question at all.
nobody cares what question is posed to Ĥ.H. How long are you going to
keep trying to get away with this strawman deception?
That you don't care about the foundations of computation is irrelevant. When-so-ever anyone or anything is asked a question that is logically impossible for them to correctly answer this places no actual limit
and anyone, anything, or computation.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
It is logically impossible
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
On 3/14/2024 8:09 PM, Richard Damon wrote:
On 3/14/24 5:54 PM, olcott wrote:
On 3/14/2024 7:23 PM, immibis wrote:
On 14/03/24 21:59, olcott wrote:In any language that you are familiar with.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct and make sure that
you explain why this element is correct and don't try to switch
to any other element outside of the above specified set.
Syntactically invalid question.
Well, since the bases of your question is a false premise, there is no
language it can be properly expresses in truthfully.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
There is no false premise above.
If there was you would have already pointed it out.
On 3/14/2024 7:22 PM, immibis wrote:
On 14/03/24 21:26, olcott wrote:
The question posed to Ĥ.H has no correct answer, thus not the
same question at all.
nobody cares what question is posed to Ĥ.H. How long are you going to
keep trying to get away with this strawman deception?
That you don't care about the foundations of computation is irrelevant. When-so-ever anyone or anything is asked a question that is logically impossible for them to correctly answer this places no actual limit
and anyone, anything, or computation.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
It is logically impossible
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
On 3/14/2024 8:06 PM, Richard Damon wrote:
On 3/14/24 4:45 PM, olcott wrote:*The strawman deception is all that you have*
On 3/14/2024 5:37 PM, Richard Damon wrote:
On 3/14/24 3:04 PM, olcott wrote:
On 3/14/2024 4:55 PM, Richard Damon wrote:
On 3/14/24 1:59 PM, olcott wrote:
On 3/14/2024 3:54 PM, Richard Damon wrote:
On 3/14/24 1:26 PM, olcott wrote:
On 3/14/2024 3:20 PM, Richard Damon wrote:
On 3/14/24 12:32 PM, olcott wrote:
On 3/14/2024 12:33 PM, Richard Damon wrote:Which isn;t the question at all, so you are just shown to be a >>>>>>>>>> stupid liar.
On 3/13/24 4:04 PM, olcott wrote:
On 3/13/2024 5:43 PM, Richard Damon wrote:
On 3/13/24 2:54 PM, olcott wrote:
On 3/13/2024 4:39 PM, Richard Damon wrote:
On 3/13/24 1:52 PM, olcott wrote:I invented it so I get to stipulate its meaning. >>>>>>>>>>>>>>> https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
On 3/13/2024 12:52 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>> On 3/13/24 10:08 AM, olcott wrote:
On 3/13/2024 11:44 AM, immibis wrote:
On 13/03/24 04:55, olcott wrote:
On 3/12/2024 10:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>
Not quite. It always gets the wrong answer, but >>>>>>>>>>>>>>>>>>>>>> only one of them for each quesiton. >>>>>>>>>>>>>>>>>>>>>>
They all gets the wrong answer on a whole class of >>>>>>>>>>>>>>>>>>>>> questions
Wrong. You said. yourself. that H1 gets the right >>>>>>>>>>>>>>>>>>>> answer for D.
Since it is a logical impossibility to determine the >>>>>>>>>>>>>>>>>>> truth
value of a self-contradictory expression the requirement >>>>>>>>>>>>>>>>>>> for H to do this is bogus.
Shows you are just a LIAR, as there IS a truth value >>>>>>>>>>>>>>>>>> to the expression that is the requirment for ANY >>>>>>>>>>>>>>>>>> SPECIFIC H.
*Lying about me being a liar may possibly cost your soul* >>>>>>>>>>>>>>>>> *Lying about me being a liar may possibly cost your soul* >>>>>>>>>>>>>>>>> *Lying about me being a liar may possibly cost your soul* >>>>>>>>>>>>>>>>>
There is no mapping from H(D,D) to Halts(D,D) that exists. >>>>>>>>>>>>>>>>> This proves that H(D,D) is being asked an incorrect >>>>>>>>>>>>>>>>> question.
Why, because it is NOT a LIE.
You don't even know the definiton of an incorrect question. >>>>>>>>>>>>>>>>
Nope, common technical term.
Cite a source.
Right, because that question include a presumption of >>>>>>>>>>>>>> something not actually present.
The fact that there DOES exist a mapping Halt(M,d) that >>>>>>>>>>>>>>>> maps all Turing Machines and there input to a result of >>>>>>>>>>>>>>>> Halting / Non-Halting for EVERY member of that input >>>>>>>>>>>>>>>> set, means tha Halts is a valid mapping to ask a decider >>>>>>>>>>>>>>>> to try to decider.That part is true.
Likewise when you ask a man that has never been married: >>>>>>>>>>>>>>> Have you stopped beating tour wife?
There are some men that have stopped beating their wife. >>>>>>>>>>>>>>
Although there is a mapping from some men to YES/NO
there is no mapping from never unmarried men to YES/NO >>>>>>>>>>>>> thus the question is incorrect for all unmarried men. >>>>>>>>>>>>>
Although there is a mapping from some TM/input pairs to YES/NO >>>>>>>>>>>>> there is no mapping from H/D to YES/NO
thus the question is incorrect for H/D
Except that the mapping requested is about the INPUTS to H, >>>>>>>>>>>> not H itsef.
In order to see that it is an incorrect question we must examine >>>>>>>>>>> the question in detail. Making sure to always ignore this key >>>>>>>>>>> detail
<is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does
not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) >>>>>>>>>>
The QUESTION is:
Does the machine and input described by this input, Halt when >>>>>>>>>> run?
The question posed to Ĥ.H has no correct answer, thus not the >>>>>>>>> same question at all.
But it DOES.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct and make sure that
you explain why this element is correct and don't try to switch
to any other element outside of the above specified set.
I didn't say there was.
Then you understand that each question posed to each Ĥ.H in the
above set has no correct answer only because each of these answers
are contradicted by the machine that H is contained within.
No, YOU don't understand that the IS a correct answer, just not the
one that H (or H^.H ) happens to give.
Then show me which contradicted answer is correct.
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy and loops so qn was
the right answer.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
*The answer must come from elements of the above set*
On 3/14/2024 8:06 PM, Richard Damon wrote:
On 3/14/24 4:45 PM, olcott wrote:*The strawman deception is all that you have*
On 3/14/2024 5:37 PM, Richard Damon wrote:
On 3/14/24 3:04 PM, olcott wrote:
On 3/14/2024 4:55 PM, Richard Damon wrote:
On 3/14/24 1:59 PM, olcott wrote:
On 3/14/2024 3:54 PM, Richard Damon wrote:
On 3/14/24 1:26 PM, olcott wrote:
On 3/14/2024 3:20 PM, Richard Damon wrote:
On 3/14/24 12:32 PM, olcott wrote:
On 3/14/2024 12:33 PM, Richard Damon wrote:Which isn;t the question at all, so you are just shown to be a >>>>>>>>>> stupid liar.
On 3/13/24 4:04 PM, olcott wrote:
On 3/13/2024 5:43 PM, Richard Damon wrote:
On 3/13/24 2:54 PM, olcott wrote:
On 3/13/2024 4:39 PM, Richard Damon wrote:
On 3/13/24 1:52 PM, olcott wrote:I invented it so I get to stipulate its meaning. >>>>>>>>>>>>>>> https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
On 3/13/2024 12:52 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>> On 3/13/24 10:08 AM, olcott wrote:
On 3/13/2024 11:44 AM, immibis wrote:
On 13/03/24 04:55, olcott wrote:
On 3/12/2024 10:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>
Not quite. It always gets the wrong answer, but >>>>>>>>>>>>>>>>>>>>>> only one of them for each quesiton. >>>>>>>>>>>>>>>>>>>>>>
They all gets the wrong answer on a whole class of >>>>>>>>>>>>>>>>>>>>> questions
Wrong. You said. yourself. that H1 gets the right >>>>>>>>>>>>>>>>>>>> answer for D.
Since it is a logical impossibility to determine the >>>>>>>>>>>>>>>>>>> truth
value of a self-contradictory expression the requirement >>>>>>>>>>>>>>>>>>> for H to do this is bogus.
Shows you are just a LIAR, as there IS a truth value >>>>>>>>>>>>>>>>>> to the expression that is the requirment for ANY >>>>>>>>>>>>>>>>>> SPECIFIC H.
*Lying about me being a liar may possibly cost your soul* >>>>>>>>>>>>>>>>> *Lying about me being a liar may possibly cost your soul* >>>>>>>>>>>>>>>>> *Lying about me being a liar may possibly cost your soul* >>>>>>>>>>>>>>>>>
There is no mapping from H(D,D) to Halts(D,D) that exists. >>>>>>>>>>>>>>>>> This proves that H(D,D) is being asked an incorrect >>>>>>>>>>>>>>>>> question.
Why, because it is NOT a LIE.
You don't even know the definiton of an incorrect question. >>>>>>>>>>>>>>>>
Nope, common technical term.
Cite a source.
Right, because that question include a presumption of >>>>>>>>>>>>>> something not actually present.
The fact that there DOES exist a mapping Halt(M,d) that >>>>>>>>>>>>>>>> maps all Turing Machines and there input to a result of >>>>>>>>>>>>>>>> Halting / Non-Halting for EVERY member of that input >>>>>>>>>>>>>>>> set, means tha Halts is a valid mapping to ask a decider >>>>>>>>>>>>>>>> to try to decider.That part is true.
Likewise when you ask a man that has never been married: >>>>>>>>>>>>>>> Have you stopped beating tour wife?
There are some men that have stopped beating their wife. >>>>>>>>>>>>>>
Although there is a mapping from some men to YES/NO
there is no mapping from never unmarried men to YES/NO >>>>>>>>>>>>> thus the question is incorrect for all unmarried men. >>>>>>>>>>>>>
Although there is a mapping from some TM/input pairs to YES/NO >>>>>>>>>>>>> there is no mapping from H/D to YES/NO
thus the question is incorrect for H/D
Except that the mapping requested is about the INPUTS to H, >>>>>>>>>>>> not H itsef.
In order to see that it is an incorrect question we must examine >>>>>>>>>>> the question in detail. Making sure to always ignore this key >>>>>>>>>>> detail
<is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does
not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) >>>>>>>>>>
The QUESTION is:
Does the machine and input described by this input, Halt when >>>>>>>>>> run?
The question posed to Ĥ.H has no correct answer, thus not the >>>>>>>>> same question at all.
But it DOES.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct and make sure that
you explain why this element is correct and don't try to switch
to any other element outside of the above specified set.
I didn't say there was.
Then you understand that each question posed to each Ĥ.H in the
above set has no correct answer only because each of these answers
are contradicted by the machine that H is contained within.
No, YOU don't understand that the IS a correct answer, just not the
one that H (or H^.H ) happens to give.
Then show me which contradicted answer is correct.
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy and loops so qn was
the right answer.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
*The answer must come from elements of the above set*
On 3/14/2024 9:15 PM, Richard Damon wrote:
On 3/14/24 6:12 PM, olcott wrote:
On 3/14/2024 8:06 PM, Richard Damon wrote:
On 3/14/24 4:45 PM, olcott wrote:*The strawman deception is all that you have*
On 3/14/2024 5:37 PM, Richard Damon wrote:
On 3/14/24 3:04 PM, olcott wrote:
On 3/14/2024 4:55 PM, Richard Damon wrote:
On 3/14/24 1:59 PM, olcott wrote:
On 3/14/2024 3:54 PM, Richard Damon wrote:
On 3/14/24 1:26 PM, olcott wrote:
On 3/14/2024 3:20 PM, Richard Damon wrote:
On 3/14/24 12:32 PM, olcott wrote:
On 3/14/2024 12:33 PM, Richard Damon wrote:Which isn;t the question at all, so you are just shown to be >>>>>>>>>>>> a stupid liar.
On 3/13/24 4:04 PM, olcott wrote:
On 3/13/2024 5:43 PM, Richard Damon wrote:
On 3/13/24 2:54 PM, olcott wrote:
On 3/13/2024 4:39 PM, Richard Damon wrote:
On 3/13/24 1:52 PM, olcott wrote:I invented it so I get to stipulate its meaning. >>>>>>>>>>>>>>>>> https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
On 3/13/2024 12:52 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/13/24 10:08 AM, olcott wrote:
On 3/13/2024 11:44 AM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>> On 13/03/24 04:55, olcott wrote:
On 3/12/2024 10:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>
Not quite. It always gets the wrong answer, but >>>>>>>>>>>>>>>>>>>>>>>> only one of them for each quesiton. >>>>>>>>>>>>>>>>>>>>>>>>
They all gets the wrong answer on a whole class >>>>>>>>>>>>>>>>>>>>>>> of questions
Wrong. You said. yourself. that H1 gets the right >>>>>>>>>>>>>>>>>>>>>> answer for D.
Since it is a logical impossibility to determine >>>>>>>>>>>>>>>>>>>>> the truth
value of a self-contradictory expression the >>>>>>>>>>>>>>>>>>>>> requirement
for H to do this is bogus.
Shows you are just a LIAR, as there IS a truth value >>>>>>>>>>>>>>>>>>>> to the expression that is the requirment for ANY >>>>>>>>>>>>>>>>>>>> SPECIFIC H.
*Lying about me being a liar may possibly cost your >>>>>>>>>>>>>>>>>>> soul*
*Lying about me being a liar may possibly cost your >>>>>>>>>>>>>>>>>>> soul*
*Lying about me being a liar may possibly cost your >>>>>>>>>>>>>>>>>>> soul*
There is no mapping from H(D,D) to Halts(D,D) that >>>>>>>>>>>>>>>>>>> exists.
This proves that H(D,D) is being asked an incorrect >>>>>>>>>>>>>>>>>>> question.
Why, because it is NOT a LIE.
You don't even know the definiton of an incorrect >>>>>>>>>>>>>>>>>> question.
Nope, common technical term.
Cite a source.
Right, because that question include a presumption of >>>>>>>>>>>>>>>> something not actually present.
The fact that there DOES exist a mapping Halt(M,d) >>>>>>>>>>>>>>>>>> that maps all Turing Machines and there input to a >>>>>>>>>>>>>>>>>> result of Halting / Non-Halting for EVERY member of >>>>>>>>>>>>>>>>>> that input set, means tha Halts is a valid mapping to >>>>>>>>>>>>>>>>>> ask a decider to try to decider.That part is true.
Likewise when you ask a man that has never been married: >>>>>>>>>>>>>>>>> Have you stopped beating tour wife?
There are some men that have stopped beating their wife. >>>>>>>>>>>>>>>>
Although there is a mapping from some men to YES/NO >>>>>>>>>>>>>>> there is no mapping from never unmarried men to YES/NO >>>>>>>>>>>>>>> thus the question is incorrect for all unmarried men. >>>>>>>>>>>>>>>
Although there is a mapping from some TM/input pairs to >>>>>>>>>>>>>>> YES/NO
there is no mapping from H/D to YES/NO
thus the question is incorrect for H/D
Except that the mapping requested is about the INPUTS to >>>>>>>>>>>>>> H, not H itsef.
In order to see that it is an incorrect question we must >>>>>>>>>>>>> examine
the question in detail. Making sure to always ignore this >>>>>>>>>>>>> key detail
<is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩
does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) >>>>>>>>>>>>
The QUESTION is:
Does the machine and input described by this input, Halt >>>>>>>>>>>> when run?
The question posed to Ĥ.H has no correct answer, thus not the >>>>>>>>>>> same question at all.
But it DOES.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct and make sure
that
you explain why this element is correct and don't try to switch >>>>>>>>> to any other element outside of the above specified set.
I didn't say there was.
Then you understand that each question posed to each Ĥ.H in the >>>>>>> above set has no correct answer only because each of these answers >>>>>>> are contradicted by the machine that H is contained within.
No, YOU don't understand that the IS a correct answer, just not
the one that H (or H^.H ) happens to give.
Then show me which contradicted answer is correct.
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy and loops so qn
was the right answer.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
*The answer must come from elements of the above set*
Nope. Not the requirement
That woould be asking which H was correct, which IS an invalid question
*Yes you are finally getting this*
Feb 20, 2015, 11:38:48 AM sci.lang
The logical law of polar questions
When posed to a man whom has never been married,
the question: Have you stopped beating your wife?
Is an incorrect polar question because neither yes nor
no is a correct answer.
Thus any H that has both of its answers contradicted
cannot possibly correctly answer that question...
(since it has a false premise, that one of them is correct).
How do you parse the question:
"Does the Machine described by your input Halt?" into you set of choices>
Remember, that part of the proof you are quoting was under the
presumption that there existed a correct H, so when you arrive at a
contradiction, the solution is that presumption was false.
On 3/14/2024 9:19 PM, Richard Damon wrote:
On 3/14/24 6:14 PM, olcott wrote:
On 3/14/2024 8:09 PM, Richard Damon wrote:
On 3/14/24 5:54 PM, olcott wrote:
On 3/14/2024 7:23 PM, immibis wrote:
On 14/03/24 21:59, olcott wrote:In any language that you are familiar with.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct and make sure that
you explain why this element is correct and don't try to switch
to any other element outside of the above specified set.
Syntactically invalid question.
Well, since the bases of your question is a false premise, there is
no language it can be properly expresses in truthfully.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
There is no false premise above.
If there was you would have already pointed it out.
We HAVE.
That there IS an H that is correct, is a false premise that all those
statements were built from.
That is not a premise, yet isomorphic to the never married
man that stopped beating his wife.
When you ask a man that has never been married:
Have you stopped beating your wife?
Feb 20, 2015, 11:38:48 AM sci.lang
*The logical law of polar questions*
When posed to a man whom has never been married,
the question: Have you stopped beating your wife?
Is an incorrect polar question because neither yes nor
no is a correct answer. https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
On 15/03/24 01:52, olcott wrote:
On 3/14/2024 7:22 PM, immibis wrote:
On 14/03/24 21:26, olcott wrote:
The question posed to Ĥ.H has no correct answer, thus not the
same question at all.
nobody cares what question is posed to Ĥ.H. How long are you going to
keep trying to get away with this strawman deception?
That you don't care about the foundations of computation is irrelevant.
When-so-ever anyone or anything is asked a question that is logically
impossible for them to correctly answer this places no actual limit
and anyone, anything, or computation.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
It is logically impossible
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
Ĥ.H isn't even asked any question. It just runs until it halts and then returns some result.
On 3/15/2024 6:13 AM, Mikko wrote:
On 2024-03-15 02:22:17 +0000, immibis said:
On 15/03/24 01:52, olcott wrote:
On 3/14/2024 7:22 PM, immibis wrote:
On 14/03/24 21:26, olcott wrote:
The question posed to Ĥ.H has no correct answer, thus not the
same question at all.
nobody cares what question is posed to Ĥ.H. How long are you going
to keep trying to get away with this strawman deception?
That you don't care about the foundations of computation is irrelevant. >>>> When-so-ever anyone or anything is asked a question that is logically
impossible for them to correctly answer this places no actual limit
and anyone, anything, or computation.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
It is logically impossible
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
Ĥ.H isn't even asked any question. It just runs until it halts and
then returns some result.
If Ĥ is as defined by Linz the only result it ever returns is "no". But
it may run forever without returning any result.
That is counter-factual: Linz never gets into this much detail.
On 3/15/2024 1:38 PM, Richard Damon wrote:
On 3/15/24 7:41 AM, olcott wrote:
On 3/15/2024 5:44 AM, Mikko wrote:
On 2024-03-15 01:12:19 +0000, olcott said:
On 3/14/2024 8:06 PM, Richard Damon wrote:
On 3/14/24 4:45 PM, olcott wrote:*The strawman deception is all that you have*
On 3/14/2024 5:37 PM, Richard Damon wrote:
On 3/14/24 3:04 PM, olcott wrote:
On 3/14/2024 4:55 PM, Richard Damon wrote:
On 3/14/24 1:59 PM, olcott wrote:
On 3/14/2024 3:54 PM, Richard Damon wrote:
On 3/14/24 1:26 PM, olcott wrote:
On 3/14/2024 3:20 PM, Richard Damon wrote:
On 3/14/24 12:32 PM, olcott wrote:
On 3/14/2024 12:33 PM, Richard Damon wrote:Which isn;t the question at all, so you are just shown to >>>>>>>>>>>>>> be a stupid liar.
On 3/13/24 4:04 PM, olcott wrote:
On 3/13/2024 5:43 PM, Richard Damon wrote:
On 3/13/24 2:54 PM, olcott wrote:
On 3/13/2024 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/13/24 1:52 PM, olcott wrote:
I invented it so I get to stipulate its meaning. >>>>>>>>>>>>>>>>>>> https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJOn 3/13/2024 12:52 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/13/24 10:08 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 11:44 AM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 13/03/24 04:55, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/2024 10:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>
Not quite. It always gets the wrong answer, >>>>>>>>>>>>>>>>>>>>>>>>>> but only one of them for each quesiton. >>>>>>>>>>>>>>>>>>>>>>>>>>
They all gets the wrong answer on a whole class >>>>>>>>>>>>>>>>>>>>>>>>> of questions
Wrong. You said. yourself. that H1 gets the >>>>>>>>>>>>>>>>>>>>>>>> right answer for D.
Since it is a logical impossibility to determine >>>>>>>>>>>>>>>>>>>>>>> the truth
value of a self-contradictory expression the >>>>>>>>>>>>>>>>>>>>>>> requirement
for H to do this is bogus.
Shows you are just a LIAR, as there IS a truth >>>>>>>>>>>>>>>>>>>>>> value to the expression that is the requirment for >>>>>>>>>>>>>>>>>>>>>> ANY SPECIFIC H.
*Lying about me being a liar may possibly cost your >>>>>>>>>>>>>>>>>>>>> soul*
*Lying about me being a liar may possibly cost your >>>>>>>>>>>>>>>>>>>>> soul*
*Lying about me being a liar may possibly cost your >>>>>>>>>>>>>>>>>>>>> soul*
There is no mapping from H(D,D) to Halts(D,D) that >>>>>>>>>>>>>>>>>>>>> exists.
This proves that H(D,D) is being asked an incorrect >>>>>>>>>>>>>>>>>>>>> question.
Why, because it is NOT a LIE.
You don't even know the definiton of an incorrect >>>>>>>>>>>>>>>>>>>> question.
Nope, common technical term.
Cite a source.
Right, because that question include a presumption of >>>>>>>>>>>>>>>>>> something not actually present.
The fact that there DOES exist a mapping Halt(M,d) >>>>>>>>>>>>>>>>>>>> that maps all Turing Machines and there input to a >>>>>>>>>>>>>>>>>>>> result of Halting / Non-Halting for EVERY member of >>>>>>>>>>>>>>>>>>>> that input set, means tha Halts is a valid mapping >>>>>>>>>>>>>>>>>>>> to ask a decider to try to decider.That part is true.
Likewise when you ask a man that has never been married: >>>>>>>>>>>>>>>>>>> Have you stopped beating tour wife?
There are some men that have stopped beating their wife. >>>>>>>>>>>>>>>>>>
Although there is a mapping from some men to YES/NO >>>>>>>>>>>>>>>>> there is no mapping from never unmarried men to YES/NO >>>>>>>>>>>>>>>>> thus the question is incorrect for all unmarried men. >>>>>>>>>>>>>>>>>
Although there is a mapping from some TM/input pairs to >>>>>>>>>>>>>>>>> YES/NO
there is no mapping from H/D to YES/NO
thus the question is incorrect for H/D
Except that the mapping requested is about the INPUTS to >>>>>>>>>>>>>>>> H, not H itsef.
In order to see that it is an incorrect question we must >>>>>>>>>>>>>>> examine
the question in detail. Making sure to always ignore this >>>>>>>>>>>>>>> key detail
<is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩
halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩
does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) >>>>>>>>>>>>>>
The QUESTION is:
Does the machine and input described by this input, Halt >>>>>>>>>>>>>> when run?
The question posed to Ĥ.H has no correct answer, thus not the >>>>>>>>>>>>> same question at all.
But it DOES.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct and make
sure that
you explain why this element is correct and don't try to switch >>>>>>>>>>> to any other element outside of the above specified set. >>>>>>>>>>>
I didn't say there was.
Then you understand that each question posed to each Ĥ.H in the >>>>>>>>> above set has no correct answer only because each of these answers >>>>>>>>> are contradicted by the machine that H is contained within.
No, YOU don't understand that the IS a correct answer, just not >>>>>>>> the one that H (or H^.H ) happens to give.
Then show me which contradicted answer is correct.
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy and loops so qn >>>>>> was the right answer.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
*The answer must come from elements of the above set*
Is a false claim about a strawman deception really the best you can
say?
The above are the program/input pairs such that every Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets the wrong answer only because whatever answer that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets is contradicted.
So?
That doesn't mean they are the set that the answer to the ACTUAL
QUESTION needs to come from.
You are just proving your stupidity and duplicity.
Objective and Subjective Specifications https://www.cs.toronto.edu/~hehner/OSS.pdf
Credit goes to you for finding the loophole in Carol's original
question: Can anyone correctly answer “no” to this question?
Carol can correctly answer that question with any word that is
synonymous with "no".
Here is the one where the loophole is closed:
Can Carol correctly answer “no” to this [yes/no] question?
The fact that anyone besides Carol can correctly answer that
question with a NO and Carol cannot possibly correctly answer
that question proves that it is a different question when posed
to Carol than when posed to anyone else.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
Carol's question posed to Carol <is> isomorphic to input ⟨Ĥ⟩ ⟨Ĥ⟩
to every Ĥ.H shown above. The fact that some other TM such as H1
(that is not contradicted) can determine a correct answer proves
that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is a different question
On 3/15/2024 2:14 PM, Richard Damon wrote:
On 3/15/24 12:00 PM, olcott wrote:
On 3/15/2024 1:38 PM, Richard Damon wrote:
On 3/15/24 7:41 AM, olcott wrote:
On 3/15/2024 5:44 AM, Mikko wrote:
On 2024-03-15 01:12:19 +0000, olcott said:
On 3/14/2024 8:06 PM, Richard Damon wrote:
On 3/14/24 4:45 PM, olcott wrote:*The strawman deception is all that you have*
On 3/14/2024 5:37 PM, Richard Damon wrote:
On 3/14/24 3:04 PM, olcott wrote:
On 3/14/2024 4:55 PM, Richard Damon wrote:
On 3/14/24 1:59 PM, olcott wrote:
On 3/14/2024 3:54 PM, Richard Damon wrote:
On 3/14/24 1:26 PM, olcott wrote:
On 3/14/2024 3:20 PM, Richard Damon wrote:
On 3/14/24 12:32 PM, olcott wrote:
On 3/14/2024 12:33 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>> On 3/13/24 4:04 PM, olcott wrote:Which isn;t the question at all, so you are just shown >>>>>>>>>>>>>>>> to be a stupid liar.
On 3/13/2024 5:43 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/13/24 2:54 PM, olcott wrote:
On 3/13/2024 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/13/24 1:52 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 12:52 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 10:08 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 11:44 AM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 13/03/24 04:55, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/2024 10:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>
I invented it so I get to stipulate its meaning. >>>>>>>>>>>>>>>>>>>>> https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJNot quite. It always gets the wrong answer, >>>>>>>>>>>>>>>>>>>>>>>>>>>> but only one of them for each quesiton. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
They all gets the wrong answer on a whole >>>>>>>>>>>>>>>>>>>>>>>>>>> class of questions
Wrong. You said. yourself. that H1 gets the >>>>>>>>>>>>>>>>>>>>>>>>>> right answer for D.
Since it is a logical impossibility to >>>>>>>>>>>>>>>>>>>>>>>>> determine the truth
value of a self-contradictory expression the >>>>>>>>>>>>>>>>>>>>>>>>> requirement
for H to do this is bogus.
Shows you are just a LIAR, as there IS a truth >>>>>>>>>>>>>>>>>>>>>>>> value to the expression that is the requirment >>>>>>>>>>>>>>>>>>>>>>>> for ANY SPECIFIC H.
*Lying about me being a liar may possibly cost >>>>>>>>>>>>>>>>>>>>>>> your soul*
*Lying about me being a liar may possibly cost >>>>>>>>>>>>>>>>>>>>>>> your soul*
*Lying about me being a liar may possibly cost >>>>>>>>>>>>>>>>>>>>>>> your soul*
There is no mapping from H(D,D) to Halts(D,D) >>>>>>>>>>>>>>>>>>>>>>> that exists.
This proves that H(D,D) is being asked an >>>>>>>>>>>>>>>>>>>>>>> incorrect question.
Why, because it is NOT a LIE.
You don't even know the definiton of an incorrect >>>>>>>>>>>>>>>>>>>>>> question.
Nope, common technical term.
Cite a source.
The fact that there DOES exist a mapping Halt(M,d) >>>>>>>>>>>>>>>>>>>>>> that maps all Turing Machines and there input to a >>>>>>>>>>>>>>>>>>>>>> result of Halting / Non-Halting for EVERY member >>>>>>>>>>>>>>>>>>>>>> of that input set, means tha Halts is a valid >>>>>>>>>>>>>>>>>>>>>> mapping to ask a decider to try to decider. >>>>>>>>>>>>>>>>>>>>>>That part is true.
Likewise when you ask a man that has never been >>>>>>>>>>>>>>>>>>>>> married:
Have you stopped beating tour wife?
There are some men that have stopped beating their >>>>>>>>>>>>>>>>>>>>> wife.
Right, because that question include a presumption >>>>>>>>>>>>>>>>>>>> of something not actually present.
Although there is a mapping from some men to YES/NO >>>>>>>>>>>>>>>>>>> there is no mapping from never unmarried men to YES/NO >>>>>>>>>>>>>>>>>>> thus the question is incorrect for all unmarried men. >>>>>>>>>>>>>>>>>>>
Although there is a mapping from some TM/input pairs >>>>>>>>>>>>>>>>>>> to YES/NO
there is no mapping from H/D to YES/NO
thus the question is incorrect for H/D
Except that the mapping requested is about the INPUTS >>>>>>>>>>>>>>>>>> to H, not H itsef.
In order to see that it is an incorrect question we >>>>>>>>>>>>>>>>> must examine
the question in detail. Making sure to always ignore >>>>>>>>>>>>>>>>> this key detail
<is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to
⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to
⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) >>>>>>>>>>>>>>>>
The QUESTION is:
Does the machine and input described by this input, Halt >>>>>>>>>>>>>>>> when run?
The question posed to Ĥ.H has no correct answer, thus not >>>>>>>>>>>>>>> the
same question at all.
But it DOES.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct and make
sure that
you explain why this element is correct and don't try to >>>>>>>>>>>>> switch
to any other element outside of the above specified set. >>>>>>>>>>>>>
I didn't say there was.
Then you understand that each question posed to each Ĥ.H in the >>>>>>>>>>> above set has no correct answer only because each of these >>>>>>>>>>> answers
are contradicted by the machine that H is contained within. >>>>>>>>>>>
No, YOU don't understand that the IS a correct answer, just >>>>>>>>>> not the one that H (or H^.H ) happens to give.
Then show me which contradicted answer is correct.
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy and loops so >>>>>>>> qn was the right answer.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not
halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
*The answer must come from elements of the above set*
Is a false claim about a strawman deception really the best you
can say?
The above are the program/input pairs such that every Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets the wrong answer only because whatever answer that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets is contradicted.
So?
That doesn't mean they are the set that the answer to the ACTUAL
QUESTION needs to come from.
You are just proving your stupidity and duplicity.
Objective and Subjective Specifications
https://www.cs.toronto.edu/~hehner/OSS.pdf
Credit goes to you for finding the loophole in Carol's original
question: Can anyone correctly answer “no” to this question?
Carol can correctly answer that question with any word that is
synonymous with "no".
Here is the one where the loophole is closed:
Can Carol correctly answer “no” to this [yes/no] question?
The fact that anyone besides Carol can correctly answer that
question with a NO and Carol cannot possibly correctly answer
that question proves that it is a different question when posed
to Carol than when posed to anyone else.
Which is IRRELEVENT to the Halting Question, as it is a purely
objective question.
The behavior of the input is INDEPENDENT of the decider looking at it.
Note, a given H^ is built on a given H, and no other, but can be given
to any decider to answer, and the correct answer will be the same
irrespective of you ask. Some will give the right answer, and some
will give the wrong answer. The fact that that H is in the latter
doesn't make the question subjective.
The only way to make the Halting Question subjective is to try to
redefine it so the input changes with who you ask, but it doesn't.
The changing H^ to match the H only happens in the Meta, where we
prove that we can find an H^ that any H will get wrong, but each of
those are SEPERATE Halting question (not all one question) and each of
those seperate questions have a correct answer.
Nope.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
Carol's question posed to Carol <is> isomorphic to input ⟨Ĥ⟩ ⟨Ĥ⟩ >>> to every Ĥ.H shown above. The fact that some other TM such as H1
(that is not contradicted) can determine a correct answer proves
that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is a different question
The Question doesn't refer to H at all.
The input ⟨Ĥ⟩ ⟨Ĥ⟩ posed to Ĥ.H
is isomorphic to this question posed to Carol:
Can Carol correctly answer “no” to this [yes/no] question?
Thus changing the meaning of the question
Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H
*in the exact same way that the meaning of this question*
Can Carol correctly answer “no” to this [yes/no] question?
is changed when it is posed to Carol.
On 3/15/2024 2:42 PM, Richard Damon wrote:
On 3/15/24 12:23 PM, olcott wrote:
On 3/15/2024 2:14 PM, Richard Damon wrote:
On 3/15/24 12:00 PM, olcott wrote:
On 3/15/2024 1:38 PM, Richard Damon wrote:
On 3/15/24 7:41 AM, olcott wrote:
On 3/15/2024 5:44 AM, Mikko wrote:
On 2024-03-15 01:12:19 +0000, olcott said:
On 3/14/2024 8:06 PM, Richard Damon wrote:
On 3/14/24 4:45 PM, olcott wrote:*The strawman deception is all that you have*
On 3/14/2024 5:37 PM, Richard Damon wrote:
On 3/14/24 3:04 PM, olcott wrote:
On 3/14/2024 4:55 PM, Richard Damon wrote:
On 3/14/24 1:59 PM, olcott wrote:
On 3/14/2024 3:54 PM, Richard Damon wrote:
On 3/14/24 1:26 PM, olcott wrote:
On 3/14/2024 3:20 PM, Richard Damon wrote:
On 3/14/24 12:32 PM, olcott wrote:
On 3/14/2024 12:33 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/13/24 4:04 PM, olcott wrote:
On 3/13/2024 5:43 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/13/24 2:54 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 1:52 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 12:52 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 10:08 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 11:44 AM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 13/03/24 04:55, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/2024 10:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
I invented it so I get to stipulate its meaning. >>>>>>>>>>>>>>>>>>>>>>> https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJShows you are just a LIAR, as there IS a truth >>>>>>>>>>>>>>>>>>>>>>>>>> value to the expression that is the requirment >>>>>>>>>>>>>>>>>>>>>>>>>> for ANY SPECIFIC H.Not quite. It always gets the wrong >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> answer, but only one of them for each >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> quesiton.
They all gets the wrong answer on a whole >>>>>>>>>>>>>>>>>>>>>>>>>>>>> class of questions
Wrong. You said. yourself. that H1 gets the >>>>>>>>>>>>>>>>>>>>>>>>>>>> right answer for D.
Since it is a logical impossibility to >>>>>>>>>>>>>>>>>>>>>>>>>>> determine the truth
value of a self-contradictory expression the >>>>>>>>>>>>>>>>>>>>>>>>>>> requirement
for H to do this is bogus. >>>>>>>>>>>>>>>>>>>>>>>>>>
*Lying about me being a liar may possibly cost >>>>>>>>>>>>>>>>>>>>>>>>> your soul*
*Lying about me being a liar may possibly cost >>>>>>>>>>>>>>>>>>>>>>>>> your soul*
*Lying about me being a liar may possibly cost >>>>>>>>>>>>>>>>>>>>>>>>> your soul*
There is no mapping from H(D,D) to Halts(D,D) >>>>>>>>>>>>>>>>>>>>>>>>> that exists.
This proves that H(D,D) is being asked an >>>>>>>>>>>>>>>>>>>>>>>>> incorrect question.
Why, because it is NOT a LIE.
You don't even know the definiton of an >>>>>>>>>>>>>>>>>>>>>>>> incorrect question.
Nope, common technical term.
Cite a source.
The fact that there DOES exist a mapping >>>>>>>>>>>>>>>>>>>>>>>> Halt(M,d) that maps all Turing Machines and >>>>>>>>>>>>>>>>>>>>>>>> there input to a result of Halting / Non-Halting >>>>>>>>>>>>>>>>>>>>>>>> for EVERY member of that input set, means tha >>>>>>>>>>>>>>>>>>>>>>>> Halts is a valid mapping to ask a decider to try >>>>>>>>>>>>>>>>>>>>>>>> to decider.That part is true.
Likewise when you ask a man that has never been >>>>>>>>>>>>>>>>>>>>>>> married:
Have you stopped beating tour wife? >>>>>>>>>>>>>>>>>>>>>>> There are some men that have stopped beating >>>>>>>>>>>>>>>>>>>>>>> their wife.
Right, because that question include a presumption >>>>>>>>>>>>>>>>>>>>>> of something not actually present. >>>>>>>>>>>>>>>>>>>>>>
Although there is a mapping from some men to YES/NO >>>>>>>>>>>>>>>>>>>>> there is no mapping from never unmarried men to YES/NO >>>>>>>>>>>>>>>>>>>>> thus the question is incorrect for all unmarried men. >>>>>>>>>>>>>>>>>>>>>
Although there is a mapping from some TM/input >>>>>>>>>>>>>>>>>>>>> pairs to YES/NO
there is no mapping from H/D to YES/NO >>>>>>>>>>>>>>>>>>>>> thus the question is incorrect for H/D >>>>>>>>>>>>>>>>>>>>>
Except that the mapping requested is about the >>>>>>>>>>>>>>>>>>>> INPUTS to H, not H itsef.
In order to see that it is an incorrect question we >>>>>>>>>>>>>>>>>>> must examine
the question in detail. Making sure to always ignore >>>>>>>>>>>>>>>>>>> this key detail
<is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to
⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to
⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
Which isn;t the question at all, so you are just shown >>>>>>>>>>>>>>>>>> to be a stupid liar.
The QUESTION is:
Does the machine and input described by this input, >>>>>>>>>>>>>>>>>> Halt when run?
The question posed to Ĥ.H has no correct answer, thus >>>>>>>>>>>>>>>>> not the
same question at all.
But it DOES.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct and make
sure that
you explain why this element is correct and don't try to >>>>>>>>>>>>>>> switch
to any other element outside of the above specified set. >>>>>>>>>>>>>>>
I didn't say there was.
Then you understand that each question posed to each Ĥ.H in >>>>>>>>>>>>> the
above set has no correct answer only because each of these >>>>>>>>>>>>> answers
are contradicted by the machine that H is contained within. >>>>>>>>>>>>>
No, YOU don't understand that the IS a correct answer, just >>>>>>>>>>>> not the one that H (or H^.H ) happens to give.
Then show me which contradicted answer is correct.
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy and loops >>>>>>>>>> so qn was the right answer.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does
not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) >>>>>>>>>
*The answer must come from elements of the above set*
Is a false claim about a strawman deception really the best you >>>>>>>> can say?
The above are the program/input pairs such that every Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets the wrong answer only because whatever answer that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets is contradicted.
So?
That doesn't mean they are the set that the answer to the ACTUAL
QUESTION needs to come from.
You are just proving your stupidity and duplicity.
Objective and Subjective Specifications
https://www.cs.toronto.edu/~hehner/OSS.pdf
Credit goes to you for finding the loophole in Carol's original
question: Can anyone correctly answer “no” to this question?
Carol can correctly answer that question with any word that is
synonymous with "no".
Here is the one where the loophole is closed:
Can Carol correctly answer “no” to this [yes/no] question?
The fact that anyone besides Carol can correctly answer that
question with a NO and Carol cannot possibly correctly answer
that question proves that it is a different question when posed
to Carol than when posed to anyone else.
Which is IRRELEVENT to the Halting Question, as it is a purely
objective question.
The behavior of the input is INDEPENDENT of the decider looking at it. >>>>
Note, a given H^ is built on a given H, and no other, but can be
given to any decider to answer, and the correct answer will be the
same irrespective of you ask. Some will give the right answer, and
some will give the wrong answer. The fact that that H is in the
latter doesn't make the question subjective.
The only way to make the Halting Question subjective is to try to
redefine it so the input changes with who you ask, but it doesn't.
The changing H^ to match the H only happens in the Meta, where we
prove that we can find an H^ that any H will get wrong, but each of
those are SEPERATE Halting question (not all one question) and each
of those seperate questions have a correct answer.
Nope.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
Carol's question posed to Carol <is> isomorphic to input ⟨Ĥ⟩ ⟨Ĥ⟩
to every Ĥ.H shown above. The fact that some other TM such as H1
(that is not contradicted) can determine a correct answer proves
that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is a different question
The Question doesn't refer to H at all.
The input ⟨Ĥ⟩ ⟨Ĥ⟩ posed to Ĥ.H
is isomorphic to this question posed to Carol:
Can Carol correctly answer “no” to this [yes/no] question?
Nope. and that LIE is a source of a lot of your ERRORS.
Carol is a volitional being.
When we hypothesize that Carol is the name of an AI machine
everything remains the same.
Thus the "correct" answer to it doesn't yet exist, and thus some
consider it just an invalid question in the present.
H is a deterministic machine, that has been stipulated to get her.
The input is a description of a deterministic machine, that has been
stupulated to get here.
The Question, does the input Halt? has a detereministic answer.
The Answer H gives has a deterministic asnwer.
The, it is deterministic as to if H was correct or not.
Thus changing the meaning of the question
Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H
*in the exact same way that the meaning of this question*
In other words, you ADMIT to changing the question!
The, admit to LYING.
(a) Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H
is isomorphic to:
(b) Can Carol correctly answer “no” to this [yes/no] question?
posed to an AI machine named Carol.
Can Carol correctly answer “no” to this [yes/no] question?
is changed when it is posed to Carol.
And that is because the question directly refers to Carol, and thus
can be subjective.
Does the machine described by (H^) (H^) doesn't refere to the decider
that we are asking, it only includes a copy of its algrothm, and not a
referance to that exact machine.
This is a key difference, the two are tied by determinism but not
identity, and thus it is not "Subjective" which refers to identity.
You just don't understand the terms.
In both cases the answer is contradicted thus making both
cases isomorphic to each other.
The only reason that:
Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H
cannot be correctly answered is that the specific Ĥ.H
is contradicted.
The only reason that:
Can Carol correctly answer “no” to this [yes/no] question?
posed to Carol
cannot be correctly answered is that the specific Carol
is contradicted.
On 3/15/2024 3:35 PM, Richard Damon wrote:
On 3/15/24 12:50 PM, olcott wrote:
On 3/15/2024 2:42 PM, Richard Damon wrote:
On 3/15/24 12:23 PM, olcott wrote:
On 3/15/2024 2:14 PM, Richard Damon wrote:
On 3/15/24 12:00 PM, olcott wrote:
On 3/15/2024 1:38 PM, Richard Damon wrote:
On 3/15/24 7:41 AM, olcott wrote:
On 3/15/2024 5:44 AM, Mikko wrote:
On 2024-03-15 01:12:19 +0000, olcott said:
On 3/14/2024 8:06 PM, Richard Damon wrote:
On 3/14/24 4:45 PM, olcott wrote:*The strawman deception is all that you have*
On 3/14/2024 5:37 PM, Richard Damon wrote:
On 3/14/24 3:04 PM, olcott wrote:
On 3/14/2024 4:55 PM, Richard Damon wrote:
On 3/14/24 1:59 PM, olcott wrote:
On 3/14/2024 3:54 PM, Richard Damon wrote:
On 3/14/24 1:26 PM, olcott wrote:
On 3/14/2024 3:20 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/14/24 12:32 PM, olcott wrote:
On 3/14/2024 12:33 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/13/24 4:04 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 5:43 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 2:54 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 1:52 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 12:52 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 10:08 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 11:44 AM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 13/03/24 04:55, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/2024 10:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
I invented it so I get to stipulate its meaning. >>>>>>>>>>>>>>>>>>>>>>>>> https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJShows you are just a LIAR, as there IS a >>>>>>>>>>>>>>>>>>>>>>>>>>>> truth value to the expression that is the >>>>>>>>>>>>>>>>>>>>>>>>>>>> requirment for ANY SPECIFIC H. >>>>>>>>>>>>>>>>>>>>>>>>>>>>Wrong. You said. yourself. that H1 gets >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> the right answer for D. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>Not quite. It always gets the wrong >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> answer, but only one of them for each >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> quesiton.
They all gets the wrong answer on a whole >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> class of questions >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Since it is a logical impossibility to >>>>>>>>>>>>>>>>>>>>>>>>>>>>> determine the truth
value of a self-contradictory expression >>>>>>>>>>>>>>>>>>>>>>>>>>>>> the requirement
for H to do this is bogus. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
*Lying about me being a liar may possibly >>>>>>>>>>>>>>>>>>>>>>>>>>> cost your soul*
*Lying about me being a liar may possibly >>>>>>>>>>>>>>>>>>>>>>>>>>> cost your soul*
*Lying about me being a liar may possibly >>>>>>>>>>>>>>>>>>>>>>>>>>> cost your soul*
There is no mapping from H(D,D) to Halts(D,D) >>>>>>>>>>>>>>>>>>>>>>>>>>> that exists.
This proves that H(D,D) is being asked an >>>>>>>>>>>>>>>>>>>>>>>>>>> incorrect question.
Why, because it is NOT a LIE. >>>>>>>>>>>>>>>>>>>>>>>>>>
You don't even know the definiton of an >>>>>>>>>>>>>>>>>>>>>>>>>> incorrect question.
Nope, common technical term.
Cite a source.
The fact that there DOES exist a mapping >>>>>>>>>>>>>>>>>>>>>>>>>> Halt(M,d) that maps all Turing Machines and >>>>>>>>>>>>>>>>>>>>>>>>>> there input to a result of Halting / >>>>>>>>>>>>>>>>>>>>>>>>>> Non-Halting for EVERY member of that input >>>>>>>>>>>>>>>>>>>>>>>>>> set, means tha Halts is a valid mapping to ask >>>>>>>>>>>>>>>>>>>>>>>>>> a decider to try to decider. >>>>>>>>>>>>>>>>>>>>>>>>>>That part is true.
Likewise when you ask a man that has never been >>>>>>>>>>>>>>>>>>>>>>>>> married:
Have you stopped beating tour wife? >>>>>>>>>>>>>>>>>>>>>>>>> There are some men that have stopped beating >>>>>>>>>>>>>>>>>>>>>>>>> their wife.
Right, because that question include a >>>>>>>>>>>>>>>>>>>>>>>> presumption of something not actually present. >>>>>>>>>>>>>>>>>>>>>>>>
Although there is a mapping from some men to YES/NO >>>>>>>>>>>>>>>>>>>>>>> there is no mapping from never unmarried men to >>>>>>>>>>>>>>>>>>>>>>> YES/NO
thus the question is incorrect for all unmarried >>>>>>>>>>>>>>>>>>>>>>> men.
Although there is a mapping from some TM/input >>>>>>>>>>>>>>>>>>>>>>> pairs to YES/NO
there is no mapping from H/D to YES/NO >>>>>>>>>>>>>>>>>>>>>>> thus the question is incorrect for H/D >>>>>>>>>>>>>>>>>>>>>>>
Except that the mapping requested is about the >>>>>>>>>>>>>>>>>>>>>> INPUTS to H, not H itsef.
In order to see that it is an incorrect question we >>>>>>>>>>>>>>>>>>>>> must examine
the question in detail. Making sure to always >>>>>>>>>>>>>>>>>>>>> ignore this key detail
<is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied
to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied
to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
Which isn;t the question at all, so you are just >>>>>>>>>>>>>>>>>>>> shown to be a stupid liar.
The QUESTION is:
Does the machine and input described by this input, >>>>>>>>>>>>>>>>>>>> Halt when run?
The question posed to Ĥ.H has no correct answer, thus >>>>>>>>>>>>>>>>>>> not the
same question at all.
But it DOES.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct and
make sure that
you explain why this element is correct and don't try >>>>>>>>>>>>>>>>> to switch
to any other element outside of the above specified set. >>>>>>>>>>>>>>>>>
I didn't say there was.
Then you understand that each question posed to each Ĥ.H >>>>>>>>>>>>>>> in the
above set has no correct answer only because each of >>>>>>>>>>>>>>> these answers
are contradicted by the machine that H is contained within. >>>>>>>>>>>>>>>
No, YOU don't understand that the IS a correct answer, >>>>>>>>>>>>>> just not the one that H (or H^.H ) happens to give. >>>>>>>>>>>>>>
Then show me which contradicted answer is correct.
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy and loops >>>>>>>>>>>> so qn was the right answer.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does
not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) >>>>>>>>>>>
*The answer must come from elements of the above set*
Is a false claim about a strawman deception really the best >>>>>>>>>> you can say?
The above are the program/input pairs such that every Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets the wrong answer only because whatever answer that Ĥ.H ⟨Ĥ⟩ >>>>>>>>> ⟨Ĥ⟩
gets is contradicted.
So?
That doesn't mean they are the set that the answer to the ACTUAL >>>>>>>> QUESTION needs to come from.
You are just proving your stupidity and duplicity.
Objective and Subjective Specifications
https://www.cs.toronto.edu/~hehner/OSS.pdf
Credit goes to you for finding the loophole in Carol's original
question: Can anyone correctly answer “no” to this question? >>>>>>>
Carol can correctly answer that question with any word that is
synonymous with "no".
Here is the one where the loophole is closed:
Can Carol correctly answer “no” to this [yes/no] question?
The fact that anyone besides Carol can correctly answer that
question with a NO and Carol cannot possibly correctly answer
that question proves that it is a different question when posed
to Carol than when posed to anyone else.
Which is IRRELEVENT to the Halting Question, as it is a purely
objective question.
The behavior of the input is INDEPENDENT of the decider looking at >>>>>> it.
Note, a given H^ is built on a given H, and no other, but can be
given to any decider to answer, and the correct answer will be the >>>>>> same irrespective of you ask. Some will give the right answer, and >>>>>> some will give the wrong answer. The fact that that H is in the
latter doesn't make the question subjective.
The only way to make the Halting Question subjective is to try to
redefine it so the input changes with who you ask, but it doesn't. >>>>>>
The changing H^ to match the H only happens in the Meta, where we
prove that we can find an H^ that any H will get wrong, but each
of those are SEPERATE Halting question (not all one question) and
each of those seperate questions have a correct answer.
Nope.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not
halt
Carol's question posed to Carol <is> isomorphic to input ⟨Ĥ⟩ ⟨Ĥ⟩
to every Ĥ.H shown above. The fact that some other TM such as H1 >>>>>>> (that is not contradicted) can determine a correct answer proves >>>>>>> that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is a different question
The Question doesn't refer to H at all.
The input ⟨Ĥ⟩ ⟨Ĥ⟩ posed to Ĥ.H
is isomorphic to this question posed to Carol:
Can Carol correctly answer “no” to this [yes/no] question?
Nope. and that LIE is a source of a lot of your ERRORS.
Carol is a volitional being.
When we hypothesize that Carol is the name of an AI machine
everything remains the same.
Nope.
Once Carol become deterministic, then the whole thing changes.
The only reason that:
Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H
cannot be correctly answered is that the specific Ĥ.H is contradicted.
The only reason that:
Can Carol correctly answer “no” to this [yes/no] question? posed to Carol cannot be correctly answered is that the specific Carol is contradicted.
On 3/15/2024 5:13 PM, Richard Damon wrote:
On 3/15/24 1:42 PM, olcott wrote:
On 3/15/2024 3:35 PM, Richard Damon wrote:
On 3/15/24 12:50 PM, olcott wrote:
On 3/15/2024 2:42 PM, Richard Damon wrote:
On 3/15/24 12:23 PM, olcott wrote:
On 3/15/2024 2:14 PM, Richard Damon wrote:
On 3/15/24 12:00 PM, olcott wrote:
On 3/15/2024 1:38 PM, Richard Damon wrote:
On 3/15/24 7:41 AM, olcott wrote:
On 3/15/2024 5:44 AM, Mikko wrote:
On 2024-03-15 01:12:19 +0000, olcott said:
On 3/14/2024 8:06 PM, Richard Damon wrote:Is a false claim about a strawman deception really the best >>>>>>>>>>>> you can say?
On 3/14/24 4:45 PM, olcott wrote:*The strawman deception is all that you have*
On 3/14/2024 5:37 PM, Richard Damon wrote:
On 3/14/24 3:04 PM, olcott wrote:
On 3/14/2024 4:55 PM, Richard Damon wrote:
On 3/14/24 1:59 PM, olcott wrote:
On 3/14/2024 3:54 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/14/24 1:26 PM, olcott wrote:
On 3/14/2024 3:20 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/14/24 12:32 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 12:33 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 4:04 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 5:43 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 2:54 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 1:52 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 12:52 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 10:08 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 11:44 AM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 13/03/24 04:55, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/2024 10:49 PM, Richard Damon >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
I invented it so I get to stipulate its meaning. >>>>>>>>>>>>>>>>>>>>>>>>>>> https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJShows you are just a LIAR, as there IS a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> truth value to the expression that is the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> requirment for ANY SPECIFIC H. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>Wrong. You said. yourself. that H1 gets >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> the right answer for D. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Not quite. It always gets the wrong >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> answer, but only one of them for each >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> quesiton.
They all gets the wrong answer on a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> whole class of questions >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Since it is a logical impossibility to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> determine the truth >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> value of a self-contradictory expression >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> the requirement
for H to do this is bogus. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*Lying about me being a liar may possibly >>>>>>>>>>>>>>>>>>>>>>>>>>>>> cost your soul*
*Lying about me being a liar may possibly >>>>>>>>>>>>>>>>>>>>>>>>>>>>> cost your soul*
*Lying about me being a liar may possibly >>>>>>>>>>>>>>>>>>>>>>>>>>>>> cost your soul*
There is no mapping from H(D,D) to >>>>>>>>>>>>>>>>>>>>>>>>>>>>> Halts(D,D) that exists. >>>>>>>>>>>>>>>>>>>>>>>>>>>>> This proves that H(D,D) is being asked an >>>>>>>>>>>>>>>>>>>>>>>>>>>>> incorrect question.
Why, because it is NOT a LIE. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
You don't even know the definiton of an >>>>>>>>>>>>>>>>>>>>>>>>>>>> incorrect question.
Nope, common technical term. >>>>>>>>>>>>>>>>>>>>>>>>>>
Cite a source.
The fact that there DOES exist a mapping >>>>>>>>>>>>>>>>>>>>>>>>>>>> Halt(M,d) that maps all Turing Machines and >>>>>>>>>>>>>>>>>>>>>>>>>>>> there input to a result of Halting / >>>>>>>>>>>>>>>>>>>>>>>>>>>> Non-Halting for EVERY member of that input >>>>>>>>>>>>>>>>>>>>>>>>>>>> set, means tha Halts is a valid mapping to >>>>>>>>>>>>>>>>>>>>>>>>>>>> ask a decider to try to decider. >>>>>>>>>>>>>>>>>>>>>>>>>>>>That part is true.
Likewise when you ask a man that has never >>>>>>>>>>>>>>>>>>>>>>>>>>> been married:
Have you stopped beating tour wife? >>>>>>>>>>>>>>>>>>>>>>>>>>> There are some men that have stopped beating >>>>>>>>>>>>>>>>>>>>>>>>>>> their wife.
Right, because that question include a >>>>>>>>>>>>>>>>>>>>>>>>>> presumption of something not actually present. >>>>>>>>>>>>>>>>>>>>>>>>>>
Although there is a mapping from some men to >>>>>>>>>>>>>>>>>>>>>>>>> YES/NO
there is no mapping from never unmarried men to >>>>>>>>>>>>>>>>>>>>>>>>> YES/NO
thus the question is incorrect for all >>>>>>>>>>>>>>>>>>>>>>>>> unmarried men.
Although there is a mapping from some TM/input >>>>>>>>>>>>>>>>>>>>>>>>> pairs to YES/NO
there is no mapping from H/D to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>> thus the question is incorrect for H/D >>>>>>>>>>>>>>>>>>>>>>>>>
Except that the mapping requested is about the >>>>>>>>>>>>>>>>>>>>>>>> INPUTS to H, not H itsef.
In order to see that it is an incorrect question >>>>>>>>>>>>>>>>>>>>>>> we must examine
the question in detail. Making sure to always >>>>>>>>>>>>>>>>>>>>>>> ignore this key detail
<is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied
to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied
to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
Which isn;t the question at all, so you are just >>>>>>>>>>>>>>>>>>>>>> shown to be a stupid liar.
The QUESTION is:
Does the machine and input described by this >>>>>>>>>>>>>>>>>>>>>> input, Halt when run?
The question posed to Ĥ.H has no correct answer, >>>>>>>>>>>>>>>>>>>>> thus not the
same question at all.
But it DOES.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct and
make sure that
you explain why this element is correct and don't try >>>>>>>>>>>>>>>>>>> to switch
to any other element outside of the above specified set. >>>>>>>>>>>>>>>>>>>
I didn't say there was.
Then you understand that each question posed to each >>>>>>>>>>>>>>>>> Ĥ.H in the
above set has no correct answer only because each of >>>>>>>>>>>>>>>>> these answers
are contradicted by the machine that H is contained >>>>>>>>>>>>>>>>> within.
No, YOU don't understand that the IS a correct answer, >>>>>>>>>>>>>>>> just not the one that H (or H^.H ) happens to give. >>>>>>>>>>>>>>>>
Then show me which contradicted answer is correct. >>>>>>>>>>>>>>>
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy and >>>>>>>>>>>>>> loops so qn was the right answer.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩
does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) >>>>>>>>>>>>>
*The answer must come from elements of the above set* >>>>>>>>>>>>
The above are the program/input pairs such that every Ĥ.H ⟨Ĥ⟩ >>>>>>>>>>> ⟨Ĥ⟩
gets the wrong answer only because whatever answer that Ĥ.H >>>>>>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩
gets is contradicted.
So?
That doesn't mean they are the set that the answer to the
ACTUAL QUESTION needs to come from.
You are just proving your stupidity and duplicity.
Objective and Subjective Specifications
https://www.cs.toronto.edu/~hehner/OSS.pdf
Credit goes to you for finding the loophole in Carol's original >>>>>>>>> question: Can anyone correctly answer “no” to this question? >>>>>>>>>
Carol can correctly answer that question with any word that is >>>>>>>>> synonymous with "no".
Here is the one where the loophole is closed:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>>
The fact that anyone besides Carol can correctly answer that >>>>>>>>> question with a NO and Carol cannot possibly correctly answer >>>>>>>>> that question proves that it is a different question when posed >>>>>>>>> to Carol than when posed to anyone else.
Which is IRRELEVENT to the Halting Question, as it is a purely >>>>>>>> objective question.
The behavior of the input is INDEPENDENT of the decider looking >>>>>>>> at it.
Note, a given H^ is built on a given H, and no other, but can be >>>>>>>> given to any decider to answer, and the correct answer will be >>>>>>>> the same irrespective of you ask. Some will give the right
answer, and some will give the wrong answer. The fact that that >>>>>>>> H is in the latter doesn't make the question subjective.
The only way to make the Halting Question subjective is to try >>>>>>>> to redefine it so the input changes with who you ask, but it
doesn't.
The changing H^ to match the H only happens in the Meta, where >>>>>>>> we prove that we can find an H^ that any H will get wrong, but >>>>>>>> each of those are SEPERATE Halting question (not all one
question) and each of those seperate questions have a correct
answer.
Nope.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does
not halt
Carol's question posed to Carol <is> isomorphic to input ⟨Ĥ⟩ ⟨Ĥ⟩
to every Ĥ.H shown above. The fact that some other TM such as H1 >>>>>>>>> (that is not contradicted) can determine a correct answer proves >>>>>>>>> that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is a different question
The Question doesn't refer to H at all.
The input ⟨Ĥ⟩ ⟨Ĥ⟩ posed to Ĥ.H
is isomorphic to this question posed to Carol:
Can Carol correctly answer “no” to this [yes/no] question?
Nope. and that LIE is a source of a lot of your ERRORS.
Carol is a volitional being.
When we hypothesize that Carol is the name of an AI machine
everything remains the same.
Nope.
Once Carol become deterministic, then the whole thing changes.
The only reason that:
Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H
cannot be correctly answered is that the specific Ĥ.H is contradicted.
The only reason that:
Can Carol correctly answer “no” to this [yes/no] question? posed to
Carol
cannot be correctly answered is that the specific Carol is contradicted. >>>
Nope.
You aren't showing any ERRORS I made but just asserting your FALSE
claims again.
Inability to show WHY my description was wrong just proves you have no
basis.
You are just demonstrating that you don't understand how logic works.
It seems you think this is just some abstract philosophy where
anything goes and rhetoric rules.
*You have provided zero correct reasoning of how*
*Carol's question posed to Carol*
*is not contradicted just like*
*Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H is contradicted*
On 3/15/2024 6:02 PM, Richard Damon wrote:
On 3/15/24 3:47 PM, olcott wrote:
On 3/15/2024 5:13 PM, Richard Damon wrote:
On 3/15/24 1:42 PM, olcott wrote:
On 3/15/2024 3:35 PM, Richard Damon wrote:
On 3/15/24 12:50 PM, olcott wrote:
On 3/15/2024 2:42 PM, Richard Damon wrote:
On 3/15/24 12:23 PM, olcott wrote:
On 3/15/2024 2:14 PM, Richard Damon wrote:Nope. and that LIE is a source of a lot of your ERRORS.
On 3/15/24 12:00 PM, olcott wrote:
On 3/15/2024 1:38 PM, Richard Damon wrote:
On 3/15/24 7:41 AM, olcott wrote:
On 3/15/2024 5:44 AM, Mikko wrote:
On 2024-03-15 01:12:19 +0000, olcott said:
On 3/14/2024 8:06 PM, Richard Damon wrote:Is a false claim about a strawman deception really the >>>>>>>>>>>>>> best you can say?
On 3/14/24 4:45 PM, olcott wrote:*The strawman deception is all that you have*
On 3/14/2024 5:37 PM, Richard Damon wrote:
On 3/14/24 3:04 PM, olcott wrote:
On 3/14/2024 4:55 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/14/24 1:59 PM, olcott wrote:
On 3/14/2024 3:54 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/14/24 1:26 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 3:20 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 12:32 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 12:33 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 4:04 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 5:43 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 2:54 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 1:52 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 12:52 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 10:08 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 11:44 AM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 13/03/24 04:55, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/2024 10:49 PM, Richard Damon >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
I invented it so I get to stipulate its >>>>>>>>>>>>>>>>>>>>>>>>>>>>> meaning.Shows you are just a LIAR, as there IS a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> truth value to the expression that is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> the requirment for ANY SPECIFIC H. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>Wrong. You said. yourself. that H1 >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> gets the right answer for D. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Not quite. It always gets the wrong >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> answer, but only one of them for >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> each quesiton. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
They all gets the wrong answer on a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> whole class of questions >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Since it is a logical impossibility to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> determine the truth >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> value of a self-contradictory >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> expression the requirement >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> for H to do this is bogus. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*Lying about me being a liar may possibly >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> cost your soul*
*Lying about me being a liar may possibly >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> cost your soul*
*Lying about me being a liar may possibly >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> cost your soul*
There is no mapping from H(D,D) to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Halts(D,D) that exists. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> This proves that H(D,D) is being asked an >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> incorrect question. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Why, because it is NOT a LIE. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
You don't even know the definiton of an >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> incorrect question. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
Nope, common technical term. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
Cite a source.
The fact that there DOES exist a mapping >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Halt(M,d) that maps all Turing Machines >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> and there input to a result of Halting / >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Non-Halting for EVERY member of that input >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> set, means tha Halts is a valid mapping to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ask a decider to try to decider. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>That part is true.
Likewise when you ask a man that has never >>>>>>>>>>>>>>>>>>>>>>>>>>>>> been married:
Have you stopped beating tour wife? >>>>>>>>>>>>>>>>>>>>>>>>>>>>> There are some men that have stopped >>>>>>>>>>>>>>>>>>>>>>>>>>>>> beating their wife.
Right, because that question include a >>>>>>>>>>>>>>>>>>>>>>>>>>>> presumption of something not actually present. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
Although there is a mapping from some men to >>>>>>>>>>>>>>>>>>>>>>>>>>> YES/NO
there is no mapping from never unmarried men >>>>>>>>>>>>>>>>>>>>>>>>>>> to YES/NO
thus the question is incorrect for all >>>>>>>>>>>>>>>>>>>>>>>>>>> unmarried men.
Although there is a mapping from some >>>>>>>>>>>>>>>>>>>>>>>>>>> TM/input pairs to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>> there is no mapping from H/D to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>> thus the question is incorrect for H/D >>>>>>>>>>>>>>>>>>>>>>>>>>>
Except that the mapping requested is about the >>>>>>>>>>>>>>>>>>>>>>>>>> INPUTS to H, not H itsef.
In order to see that it is an incorrect >>>>>>>>>>>>>>>>>>>>>>>>> question we must examine
the question in detail. Making sure to always >>>>>>>>>>>>>>>>>>>>>>>>> ignore this key detail
<is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ
applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ
applied to ⟨Ĥ⟩ does not halt >>>>>>>>>>>>>>>>>>>>>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
Which isn;t the question at all, so you are just >>>>>>>>>>>>>>>>>>>>>>>> shown to be a stupid liar.
The QUESTION is:
Does the machine and input described by this >>>>>>>>>>>>>>>>>>>>>>>> input, Halt when run?
The question posed to Ĥ.H has no correct answer, >>>>>>>>>>>>>>>>>>>>>>> thus not the
same question at all.
But it DOES.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct
and make sure that
you explain why this element is correct and don't >>>>>>>>>>>>>>>>>>>>> try to switch
to any other element outside of the above specified >>>>>>>>>>>>>>>>>>>>> set.
I didn't say there was.
Then you understand that each question posed to each >>>>>>>>>>>>>>>>>>> Ĥ.H in the
above set has no correct answer only because each of >>>>>>>>>>>>>>>>>>> these answers
are contradicted by the machine that H is contained >>>>>>>>>>>>>>>>>>> within.
No, YOU don't understand that the IS a correct answer, >>>>>>>>>>>>>>>>>> just not the one that H (or H^.H ) happens to give. >>>>>>>>>>>>>>>>>>
Then show me which contradicted answer is correct. >>>>>>>>>>>>>>>>>
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy and >>>>>>>>>>>>>>>> loops so qn was the right answer.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩
halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩
does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) >>>>>>>>>>>>>>>
*The answer must come from elements of the above set* >>>>>>>>>>>>>>
The above are the program/input pairs such that every Ĥ.H >>>>>>>>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩
gets the wrong answer only because whatever answer that Ĥ.H >>>>>>>>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩
gets is contradicted.
So?
That doesn't mean they are the set that the answer to the >>>>>>>>>>>> ACTUAL QUESTION needs to come from.
You are just proving your stupidity and duplicity.
Objective and Subjective Specifications
https://www.cs.toronto.edu/~hehner/OSS.pdf
Credit goes to you for finding the loophole in Carol's original >>>>>>>>>>> question: Can anyone correctly answer “no” to this question? >>>>>>>>>>>
Carol can correctly answer that question with any word that is >>>>>>>>>>> synonymous with "no".
Here is the one where the loophole is closed:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>>>>
The fact that anyone besides Carol can correctly answer that >>>>>>>>>>> question with a NO and Carol cannot possibly correctly answer >>>>>>>>>>> that question proves that it is a different question when posed >>>>>>>>>>> to Carol than when posed to anyone else.
Which is IRRELEVENT to the Halting Question, as it is a purely >>>>>>>>>> objective question.
The behavior of the input is INDEPENDENT of the decider
looking at it.
Note, a given H^ is built on a given H, and no other, but can >>>>>>>>>> be given to any decider to answer, and the correct answer will >>>>>>>>>> be the same irrespective of you ask. Some will give the right >>>>>>>>>> answer, and some will give the wrong answer. The fact that >>>>>>>>>> that H is in the latter doesn't make the question subjective. >>>>>>>>>>
The only way to make the Halting Question subjective is to try >>>>>>>>>> to redefine it so the input changes with who you ask, but it >>>>>>>>>> doesn't.
The changing H^ to match the H only happens in the Meta, where >>>>>>>>>> we prove that we can find an H^ that any H will get wrong, but >>>>>>>>>> each of those are SEPERATE Halting question (not all one
question) and each of those seperate questions have a correct >>>>>>>>>> answer.
Nope.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does
not halt
Carol's question posed to Carol <is> isomorphic to input ⟨Ĥ⟩ ⟨Ĥ⟩
to every Ĥ.H shown above. The fact that some other TM such as H1 >>>>>>>>>>> (that is not contradicted) can determine a correct answer proves >>>>>>>>>>> that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is a different question
The Question doesn't refer to H at all.
The input ⟨Ĥ⟩ ⟨Ĥ⟩ posed to Ĥ.H
is isomorphic to this question posed to Carol:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>
Carol is a volitional being.
When we hypothesize that Carol is the name of an AI machine
everything remains the same.
Nope.
Once Carol become deterministic, then the whole thing changes.
The only reason that:
Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H
cannot be correctly answered is that the specific Ĥ.H is contradicted. >>>>>
The only reason that:
Can Carol correctly answer “no” to this [yes/no] question? posed to >>>>> Carol
cannot be correctly answered is that the specific Carol is
contradicted.
Nope.
You aren't showing any ERRORS I made but just asserting your FALSE
claims again.
Inability to show WHY my description was wrong just proves you have
no basis.
You are just demonstrating that you don't understand how logic works.
It seems you think this is just some abstract philosophy where
anything goes and rhetoric rules.
*You have provided zero correct reasoning of how*
*Carol's question posed to Carol*
*is not contradicted just like*
*Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H is contradicted*
Yes, I have.
YOU have provided ZERO reasoning how they are.
Dos H^ (H^) Halt? even when posed to H^.H has an answer!
When posed to each entity (Carol/Ĥ.H)
their respective question (a)/(b):
(a) Carol correctly answer “no” to this [yes/no] question?
(b) Does Ĥ ⟨Ĥ⟩ halt?
lacks a correct answer because this answer is contradicted.
*Incorrect questions do not lack answers they lack correct answers*
Carol could answer by flipping the bird.
On 3/15/2024 1:38 PM, Richard Damon wrote:
On 3/15/24 7:41 AM, olcott wrote:
On 3/15/2024 5:44 AM, Mikko wrote:
On 2024-03-15 01:12:19 +0000, olcott said:
On 3/14/2024 8:06 PM, Richard Damon wrote:
On 3/14/24 4:45 PM, olcott wrote:*The strawman deception is all that you have*
On 3/14/2024 5:37 PM, Richard Damon wrote:
On 3/14/24 3:04 PM, olcott wrote:
On 3/14/2024 4:55 PM, Richard Damon wrote:
On 3/14/24 1:59 PM, olcott wrote:
On 3/14/2024 3:54 PM, Richard Damon wrote:
On 3/14/24 1:26 PM, olcott wrote:
On 3/14/2024 3:20 PM, Richard Damon wrote:
On 3/14/24 12:32 PM, olcott wrote:
On 3/14/2024 12:33 PM, Richard Damon wrote:Which isn;t the question at all, so you are just shown to >>>>>>>>>>>>>> be a stupid liar.
On 3/13/24 4:04 PM, olcott wrote:
On 3/13/2024 5:43 PM, Richard Damon wrote:
On 3/13/24 2:54 PM, olcott wrote:
On 3/13/2024 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/13/24 1:52 PM, olcott wrote:
I invented it so I get to stipulate its meaning. >>>>>>>>>>>>>>>>>>> https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJOn 3/13/2024 12:52 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/13/24 10:08 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 11:44 AM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 13/03/24 04:55, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/2024 10:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>
Not quite. It always gets the wrong answer, >>>>>>>>>>>>>>>>>>>>>>>>>> but only one of them for each quesiton. >>>>>>>>>>>>>>>>>>>>>>>>>>
They all gets the wrong answer on a whole class >>>>>>>>>>>>>>>>>>>>>>>>> of questions
Wrong. You said. yourself. that H1 gets the >>>>>>>>>>>>>>>>>>>>>>>> right answer for D.
Since it is a logical impossibility to determine >>>>>>>>>>>>>>>>>>>>>>> the truth
value of a self-contradictory expression the >>>>>>>>>>>>>>>>>>>>>>> requirement
for H to do this is bogus.
Shows you are just a LIAR, as there IS a truth >>>>>>>>>>>>>>>>>>>>>> value to the expression that is the requirment for >>>>>>>>>>>>>>>>>>>>>> ANY SPECIFIC H.
*Lying about me being a liar may possibly cost your >>>>>>>>>>>>>>>>>>>>> soul*
*Lying about me being a liar may possibly cost your >>>>>>>>>>>>>>>>>>>>> soul*
*Lying about me being a liar may possibly cost your >>>>>>>>>>>>>>>>>>>>> soul*
There is no mapping from H(D,D) to Halts(D,D) that >>>>>>>>>>>>>>>>>>>>> exists.
This proves that H(D,D) is being asked an incorrect >>>>>>>>>>>>>>>>>>>>> question.
Why, because it is NOT a LIE.
You don't even know the definiton of an incorrect >>>>>>>>>>>>>>>>>>>> question.
Nope, common technical term.
Cite a source.
Right, because that question include a presumption of >>>>>>>>>>>>>>>>>> something not actually present.
The fact that there DOES exist a mapping Halt(M,d) >>>>>>>>>>>>>>>>>>>> that maps all Turing Machines and there input to a >>>>>>>>>>>>>>>>>>>> result of Halting / Non-Halting for EVERY member of >>>>>>>>>>>>>>>>>>>> that input set, means tha Halts is a valid mapping >>>>>>>>>>>>>>>>>>>> to ask a decider to try to decider.That part is true.
Likewise when you ask a man that has never been married: >>>>>>>>>>>>>>>>>>> Have you stopped beating tour wife?
There are some men that have stopped beating their wife. >>>>>>>>>>>>>>>>>>
Although there is a mapping from some men to YES/NO >>>>>>>>>>>>>>>>> there is no mapping from never unmarried men to YES/NO >>>>>>>>>>>>>>>>> thus the question is incorrect for all unmarried men. >>>>>>>>>>>>>>>>>
Although there is a mapping from some TM/input pairs to >>>>>>>>>>>>>>>>> YES/NO
there is no mapping from H/D to YES/NO
thus the question is incorrect for H/D
Except that the mapping requested is about the INPUTS to >>>>>>>>>>>>>>>> H, not H itsef.
In order to see that it is an incorrect question we must >>>>>>>>>>>>>>> examine
the question in detail. Making sure to always ignore this >>>>>>>>>>>>>>> key detail
<is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩
halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩
does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) >>>>>>>>>>>>>>
The QUESTION is:
Does the machine and input described by this input, Halt >>>>>>>>>>>>>> when run?
The question posed to Ĥ.H has no correct answer, thus not the >>>>>>>>>>>>> same question at all.
But it DOES.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct and make
sure that
you explain why this element is correct and don't try to switch >>>>>>>>>>> to any other element outside of the above specified set. >>>>>>>>>>>
I didn't say there was.
Then you understand that each question posed to each Ĥ.H in the >>>>>>>>> above set has no correct answer only because each of these answers >>>>>>>>> are contradicted by the machine that H is contained within.
No, YOU don't understand that the IS a correct answer, just not >>>>>>>> the one that H (or H^.H ) happens to give.
Then show me which contradicted answer is correct.
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy and loops so qn >>>>>> was the right answer.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
*The answer must come from elements of the above set*
Is a false claim about a strawman deception really the best you can
say?
The above are the program/input pairs such that every Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets the wrong answer only because whatever answer that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets is contradicted.
So?
That doesn't mean they are the set that the answer to the ACTUAL
QUESTION needs to come from.
You are just proving your stupidity and duplicity.
Objective and Subjective Specifications https://www.cs.toronto.edu/~hehner/OSS.pdf
Credit goes to you for finding the loophole in Carol's original
question: Can anyone correctly answer “no” to this question?
Carol can correctly answer that question with any word that is
synonymous with "no".
Here is the one where the loophole is closed:
Can Carol correctly answer “no” to this [yes/no] question?
The fact that anyone besides Carol can correctly answer that
question with a NO and Carol cannot possibly correctly answer
that question proves that it is a different question when posed
to Carol than when posed to anyone else.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
Carol's question posed to Carol <is> isomorphic to input ⟨Ĥ⟩ ⟨Ĥ⟩
to every Ĥ.H shown above. The fact that some other TM such as H1
(that is not contradicted) can determine a correct answer proves
that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is a different question
On 3/15/2024 2:14 PM, Richard Damon wrote:
On 3/15/24 12:00 PM, olcott wrote:
On 3/15/2024 1:38 PM, Richard Damon wrote:
On 3/15/24 7:41 AM, olcott wrote:
On 3/15/2024 5:44 AM, Mikko wrote:
On 2024-03-15 01:12:19 +0000, olcott said:
On 3/14/2024 8:06 PM, Richard Damon wrote:
On 3/14/24 4:45 PM, olcott wrote:*The strawman deception is all that you have*
On 3/14/2024 5:37 PM, Richard Damon wrote:
On 3/14/24 3:04 PM, olcott wrote:
On 3/14/2024 4:55 PM, Richard Damon wrote:
On 3/14/24 1:59 PM, olcott wrote:
On 3/14/2024 3:54 PM, Richard Damon wrote:
On 3/14/24 1:26 PM, olcott wrote:
On 3/14/2024 3:20 PM, Richard Damon wrote:
On 3/14/24 12:32 PM, olcott wrote:
On 3/14/2024 12:33 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>> On 3/13/24 4:04 PM, olcott wrote:Which isn;t the question at all, so you are just shown >>>>>>>>>>>>>>>> to be a stupid liar.
On 3/13/2024 5:43 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/13/24 2:54 PM, olcott wrote:
On 3/13/2024 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/13/24 1:52 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 12:52 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 10:08 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 11:44 AM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 13/03/24 04:55, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/2024 10:49 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>
I invented it so I get to stipulate its meaning. >>>>>>>>>>>>>>>>>>>>> https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJNot quite. It always gets the wrong answer, >>>>>>>>>>>>>>>>>>>>>>>>>>>> but only one of them for each quesiton. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
They all gets the wrong answer on a whole >>>>>>>>>>>>>>>>>>>>>>>>>>> class of questions
Wrong. You said. yourself. that H1 gets the >>>>>>>>>>>>>>>>>>>>>>>>>> right answer for D.
Since it is a logical impossibility to >>>>>>>>>>>>>>>>>>>>>>>>> determine the truth
value of a self-contradictory expression the >>>>>>>>>>>>>>>>>>>>>>>>> requirement
for H to do this is bogus.
Shows you are just a LIAR, as there IS a truth >>>>>>>>>>>>>>>>>>>>>>>> value to the expression that is the requirment >>>>>>>>>>>>>>>>>>>>>>>> for ANY SPECIFIC H.
*Lying about me being a liar may possibly cost >>>>>>>>>>>>>>>>>>>>>>> your soul*
*Lying about me being a liar may possibly cost >>>>>>>>>>>>>>>>>>>>>>> your soul*
*Lying about me being a liar may possibly cost >>>>>>>>>>>>>>>>>>>>>>> your soul*
There is no mapping from H(D,D) to Halts(D,D) >>>>>>>>>>>>>>>>>>>>>>> that exists.
This proves that H(D,D) is being asked an >>>>>>>>>>>>>>>>>>>>>>> incorrect question.
Why, because it is NOT a LIE.
You don't even know the definiton of an incorrect >>>>>>>>>>>>>>>>>>>>>> question.
Nope, common technical term.
Cite a source.
The fact that there DOES exist a mapping Halt(M,d) >>>>>>>>>>>>>>>>>>>>>> that maps all Turing Machines and there input to a >>>>>>>>>>>>>>>>>>>>>> result of Halting / Non-Halting for EVERY member >>>>>>>>>>>>>>>>>>>>>> of that input set, means tha Halts is a valid >>>>>>>>>>>>>>>>>>>>>> mapping to ask a decider to try to decider. >>>>>>>>>>>>>>>>>>>>>>That part is true.
Likewise when you ask a man that has never been >>>>>>>>>>>>>>>>>>>>> married:
Have you stopped beating tour wife?
There are some men that have stopped beating their >>>>>>>>>>>>>>>>>>>>> wife.
Right, because that question include a presumption >>>>>>>>>>>>>>>>>>>> of something not actually present.
Although there is a mapping from some men to YES/NO >>>>>>>>>>>>>>>>>>> there is no mapping from never unmarried men to YES/NO >>>>>>>>>>>>>>>>>>> thus the question is incorrect for all unmarried men. >>>>>>>>>>>>>>>>>>>
Although there is a mapping from some TM/input pairs >>>>>>>>>>>>>>>>>>> to YES/NO
there is no mapping from H/D to YES/NO
thus the question is incorrect for H/D
Except that the mapping requested is about the INPUTS >>>>>>>>>>>>>>>>>> to H, not H itsef.
In order to see that it is an incorrect question we >>>>>>>>>>>>>>>>> must examine
the question in detail. Making sure to always ignore >>>>>>>>>>>>>>>>> this key detail
<is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to
⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to
⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) >>>>>>>>>>>>>>>>
The QUESTION is:
Does the machine and input described by this input, Halt >>>>>>>>>>>>>>>> when run?
The question posed to Ĥ.H has no correct answer, thus not >>>>>>>>>>>>>>> the
same question at all.
But it DOES.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct and make
sure that
you explain why this element is correct and don't try to >>>>>>>>>>>>> switch
to any other element outside of the above specified set. >>>>>>>>>>>>>
I didn't say there was.
Then you understand that each question posed to each Ĥ.H in the >>>>>>>>>>> above set has no correct answer only because each of these >>>>>>>>>>> answers
are contradicted by the machine that H is contained within. >>>>>>>>>>>
No, YOU don't understand that the IS a correct answer, just >>>>>>>>>> not the one that H (or H^.H ) happens to give.
Then show me which contradicted answer is correct.
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy and loops so >>>>>>>> qn was the right answer.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not
halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
*The answer must come from elements of the above set*
Is a false claim about a strawman deception really the best you
can say?
The above are the program/input pairs such that every Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets the wrong answer only because whatever answer that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets is contradicted.
So?
That doesn't mean they are the set that the answer to the ACTUAL
QUESTION needs to come from.
You are just proving your stupidity and duplicity.
Objective and Subjective Specifications
https://www.cs.toronto.edu/~hehner/OSS.pdf
Credit goes to you for finding the loophole in Carol's original
question: Can anyone correctly answer “no” to this question?
Carol can correctly answer that question with any word that is
synonymous with "no".
Here is the one where the loophole is closed:
Can Carol correctly answer “no” to this [yes/no] question?
The fact that anyone besides Carol can correctly answer that
question with a NO and Carol cannot possibly correctly answer
that question proves that it is a different question when posed
to Carol than when posed to anyone else.
Which is IRRELEVENT to the Halting Question, as it is a purely
objective question.
The behavior of the input is INDEPENDENT of the decider looking at it.
Note, a given H^ is built on a given H, and no other, but can be given
to any decider to answer, and the correct answer will be the same
irrespective of you ask. Some will give the right answer, and some
will give the wrong answer. The fact that that H is in the latter
doesn't make the question subjective.
The only way to make the Halting Question subjective is to try to
redefine it so the input changes with who you ask, but it doesn't.
The changing H^ to match the H only happens in the Meta, where we
prove that we can find an H^ that any H will get wrong, but each of
those are SEPERATE Halting question (not all one question) and each of
those seperate questions have a correct answer.
Nope.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
Carol's question posed to Carol <is> isomorphic to input ⟨Ĥ⟩ ⟨Ĥ⟩ >>> to every Ĥ.H shown above. The fact that some other TM such as H1
(that is not contradicted) can determine a correct answer proves
that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is a different question
The Question doesn't refer to H at all.
The input ⟨Ĥ⟩ ⟨Ĥ⟩ posed to Ĥ.H
is isomorphic to this question posed to Carol:
Can Carol correctly answer “no” to this [yes/no] question?
Thus changing the meaning of the question
Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H
*in the exact same way that the meaning of this question*
Can Carol correctly answer “no” to this [yes/no] question?
is changed when it is posed to Carol.
On 3/15/2024 6:02 PM, Richard Damon wrote:
On 3/15/24 3:47 PM, olcott wrote:
On 3/15/2024 5:13 PM, Richard Damon wrote:
On 3/15/24 1:42 PM, olcott wrote:
On 3/15/2024 3:35 PM, Richard Damon wrote:
On 3/15/24 12:50 PM, olcott wrote:
On 3/15/2024 2:42 PM, Richard Damon wrote:
On 3/15/24 12:23 PM, olcott wrote:
On 3/15/2024 2:14 PM, Richard Damon wrote:Nope. and that LIE is a source of a lot of your ERRORS.
On 3/15/24 12:00 PM, olcott wrote:
On 3/15/2024 1:38 PM, Richard Damon wrote:
On 3/15/24 7:41 AM, olcott wrote:
On 3/15/2024 5:44 AM, Mikko wrote:
On 2024-03-15 01:12:19 +0000, olcott said:
On 3/14/2024 8:06 PM, Richard Damon wrote:Is a false claim about a strawman deception really the >>>>>>>>>>>>>> best you can say?
On 3/14/24 4:45 PM, olcott wrote:*The strawman deception is all that you have*
On 3/14/2024 5:37 PM, Richard Damon wrote:
On 3/14/24 3:04 PM, olcott wrote:
On 3/14/2024 4:55 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/14/24 1:59 PM, olcott wrote:
On 3/14/2024 3:54 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/14/24 1:26 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 3:20 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 12:32 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 12:33 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 4:04 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 5:43 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 2:54 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 1:52 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 12:52 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 10:08 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 11:44 AM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 13/03/24 04:55, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/2024 10:49 PM, Richard Damon >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
I invented it so I get to stipulate its >>>>>>>>>>>>>>>>>>>>>>>>>>>>> meaning.Shows you are just a LIAR, as there IS a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> truth value to the expression that is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> the requirment for ANY SPECIFIC H. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>Wrong. You said. yourself. that H1 >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> gets the right answer for D. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Not quite. It always gets the wrong >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> answer, but only one of them for >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> each quesiton. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
They all gets the wrong answer on a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> whole class of questions >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Since it is a logical impossibility to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> determine the truth >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> value of a self-contradictory >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> expression the requirement >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> for H to do this is bogus. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*Lying about me being a liar may possibly >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> cost your soul*
*Lying about me being a liar may possibly >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> cost your soul*
*Lying about me being a liar may possibly >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> cost your soul*
There is no mapping from H(D,D) to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Halts(D,D) that exists. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> This proves that H(D,D) is being asked an >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> incorrect question. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Why, because it is NOT a LIE. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
You don't even know the definiton of an >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> incorrect question. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
Nope, common technical term. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
Cite a source.
The fact that there DOES exist a mapping >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Halt(M,d) that maps all Turing Machines >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> and there input to a result of Halting / >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Non-Halting for EVERY member of that input >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> set, means tha Halts is a valid mapping to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ask a decider to try to decider. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>That part is true.
Likewise when you ask a man that has never >>>>>>>>>>>>>>>>>>>>>>>>>>>>> been married:
Have you stopped beating tour wife? >>>>>>>>>>>>>>>>>>>>>>>>>>>>> There are some men that have stopped >>>>>>>>>>>>>>>>>>>>>>>>>>>>> beating their wife.
Right, because that question include a >>>>>>>>>>>>>>>>>>>>>>>>>>>> presumption of something not actually present. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
Although there is a mapping from some men to >>>>>>>>>>>>>>>>>>>>>>>>>>> YES/NO
there is no mapping from never unmarried men >>>>>>>>>>>>>>>>>>>>>>>>>>> to YES/NO
thus the question is incorrect for all >>>>>>>>>>>>>>>>>>>>>>>>>>> unmarried men.
Although there is a mapping from some >>>>>>>>>>>>>>>>>>>>>>>>>>> TM/input pairs to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>> there is no mapping from H/D to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>> thus the question is incorrect for H/D >>>>>>>>>>>>>>>>>>>>>>>>>>>
Except that the mapping requested is about the >>>>>>>>>>>>>>>>>>>>>>>>>> INPUTS to H, not H itsef.
In order to see that it is an incorrect >>>>>>>>>>>>>>>>>>>>>>>>> question we must examine
the question in detail. Making sure to always >>>>>>>>>>>>>>>>>>>>>>>>> ignore this key detail
<is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ
applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ
applied to ⟨Ĥ⟩ does not halt >>>>>>>>>>>>>>>>>>>>>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
Which isn;t the question at all, so you are just >>>>>>>>>>>>>>>>>>>>>>>> shown to be a stupid liar.
The QUESTION is:
Does the machine and input described by this >>>>>>>>>>>>>>>>>>>>>>>> input, Halt when run?
The question posed to Ĥ.H has no correct answer, >>>>>>>>>>>>>>>>>>>>>>> thus not the
same question at all.
But it DOES.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct
and make sure that
you explain why this element is correct and don't >>>>>>>>>>>>>>>>>>>>> try to switch
to any other element outside of the above specified >>>>>>>>>>>>>>>>>>>>> set.
I didn't say there was.
Then you understand that each question posed to each >>>>>>>>>>>>>>>>>>> Ĥ.H in the
above set has no correct answer only because each of >>>>>>>>>>>>>>>>>>> these answers
are contradicted by the machine that H is contained >>>>>>>>>>>>>>>>>>> within.
No, YOU don't understand that the IS a correct answer, >>>>>>>>>>>>>>>>>> just not the one that H (or H^.H ) happens to give. >>>>>>>>>>>>>>>>>>
Then show me which contradicted answer is correct. >>>>>>>>>>>>>>>>>
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy and >>>>>>>>>>>>>>>> loops so qn was the right answer.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩
halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩
does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) >>>>>>>>>>>>>>>
*The answer must come from elements of the above set* >>>>>>>>>>>>>>
The above are the program/input pairs such that every Ĥ.H >>>>>>>>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩
gets the wrong answer only because whatever answer that Ĥ.H >>>>>>>>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩
gets is contradicted.
So?
That doesn't mean they are the set that the answer to the >>>>>>>>>>>> ACTUAL QUESTION needs to come from.
You are just proving your stupidity and duplicity.
Objective and Subjective Specifications
https://www.cs.toronto.edu/~hehner/OSS.pdf
Credit goes to you for finding the loophole in Carol's original >>>>>>>>>>> question: Can anyone correctly answer “no” to this question? >>>>>>>>>>>
Carol can correctly answer that question with any word that is >>>>>>>>>>> synonymous with "no".
Here is the one where the loophole is closed:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>>>>
The fact that anyone besides Carol can correctly answer that >>>>>>>>>>> question with a NO and Carol cannot possibly correctly answer >>>>>>>>>>> that question proves that it is a different question when posed >>>>>>>>>>> to Carol than when posed to anyone else.
Which is IRRELEVENT to the Halting Question, as it is a purely >>>>>>>>>> objective question.
The behavior of the input is INDEPENDENT of the decider
looking at it.
Note, a given H^ is built on a given H, and no other, but can >>>>>>>>>> be given to any decider to answer, and the correct answer will >>>>>>>>>> be the same irrespective of you ask. Some will give the right >>>>>>>>>> answer, and some will give the wrong answer. The fact that >>>>>>>>>> that H is in the latter doesn't make the question subjective. >>>>>>>>>>
The only way to make the Halting Question subjective is to try >>>>>>>>>> to redefine it so the input changes with who you ask, but it >>>>>>>>>> doesn't.
The changing H^ to match the H only happens in the Meta, where >>>>>>>>>> we prove that we can find an H^ that any H will get wrong, but >>>>>>>>>> each of those are SEPERATE Halting question (not all one
question) and each of those seperate questions have a correct >>>>>>>>>> answer.
Nope.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does
not halt
Carol's question posed to Carol <is> isomorphic to input ⟨Ĥ⟩ ⟨Ĥ⟩
to every Ĥ.H shown above. The fact that some other TM such as H1 >>>>>>>>>>> (that is not contradicted) can determine a correct answer proves >>>>>>>>>>> that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is a different question
The Question doesn't refer to H at all.
The input ⟨Ĥ⟩ ⟨Ĥ⟩ posed to Ĥ.H
is isomorphic to this question posed to Carol:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>
Carol is a volitional being.
When we hypothesize that Carol is the name of an AI machine
everything remains the same.
Nope.
Once Carol become deterministic, then the whole thing changes.
The only reason that:
Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H
cannot be correctly answered is that the specific Ĥ.H is contradicted. >>>>>
The only reason that:
Can Carol correctly answer “no” to this [yes/no] question? posed to >>>>> Carol
cannot be correctly answered is that the specific Carol is
contradicted.
Nope.
You aren't showing any ERRORS I made but just asserting your FALSE
claims again.
Inability to show WHY my description was wrong just proves you have
no basis.
You are just demonstrating that you don't understand how logic works.
It seems you think this is just some abstract philosophy where
anything goes and rhetoric rules.
*You have provided zero correct reasoning of how*
*Carol's question posed to Carol*
*is not contradicted just like*
*Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H is contradicted*
Yes, I have.
YOU have provided ZERO reasoning how they are.
Dos H^ (H^) Halt? even when posed to H^.H has an answer!
When posed to each entity (Carol/Ĥ.H)
their respective question (a)/(b):
(a) Carol correctly answer “no” to this [yes/no] question?
(b) Does Ĥ ⟨Ĥ⟩ halt?
lacks a correct answer because this answer is contradicted.
*Incorrect questions do not lack answers they lack correct answers*
Carol could answer by flipping the bird.
On 3/15/2024 10:15 PM, immibis wrote:
On 16/03/24 00:17, olcott wrote:
On 3/15/2024 6:02 PM, Richard Damon wrote:
On 3/15/24 3:47 PM, olcott wrote:
On 3/15/2024 5:13 PM, Richard Damon wrote:
On 3/15/24 1:42 PM, olcott wrote:
On 3/15/2024 3:35 PM, Richard Damon wrote:
On 3/15/24 12:50 PM, olcott wrote:
On 3/15/2024 2:42 PM, Richard Damon wrote:
On 3/15/24 12:23 PM, olcott wrote:
On 3/15/2024 2:14 PM, Richard Damon wrote:Nope. and that LIE is a source of a lot of your ERRORS.
On 3/15/24 12:00 PM, olcott wrote:
On 3/15/2024 1:38 PM, Richard Damon wrote:
On 3/15/24 7:41 AM, olcott wrote:
On 3/15/2024 5:44 AM, Mikko wrote:
On 2024-03-15 01:12:19 +0000, olcott said:
On 3/14/2024 8:06 PM, Richard Damon wrote:Is a false claim about a strawman deception really the >>>>>>>>>>>>>>>> best you can say?
On 3/14/24 4:45 PM, olcott wrote:*The strawman deception is all that you have* >>>>>>>>>>>>>>>>>
On 3/14/2024 5:37 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/14/24 3:04 PM, olcott wrote:
On 3/14/2024 4:55 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/14/24 1:59 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 3:54 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 1:26 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 3:20 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 12:32 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 12:33 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 4:04 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 5:43 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 2:54 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 1:52 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 12:52 PM, Richard Damon >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
I invented it so I get to stipulate its >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> meaning.On 3/13/24 10:08 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 11:44 AM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 13/03/24 04:55, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/2024 10:49 PM, Richard >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Shows you are just a LIAR, as there IS >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> a truth value to the expression that >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> is the requirment for ANY SPECIFIC H. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>Wrong. You said. yourself. that H1 >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> gets the right answer for D. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>Not quite. It always gets the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrong answer, but only one of them >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> for each quesiton. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
They all gets the wrong answer on a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> whole class of questions >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Since it is a logical impossibility >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> to determine the truth >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> value of a self-contradictory >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> expression the requirement >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> for H to do this is bogus. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> *Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> *Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
There is no mapping from H(D,D) to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Halts(D,D) that exists. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> This proves that H(D,D) is being asked >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> an incorrect question. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Why, because it is NOT a LIE. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
You don't even know the definiton of an >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> incorrect question. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
Nope, common technical term. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Cite a source.
Right, because that question include a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> presumption of something not actually >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> present.
The fact that there DOES exist a mapping >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Halt(M,d) that maps all Turing Machines >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> and there input to a result of Halting / >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Non-Halting for EVERY member of that >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> input set, means tha Halts is a valid >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> mapping to ask a decider to try to decider. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>That part is true. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Likewise when you ask a man that has >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> never been married: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Have you stopped beating tour wife? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> There are some men that have stopped >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> beating their wife. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Although there is a mapping from some men >>>>>>>>>>>>>>>>>>>>>>>>>>>>> to YES/NO
there is no mapping from never unmarried >>>>>>>>>>>>>>>>>>>>>>>>>>>>> men to YES/NO
thus the question is incorrect for all >>>>>>>>>>>>>>>>>>>>>>>>>>>>> unmarried men.
Although there is a mapping from some >>>>>>>>>>>>>>>>>>>>>>>>>>>>> TM/input pairs to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>> there is no mapping from H/D to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>> thus the question is incorrect for H/D >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Except that the mapping requested is about >>>>>>>>>>>>>>>>>>>>>>>>>>>> the INPUTS to H, not H itsef. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
In order to see that it is an incorrect >>>>>>>>>>>>>>>>>>>>>>>>>>> question we must examine >>>>>>>>>>>>>>>>>>>>>>>>>>> the question in detail. Making sure to always >>>>>>>>>>>>>>>>>>>>>>>>>>> ignore this key detail
<is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ
applied to ⟨Ĥ⟩ halts >>>>>>>>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ
applied to ⟨Ĥ⟩ does not halt >>>>>>>>>>>>>>>>>>>>>>>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
Which isn;t the question at all, so you are >>>>>>>>>>>>>>>>>>>>>>>>>> just shown to be a stupid liar. >>>>>>>>>>>>>>>>>>>>>>>>>>
The QUESTION is:
Does the machine and input described by this >>>>>>>>>>>>>>>>>>>>>>>>>> input, Halt when run?
The question posed to Ĥ.H has no correct >>>>>>>>>>>>>>>>>>>>>>>>> answer, thus not the
same question at all.
But it DOES.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct
and make sure that
you explain why this element is correct and don't >>>>>>>>>>>>>>>>>>>>>>> try to switch
to any other element outside of the above >>>>>>>>>>>>>>>>>>>>>>> specified set.
I didn't say there was.
Then you understand that each question posed to >>>>>>>>>>>>>>>>>>>>> each Ĥ.H in the
above set has no correct answer only because each >>>>>>>>>>>>>>>>>>>>> of these answers
are contradicted by the machine that H is contained >>>>>>>>>>>>>>>>>>>>> within.
No, YOU don't understand that the IS a correct >>>>>>>>>>>>>>>>>>>> answer, just not the one that H (or H^.H ) happens >>>>>>>>>>>>>>>>>>>> to give.
Then show me which contradicted answer is correct. >>>>>>>>>>>>>>>>>>>
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy and >>>>>>>>>>>>>>>>>> loops so qn was the right answer.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to
⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to
⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) >>>>>>>>>>>>>>>>>
*The answer must come from elements of the above set* >>>>>>>>>>>>>>>>
The above are the program/input pairs such that every Ĥ.H >>>>>>>>>>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩
gets the wrong answer only because whatever answer that >>>>>>>>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets is contradicted.
So?
That doesn't mean they are the set that the answer to the >>>>>>>>>>>>>> ACTUAL QUESTION needs to come from.
You are just proving your stupidity and duplicity.
Objective and Subjective Specifications
https://www.cs.toronto.edu/~hehner/OSS.pdf
Credit goes to you for finding the loophole in Carol's >>>>>>>>>>>>> original
question: Can anyone correctly answer “no” to this question? >>>>>>>>>>>>>
Carol can correctly answer that question with any word that is >>>>>>>>>>>>> synonymous with "no".
Here is the one where the loophole is closed:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>>>>>>
The fact that anyone besides Carol can correctly answer that >>>>>>>>>>>>> question with a NO and Carol cannot possibly correctly answer >>>>>>>>>>>>> that question proves that it is a different question when >>>>>>>>>>>>> posed
to Carol than when posed to anyone else.
Which is IRRELEVENT to the Halting Question, as it is a >>>>>>>>>>>> purely objective question.
The behavior of the input is INDEPENDENT of the decider >>>>>>>>>>>> looking at it.
Note, a given H^ is built on a given H, and no other, but >>>>>>>>>>>> can be given to any decider to answer, and the correct >>>>>>>>>>>> answer will be the same irrespective of you ask. Some will >>>>>>>>>>>> give the right answer, and some will give the wrong answer. >>>>>>>>>>>> The fact that that H is in the latter doesn't make the >>>>>>>>>>>> question subjective.
The only way to make the Halting Question subjective is to >>>>>>>>>>>> try to redefine it so the input changes with who you ask, >>>>>>>>>>>> but it doesn't.
The changing H^ to match the H only happens in the Meta, >>>>>>>>>>>> where we prove that we can find an H^ that any H will get >>>>>>>>>>>> wrong, but each of those are SEPERATE Halting question (not >>>>>>>>>>>> all one question) and each of those seperate questions have >>>>>>>>>>>> a correct answer.
Nope.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩
does not halt
Carol's question posed to Carol <is> isomorphic to input >>>>>>>>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩
to every Ĥ.H shown above. The fact that some other TM such >>>>>>>>>>>>> as H1
(that is not contradicted) can determine a correct answer >>>>>>>>>>>>> proves
that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is a different question
The Question doesn't refer to H at all.
The input ⟨Ĥ⟩ ⟨Ĥ⟩ posed to Ĥ.H
is isomorphic to this question posed to Carol:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>>>
Carol is a volitional being.
When we hypothesize that Carol is the name of an AI machine
everything remains the same.
Nope.
Once Carol become deterministic, then the whole thing changes. >>>>>>>>
The only reason that:
Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H
cannot be correctly answered is that the specific Ĥ.H is
contradicted.
The only reason that:
Can Carol correctly answer “no” to this [yes/no] question? posed >>>>>>> to Carol
cannot be correctly answered is that the specific Carol is
contradicted.
Nope.
You aren't showing any ERRORS I made but just asserting your FALSE >>>>>> claims again.
Inability to show WHY my description was wrong just proves you
have no basis.
You are just demonstrating that you don't understand how logic works. >>>>>>
It seems you think this is just some abstract philosophy where
anything goes and rhetoric rules.
*You have provided zero correct reasoning of how*
*Carol's question posed to Carol*
*is not contradicted just like*
*Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H is contradicted*
Yes, I have.
YOU have provided ZERO reasoning how they are.
Dos H^ (H^) Halt? even when posed to H^.H has an answer!
When posed to each entity (Carol/Ĥ.H)
their respective question (a)/(b):
(a) Carol correctly answer “no” to this [yes/no] question?
(b) Does Ĥ ⟨Ĥ⟩ halt?
lacks a correct answer because this answer is contradicted.
(b) has a correct answer, which is "yes"
When Ĥ gives that answer it is contradicted by Ĥ,
thus it is the wrong answer.
On 3/15/2024 11:24 PM, immibis wrote:
On 16/03/24 04:52, olcott wrote:The possible answers that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gives are:
On 3/15/2024 10:15 PM, immibis wrote:
On 16/03/24 00:17, olcott wrote:
On 3/15/2024 6:02 PM, Richard Damon wrote:
On 3/15/24 3:47 PM, olcott wrote:
On 3/15/2024 5:13 PM, Richard Damon wrote:
On 3/15/24 1:42 PM, olcott wrote:
On 3/15/2024 3:35 PM, Richard Damon wrote:
On 3/15/24 12:50 PM, olcott wrote:
On 3/15/2024 2:42 PM, Richard Damon wrote:
On 3/15/24 12:23 PM, olcott wrote:
On 3/15/2024 2:14 PM, Richard Damon wrote:Nope. and that LIE is a source of a lot of your ERRORS. >>>>>>>>>>>>
On 3/15/24 12:00 PM, olcott wrote:
On 3/15/2024 1:38 PM, Richard Damon wrote:
On 3/15/24 7:41 AM, olcott wrote:Objective and Subjective Specifications
On 3/15/2024 5:44 AM, Mikko wrote:
On 2024-03-15 01:12:19 +0000, olcott said: >>>>>>>>>>>>>>>>>>
On 3/14/2024 8:06 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/14/24 4:45 PM, olcott wrote:Is a false claim about a strawman deception really the >>>>>>>>>>>>>>>>>> best you can say?
*The strawman deception is all that you have* >>>>>>>>>>>>>>>>>>>On 3/14/2024 5:37 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/14/24 3:04 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 4:55 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 1:59 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 3:54 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 1:26 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 3:20 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 12:32 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 12:33 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 4:04 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 5:43 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 2:54 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 1:52 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 12:52 PM, Richard Damon >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
I invented it so I get to stipulate its >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> meaning.On 3/13/24 10:08 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 11:44 AM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 13/03/24 04:55, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/2024 10:49 PM, Richard >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Not quite. It always gets the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrong answer, but only one of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> them for each quesiton. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Shows you are just a LIAR, as there >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> IS a truth value to the expression >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that is the requirment for ANY >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> SPECIFIC H. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>They all gets the wrong answer on >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> a whole class of questions >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>Wrong. You said. yourself. that H1 >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> gets the right answer for D. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Since it is a logical impossibility >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> to determine the truth >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> value of a self-contradictory >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> expression the requirement >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> for H to do this is bogus. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> *Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> *Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
There is no mapping from H(D,D) to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Halts(D,D) that exists. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> This proves that H(D,D) is being >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> asked an incorrect question. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Why, because it is NOT a LIE. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
You don't even know the definiton of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> an incorrect question. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
Nope, common technical term. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Cite a source.
Right, because that question include a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> presumption of something not actually >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> present.
The fact that there DOES exist a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> mapping Halt(M,d) that maps all Turing >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Machines and there input to a result >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> of Halting / Non-Halting for EVERY >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> member of that input set, means tha >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Halts is a valid mapping to ask a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> decider to try to decider. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>That part is true. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Likewise when you ask a man that has >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> never been married: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Have you stopped beating tour wife? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> There are some men that have stopped >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> beating their wife. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Although there is a mapping from some men >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> to YES/NO
there is no mapping from never unmarried >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> men to YES/NO
thus the question is incorrect for all >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unmarried men.
Although there is a mapping from some >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> TM/input pairs to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> there is no mapping from H/D to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> thus the question is incorrect for H/D >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Except that the mapping requested is about >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> the INPUTS to H, not H itsef. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
In order to see that it is an incorrect >>>>>>>>>>>>>>>>>>>>>>>>>>>>> question we must examine >>>>>>>>>>>>>>>>>>>>>>>>>>>>> the question in detail. Making sure to >>>>>>>>>>>>>>>>>>>>>>>>>>>>> always ignore this key detail >>>>>>>>>>>>>>>>>>>>>>>>>>>>> <is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ
applied to ⟨Ĥ⟩ halts >>>>>>>>>>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ
applied to ⟨Ĥ⟩ does not halt >>>>>>>>>>>>>>>>>>>>>>>>>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
Which isn;t the question at all, so you are >>>>>>>>>>>>>>>>>>>>>>>>>>>> just shown to be a stupid liar. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
The QUESTION is:
Does the machine and input described by this >>>>>>>>>>>>>>>>>>>>>>>>>>>> input, Halt when run?
The question posed to Ĥ.H has no correct >>>>>>>>>>>>>>>>>>>>>>>>>>> answer, thus not the
same question at all.
But it DOES.
Then tell me which element of: >>>>>>>>>>>>>>>>>>>>>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is
correct and make sure that
you explain why this element is correct and >>>>>>>>>>>>>>>>>>>>>>>>> don't try to switch
to any other element outside of the above >>>>>>>>>>>>>>>>>>>>>>>>> specified set.
I didn't say there was.
Then you understand that each question posed to >>>>>>>>>>>>>>>>>>>>>>> each Ĥ.H in the
above set has no correct answer only because each >>>>>>>>>>>>>>>>>>>>>>> of these answers
are contradicted by the machine that H is >>>>>>>>>>>>>>>>>>>>>>> contained within.
No, YOU don't understand that the IS a correct >>>>>>>>>>>>>>>>>>>>>> answer, just not the one that H (or H^.H ) happens >>>>>>>>>>>>>>>>>>>>>> to give.
Then show me which contradicted answer is correct. >>>>>>>>>>>>>>>>>>>>>
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy >>>>>>>>>>>>>>>>>>>> and loops so qn was the right answer.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to
⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to
⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
*The answer must come from elements of the above set* >>>>>>>>>>>>>>>>>>
The above are the program/input pairs such that every >>>>>>>>>>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets the wrong answer only because whatever answer that >>>>>>>>>>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets is contradicted.
So?
That doesn't mean they are the set that the answer to >>>>>>>>>>>>>>>> the ACTUAL QUESTION needs to come from.
You are just proving your stupidity and duplicity. >>>>>>>>>>>>>>>
https://www.cs.toronto.edu/~hehner/OSS.pdf
Credit goes to you for finding the loophole in Carol's >>>>>>>>>>>>>>> original
question: Can anyone correctly answer “no” to this question?
Carol can correctly answer that question with any word >>>>>>>>>>>>>>> that is
synonymous with "no".
Here is the one where the loophole is closed:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>>>>>>>>
The fact that anyone besides Carol can correctly answer that >>>>>>>>>>>>>>> question with a NO and Carol cannot possibly correctly >>>>>>>>>>>>>>> answer
that question proves that it is a different question when >>>>>>>>>>>>>>> posed
to Carol than when posed to anyone else.
Which is IRRELEVENT to the Halting Question, as it is a >>>>>>>>>>>>>> purely objective question.
The behavior of the input is INDEPENDENT of the decider >>>>>>>>>>>>>> looking at it.
Note, a given H^ is built on a given H, and no other, but >>>>>>>>>>>>>> can be given to any decider to answer, and the correct >>>>>>>>>>>>>> answer will be the same irrespective of you ask. Some will >>>>>>>>>>>>>> give the right answer, and some will give the wrong >>>>>>>>>>>>>> answer. The fact that that H is in the latter doesn't make >>>>>>>>>>>>>> the question subjective.
The only way to make the Halting Question subjective is to >>>>>>>>>>>>>> try to redefine it so the input changes with who you ask, >>>>>>>>>>>>>> but it doesn't.
The changing H^ to match the H only happens in the Meta, >>>>>>>>>>>>>> where we prove that we can find an H^ that any H will get >>>>>>>>>>>>>> wrong, but each of those are SEPERATE Halting question >>>>>>>>>>>>>> (not all one question) and each of those seperate
questions have a correct answer.
Nope.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩
halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩
does not halt
Carol's question posed to Carol <is> isomorphic to input >>>>>>>>>>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩
to every Ĥ.H shown above. The fact that some other TM >>>>>>>>>>>>>>> such as H1
(that is not contradicted) can determine a correct answer >>>>>>>>>>>>>>> proves
that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is a different question >>>>>>>>>>>>>>>
The Question doesn't refer to H at all.
The input ⟨Ĥ⟩ ⟨Ĥ⟩ posed to Ĥ.H
is isomorphic to this question posed to Carol:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>>>>>
Carol is a volitional being.
When we hypothesize that Carol is the name of an AI machine >>>>>>>>>>> everything remains the same.
Nope.
Once Carol become deterministic, then the whole thing changes. >>>>>>>>>>
The only reason that:
Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H
cannot be correctly answered is that the specific Ĥ.H is
contradicted.
The only reason that:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>> posed to Carol
cannot be correctly answered is that the specific Carol is
contradicted.
Nope.
You aren't showing any ERRORS I made but just asserting your
FALSE claims again.
Inability to show WHY my description was wrong just proves you >>>>>>>> have no basis.
You are just demonstrating that you don't understand how logic >>>>>>>> works.
It seems you think this is just some abstract philosophy where >>>>>>>> anything goes and rhetoric rules.
*You have provided zero correct reasoning of how*
*Carol's question posed to Carol*
*is not contradicted just like*
*Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H is contradicted*
Yes, I have.
YOU have provided ZERO reasoning how they are.
Dos H^ (H^) Halt? even when posed to H^.H has an answer!
When posed to each entity (Carol/Ĥ.H)
their respective question (a)/(b):
(a) Carol correctly answer “no” to this [yes/no] question?
(b) Does Ĥ halt?
lacks a correct answer because this answer is contradicted.
(b) has a correct answer, which is "yes"
When Ĥ gives that answer it is contradicted by Ĥ,
thus it is the wrong answer.
What does "When Ĥ gives that answer" mean?
(a) Ĥ.Hqy then loop (always does the opposite of what it says).
(b) Ĥ.Hqn then halt (always does the opposite of what it says).
Ĥ is a program which can only do what it is programmed, and it is
programmed to answer "no" even though the correct answer is "yes".
On 3/15/2024 11:24 PM, immibis wrote:
On 16/03/24 04:52, olcott wrote:The possible answers that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gives are:
On 3/15/2024 10:15 PM, immibis wrote:
On 16/03/24 00:17, olcott wrote:
On 3/15/2024 6:02 PM, Richard Damon wrote:
On 3/15/24 3:47 PM, olcott wrote:
On 3/15/2024 5:13 PM, Richard Damon wrote:
On 3/15/24 1:42 PM, olcott wrote:
On 3/15/2024 3:35 PM, Richard Damon wrote:
On 3/15/24 12:50 PM, olcott wrote:
On 3/15/2024 2:42 PM, Richard Damon wrote:
On 3/15/24 12:23 PM, olcott wrote:
On 3/15/2024 2:14 PM, Richard Damon wrote:Nope. and that LIE is a source of a lot of your ERRORS. >>>>>>>>>>>>
On 3/15/24 12:00 PM, olcott wrote:
On 3/15/2024 1:38 PM, Richard Damon wrote:
On 3/15/24 7:41 AM, olcott wrote:Objective and Subjective Specifications
On 3/15/2024 5:44 AM, Mikko wrote:
On 2024-03-15 01:12:19 +0000, olcott said: >>>>>>>>>>>>>>>>>>
On 3/14/2024 8:06 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/14/24 4:45 PM, olcott wrote:Is a false claim about a strawman deception really the >>>>>>>>>>>>>>>>>> best you can say?
*The strawman deception is all that you have* >>>>>>>>>>>>>>>>>>>On 3/14/2024 5:37 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/14/24 3:04 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 4:55 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 1:59 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 3:54 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 1:26 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 3:20 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 12:32 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 12:33 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 4:04 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 5:43 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 2:54 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 1:52 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 12:52 PM, Richard Damon >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
I invented it so I get to stipulate its >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> meaning.On 3/13/24 10:08 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 11:44 AM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 13/03/24 04:55, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/2024 10:49 PM, Richard >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Not quite. It always gets the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrong answer, but only one of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> them for each quesiton. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Shows you are just a LIAR, as there >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> IS a truth value to the expression >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> that is the requirment for ANY >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> SPECIFIC H. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>They all gets the wrong answer on >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> a whole class of questions >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>Wrong. You said. yourself. that H1 >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> gets the right answer for D. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Since it is a logical impossibility >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> to determine the truth >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> value of a self-contradictory >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> expression the requirement >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> for H to do this is bogus. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> *Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> *Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
There is no mapping from H(D,D) to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Halts(D,D) that exists. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> This proves that H(D,D) is being >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> asked an incorrect question. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Why, because it is NOT a LIE. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
You don't even know the definiton of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> an incorrect question. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
Nope, common technical term. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Cite a source.
Right, because that question include a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> presumption of something not actually >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> present.
The fact that there DOES exist a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> mapping Halt(M,d) that maps all Turing >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Machines and there input to a result >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> of Halting / Non-Halting for EVERY >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> member of that input set, means tha >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Halts is a valid mapping to ask a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> decider to try to decider. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>That part is true. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Likewise when you ask a man that has >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> never been married: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Have you stopped beating tour wife? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> There are some men that have stopped >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> beating their wife. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Although there is a mapping from some men >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> to YES/NO
there is no mapping from never unmarried >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> men to YES/NO
thus the question is incorrect for all >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unmarried men.
Although there is a mapping from some >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> TM/input pairs to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> there is no mapping from H/D to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> thus the question is incorrect for H/D >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Except that the mapping requested is about >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> the INPUTS to H, not H itsef. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
In order to see that it is an incorrect >>>>>>>>>>>>>>>>>>>>>>>>>>>>> question we must examine >>>>>>>>>>>>>>>>>>>>>>>>>>>>> the question in detail. Making sure to >>>>>>>>>>>>>>>>>>>>>>>>>>>>> always ignore this key detail >>>>>>>>>>>>>>>>>>>>>>>>>>>>> <is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ
applied to ⟨Ĥ⟩ halts >>>>>>>>>>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ
applied to ⟨Ĥ⟩ does not halt >>>>>>>>>>>>>>>>>>>>>>>>>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
Which isn;t the question at all, so you are >>>>>>>>>>>>>>>>>>>>>>>>>>>> just shown to be a stupid liar. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
The QUESTION is:
Does the machine and input described by this >>>>>>>>>>>>>>>>>>>>>>>>>>>> input, Halt when run?
The question posed to Ĥ.H has no correct >>>>>>>>>>>>>>>>>>>>>>>>>>> answer, thus not the
same question at all.
But it DOES.
Then tell me which element of: >>>>>>>>>>>>>>>>>>>>>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is
correct and make sure that
you explain why this element is correct and >>>>>>>>>>>>>>>>>>>>>>>>> don't try to switch
to any other element outside of the above >>>>>>>>>>>>>>>>>>>>>>>>> specified set.
I didn't say there was.
Then you understand that each question posed to >>>>>>>>>>>>>>>>>>>>>>> each Ĥ.H in the
above set has no correct answer only because each >>>>>>>>>>>>>>>>>>>>>>> of these answers
are contradicted by the machine that H is >>>>>>>>>>>>>>>>>>>>>>> contained within.
No, YOU don't understand that the IS a correct >>>>>>>>>>>>>>>>>>>>>> answer, just not the one that H (or H^.H ) happens >>>>>>>>>>>>>>>>>>>>>> to give.
Then show me which contradicted answer is correct. >>>>>>>>>>>>>>>>>>>>>
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy >>>>>>>>>>>>>>>>>>>> and loops so qn was the right answer.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to
⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to
⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
*The answer must come from elements of the above set* >>>>>>>>>>>>>>>>>>
The above are the program/input pairs such that every >>>>>>>>>>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets the wrong answer only because whatever answer that >>>>>>>>>>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets is contradicted.
So?
That doesn't mean they are the set that the answer to >>>>>>>>>>>>>>>> the ACTUAL QUESTION needs to come from.
You are just proving your stupidity and duplicity. >>>>>>>>>>>>>>>
https://www.cs.toronto.edu/~hehner/OSS.pdf
Credit goes to you for finding the loophole in Carol's >>>>>>>>>>>>>>> original
question: Can anyone correctly answer “no” to this question?
Carol can correctly answer that question with any word >>>>>>>>>>>>>>> that is
synonymous with "no".
Here is the one where the loophole is closed:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>>>>>>>>
The fact that anyone besides Carol can correctly answer that >>>>>>>>>>>>>>> question with a NO and Carol cannot possibly correctly >>>>>>>>>>>>>>> answer
that question proves that it is a different question when >>>>>>>>>>>>>>> posed
to Carol than when posed to anyone else.
Which is IRRELEVENT to the Halting Question, as it is a >>>>>>>>>>>>>> purely objective question.
The behavior of the input is INDEPENDENT of the decider >>>>>>>>>>>>>> looking at it.
Note, a given H^ is built on a given H, and no other, but >>>>>>>>>>>>>> can be given to any decider to answer, and the correct >>>>>>>>>>>>>> answer will be the same irrespective of you ask. Some will >>>>>>>>>>>>>> give the right answer, and some will give the wrong >>>>>>>>>>>>>> answer. The fact that that H is in the latter doesn't make >>>>>>>>>>>>>> the question subjective.
The only way to make the Halting Question subjective is to >>>>>>>>>>>>>> try to redefine it so the input changes with who you ask, >>>>>>>>>>>>>> but it doesn't.
The changing H^ to match the H only happens in the Meta, >>>>>>>>>>>>>> where we prove that we can find an H^ that any H will get >>>>>>>>>>>>>> wrong, but each of those are SEPERATE Halting question >>>>>>>>>>>>>> (not all one question) and each of those seperate
questions have a correct answer.
Nope.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩
halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩
does not halt
Carol's question posed to Carol <is> isomorphic to input >>>>>>>>>>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩
to every Ĥ.H shown above. The fact that some other TM >>>>>>>>>>>>>>> such as H1
(that is not contradicted) can determine a correct answer >>>>>>>>>>>>>>> proves
that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is a different question >>>>>>>>>>>>>>>
The Question doesn't refer to H at all.
The input ⟨Ĥ⟩ ⟨Ĥ⟩ posed to Ĥ.H
is isomorphic to this question posed to Carol:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>>>>>
Carol is a volitional being.
When we hypothesize that Carol is the name of an AI machine >>>>>>>>>>> everything remains the same.
Nope.
Once Carol become deterministic, then the whole thing changes. >>>>>>>>>>
The only reason that:
Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H
cannot be correctly answered is that the specific Ĥ.H is
contradicted.
The only reason that:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>> posed to Carol
cannot be correctly answered is that the specific Carol is
contradicted.
Nope.
You aren't showing any ERRORS I made but just asserting your
FALSE claims again.
Inability to show WHY my description was wrong just proves you >>>>>>>> have no basis.
You are just demonstrating that you don't understand how logic >>>>>>>> works.
It seems you think this is just some abstract philosophy where >>>>>>>> anything goes and rhetoric rules.
*You have provided zero correct reasoning of how*
*Carol's question posed to Carol*
*is not contradicted just like*
*Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H is contradicted*
Yes, I have.
YOU have provided ZERO reasoning how they are.
Dos H^ (H^) Halt? even when posed to H^.H has an answer!
When posed to each entity (Carol/Ĥ.H)
their respective question (a)/(b):
(a) Carol correctly answer “no” to this [yes/no] question?
(b) Does Ĥ halt?
lacks a correct answer because this answer is contradicted.
(b) has a correct answer, which is "yes"
When Ĥ gives that answer it is contradicted by Ĥ,
thus it is the wrong answer.
What does "When Ĥ gives that answer" mean?
(a) Ĥ.Hqy then loop (always does the opposite of what it says).
(b) Ĥ.Hqn then halt (always does the opposite of what it says).
Ĥ is a program which can only do what it is programmed, and it is
programmed to answer "no" even though the correct answer is "yes".
On 3/16/2024 11:33 AM, immibis wrote:
On 16/03/24 06:21, olcott wrote:
On 3/15/2024 11:24 PM, immibis wrote:
On 16/03/24 04:52, olcott wrote:The possible answers that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gives are:
On 3/15/2024 10:15 PM, immibis wrote:
On 16/03/24 00:17, olcott wrote:
On 3/15/2024 6:02 PM, Richard Damon wrote:
On 3/15/24 3:47 PM, olcott wrote:
On 3/15/2024 5:13 PM, Richard Damon wrote:
On 3/15/24 1:42 PM, olcott wrote:
On 3/15/2024 3:35 PM, Richard Damon wrote:
On 3/15/24 12:50 PM, olcott wrote:
On 3/15/2024 2:42 PM, Richard Damon wrote:
On 3/15/24 12:23 PM, olcott wrote:
On 3/15/2024 2:14 PM, Richard Damon wrote:Nope. and that LIE is a source of a lot of your ERRORS. >>>>>>>>>>>>>>
On 3/15/24 12:00 PM, olcott wrote:
On 3/15/2024 1:38 PM, Richard Damon wrote:
On 3/15/24 7:41 AM, olcott wrote:Objective and Subjective Specifications
On 3/15/2024 5:44 AM, Mikko wrote:
On 2024-03-15 01:12:19 +0000, olcott said: >>>>>>>>>>>>>>>>>>>>
On 3/14/2024 8:06 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/14/24 4:45 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 5:37 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 3:04 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 4:55 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 1:59 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 3:54 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 1:26 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 3:20 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 12:32 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 12:33 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 4:04 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 5:43 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 2:54 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 4:39 PM, Richard Damon >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:Is a false claim about a strawman deception really >>>>>>>>>>>>>>>>>>>> the best you can say?
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ appliedBut it DOES.The question posed to Ĥ.H has no correct >>>>>>>>>>>>>>>>>>>>>>>>>>>>> answer, thus not the >>>>>>>>>>>>>>>>>>>>>>>>>>>>> same question at all. >>>>>>>>>>>>>>>>>>>>>>>>>>>>On 3/13/24 1:52 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 12:52 PM, Richard >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 10:08 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 11:44 AM, immibis >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 13/03/24 04:55, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/2024 10:49 PM, Richard >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Not quite. It always gets the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrong answer, but only one of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> them for each quesiton. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> They all gets the wrong answer >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> on a whole class of questions >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Wrong. You said. yourself. that >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> H1 gets the right answer for D. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>I invented it so I get to stipulate >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> its meaning. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
Since it is a logical >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> impossibility to determine the truth >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> value of a self-contradictory >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> expression the requirement >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> for H to do this is bogus. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>Shows you are just a LIAR, as >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> there IS a truth value to the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> expression that is the requirment >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> for ANY SPECIFIC H. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> *Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> *Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
There is no mapping from H(D,D) to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Halts(D,D) that exists. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> This proves that H(D,D) is being >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> asked an incorrect question. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Why, because it is NOT a LIE. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
You don't even know the definiton of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> an incorrect question. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Nope, common technical term. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Cite a source. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Right, because that question include a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> presumption of something not actually >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> present.
The fact that there DOES exist a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> mapping Halt(M,d) that maps all >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Turing Machines and there input to a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> result of Halting / Non-Halting for >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> EVERY member of that input set, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> means tha Halts is a valid mapping >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> to ask a decider to try to decider. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>That part is true. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Likewise when you ask a man that has >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> never been married: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Have you stopped beating tour wife? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> There are some men that have stopped >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> beating their wife. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Although there is a mapping from some >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> men to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> there is no mapping from never >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unmarried men to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> thus the question is incorrect for all >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unmarried men. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Although there is a mapping from some >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> TM/input pairs to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> there is no mapping from H/D to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> thus the question is incorrect for H/D >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Except that the mapping requested is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> about the INPUTS to H, not H itsef. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
In order to see that it is an incorrect >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> question we must examine >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> the question in detail. Making sure to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> always ignore this key detail >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> <is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ
applied to ⟨Ĥ⟩ halts >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ
applied to ⟨Ĥ⟩ does not halt >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
Which isn;t the question at all, so you >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> are just shown to be a stupid liar. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
The QUESTION is:
Does the machine and input described by >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> this input, Halt when run? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Then tell me which element of: >>>>>>>>>>>>>>>>>>>>>>>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is
correct and make sure that >>>>>>>>>>>>>>>>>>>>>>>>>>> you explain why this element is correct and >>>>>>>>>>>>>>>>>>>>>>>>>>> don't try to switch
to any other element outside of the above >>>>>>>>>>>>>>>>>>>>>>>>>>> specified set.
I didn't say there was.
Then you understand that each question posed to >>>>>>>>>>>>>>>>>>>>>>>>> each Ĥ.H in the
above set has no correct answer only because >>>>>>>>>>>>>>>>>>>>>>>>> each of these answers
are contradicted by the machine that H is >>>>>>>>>>>>>>>>>>>>>>>>> contained within.
No, YOU don't understand that the IS a correct >>>>>>>>>>>>>>>>>>>>>>>> answer, just not the one that H (or H^.H ) >>>>>>>>>>>>>>>>>>>>>>>> happens to give.
Then show me which contradicted answer is correct. >>>>>>>>>>>>>>>>>>>>>>>
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy >>>>>>>>>>>>>>>>>>>>>> and loops so qn was the right answer. >>>>>>>>>>>>>>>>>>>>> *The strawman deception is all that you have* >>>>>>>>>>>>>>>>>>>>>
to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied
to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
*The answer must come from elements of the above set* >>>>>>>>>>>>>>>>>>>>
The above are the program/input pairs such that every >>>>>>>>>>>>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets the wrong answer only because whatever answer >>>>>>>>>>>>>>>>>>> that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets is contradicted.
So?
That doesn't mean they are the set that the answer to >>>>>>>>>>>>>>>>>> the ACTUAL QUESTION needs to come from.
You are just proving your stupidity and duplicity. >>>>>>>>>>>>>>>>>
https://www.cs.toronto.edu/~hehner/OSS.pdf
Credit goes to you for finding the loophole in Carol's >>>>>>>>>>>>>>>>> original
question: Can anyone correctly answer “no” to this >>>>>>>>>>>>>>>>> question?
Carol can correctly answer that question with any word >>>>>>>>>>>>>>>>> that is
synonymous with "no".
Here is the one where the loophole is closed: >>>>>>>>>>>>>>>>> Can Carol correctly answer “no” to this [yes/no] question?
The fact that anyone besides Carol can correctly answer >>>>>>>>>>>>>>>>> that
question with a NO and Carol cannot possibly correctly >>>>>>>>>>>>>>>>> answer
that question proves that it is a different question >>>>>>>>>>>>>>>>> when posed
to Carol than when posed to anyone else.
Which is IRRELEVENT to the Halting Question, as it is a >>>>>>>>>>>>>>>> purely objective question.
The behavior of the input is INDEPENDENT of the decider >>>>>>>>>>>>>>>> looking at it.
Note, a given H^ is built on a given H, and no other, >>>>>>>>>>>>>>>> but can be given to any decider to answer, and the >>>>>>>>>>>>>>>> correct answer will be the same irrespective of you ask. >>>>>>>>>>>>>>>> Some will give the right answer, and some will give the >>>>>>>>>>>>>>>> wrong answer. The fact that that H is in the latter >>>>>>>>>>>>>>>> doesn't make the question subjective.
The only way to make the Halting Question subjective is >>>>>>>>>>>>>>>> to try to redefine it so the input changes with who you >>>>>>>>>>>>>>>> ask, but it doesn't.
The changing H^ to match the H only happens in the Meta, >>>>>>>>>>>>>>>> where we prove that we can find an H^ that any H will >>>>>>>>>>>>>>>> get wrong, but each of those are SEPERATE Halting >>>>>>>>>>>>>>>> question (not all one question) and each of those >>>>>>>>>>>>>>>> seperate questions have a correct answer.
Nope.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to
⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to
⟨Ĥ⟩ does not halt
Carol's question posed to Carol <is> isomorphic to >>>>>>>>>>>>>>>>> input ⟨Ĥ⟩ ⟨Ĥ⟩
to every Ĥ.H shown above. The fact that some other TM >>>>>>>>>>>>>>>>> such as H1
(that is not contradicted) can determine a correct >>>>>>>>>>>>>>>>> answer proves
that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is a different question >>>>>>>>>>>>>>>>>
The Question doesn't refer to H at all.
The input ⟨Ĥ⟩ ⟨Ĥ⟩ posed to Ĥ.H
is isomorphic to this question posed to Carol:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>>>>>>>
Carol is a volitional being.
When we hypothesize that Carol is the name of an AI machine >>>>>>>>>>>>> everything remains the same.
Nope.
Once Carol become deterministic, then the whole thing changes. >>>>>>>>>>>>
The only reason that:
Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H
cannot be correctly answered is that the specific Ĥ.H is >>>>>>>>>>> contradicted.
The only reason that:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>>>> posed to Carol
cannot be correctly answered is that the specific Carol is >>>>>>>>>>> contradicted.
Nope.
You aren't showing any ERRORS I made but just asserting your >>>>>>>>>> FALSE claims again.
Inability to show WHY my description was wrong just proves you >>>>>>>>>> have no basis.
You are just demonstrating that you don't understand how logic >>>>>>>>>> works.
It seems you think this is just some abstract philosophy where >>>>>>>>>> anything goes and rhetoric rules.
*You have provided zero correct reasoning of how*
*Carol's question posed to Carol*
*is not contradicted just like*
*Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H is contradicted*
Yes, I have.
YOU have provided ZERO reasoning how they are.
Dos H^ (H^) Halt? even when posed to H^.H has an answer!
When posed to each entity (Carol/Ĥ.H)
their respective question (a)/(b):
(a) Carol correctly answer “no” to this [yes/no] question?
(b) Does Ĥ halt?
lacks a correct answer because this answer is contradicted.
(b) has a correct answer, which is "yes"
When Ĥ gives that answer it is contradicted by Ĥ,
thus it is the wrong answer.
What does "When Ĥ gives that answer" mean?
(a) Ĥ.Hqy then loop (always does the opposite of what it says).
(b) Ĥ.Hqn then halt (always does the opposite of what it says).
No, only one answer is possible, which is the one that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is
programmed to give.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
Because every Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is contradicted.
Ĥ is a program which can only do what it is programmed, and it is
programmed to answer "no" even though the correct answer is "yes".
On 3/16/2024 10:37 AM, Richard Damon wrote:
On 3/15/24 10:21 PM, olcott wrote:
On 3/15/2024 11:24 PM, immibis wrote:
On 16/03/24 04:52, olcott wrote:The possible answers that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gives are:
On 3/15/2024 10:15 PM, immibis wrote:
On 16/03/24 00:17, olcott wrote:
On 3/15/2024 6:02 PM, Richard Damon wrote:
On 3/15/24 3:47 PM, olcott wrote:
On 3/15/2024 5:13 PM, Richard Damon wrote:
On 3/15/24 1:42 PM, olcott wrote:
On 3/15/2024 3:35 PM, Richard Damon wrote:
On 3/15/24 12:50 PM, olcott wrote:
On 3/15/2024 2:42 PM, Richard Damon wrote:
On 3/15/24 12:23 PM, olcott wrote:
On 3/15/2024 2:14 PM, Richard Damon wrote:Nope. and that LIE is a source of a lot of your ERRORS. >>>>>>>>>>>>>>
On 3/15/24 12:00 PM, olcott wrote:
On 3/15/2024 1:38 PM, Richard Damon wrote:
On 3/15/24 7:41 AM, olcott wrote:Objective and Subjective Specifications
On 3/15/2024 5:44 AM, Mikko wrote:
On 2024-03-15 01:12:19 +0000, olcott said: >>>>>>>>>>>>>>>>>>>>
On 3/14/2024 8:06 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/14/24 4:45 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 5:37 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 3:04 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 4:55 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 1:59 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 3:54 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 1:26 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 3:20 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 12:32 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 12:33 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 4:04 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 5:43 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 2:54 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 4:39 PM, Richard Damon >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:Is a false claim about a strawman deception really >>>>>>>>>>>>>>>>>>>> the best you can say?
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ appliedBut it DOES.The question posed to Ĥ.H has no correct >>>>>>>>>>>>>>>>>>>>>>>>>>>>> answer, thus not the >>>>>>>>>>>>>>>>>>>>>>>>>>>>> same question at all. >>>>>>>>>>>>>>>>>>>>>>>>>>>>On 3/13/24 1:52 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 12:52 PM, Richard >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 10:08 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 11:44 AM, immibis >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 13/03/24 04:55, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/2024 10:49 PM, Richard >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Not quite. It always gets the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrong answer, but only one of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> them for each quesiton. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> They all gets the wrong answer >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> on a whole class of questions >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Wrong. You said. yourself. that >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> H1 gets the right answer for D. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>I invented it so I get to stipulate >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> its meaning. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
Since it is a logical >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> impossibility to determine the truth >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> value of a self-contradictory >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> expression the requirement >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> for H to do this is bogus. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>Shows you are just a LIAR, as >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> there IS a truth value to the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> expression that is the requirment >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> for ANY SPECIFIC H. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> *Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> *Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
There is no mapping from H(D,D) to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Halts(D,D) that exists. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> This proves that H(D,D) is being >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> asked an incorrect question. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Why, because it is NOT a LIE. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
You don't even know the definiton of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> an incorrect question. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Nope, common technical term. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Cite a source. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Right, because that question include a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> presumption of something not actually >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> present.
The fact that there DOES exist a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> mapping Halt(M,d) that maps all >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Turing Machines and there input to a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> result of Halting / Non-Halting for >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> EVERY member of that input set, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> means tha Halts is a valid mapping >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> to ask a decider to try to decider. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>That part is true. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Likewise when you ask a man that has >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> never been married: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Have you stopped beating tour wife? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> There are some men that have stopped >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> beating their wife. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Although there is a mapping from some >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> men to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> there is no mapping from never >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unmarried men to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> thus the question is incorrect for all >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unmarried men. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Although there is a mapping from some >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> TM/input pairs to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> there is no mapping from H/D to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> thus the question is incorrect for H/D >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Except that the mapping requested is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> about the INPUTS to H, not H itsef. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
In order to see that it is an incorrect >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> question we must examine >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> the question in detail. Making sure to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> always ignore this key detail >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> <is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ
applied to ⟨Ĥ⟩ halts >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ
applied to ⟨Ĥ⟩ does not halt >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
Which isn;t the question at all, so you >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> are just shown to be a stupid liar. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
The QUESTION is:
Does the machine and input described by >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> this input, Halt when run? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Then tell me which element of: >>>>>>>>>>>>>>>>>>>>>>>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is
correct and make sure that >>>>>>>>>>>>>>>>>>>>>>>>>>> you explain why this element is correct and >>>>>>>>>>>>>>>>>>>>>>>>>>> don't try to switch
to any other element outside of the above >>>>>>>>>>>>>>>>>>>>>>>>>>> specified set.
I didn't say there was.
Then you understand that each question posed to >>>>>>>>>>>>>>>>>>>>>>>>> each Ĥ.H in the
above set has no correct answer only because >>>>>>>>>>>>>>>>>>>>>>>>> each of these answers
are contradicted by the machine that H is >>>>>>>>>>>>>>>>>>>>>>>>> contained within.
No, YOU don't understand that the IS a correct >>>>>>>>>>>>>>>>>>>>>>>> answer, just not the one that H (or H^.H ) >>>>>>>>>>>>>>>>>>>>>>>> happens to give.
Then show me which contradicted answer is correct. >>>>>>>>>>>>>>>>>>>>>>>
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy >>>>>>>>>>>>>>>>>>>>>> and loops so qn was the right answer. >>>>>>>>>>>>>>>>>>>>> *The strawman deception is all that you have* >>>>>>>>>>>>>>>>>>>>>
to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied
to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
*The answer must come from elements of the above set* >>>>>>>>>>>>>>>>>>>>
The above are the program/input pairs such that every >>>>>>>>>>>>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets the wrong answer only because whatever answer >>>>>>>>>>>>>>>>>>> that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets is contradicted.
So?
That doesn't mean they are the set that the answer to >>>>>>>>>>>>>>>>>> the ACTUAL QUESTION needs to come from.
You are just proving your stupidity and duplicity. >>>>>>>>>>>>>>>>>
https://www.cs.toronto.edu/~hehner/OSS.pdf
Credit goes to you for finding the loophole in Carol's >>>>>>>>>>>>>>>>> original
question: Can anyone correctly answer “no” to this >>>>>>>>>>>>>>>>> question?
Carol can correctly answer that question with any word >>>>>>>>>>>>>>>>> that is
synonymous with "no".
Here is the one where the loophole is closed: >>>>>>>>>>>>>>>>> Can Carol correctly answer “no” to this [yes/no] question?
The fact that anyone besides Carol can correctly answer >>>>>>>>>>>>>>>>> that
question with a NO and Carol cannot possibly correctly >>>>>>>>>>>>>>>>> answer
that question proves that it is a different question >>>>>>>>>>>>>>>>> when posed
to Carol than when posed to anyone else.
Which is IRRELEVENT to the Halting Question, as it is a >>>>>>>>>>>>>>>> purely objective question.
The behavior of the input is INDEPENDENT of the decider >>>>>>>>>>>>>>>> looking at it.
Note, a given H^ is built on a given H, and no other, >>>>>>>>>>>>>>>> but can be given to any decider to answer, and the >>>>>>>>>>>>>>>> correct answer will be the same irrespective of you ask. >>>>>>>>>>>>>>>> Some will give the right answer, and some will give the >>>>>>>>>>>>>>>> wrong answer. The fact that that H is in the latter >>>>>>>>>>>>>>>> doesn't make the question subjective.
The only way to make the Halting Question subjective is >>>>>>>>>>>>>>>> to try to redefine it so the input changes with who you >>>>>>>>>>>>>>>> ask, but it doesn't.
The changing H^ to match the H only happens in the Meta, >>>>>>>>>>>>>>>> where we prove that we can find an H^ that any H will >>>>>>>>>>>>>>>> get wrong, but each of those are SEPERATE Halting >>>>>>>>>>>>>>>> question (not all one question) and each of those >>>>>>>>>>>>>>>> seperate questions have a correct answer.
Nope.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to
⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to
⟨Ĥ⟩ does not halt
Carol's question posed to Carol <is> isomorphic to >>>>>>>>>>>>>>>>> input ⟨Ĥ⟩ ⟨Ĥ⟩
to every Ĥ.H shown above. The fact that some other TM >>>>>>>>>>>>>>>>> such as H1
(that is not contradicted) can determine a correct >>>>>>>>>>>>>>>>> answer proves
that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is a different question >>>>>>>>>>>>>>>>>
The Question doesn't refer to H at all.
The input ⟨Ĥ⟩ ⟨Ĥ⟩ posed to Ĥ.H
is isomorphic to this question posed to Carol:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>>>>>>>
Carol is a volitional being.
When we hypothesize that Carol is the name of an AI machine >>>>>>>>>>>>> everything remains the same.
Nope.
Once Carol become deterministic, then the whole thing changes. >>>>>>>>>>>>
The only reason that:
Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H
cannot be correctly answered is that the specific Ĥ.H is >>>>>>>>>>> contradicted.
The only reason that:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>>>> posed to Carol
cannot be correctly answered is that the specific Carol is >>>>>>>>>>> contradicted.
Nope.
You aren't showing any ERRORS I made but just asserting your >>>>>>>>>> FALSE claims again.
Inability to show WHY my description was wrong just proves you >>>>>>>>>> have no basis.
You are just demonstrating that you don't understand how logic >>>>>>>>>> works.
It seems you think this is just some abstract philosophy where >>>>>>>>>> anything goes and rhetoric rules.
*You have provided zero correct reasoning of how*
*Carol's question posed to Carol*
*is not contradicted just like*
*Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H is contradicted*
Yes, I have.
YOU have provided ZERO reasoning how they are.
Dos H^ (H^) Halt? even when posed to H^.H has an answer!
When posed to each entity (Carol/Ĥ.H)
their respective question (a)/(b):
(a) Carol correctly answer “no” to this [yes/no] question?
(b) Does Ĥ halt?
lacks a correct answer because this answer is contradicted.
(b) has a correct answer, which is "yes"
When Ĥ gives that answer it is contradicted by Ĥ,
thus it is the wrong answer.
What does "When Ĥ gives that answer" mean?
(a) Ĥ.Hqy then loop (always does the opposite of what it says).
(b) Ĥ.Hqn then halt (always does the opposite of what it says).
Ĥ is a program which can only do what it is programmed, and it is
programmed to answer "no" even though the correct answer is "yes".
Nut (a) isn't AN ANSWER, as it isn't given to any machine that uses it.
You don't seem to understand what answer is.
And, H^ is not defined to apply any semantic to its return, so you
can't assume any.
H is defined to give an answer, but all H's will give the wrong answer
for the H^ built from it.
*Only because every answer that they give is contradicted*
On 3/16/2024 10:37 AM, Richard Damon wrote:
On 3/15/24 10:21 PM, olcott wrote:
On 3/15/2024 11:24 PM, immibis wrote:
On 16/03/24 04:52, olcott wrote:The possible answers that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gives are:
On 3/15/2024 10:15 PM, immibis wrote:
On 16/03/24 00:17, olcott wrote:
On 3/15/2024 6:02 PM, Richard Damon wrote:
On 3/15/24 3:47 PM, olcott wrote:
On 3/15/2024 5:13 PM, Richard Damon wrote:
On 3/15/24 1:42 PM, olcott wrote:
On 3/15/2024 3:35 PM, Richard Damon wrote:
On 3/15/24 12:50 PM, olcott wrote:
On 3/15/2024 2:42 PM, Richard Damon wrote:
On 3/15/24 12:23 PM, olcott wrote:
On 3/15/2024 2:14 PM, Richard Damon wrote:Nope. and that LIE is a source of a lot of your ERRORS. >>>>>>>>>>>>>>
On 3/15/24 12:00 PM, olcott wrote:
On 3/15/2024 1:38 PM, Richard Damon wrote:
On 3/15/24 7:41 AM, olcott wrote:Objective and Subjective Specifications
On 3/15/2024 5:44 AM, Mikko wrote:
On 2024-03-15 01:12:19 +0000, olcott said: >>>>>>>>>>>>>>>>>>>>
On 3/14/2024 8:06 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/14/24 4:45 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 5:37 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 3:04 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 4:55 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 1:59 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 3:54 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 1:26 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 3:20 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 12:32 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 12:33 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 4:04 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 5:43 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 2:54 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 4:39 PM, Richard Damon >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:Is a false claim about a strawman deception really >>>>>>>>>>>>>>>>>>>> the best you can say?
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ appliedBut it DOES.The question posed to Ĥ.H has no correct >>>>>>>>>>>>>>>>>>>>>>>>>>>>> answer, thus not the >>>>>>>>>>>>>>>>>>>>>>>>>>>>> same question at all. >>>>>>>>>>>>>>>>>>>>>>>>>>>>On 3/13/24 1:52 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 12:52 PM, Richard >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 10:08 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 11:44 AM, immibis >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 13/03/24 04:55, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/2024 10:49 PM, Richard >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Not quite. It always gets the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrong answer, but only one of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> them for each quesiton. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> They all gets the wrong answer >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> on a whole class of questions >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Wrong. You said. yourself. that >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> H1 gets the right answer for D. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>I invented it so I get to stipulate >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> its meaning. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
Since it is a logical >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> impossibility to determine the truth >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> value of a self-contradictory >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> expression the requirement >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> for H to do this is bogus. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>Shows you are just a LIAR, as >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> there IS a truth value to the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> expression that is the requirment >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> for ANY SPECIFIC H. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> *Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> *Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
There is no mapping from H(D,D) to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Halts(D,D) that exists. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> This proves that H(D,D) is being >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> asked an incorrect question. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Why, because it is NOT a LIE. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
You don't even know the definiton of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> an incorrect question. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Nope, common technical term. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Cite a source. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Right, because that question include a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> presumption of something not actually >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> present.
The fact that there DOES exist a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> mapping Halt(M,d) that maps all >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Turing Machines and there input to a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> result of Halting / Non-Halting for >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> EVERY member of that input set, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> means tha Halts is a valid mapping >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> to ask a decider to try to decider. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>That part is true. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Likewise when you ask a man that has >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> never been married: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Have you stopped beating tour wife? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> There are some men that have stopped >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> beating their wife. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Although there is a mapping from some >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> men to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> there is no mapping from never >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unmarried men to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> thus the question is incorrect for all >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unmarried men. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Although there is a mapping from some >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> TM/input pairs to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> there is no mapping from H/D to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> thus the question is incorrect for H/D >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Except that the mapping requested is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> about the INPUTS to H, not H itsef. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
In order to see that it is an incorrect >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> question we must examine >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> the question in detail. Making sure to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> always ignore this key detail >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> <is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ
applied to ⟨Ĥ⟩ halts >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ
applied to ⟨Ĥ⟩ does not halt >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
Which isn;t the question at all, so you >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> are just shown to be a stupid liar. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
The QUESTION is:
Does the machine and input described by >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> this input, Halt when run? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Then tell me which element of: >>>>>>>>>>>>>>>>>>>>>>>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is
correct and make sure that >>>>>>>>>>>>>>>>>>>>>>>>>>> you explain why this element is correct and >>>>>>>>>>>>>>>>>>>>>>>>>>> don't try to switch
to any other element outside of the above >>>>>>>>>>>>>>>>>>>>>>>>>>> specified set.
I didn't say there was.
Then you understand that each question posed to >>>>>>>>>>>>>>>>>>>>>>>>> each Ĥ.H in the
above set has no correct answer only because >>>>>>>>>>>>>>>>>>>>>>>>> each of these answers
are contradicted by the machine that H is >>>>>>>>>>>>>>>>>>>>>>>>> contained within.
No, YOU don't understand that the IS a correct >>>>>>>>>>>>>>>>>>>>>>>> answer, just not the one that H (or H^.H ) >>>>>>>>>>>>>>>>>>>>>>>> happens to give.
Then show me which contradicted answer is correct. >>>>>>>>>>>>>>>>>>>>>>>
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy >>>>>>>>>>>>>>>>>>>>>> and loops so qn was the right answer. >>>>>>>>>>>>>>>>>>>>> *The strawman deception is all that you have* >>>>>>>>>>>>>>>>>>>>>
to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied
to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
*The answer must come from elements of the above set* >>>>>>>>>>>>>>>>>>>>
The above are the program/input pairs such that every >>>>>>>>>>>>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets the wrong answer only because whatever answer >>>>>>>>>>>>>>>>>>> that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets is contradicted.
So?
That doesn't mean they are the set that the answer to >>>>>>>>>>>>>>>>>> the ACTUAL QUESTION needs to come from.
You are just proving your stupidity and duplicity. >>>>>>>>>>>>>>>>>
https://www.cs.toronto.edu/~hehner/OSS.pdf
Credit goes to you for finding the loophole in Carol's >>>>>>>>>>>>>>>>> original
question: Can anyone correctly answer “no” to this >>>>>>>>>>>>>>>>> question?
Carol can correctly answer that question with any word >>>>>>>>>>>>>>>>> that is
synonymous with "no".
Here is the one where the loophole is closed: >>>>>>>>>>>>>>>>> Can Carol correctly answer “no” to this [yes/no] question?
The fact that anyone besides Carol can correctly answer >>>>>>>>>>>>>>>>> that
question with a NO and Carol cannot possibly correctly >>>>>>>>>>>>>>>>> answer
that question proves that it is a different question >>>>>>>>>>>>>>>>> when posed
to Carol than when posed to anyone else.
Which is IRRELEVENT to the Halting Question, as it is a >>>>>>>>>>>>>>>> purely objective question.
The behavior of the input is INDEPENDENT of the decider >>>>>>>>>>>>>>>> looking at it.
Note, a given H^ is built on a given H, and no other, >>>>>>>>>>>>>>>> but can be given to any decider to answer, and the >>>>>>>>>>>>>>>> correct answer will be the same irrespective of you ask. >>>>>>>>>>>>>>>> Some will give the right answer, and some will give the >>>>>>>>>>>>>>>> wrong answer. The fact that that H is in the latter >>>>>>>>>>>>>>>> doesn't make the question subjective.
The only way to make the Halting Question subjective is >>>>>>>>>>>>>>>> to try to redefine it so the input changes with who you >>>>>>>>>>>>>>>> ask, but it doesn't.
The changing H^ to match the H only happens in the Meta, >>>>>>>>>>>>>>>> where we prove that we can find an H^ that any H will >>>>>>>>>>>>>>>> get wrong, but each of those are SEPERATE Halting >>>>>>>>>>>>>>>> question (not all one question) and each of those >>>>>>>>>>>>>>>> seperate questions have a correct answer.
Nope.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to
⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to
⟨Ĥ⟩ does not halt
Carol's question posed to Carol <is> isomorphic to >>>>>>>>>>>>>>>>> input ⟨Ĥ⟩ ⟨Ĥ⟩
to every Ĥ.H shown above. The fact that some other TM >>>>>>>>>>>>>>>>> such as H1
(that is not contradicted) can determine a correct >>>>>>>>>>>>>>>>> answer proves
that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is a different question >>>>>>>>>>>>>>>>>
The Question doesn't refer to H at all.
The input ⟨Ĥ⟩ ⟨Ĥ⟩ posed to Ĥ.H
is isomorphic to this question posed to Carol:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>>>>>>>
Carol is a volitional being.
When we hypothesize that Carol is the name of an AI machine >>>>>>>>>>>>> everything remains the same.
Nope.
Once Carol become deterministic, then the whole thing changes. >>>>>>>>>>>>
The only reason that:
Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H
cannot be correctly answered is that the specific Ĥ.H is >>>>>>>>>>> contradicted.
The only reason that:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>>>> posed to Carol
cannot be correctly answered is that the specific Carol is >>>>>>>>>>> contradicted.
Nope.
You aren't showing any ERRORS I made but just asserting your >>>>>>>>>> FALSE claims again.
Inability to show WHY my description was wrong just proves you >>>>>>>>>> have no basis.
You are just demonstrating that you don't understand how logic >>>>>>>>>> works.
It seems you think this is just some abstract philosophy where >>>>>>>>>> anything goes and rhetoric rules.
*You have provided zero correct reasoning of how*
*Carol's question posed to Carol*
*is not contradicted just like*
*Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H is contradicted*
Yes, I have.
YOU have provided ZERO reasoning how they are.
Dos H^ (H^) Halt? even when posed to H^.H has an answer!
When posed to each entity (Carol/Ĥ.H)
their respective question (a)/(b):
(a) Carol correctly answer “no” to this [yes/no] question?
(b) Does Ĥ halt?
lacks a correct answer because this answer is contradicted.
(b) has a correct answer, which is "yes"
When Ĥ gives that answer it is contradicted by Ĥ,
thus it is the wrong answer.
What does "When Ĥ gives that answer" mean?
(a) Ĥ.Hqy then loop (always does the opposite of what it says).
(b) Ĥ.Hqn then halt (always does the opposite of what it says).
Ĥ is a program which can only do what it is programmed, and it is
programmed to answer "no" even though the correct answer is "yes".
Nut (a) isn't AN ANSWER, as it isn't given to any machine that uses it.
You don't seem to understand what answer is.
And, H^ is not defined to apply any semantic to its return, so you
can't assume any.
H is defined to give an answer, but all H's will give the wrong answer
for the H^ built from it.
*Only because every answer that they give is contradicted*
Carol's Question posed to Carol:
Can Carol correctly answer “no” to this [yes/no] question?
and
Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
lack a correct answer because this answer is contradicted.
*Incorrect questions lack correct answers*
On 3/16/2024 11:33 AM, immibis wrote:
On 16/03/24 06:21, olcott wrote:
On 3/15/2024 11:24 PM, immibis wrote:
On 16/03/24 04:52, olcott wrote:The possible answers that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gives are:
On 3/15/2024 10:15 PM, immibis wrote:
On 16/03/24 00:17, olcott wrote:
On 3/15/2024 6:02 PM, Richard Damon wrote:
On 3/15/24 3:47 PM, olcott wrote:
On 3/15/2024 5:13 PM, Richard Damon wrote:
On 3/15/24 1:42 PM, olcott wrote:
On 3/15/2024 3:35 PM, Richard Damon wrote:
On 3/15/24 12:50 PM, olcott wrote:
On 3/15/2024 2:42 PM, Richard Damon wrote:
On 3/15/24 12:23 PM, olcott wrote:
On 3/15/2024 2:14 PM, Richard Damon wrote:Nope. and that LIE is a source of a lot of your ERRORS. >>>>>>>>>>>>>>
On 3/15/24 12:00 PM, olcott wrote:
On 3/15/2024 1:38 PM, Richard Damon wrote:
On 3/15/24 7:41 AM, olcott wrote:Objective and Subjective Specifications
On 3/15/2024 5:44 AM, Mikko wrote:
On 2024-03-15 01:12:19 +0000, olcott said: >>>>>>>>>>>>>>>>>>>>
On 3/14/2024 8:06 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/14/24 4:45 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 5:37 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 3:04 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 4:55 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 1:59 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 3:54 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 1:26 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 3:20 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 12:32 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 12:33 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 4:04 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 5:43 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 2:54 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 4:39 PM, Richard Damon >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:Is a false claim about a strawman deception really >>>>>>>>>>>>>>>>>>>> the best you can say?
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ appliedBut it DOES.The question posed to Ĥ.H has no correct >>>>>>>>>>>>>>>>>>>>>>>>>>>>> answer, thus not the >>>>>>>>>>>>>>>>>>>>>>>>>>>>> same question at all. >>>>>>>>>>>>>>>>>>>>>>>>>>>>On 3/13/24 1:52 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 12:52 PM, Richard >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 10:08 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 11:44 AM, immibis >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 13/03/24 04:55, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/2024 10:49 PM, Richard >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Not quite. It always gets the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrong answer, but only one of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> them for each quesiton. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> They all gets the wrong answer >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> on a whole class of questions >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Wrong. You said. yourself. that >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> H1 gets the right answer for D. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>I invented it so I get to stipulate >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> its meaning. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
Since it is a logical >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> impossibility to determine the truth >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> value of a self-contradictory >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> expression the requirement >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> for H to do this is bogus. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>Shows you are just a LIAR, as >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> there IS a truth value to the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> expression that is the requirment >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> for ANY SPECIFIC H. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> *Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> *Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
There is no mapping from H(D,D) to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Halts(D,D) that exists. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> This proves that H(D,D) is being >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> asked an incorrect question. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Why, because it is NOT a LIE. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
You don't even know the definiton of >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> an incorrect question. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Nope, common technical term. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Cite a source. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Right, because that question include a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> presumption of something not actually >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> present.
The fact that there DOES exist a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> mapping Halt(M,d) that maps all >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Turing Machines and there input to a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> result of Halting / Non-Halting for >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> EVERY member of that input set, >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> means tha Halts is a valid mapping >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> to ask a decider to try to decider. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>That part is true. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Likewise when you ask a man that has >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> never been married: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Have you stopped beating tour wife? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> There are some men that have stopped >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> beating their wife. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Although there is a mapping from some >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> men to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> there is no mapping from never >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unmarried men to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> thus the question is incorrect for all >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> unmarried men. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Although there is a mapping from some >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> TM/input pairs to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> there is no mapping from H/D to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> thus the question is incorrect for H/D >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Except that the mapping requested is >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> about the INPUTS to H, not H itsef. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
In order to see that it is an incorrect >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> question we must examine >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> the question in detail. Making sure to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> always ignore this key detail >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> <is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ
applied to ⟨Ĥ⟩ halts >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ
applied to ⟨Ĥ⟩ does not halt >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
Which isn;t the question at all, so you >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> are just shown to be a stupid liar. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
The QUESTION is:
Does the machine and input described by >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> this input, Halt when run? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Then tell me which element of: >>>>>>>>>>>>>>>>>>>>>>>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is
correct and make sure that >>>>>>>>>>>>>>>>>>>>>>>>>>> you explain why this element is correct and >>>>>>>>>>>>>>>>>>>>>>>>>>> don't try to switch
to any other element outside of the above >>>>>>>>>>>>>>>>>>>>>>>>>>> specified set.
I didn't say there was.
Then you understand that each question posed to >>>>>>>>>>>>>>>>>>>>>>>>> each Ĥ.H in the
above set has no correct answer only because >>>>>>>>>>>>>>>>>>>>>>>>> each of these answers
are contradicted by the machine that H is >>>>>>>>>>>>>>>>>>>>>>>>> contained within.
No, YOU don't understand that the IS a correct >>>>>>>>>>>>>>>>>>>>>>>> answer, just not the one that H (or H^.H ) >>>>>>>>>>>>>>>>>>>>>>>> happens to give.
Then show me which contradicted answer is correct. >>>>>>>>>>>>>>>>>>>>>>>
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy >>>>>>>>>>>>>>>>>>>>>> and loops so qn was the right answer. >>>>>>>>>>>>>>>>>>>>> *The strawman deception is all that you have* >>>>>>>>>>>>>>>>>>>>>
to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied
to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
*The answer must come from elements of the above set* >>>>>>>>>>>>>>>>>>>>
The above are the program/input pairs such that every >>>>>>>>>>>>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets the wrong answer only because whatever answer >>>>>>>>>>>>>>>>>>> that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets is contradicted.
So?
That doesn't mean they are the set that the answer to >>>>>>>>>>>>>>>>>> the ACTUAL QUESTION needs to come from.
You are just proving your stupidity and duplicity. >>>>>>>>>>>>>>>>>
https://www.cs.toronto.edu/~hehner/OSS.pdf
Credit goes to you for finding the loophole in Carol's >>>>>>>>>>>>>>>>> original
question: Can anyone correctly answer “no” to this >>>>>>>>>>>>>>>>> question?
Carol can correctly answer that question with any word >>>>>>>>>>>>>>>>> that is
synonymous with "no".
Here is the one where the loophole is closed: >>>>>>>>>>>>>>>>> Can Carol correctly answer “no” to this [yes/no] question?
The fact that anyone besides Carol can correctly answer >>>>>>>>>>>>>>>>> that
question with a NO and Carol cannot possibly correctly >>>>>>>>>>>>>>>>> answer
that question proves that it is a different question >>>>>>>>>>>>>>>>> when posed
to Carol than when posed to anyone else.
Which is IRRELEVENT to the Halting Question, as it is a >>>>>>>>>>>>>>>> purely objective question.
The behavior of the input is INDEPENDENT of the decider >>>>>>>>>>>>>>>> looking at it.
Note, a given H^ is built on a given H, and no other, >>>>>>>>>>>>>>>> but can be given to any decider to answer, and the >>>>>>>>>>>>>>>> correct answer will be the same irrespective of you ask. >>>>>>>>>>>>>>>> Some will give the right answer, and some will give the >>>>>>>>>>>>>>>> wrong answer. The fact that that H is in the latter >>>>>>>>>>>>>>>> doesn't make the question subjective.
The only way to make the Halting Question subjective is >>>>>>>>>>>>>>>> to try to redefine it so the input changes with who you >>>>>>>>>>>>>>>> ask, but it doesn't.
The changing H^ to match the H only happens in the Meta, >>>>>>>>>>>>>>>> where we prove that we can find an H^ that any H will >>>>>>>>>>>>>>>> get wrong, but each of those are SEPERATE Halting >>>>>>>>>>>>>>>> question (not all one question) and each of those >>>>>>>>>>>>>>>> seperate questions have a correct answer.
Nope.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to
⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to
⟨Ĥ⟩ does not halt
Carol's question posed to Carol <is> isomorphic to >>>>>>>>>>>>>>>>> input ⟨Ĥ⟩ ⟨Ĥ⟩
to every Ĥ.H shown above. The fact that some other TM >>>>>>>>>>>>>>>>> such as H1
(that is not contradicted) can determine a correct >>>>>>>>>>>>>>>>> answer proves
that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is a different question >>>>>>>>>>>>>>>>>
The Question doesn't refer to H at all.
The input ⟨Ĥ⟩ ⟨Ĥ⟩ posed to Ĥ.H
is isomorphic to this question posed to Carol:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>>>>>>>
Carol is a volitional being.
When we hypothesize that Carol is the name of an AI machine >>>>>>>>>>>>> everything remains the same.
Nope.
Once Carol become deterministic, then the whole thing changes. >>>>>>>>>>>>
The only reason that:
Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H
cannot be correctly answered is that the specific Ĥ.H is >>>>>>>>>>> contradicted.
The only reason that:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>>>> posed to Carol
cannot be correctly answered is that the specific Carol is >>>>>>>>>>> contradicted.
Nope.
You aren't showing any ERRORS I made but just asserting your >>>>>>>>>> FALSE claims again.
Inability to show WHY my description was wrong just proves you >>>>>>>>>> have no basis.
You are just demonstrating that you don't understand how logic >>>>>>>>>> works.
It seems you think this is just some abstract philosophy where >>>>>>>>>> anything goes and rhetoric rules.
*You have provided zero correct reasoning of how*
*Carol's question posed to Carol*
*is not contradicted just like*
*Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H is contradicted*
Yes, I have.
YOU have provided ZERO reasoning how they are.
Dos H^ (H^) Halt? even when posed to H^.H has an answer!
When posed to each entity (Carol/Ĥ.H)
their respective question (a)/(b):
(a) Carol correctly answer “no” to this [yes/no] question?
(b) Does Ĥ halt?
lacks a correct answer because this answer is contradicted.
(b) has a correct answer, which is "yes"
When Ĥ gives that answer it is contradicted by Ĥ,
thus it is the wrong answer.
What does "When Ĥ gives that answer" mean?
(a) Ĥ.Hqy then loop (always does the opposite of what it says).
(b) Ĥ.Hqn then halt (always does the opposite of what it says).
No, only one answer is possible, which is the one that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is
programmed to give.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
Because every Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is contradicted.
Ĥ is a program which can only do what it is programmed, and it is
programmed to answer "no" even though the correct answer is "yes".
On 3/15/2024 10:15 PM, immibis wrote:
On 16/03/24 00:17, olcott wrote:
On 3/15/2024 6:02 PM, Richard Damon wrote:
On 3/15/24 3:47 PM, olcott wrote:
On 3/15/2024 5:13 PM, Richard Damon wrote:
On 3/15/24 1:42 PM, olcott wrote:
On 3/15/2024 3:35 PM, Richard Damon wrote:
On 3/15/24 12:50 PM, olcott wrote:
On 3/15/2024 2:42 PM, Richard Damon wrote:
On 3/15/24 12:23 PM, olcott wrote:
On 3/15/2024 2:14 PM, Richard Damon wrote:Nope. and that LIE is a source of a lot of your ERRORS.
On 3/15/24 12:00 PM, olcott wrote:
On 3/15/2024 1:38 PM, Richard Damon wrote:
On 3/15/24 7:41 AM, olcott wrote:
On 3/15/2024 5:44 AM, Mikko wrote:
On 2024-03-15 01:12:19 +0000, olcott said:
On 3/14/2024 8:06 PM, Richard Damon wrote:Is a false claim about a strawman deception really the >>>>>>>>>>>>>>>> best you can say?
On 3/14/24 4:45 PM, olcott wrote:*The strawman deception is all that you have* >>>>>>>>>>>>>>>>>
On 3/14/2024 5:37 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>> On 3/14/24 3:04 PM, olcott wrote:
On 3/14/2024 4:55 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>> On 3/14/24 1:59 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 3:54 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 1:26 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 3:20 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/24 12:32 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/14/2024 12:33 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 4:04 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 5:43 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 2:54 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 4:39 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/24 1:52 PM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 12:52 PM, Richard Damon >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrote:
I invented it so I get to stipulate its >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> meaning.On 3/13/24 10:08 AM, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/13/2024 11:44 AM, immibis wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 13/03/24 04:55, olcott wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> On 3/12/2024 10:49 PM, Richard >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Damon wrote: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Shows you are just a LIAR, as there IS >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> a truth value to the expression that >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> is the requirment for ANY SPECIFIC H. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>Wrong. You said. yourself. that H1 >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> gets the right answer for D. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>Not quite. It always gets the >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> wrong answer, but only one of them >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> for each quesiton. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
They all gets the wrong answer on a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> whole class of questions >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Since it is a logical impossibility >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> to determine the truth >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> value of a self-contradictory >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> expression the requirement >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> for H to do this is bogus. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
*Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> *Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> *Lying about me being a liar may >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> possibly cost your soul* >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
There is no mapping from H(D,D) to >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Halts(D,D) that exists. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> This proves that H(D,D) is being asked >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> an incorrect question. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Why, because it is NOT a LIE. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
You don't even know the definiton of an >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> incorrect question. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
https://groups.google.com/g/sci.lang/c/AO5Vlupeelo/m/nxJy7N2vULwJ
Nope, common technical term. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Cite a source.
Right, because that question include a >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> presumption of something not actually >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> present.
The fact that there DOES exist a mapping >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Halt(M,d) that maps all Turing Machines >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> and there input to a result of Halting / >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Non-Halting for EVERY member of that >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> input set, means tha Halts is a valid >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> mapping to ask a decider to try to decider. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>That part is true. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Likewise when you ask a man that has >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> never been married: >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Have you stopped beating tour wife? >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> There are some men that have stopped >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> beating their wife. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Although there is a mapping from some men >>>>>>>>>>>>>>>>>>>>>>>>>>>>> to YES/NO
there is no mapping from never unmarried >>>>>>>>>>>>>>>>>>>>>>>>>>>>> men to YES/NO
thus the question is incorrect for all >>>>>>>>>>>>>>>>>>>>>>>>>>>>> unmarried men.
Although there is a mapping from some >>>>>>>>>>>>>>>>>>>>>>>>>>>>> TM/input pairs to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>> there is no mapping from H/D to YES/NO >>>>>>>>>>>>>>>>>>>>>>>>>>>>> thus the question is incorrect for H/D >>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Except that the mapping requested is about >>>>>>>>>>>>>>>>>>>>>>>>>>>> the INPUTS to H, not H itsef. >>>>>>>>>>>>>>>>>>>>>>>>>>>>
In order to see that it is an incorrect >>>>>>>>>>>>>>>>>>>>>>>>>>> question we must examine >>>>>>>>>>>>>>>>>>>>>>>>>>> the question in detail. Making sure to always >>>>>>>>>>>>>>>>>>>>>>>>>>> ignore this key detail
<is> cheating.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ
applied to ⟨Ĥ⟩ halts >>>>>>>>>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ
applied to ⟨Ĥ⟩ does not halt >>>>>>>>>>>>>>>>>>>>>>>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
Which isn;t the question at all, so you are >>>>>>>>>>>>>>>>>>>>>>>>>> just shown to be a stupid liar. >>>>>>>>>>>>>>>>>>>>>>>>>>
The QUESTION is:
Does the machine and input described by this >>>>>>>>>>>>>>>>>>>>>>>>>> input, Halt when run?
The question posed to Ĥ.H has no correct >>>>>>>>>>>>>>>>>>>>>>>>> answer, thus not the
same question at all.
But it DOES.
Then tell me which element of:
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) is correct
and make sure that
you explain why this element is correct and don't >>>>>>>>>>>>>>>>>>>>>>> try to switch
to any other element outside of the above >>>>>>>>>>>>>>>>>>>>>>> specified set.
I didn't say there was.
Then you understand that each question posed to >>>>>>>>>>>>>>>>>>>>> each Ĥ.H in the
above set has no correct answer only because each >>>>>>>>>>>>>>>>>>>>> of these answers
are contradicted by the machine that H is contained >>>>>>>>>>>>>>>>>>>>> within.
No, YOU don't understand that the IS a correct >>>>>>>>>>>>>>>>>>>> answer, just not the one that H (or H^.H ) happens >>>>>>>>>>>>>>>>>>>> to give.
Then show me which contradicted answer is correct. >>>>>>>>>>>>>>>>>>>
If H (H^) (H^) goes to qy, then H^ (H^) goes to qy and >>>>>>>>>>>>>>>>>> loops so qn was the right answer.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to
⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to
⟨Ĥ⟩ does not halt
∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩)) >>>>>>>>>>>>>>>>>
*The answer must come from elements of the above set* >>>>>>>>>>>>>>>>
The above are the program/input pairs such that every Ĥ.H >>>>>>>>>>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩
gets the wrong answer only because whatever answer that >>>>>>>>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
gets is contradicted.
So?
That doesn't mean they are the set that the answer to the >>>>>>>>>>>>>> ACTUAL QUESTION needs to come from.
You are just proving your stupidity and duplicity.
Objective and Subjective Specifications
https://www.cs.toronto.edu/~hehner/OSS.pdf
Credit goes to you for finding the loophole in Carol's >>>>>>>>>>>>> original
question: Can anyone correctly answer “no” to this question? >>>>>>>>>>>>>
Carol can correctly answer that question with any word that is >>>>>>>>>>>>> synonymous with "no".
Here is the one where the loophole is closed:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>>>>>>
The fact that anyone besides Carol can correctly answer that >>>>>>>>>>>>> question with a NO and Carol cannot possibly correctly answer >>>>>>>>>>>>> that question proves that it is a different question when >>>>>>>>>>>>> posed
to Carol than when posed to anyone else.
Which is IRRELEVENT to the Halting Question, as it is a >>>>>>>>>>>> purely objective question.
The behavior of the input is INDEPENDENT of the decider >>>>>>>>>>>> looking at it.
Note, a given H^ is built on a given H, and no other, but >>>>>>>>>>>> can be given to any decider to answer, and the correct >>>>>>>>>>>> answer will be the same irrespective of you ask. Some will >>>>>>>>>>>> give the right answer, and some will give the wrong answer. >>>>>>>>>>>> The fact that that H is in the latter doesn't make the >>>>>>>>>>>> question subjective.
The only way to make the Halting Question subjective is to >>>>>>>>>>>> try to redefine it so the input changes with who you ask, >>>>>>>>>>>> but it doesn't.
The changing H^ to match the H only happens in the Meta, >>>>>>>>>>>> where we prove that we can find an H^ that any H will get >>>>>>>>>>>> wrong, but each of those are SEPERATE Halting question (not >>>>>>>>>>>> all one question) and each of those seperate questions have >>>>>>>>>>>> a correct answer.
Nope.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩
does not halt
Carol's question posed to Carol <is> isomorphic to input >>>>>>>>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩
to every Ĥ.H shown above. The fact that some other TM such >>>>>>>>>>>>> as H1
(that is not contradicted) can determine a correct answer >>>>>>>>>>>>> proves
that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is a different question
The Question doesn't refer to H at all.
The input ⟨Ĥ⟩ ⟨Ĥ⟩ posed to Ĥ.H
is isomorphic to this question posed to Carol:
Can Carol correctly answer “no” to this [yes/no] question? >>>>>>>>>>
Carol is a volitional being.
When we hypothesize that Carol is the name of an AI machine
everything remains the same.
Nope.
Once Carol become deterministic, then the whole thing changes. >>>>>>>>
The only reason that:
Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H
cannot be correctly answered is that the specific Ĥ.H is
contradicted.
The only reason that:
Can Carol correctly answer “no” to this [yes/no] question? posed >>>>>>> to Carol
cannot be correctly answered is that the specific Carol is
contradicted.
Nope.
You aren't showing any ERRORS I made but just asserting your FALSE >>>>>> claims again.
Inability to show WHY my description was wrong just proves you
have no basis.
You are just demonstrating that you don't understand how logic works. >>>>>>
It seems you think this is just some abstract philosophy where
anything goes and rhetoric rules.
*You have provided zero correct reasoning of how*
*Carol's question posed to Carol*
*is not contradicted just like*
*Does Ĥ ⟨Ĥ⟩ halt? posed to Ĥ.H is contradicted*
Yes, I have.
YOU have provided ZERO reasoning how they are.
Dos H^ (H^) Halt? even when posed to H^.H has an answer!
When posed to each entity (Carol/Ĥ.H)
their respective question (a)/(b):
(a) Carol correctly answer “no” to this [yes/no] question?
(b) Does Ĥ ⟨Ĥ⟩ halt?
lacks a correct answer because this answer is contradicted.
(b) has a correct answer, which is "yes"
When Ĥ gives that answer it is contradicted by Ĥ,
thus it is the wrong answer.
*Incorrect questions do not lack answers they lack correct answers*
Carol could answer by flipping the bird.
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