https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents
the entire infinite set of what would otherwise be an infinite set of
H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
Any yes/no question that lacks a correct yes/no answer is an incorrect question. The inability to correctly answer an incorrect question places
no limit on anyone or anything.
On the other hand when embedded_H is trying to answer the question:
Could my input terminate normally?
this question can be answered with a “no”.
https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents
the entire infinite set of what would otherwise be an infinite set of
H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
On 10/25/2023 5:25 PM, olcott wrote:
https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents
the entire infinite set of what would otherwise be an infinite set of
H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
When we accept the idea of a pure function and that computations
must be pure functions then we know that it would be incorrect
for H to report on the computation that itself is contained within.
embedded_H cannot even see the computation that itself is contained
within it can only see the behavior of its actual input.
On 10/25/2023 5:25 PM, olcott wrote:
https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents
the entire infinite set of what would otherwise be an infinite set of
H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
When we accept the idea of a pure function and that computations
must be pure functions then we know that it would be incorrect
for H to report on the computation that itself is contained within.
embedded_H cannot even see the computation that itself is contained
within it can only see the behavior of its actual input.
On 10/25/2023 6:01 PM, olcott wrote:
On 10/25/2023 5:25 PM, olcott wrote:
https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents
the entire infinite set of what would otherwise be an infinite set of
H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
When we accept the idea of a pure function and that computations
must be pure functions then we know that it would be incorrect
for H to report on the computation that itself is contained within.
embedded_H cannot even see the computation that itself is contained
within it can only see the behavior of its actual input.
Right, it needs to answer about the computataion specified by the input.
Thus embedded_H is not allowed to report on the behavior of the
directly executed Ĥ ⟨Ĥ⟩ because that <is> the computation that it
is contained within.
On 10/25/2023 6:01 PM, olcott wrote:
On 10/25/2023 5:25 PM, olcott wrote:
https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents
the entire infinite set of what would otherwise be an infinite set of
H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
When we accept the idea of a pure function and that computations
must be pure functions then we know that it would be incorrect
for H to report on the computation that itself is contained within.
embedded_H cannot even see the computation that itself is contained
within it can only see the behavior of its actual input.
Right, it needs to answer about the computataion specified by the input.
Thus embedded_H is not allowed to report on the behavior of the
directly executed Ĥ ⟨Ĥ⟩ because that <is> the computation that it
is contained within.
On 10/25/2023 6:01 PM, olcott wrote:
On 10/25/2023 5:25 PM, olcott wrote:
https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents
the entire infinite set of what would otherwise be an infinite set of
H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
When we accept the idea of a pure function and that computations
must be pure functions then we know that it would be incorrect
for H to report on the computation that itself is contained within.
embedded_H cannot even see the computation that itself is contained
within it can only see the behavior of its actual input.
Right, it needs to answer about the computataion specified by the input.
Thus embedded_H is not allowed to report on the behavior of the
directly executed Ĥ ⟨Ĥ⟩ because that <is> the computation that it
is contained within.
On 10/25/2023 7:55 PM, olcott wrote:
On 10/25/2023 6:01 PM, olcott wrote:
On 10/25/2023 5:25 PM, olcott wrote:
https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents
the entire infinite set of what would otherwise be an infinite set of
H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
When we accept the idea of a pure function and that computations
must be pure functions then we know that it would be incorrect
for H to report on the computation that itself is contained within.
embedded_H cannot even see the computation that itself is contained
within it can only see the behavior of its actual input.
Right, it needs to answer about the computataion specified by the input.
Thus embedded_H is not allowed to report on the behavior of the
directly executed Ĥ ⟨Ĥ⟩ because that <is> the computation that it
is contained within.
It <is> not the computation that is specified by the
input because the computation specified by the input
includes that Ĥ is invoking embedded_H recursively.
On 10/25/2023 7:55 PM, olcott wrote:
On 10/25/2023 6:01 PM, olcott wrote:
On 10/25/2023 5:25 PM, olcott wrote:
https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents
the entire infinite set of what would otherwise be an infinite set of
H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
When we accept the idea of a pure function and that computations
must be pure functions then we know that it would be incorrect
for H to report on the computation that itself is contained within.
embedded_H cannot even see the computation that itself is contained
within it can only see the behavior of its actual input.
Right, it needs to answer about the computataion specified by the input.
Thus embedded_H is not allowed to report on the behavior of the
directly executed Ĥ ⟨Ĥ⟩ because that <is> the computation that it
is contained within.
Since it is WRONG for embedded_H to report on the behavior of the direct execution of Ĥ ⟨Ĥ⟩, that means that embedded_H is not allowed to report that the direct execution of Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.
On 10/25/2023 7:55 PM, olcott wrote:
On 10/25/2023 6:01 PM, olcott wrote:
On 10/25/2023 5:25 PM, olcott wrote:
https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents
the entire infinite set of what would otherwise be an infinite set of
H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
When we accept the idea of a pure function and that computations
must be pure functions then we know that it would be incorrect
for H to report on the computation that itself is contained within.
embedded_H cannot even see the computation that itself is contained
within it can only see the behavior of its actual input.
Right, it needs to answer about the computataion specified by the input.
Thus embedded_H is not allowed to report on the behavior of the
directly executed Ĥ ⟨Ĥ⟩ because that <is> the computation that it
is contained within.
Since it is WRONG for embedded_H to report on the behavior of the direct execution of Ĥ ⟨Ĥ⟩, that means that embedded_H is not allowed to report that the direct execution of Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.
On 10/25/2023 6:01 PM, olcott wrote:
On 10/25/2023 5:25 PM, olcott wrote:
https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents
the entire infinite set of what would otherwise be an infinite set of
H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
When we accept the idea of a pure function and that computations
must be pure functions then we know that it would be incorrect
for H to report on the computation that itself is contained within.
embedded_H cannot even see the computation that itself is contained
within it can only see the behavior of its actual input.
Right, it needs to answer about the computataion specified by the input.
Thus embedded_H is not allowed to report on the behavior of the
directly executed Ĥ ⟨Ĥ⟩ because that <is> the computation that it
is contained within.
On 10/25/2023 7:55 PM, olcott wrote:
On 10/25/2023 6:01 PM, olcott wrote:
On 10/25/2023 5:25 PM, olcott wrote:
https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents
the entire infinite set of what would otherwise be an infinite set of
H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
When we accept the idea of a pure function and that computations
must be pure functions then we know that it would be incorrect
for H to report on the computation that itself is contained within.
embedded_H cannot even see the computation that itself is contained
within it can only see the behavior of its actual input.
Right, it needs to answer about the computataion specified by the input.
Thus embedded_H is not allowed to report on the behavior of the
directly executed Ĥ ⟨Ĥ⟩ because that <is> the computation that it
is contained within.
Since it is WRONG for embedded_H to report on the behavior of the direct execution of Ĥ ⟨Ĥ⟩, that means that embedded_H is not allowed to report that the direct execution of Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.
On 10/25/2023 6:01 PM, olcott wrote:
On 10/25/2023 5:25 PM, olcott wrote:
https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents
the entire infinite set of what would otherwise be an infinite set of
H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
When we accept the idea of a pure function and that computations
must be pure functions then we know that it would be incorrect
for H to report on the computation that itself is contained within.
embedded_H cannot even see the computation that itself is contained
within it can only see the behavior of its actual input.
Right, it needs to answer about the computataion specified by the input.
Thus embedded_H is not allowed to report on the behavior of the
directly executed Ĥ ⟨Ĥ⟩ because that <is> the computation that it
is contained within.
On 10/25/2023 10:01 PM, olcott wrote:
On 10/25/2023 7:55 PM, olcott wrote:
On 10/25/2023 6:01 PM, olcott wrote:
On 10/25/2023 5:25 PM, olcott wrote:
https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents >>>>> the entire infinite set of what would otherwise be an infinite set of >>>>> H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
When we accept the idea of a pure function and that computations
must be pure functions then we know that it would be incorrect
for H to report on the computation that itself is contained within.
embedded_H cannot even see the computation that itself is contained
within it can only see the behavior of its actual input.
Right, it needs to answer about the computataion specified by the input. >>>
Thus embedded_H is not allowed to report on the behavior of the
directly executed Ĥ ⟨Ĥ⟩ because that <is> the computation that it
is contained within.
Since it is WRONG for embedded_H to report on the behavior of the direct
execution of Ĥ ⟨Ĥ⟩, that means that embedded_H is not allowed to report
that the direct execution of Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.
Only stupid idiots call other people stupid idiots.
People that are not stupid idiots realize that ad
hominem attacks from them directed at others make
them look very foolish.
AD Hominem attacks on others is a direct admission
that one has run out of reasoning.
On 10/25/2023 7:55 PM, olcott wrote:
On 10/25/2023 6:01 PM, olcott wrote:
On 10/25/2023 5:25 PM, olcott wrote:
https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents
the entire infinite set of what would otherwise be an infinite set of
H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
When we accept the idea of a pure function and that computations
must be pure functions then we know that it would be incorrect
for H to report on the computation that itself is contained within.
embedded_H cannot even see the computation that itself is contained
within it can only see the behavior of its actual input.
Right, it needs to answer about the computataion specified by the input.
Thus embedded_H is not allowed to report on the behavior of the
directly executed Ĥ ⟨Ĥ⟩ because that <is> the computation that it
is contained within.
Since it is WRONG for embedded_H to report on the behavior of the direct execution of Ĥ ⟨Ĥ⟩, that means that embedded_H is not allowed to report that the direct execution of Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.
The computation that embedded_H is contained within is
not the input and does not have the behavior of the input.
On 10/25/2023 10:01 PM, olcott wrote:
On 10/25/2023 7:55 PM, olcott wrote:
On 10/25/2023 6:01 PM, olcott wrote:
On 10/25/2023 5:25 PM, olcott wrote:
https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents >>>>> the entire infinite set of what would otherwise be an infinite set of >>>>> H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
When we accept the idea of a pure function and that computations
must be pure functions then we know that it would be incorrect
for H to report on the computation that itself is contained within.
embedded_H cannot even see the computation that itself is contained
within it can only see the behavior of its actual input.
Right, it needs to answer about the computataion specified by the input. >>>
Thus embedded_H is not allowed to report on the behavior of the
directly executed Ĥ ⟨Ĥ⟩ because that <is> the computation that it
is contained within.
Since it is WRONG for embedded_H to report on the behavior of the direct
execution of Ĥ ⟨Ĥ⟩, that means that embedded_H is not allowed to report
that the direct execution of Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.
Only stupid idiots call other people stupid idiots.
People that are not stupid idiots realize that ad
hominem attacks from them directed at others make
them look very foolish.
AD Hominem attacks on others is a direct admission
that one has run out of reasoning.
On 10/25/2023 10:50 PM, olcott wrote:
On 10/25/2023 10:01 PM, olcott wrote:
On 10/25/2023 7:55 PM, olcott wrote:
On 10/25/2023 6:01 PM, olcott wrote:
On 10/25/2023 5:25 PM, olcott wrote:
https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents >>>>>> the entire infinite set of what would otherwise be an infinite set of >>>>>> H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
When we accept the idea of a pure function and that computations
must be pure functions then we know that it would be incorrect
for H to report on the computation that itself is contained within.
embedded_H cannot even see the computation that itself is contained
within it can only see the behavior of its actual input.
Right, it needs to answer about the computataion specified by the
input.
Thus embedded_H is not allowed to report on the behavior of the
directly executed Ĥ ⟨Ĥ⟩ because that <is> the computation that it >>>> is contained within.
Since it is WRONG for embedded_H to report on the behavior of the direct >>> execution of Ĥ ⟨Ĥ⟩, that means that embedded_H is not allowed to report
that the direct execution of Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.
Only stupid idiots call other people stupid idiots.
People that are not stupid idiots realize that ad
hominem attacks from them directed at others make
them look very foolish.
AD Hominem attacks on others is a direct admission
that one has run out of reasoning.
Nothing but insults without any reasoning is one
of the stupidest things to do within the context
of fields having professional decorum.
On 10/25/2023 7:55 PM, olcott wrote:
On 10/25/2023 6:01 PM, olcott wrote:
On 10/25/2023 5:25 PM, olcott wrote:
https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents
the entire infinite set of what would otherwise be an infinite set of
H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
When we accept the idea of a pure function and that computations
must be pure functions then we know that it would be incorrect
for H to report on the computation that itself is contained within.
embedded_H cannot even see the computation that itself is contained
within it can only see the behavior of its actual input.
Right, it needs to answer about the computataion specified by the input.
Thus embedded_H is not allowed to report on the behavior of the
directly executed Ĥ ⟨Ĥ⟩ because that <is> the computation that it
is contained within.
Since it is WRONG for embedded_H to report on the behavior of the direct execution of Ĥ ⟨Ĥ⟩, that means that embedded_H is not allowed to report that the direct execution of Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.
On 10/25/2023 9:13 PM, olcott wrote:
On 10/25/2023 7:55 PM, olcott wrote:
On 10/25/2023 6:01 PM, olcott wrote:
On 10/25/2023 5:25 PM, olcott wrote:
https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents >>>>> the entire infinite set of what would otherwise be an infinite set of >>>>> H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
When we accept the idea of a pure function and that computations
must be pure functions then we know that it would be incorrect
for H to report on the computation that itself is contained within.
embedded_H cannot even see the computation that itself is contained
within it can only see the behavior of its actual input.
Right, it needs to answer about the computataion specified by the input. >>>
Thus embedded_H is not allowed to report on the behavior of the
directly executed Ĥ ⟨Ĥ⟩ because that <is> the computation that it
is contained within.
Since it is WRONG for embedded_H to report on the behavior of the direct
execution of Ĥ ⟨Ĥ⟩, that means that embedded_H is not allowed to report
that the direct execution of Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.
People that understand that computable functions must be a pure
function of their inputs and thus are not allowed to report on
the computation that they are contained within also understand
that embedded_H is not allowed to report that the direct execution
of Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.
We a few repetitions people that are not quite smart enough
eventually get this.
On 10/25/2023 9:13 PM, olcott wrote:
On 10/25/2023 7:55 PM, olcott wrote:
On 10/25/2023 6:01 PM, olcott wrote:
On 10/25/2023 5:25 PM, olcott wrote:
https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents >>>>> the entire infinite set of what would otherwise be an infinite set of >>>>> H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
When we accept the idea of a pure function and that computations
must be pure functions then we know that it would be incorrect
for H to report on the computation that itself is contained within.
embedded_H cannot even see the computation that itself is contained
within it can only see the behavior of its actual input.
Right, it needs to answer about the computataion specified by the input. >>>
Thus embedded_H is not allowed to report on the behavior of the
directly executed Ĥ ⟨Ĥ⟩ because that <is> the computation that it
is contained within.
Since it is WRONG for embedded_H to report on the behavior of the direct
execution of Ĥ ⟨Ĥ⟩, that means that embedded_H is not allowed to report
that the direct execution of Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.
People that understand that computable functions must be a pure
function of their inputs and thus are not allowed to report on
the computation that they are contained within also understand
that embedded_H is not allowed to report that the direct execution
of Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.
We a few repetitions people that are not quite smart enough
eventually get this.
On 10/25/2023 11:34 PM, olcott wrote:
On 10/25/2023 9:13 PM, olcott wrote:
On 10/25/2023 7:55 PM, olcott wrote:
On 10/25/2023 6:01 PM, olcott wrote:
On 10/25/2023 5:25 PM, olcott wrote:
https://www.liarparadox.org/Linz_Proof.pdf
This Turing Machine description at the top of page 3
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ ∞
q0 WM ⊢* Ĥq0 WM WM ⊢* Ĥ y1 qn y2
Is simplified and clarified to this:
when Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The advantage of the Linz proof is that it simultaneously represents >>>>>> the entire infinite set of what would otherwise be an infinite set of >>>>>> H/D pairs by using a single TM template for Ĥ.
*We can say that both yes and no are incorrect answers for every*
*embedded_H that attempts to answer the question*
Does the computation that I am contained within halt?
When we accept the idea of a pure function and that computations
must be pure functions then we know that it would be incorrect
for H to report on the computation that itself is contained within.
embedded_H cannot even see the computation that itself is contained
within it can only see the behavior of its actual input.
Right, it needs to answer about the computataion specified by the
input.
Thus embedded_H is not allowed to report on the behavior of the
directly executed Ĥ ⟨Ĥ⟩ because that <is> the computation that it >>>> is contained within.
Since it is WRONG for embedded_H to report on the behavior of the direct >>> execution of Ĥ ⟨Ĥ⟩, that means that embedded_H is not allowed to report
that the direct execution of Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.
People that understand that computable functions must be a pure
function of their inputs and thus are not allowed to report on
the computation that they are contained within also understand
that embedded_H is not allowed to report that the direct execution
of Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.
We a few repetitions people that are not quite smart enough
eventually get this.
IF H / embedded_H is a pure function, how can it tell that it is being
called by Ĥ to change its behavior on the input (Ĥ) (Ĥ)?
It cannot, that is why it must rely on ⟨Ĥ⟩ correctly
simulated by embedded_H as its only valid measure.
embedded_H is expressly not allowed to consider that
Ĥ ⟨Ĥ⟩ reaches Ĥ.qn and halts.
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