From Newsgroup: comp.ai.philosophy

On 8/4/2025 12:33 PM, Mr Flibble wrote:

On Mon, 04 Aug 2025 11:55:41 -0500, olcott wrote:

Until you understand that there never was any actual finite string
machine description input that does the opposite of whatever the decider
decides.

What makes you think that? You know quines are a thing, right?

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.

Ĥ.embedded_H cannot report on its own behavior as the Linz proof
incorrectly presumes in its "if" statements.

TM deciders only compute the mapping from their actual finite string
inputs they never compute any mapping from non-inputs such as actual
Turing machines themselves.

Again an ENCODING of a TM is a valid input.

/Flibble

*From the bottom of page 319 has been adapted to this* https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.

The contradiction in Linz's (or Turing's) self-referential halting construction only appears if one insists that the machine can and must
decide on its own behavior, which is neither possible nor required.

https://chatgpt.com/share/6890ee5a-52bc-8011-852e-3d9f97bcfbd8
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
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From Newsgroup: comp.ai.philosophy

Op 04.aug.2025 om 19:41 schreef olcott:

On 8/4/2025 12:33 PM, Mr Flibble wrote:

On Mon, 04 Aug 2025 11:55:41 -0500, olcott wrote:

Until you understand that there never was any actual finite string
machine description input that does the opposite of whatever the decider >>> decides.

What makes you think that? You know quines are a thing, right?

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
     if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
     if Ĥ applied to ⟨Ĥ⟩ does not halt.

Ĥ.embedded_H cannot report on its own behavior as the Linz proof
incorrectly presumes in its "if" statements.

TM deciders only compute the mapping from their actual finite string
inputs they never compute any mapping from non-inputs such as actual
Turing machines themselves.

Again an ENCODING of a TM is a valid input.

/Flibble

*From the bottom of page 319 has been adapted to this* https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
   if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
   if Ĥ applied to ⟨Ĥ⟩ does not halt.

The contradiction in Linz's (or Turing's) self-referential halting construction only appears if one insists that the machine can and must decide on its own behavior, which is neither possible nor required.

That is a huge mistake. Nobody insists on that. We only insists that it
must decide on its input, which specifies an aborting and halting
program, not your hypothetical non-input that does not abort.
That the input specifies similar behaviour as the simulating program is irrelevant.
Te only conclusion is that simulation is not the right tool to solve
this problem, because it causes recursive simulation, requiring a
premature abort.

https://chatgpt.com/share/6890ee5a-52bc-8011-852e-3d9f97bcfbd8

--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: comp.ai.philosophy

On 8/5/2025 3:33 AM, Fred. Zwarts wrote:

Op 04.aug.2025 om 19:41 schreef olcott:

On 8/4/2025 12:33 PM, Mr Flibble wrote:

On Mon, 04 Aug 2025 11:55:41 -0500, olcott wrote:

Until you understand that there never was any actual finite string
machine description input that does the opposite of whatever the
decider
decides.

What makes you think that? You know quines are a thing, right?

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
     if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
     if Ĥ applied to ⟨Ĥ⟩ does not halt.

Ĥ.embedded_H cannot report on its own behavior as the Linz proof
incorrectly presumes in its "if" statements.

TM deciders only compute the mapping from their actual finite string
inputs they never compute any mapping from non-inputs such as actual
Turing machines themselves.

Again an ENCODING of a TM is a valid input.

/Flibble

*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
    if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
    if Ĥ applied to ⟨Ĥ⟩ does not halt.

The contradiction in Linz's (or Turing's) self-referential halting
construction only appears if one insists that the machine can and must
decide on its own behavior, which is neither possible nor required.

That is a huge mistake. Nobody insists on that. We only insists that it
must decide on its input, which specifies an aborting and halting
program, not your hypothetical non-input that does not abort.

*From the bottom of page 319 has been adapted to this* https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Definition of Turing Machine Ĥ applied
to its own machine description ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.

*Lines 2 and 4 above*
*Does insist that Ĥ.embedded_H report on its own behavior*

"The contradiction in Linz's (or Turing's) self-referential
halting construction only appears if one insists that the
machine can and must decide on its own behavior, which is
neither possible nor required."

https://chatgpt.com/share/6890ee5a-52bc-8011-852e-3d9f97bcfbd8
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: comp.ai.philosophy

On 8/5/25 11:34 AM, olcott wrote:

On 8/5/2025 3:33 AM, Fred. Zwarts wrote:

Op 04.aug.2025 om 19:41 schreef olcott:

On 8/4/2025 12:33 PM, Mr Flibble wrote:

On Mon, 04 Aug 2025 11:55:41 -0500, olcott wrote:

Until you understand that there never was any actual finite string
machine description input that does the opposite of whatever the
decider
decides.

What makes you think that? You know quines are a thing, right?

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
     if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
     if Ĥ applied to ⟨Ĥ⟩ does not halt.

Ĥ.embedded_H cannot report on its own behavior as the Linz proof
incorrectly presumes in its "if" statements.

TM deciders only compute the mapping from their actual finite string >>>>> inputs they never compute any mapping from non-inputs such as actual >>>>> Turing machines themselves.

Again an ENCODING of a TM is a valid input.

/Flibble

*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
    if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
    if Ĥ applied to ⟨Ĥ⟩ does not halt.

The contradiction in Linz's (or Turing's) self-referential halting
construction only appears if one insists that the machine can and
must decide on its own behavior, which is neither possible nor required.

That is a huge mistake. Nobody insists on that. We only insists that
it must decide on its input, which specifies an aborting and halting
program, not your hypothetical non-input that does not abort.

*From the bottom of page 319 has been adapted to this* https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Definition of Turing Machine Ĥ applied
to its own machine description ⟨Ĥ⟩
   Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
if Ĥ applied to ⟨Ĥ⟩ halts, and
   Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.

*Lines 2 and 4 above*
*Does insist that Ĥ.embedded_H report on its own behavior*

Only because that is the behavior of its input.

   "The contradiction in Linz's (or Turing's) self-referential
    halting construction only appears if one insists that the
    machine can and must decide on its own behavior, which is
    neither possible nor required."

But there is no "self-reference" just something being given its own representation.

https://chatgpt.com/share/6890ee5a-52bc-8011-852e-3d9f97bcfbd8

And you LIED to it by saying

Do you understand that no Turing machine decider can directly report on
its own behavior because Turing machine deciders to not take other
actual Turing machines as inputs. Turing machine deciders can at best
only report on the behavior that a finite string machine description specifies.

Since it IS possible for the finite string machine description to
specify the machine do9ng the deciding, and thus its behavior.
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From Newsgroup: comp.ai.philosophy

On 8/5/2025 5:47 PM, Richard Damon wrote:

On 8/5/25 11:34 AM, olcott wrote:

On 8/5/2025 3:33 AM, Fred. Zwarts wrote:

Op 04.aug.2025 om 19:41 schreef olcott:

On 8/4/2025 12:33 PM, Mr Flibble wrote:

On Mon, 04 Aug 2025 11:55:41 -0500, olcott wrote:

Until you understand that there never was any actual finite string >>>>>> machine description input that does the opposite of whatever the
decider
decides.

What makes you think that? You know quines are a thing, right?

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
     if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
     if Ĥ applied to ⟨Ĥ⟩ does not halt.

Ĥ.embedded_H cannot report on its own behavior as the Linz proof
incorrectly presumes in its "if" statements.

TM deciders only compute the mapping from their actual finite string >>>>>> inputs they never compute any mapping from non-inputs such as actual >>>>>> Turing machines themselves.

Again an ENCODING of a TM is a valid input.

/Flibble

*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
    if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
    if Ĥ applied to ⟨Ĥ⟩ does not halt.

The contradiction in Linz's (or Turing's) self-referential halting
construction only appears if one insists that the machine can and
must decide on its own behavior, which is neither possible nor
required.

That is a huge mistake. Nobody insists on that. We only insists that
it must decide on its input, which specifies an aborting and halting
program, not your hypothetical non-input that does not abort.

*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Definition of Turing Machine Ĥ applied
to its own machine description ⟨Ĥ⟩
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
if Ĥ applied to ⟨Ĥ⟩ halts, and
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.

*Lines 2 and 4 above*
*Does insist that Ĥ.embedded_H report on its own behavior*

Only because that is the behavior of its input.

That is not the behavior of its input.

    "The contradiction in Linz's (or Turing's) self-referential
     halting construction only appears if one insists that the
     machine can and must decide on its own behavior, which is
     neither possible nor required."

But there is no "self-reference" just something being given its own representation.

The actual *self-reference* is not Ĥ applied to ⟨Ĥ⟩.
It is the next line: *if Ĥ applied to ⟨Ĥ⟩ halts*
It is this line that incorrectly requires Ĥ to
report on its own behavior.

https://chatgpt.com/share/6890ee5a-52bc-8011-852e-3d9f97bcfbd8

And you LIED to it by saying

So you don't understand that what it said *IS A TRUISM* ???
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: comp.ai.philosophy

On Tue, 05 Aug 2025 19:54:42 -0500, olcott wrote:

On 8/5/2025 5:47 PM, Richard Damon wrote:

On 8/5/25 11:34 AM, olcott wrote:

On 8/5/2025 3:33 AM, Fred. Zwarts wrote:

Op 04.aug.2025 om 19:41 schreef olcott:

On 8/4/2025 12:33 PM, Mr Flibble wrote:

On Mon, 04 Aug 2025 11:55:41 -0500, olcott wrote:

Until you understand that there never was any actual finite string >>>>>>> machine description input that does the opposite of whatever the >>>>>>> decider decides.

What makes you think that? You know quines are a thing, right?

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞, >>>>>>>      if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
     if Ĥ applied to ⟨Ĥ⟩ does not halt.

Ĥ.embedded_H cannot report on its own behavior as the Linz proof >>>>>>> incorrectly presumes in its "if" statements.

TM deciders only compute the mapping from their actual finite
string inputs they never compute any mapping from non-inputs such >>>>>>> as actual Turing machines themselves.

Again an ENCODING of a TM is a valid input.

/Flibble

*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
    if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
    if Ĥ applied to ⟨Ĥ⟩ does not halt.

The contradiction in Linz's (or Turing's) self-referential halting
construction only appears if one insists that the machine can and
must decide on its own behavior, which is neither possible nor
required.

That is a huge mistake. Nobody insists on that. We only insists that
it must decide on its input, which specifies an aborting and halting
program, not your hypothetical non-input that does not abort.

*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Definition of Turing Machine Ĥ applied to its own machine description
⟨Ĥ⟩
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
if Ĥ applied to ⟨Ĥ⟩ halts, and
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.

*Lines 2 and 4 above*
*Does insist that Ĥ.embedded_H report on its own behavior*

Only because that is the behavior of its input.

That is not the behavior of its input.

    "The contradiction in Linz's (or Turing's) self-referential
     halting construction only appears if one insists that the
     machine can and must decide on its own behavior, which is
     neither possible nor required."

But there is no "self-reference" just something being given its own
representation.

The actual *self-reference* is not Ĥ applied to ⟨Ĥ⟩. It is the next line: *if Ĥ applied to ⟨Ĥ⟩ halts*
It is this line that incorrectly requires Ĥ to report on its own
behavior.

https://chatgpt.com/share/6890ee5a-52bc-8011-852e-3d9f97bcfbd8

And you LIED to it by saying

So you don't understand that what it said *IS A TRUISM* ???

Olcott spouting bollocks is a truism.

/Flibble
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: comp.ai.philosophy

On 8/5/25 8:54 PM, olcott wrote:

On 8/5/2025 5:47 PM, Richard Damon wrote:

On 8/5/25 11:34 AM, olcott wrote:

On 8/5/2025 3:33 AM, Fred. Zwarts wrote:

Op 04.aug.2025 om 19:41 schreef olcott:

On 8/4/2025 12:33 PM, Mr Flibble wrote:

On Mon, 04 Aug 2025 11:55:41 -0500, olcott wrote:

Until you understand that there never was any actual finite string >>>>>>> machine description input that does the opposite of whatever the >>>>>>> decider
decides.

What makes you think that? You know quines are a thing, right?

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞, >>>>>>>      if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
     if Ĥ applied to ⟨Ĥ⟩ does not halt.

Ĥ.embedded_H cannot report on its own behavior as the Linz proof >>>>>>> incorrectly presumes in its "if" statements.

TM deciders only compute the mapping from their actual finite string >>>>>>> inputs they never compute any mapping from non-inputs such as actual >>>>>>> Turing machines themselves.

Again an ENCODING of a TM is a valid input.

/Flibble

*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
    if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
    if Ĥ applied to ⟨Ĥ⟩ does not halt.

The contradiction in Linz's (or Turing's) self-referential halting
construction only appears if one insists that the machine can and
must decide on its own behavior, which is neither possible nor
required.

That is a huge mistake. Nobody insists on that. We only insists that
it must decide on its input, which specifies an aborting and halting
program, not your hypothetical non-input that does not abort.

*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Definition of Turing Machine Ĥ applied
to its own machine description ⟨Ĥ⟩
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
if Ĥ applied to ⟨Ĥ⟩ halts, and
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.

*Lines 2 and 4 above*
*Does insist that Ĥ.embedded_H report on its own behavior*

Only because that is the behavior of its input.

That is not the behavior of its input.

    "The contradiction in Linz's (or Turing's) self-referential
     halting construction only appears if one insists that the
     machine can and must decide on its own behavior, which is
     neither possible nor required."

But there is no "self-reference" just something being given its own
representation.

The actual *self-reference* is not Ĥ applied to ⟨Ĥ⟩.
It is the next line: *if Ĥ applied to ⟨Ĥ⟩ halts*
It is this line that incorrectly requires Ĥ to
report on its own behavior.

There is no "self-reference".

And the line *if Ĥ applied to ⟨Ĥ⟩ halts* is a requirement/specification on H, not Ĥ, so there is not "self" here.

Note, there isn't even a "reference" in any real sense, unless you think
your DNA "refers" to you.

(Ĥ) doesn't "refer" to Ĥ, but is a full description that fully defines
that machine, but doesn't "name" it.

And in fact, there should be no way for H to determine that (Ĥ) includes
a description of itself except for the limitation you put into your
system that makes it non-Turing Complete.

It can be shown that in a Turing Complete system, it is, in general, impossible to prove that two descriptions are of the same machine, or
that a given description matches a given machine. You might get lucky,
and find the match as a needle in the haystack, but you can't prove that
they don't match, and it can take unbounded time to show they match.

Thus, your "algorithm" fails to be relaible in a Turing Complete
environment. This shows in your rule that DDD can't make a copy of HHH,
but is forced to use one particular version/representation of it so that
HHH could make that detection.

https://chatgpt.com/share/6890ee5a-52bc-8011-852e-3d9f97bcfbd8

And you LIED to it by saying

So you don't understand that what it said *IS A TRUISM* ???

--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: comp.ai.philosophy

On 8/5/2025 9:15 PM, Richard Damon wrote:

On 8/5/25 8:54 PM, olcott wrote:

On 8/5/2025 5:47 PM, Richard Damon wrote:

On 8/5/25 11:34 AM, olcott wrote:

On 8/5/2025 3:33 AM, Fred. Zwarts wrote:

Op 04.aug.2025 om 19:41 schreef olcott:

On 8/4/2025 12:33 PM, Mr Flibble wrote:

On Mon, 04 Aug 2025 11:55:41 -0500, olcott wrote:

Until you understand that there never was any actual finite string >>>>>>>> machine description input that does the opposite of whatever the >>>>>>>> decider
decides.

What makes you think that? You know quines are a thing, right?

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞, >>>>>>>>      if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn >>>>>>>>      if Ĥ applied to ⟨Ĥ⟩ does not halt.

Ĥ.embedded_H cannot report on its own behavior as the Linz proof >>>>>>>> incorrectly presumes in its "if" statements.

TM deciders only compute the mapping from their actual finite >>>>>>>> string
inputs they never compute any mapping from non-inputs such as >>>>>>>> actual
Turing machines themselves.

Again an ENCODING of a TM is a valid input.

/Flibble

*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
    if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
    if Ĥ applied to ⟨Ĥ⟩ does not halt.

The contradiction in Linz's (or Turing's) self-referential halting >>>>>> construction only appears if one insists that the machine can and >>>>>> must decide on its own behavior, which is neither possible nor
required.

That is a huge mistake. Nobody insists on that. We only insists
that it must decide on its input, which specifies an aborting and
halting program, not your hypothetical non-input that does not abort. >>>>

*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Definition of Turing Machine Ĥ applied
to its own machine description ⟨Ĥ⟩
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞, >>>> if Ĥ applied to ⟨Ĥ⟩ halts, and
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.

*Lines 2 and 4 above*
*Does insist that Ĥ.embedded_H report on its own behavior*

Only because that is the behavior of its input.

That is not the behavior of its input.

    "The contradiction in Linz's (or Turing's) self-referential
     halting construction only appears if one insists that the
     machine can and must decide on its own behavior, which is
     neither possible nor required."

But there is no "self-reference" just something being given its own
representation.

The actual *self-reference* is not Ĥ applied to ⟨Ĥ⟩.
It is the next line: *if Ĥ applied to ⟨Ĥ⟩ halts*
It is this line that incorrectly requires Ĥ to
report on its own behavior.

There is no "self-reference".

And the line  *if Ĥ applied to ⟨Ĥ⟩ halts* is a requirement/specification
on H, not Ĥ, so there is not "self" here.

It is a requirement of the aspect of Ĥ named Ĥ.embedded_H.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: comp.ai.philosophy

On 8/5/25 10:56 PM, olcott wrote:

On 8/5/2025 9:15 PM, Richard Damon wrote:

On 8/5/25 8:54 PM, olcott wrote:

On 8/5/2025 5:47 PM, Richard Damon wrote:

On 8/5/25 11:34 AM, olcott wrote:

On 8/5/2025 3:33 AM, Fred. Zwarts wrote:

Op 04.aug.2025 om 19:41 schreef olcott:

On 8/4/2025 12:33 PM, Mr Flibble wrote:

On Mon, 04 Aug 2025 11:55:41 -0500, olcott wrote:

Until you understand that there never was any actual finite string >>>>>>>>> machine description input that does the opposite of whatever >>>>>>>>> the decider
decides.

What makes you think that? You know quines are a thing, right? >>>>>>>>

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞, >>>>>>>>>      if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn >>>>>>>>>      if Ĥ applied to ⟨Ĥ⟩ does not halt.

Ĥ.embedded_H cannot report on its own behavior as the Linz proof >>>>>>>>> incorrectly presumes in its "if" statements.

TM deciders only compute the mapping from their actual finite >>>>>>>>> string
inputs they never compute any mapping from non-inputs such as >>>>>>>>> actual
Turing machines themselves.

Again an ENCODING of a TM is a valid input.

/Flibble

*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞, >>>>>>>     if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
    if Ĥ applied to ⟨Ĥ⟩ does not halt.

The contradiction in Linz's (or Turing's) self-referential
halting construction only appears if one insists that the machine >>>>>>> can and must decide on its own behavior, which is neither
possible nor required.

That is a huge mistake. Nobody insists on that. We only insists
that it must decide on its input, which specifies an aborting and >>>>>> halting program, not your hypothetical non-input that does not abort. >>>>>

*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

Definition of Turing Machine Ĥ applied
to its own machine description ⟨Ĥ⟩
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞, >>>>> if Ĥ applied to ⟨Ĥ⟩ halts, and
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn >>>>> if Ĥ applied to ⟨Ĥ⟩ does not halt.

*Lines 2 and 4 above*
*Does insist that Ĥ.embedded_H report on its own behavior*

Only because that is the behavior of its input.

That is not the behavior of its input.

    "The contradiction in Linz's (or Turing's) self-referential
     halting construction only appears if one insists that the
     machine can and must decide on its own behavior, which is
     neither possible nor required."

But there is no "self-reference" just something being given its own
representation.

The actual *self-reference* is not Ĥ applied to ⟨Ĥ⟩.
It is the next line: *if Ĥ applied to ⟨Ĥ⟩ halts*
It is this line that incorrectly requires Ĥ to
report on its own behavior.

There is no "self-reference".

And the line  *if Ĥ applied to ⟨Ĥ⟩ halts* is a requirement/
specification on H, not Ĥ, so there is not "self" here.

It is a requirement of the aspect of Ĥ named Ĥ.embedded_H.

Depends on which system you are working in.

IF you are talking about Linz system, where H *IS* a Halt Decider, and
MUST obey that requirement, but just might not exist, then yes, that requirement is inherited by Ĥ, but Ĥ not being able to do that just
proves that Ĥ doesn't exist, but since, by the method of construction,
if H exsits Ĥ exists, and thus if Ĥ doesn't exists, H doesn't exist, all
you are doing is proving that there is no H that meets the requirements,
and thus the set of Halt Deciders is empty.

That world is built on the concept of proof by contradiction, which it
seems you don't understand because it doesn't meet your very limited
concept of logic.

In the world where we can say that H exists, but are asking if it is
correct, the criteria does not transfer to Ĥ. You can say that it needs
to be meet for *H* to be correct, but that still isn't a requirement on Ĥ.


Sorry, you are just showing that you don't understand what you are
talking about, but are just repeating phrases by rote that you don't understand their meaning, because you chose to be ignorant.
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: comp.ai.philosophy

On 8/6/2025 6:01 AM, Richard Damon wrote:

On 8/5/25 10:56 PM, olcott wrote:

It is a requirement of the aspect of Ĥ named Ĥ.embedded_H.

Depends on which system you are working in.

IF you are talking about Linz system, where H *IS* a Halt Decider,

If H *is* a halt decider then the proof that no halt
decider exists is done.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.

*Lines 2 and 4 above*
*Does insist that Ĥ.embedded_H report on its own behavior*

"The contradiction in Linz's (or Turing's) self-referential
halting construction only appears if one insists that the
machine can and must decide on its own behavior, which is
neither possible nor required."

https://chatgpt.com/share/6890ee5a-52bc-8011-852e-3d9f97bcfbd8
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: comp.ai.philosophy

On 8/6/25 7:15 AM, olcott wrote:

On 8/6/2025 6:01 AM, Richard Damon wrote:

On 8/5/25 10:56 PM, olcott wrote:

It is a requirement of the aspect of Ĥ named Ĥ.embedded_H.

Depends on which system you are working in.

IF you are talking about Linz system, where H *IS* a Halt Decider,

If H *is* a halt decider then the proof that no halt
decider exists is done.

Nope, because while in that system, H *IS* a Halt Decider, but might not actually exist.

It seems you don't undertstand that concept,

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.

*Lines 2 and 4 above*
*Does insist that Ĥ.embedded_H report on its own behavior*

Which is perfectly valid, if done by giving the decider an input
representing a program that uses that decider.

Note, the input doesn't "refer" to the decider, it just has a
description of that decider within the description of the program.

   "The contradiction in Linz's (or Turing's) self-referential
    halting construction only appears if one insists that the
    machine can and must decide on its own behavior, which is
    neither possible nor required."

https://chatgpt.com/share/6890ee5a-52bc-8011-852e-3d9f97bcfbd8

Which since you LIED and said deciders aren't responsible for all input
in the domain (the descriptions of all programs) that answer is just meaningless.

Sorry, you are just proving you are just a stupid liar.

--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: comp.ai.philosophy

On 8/6/2025 6:29 AM, Richard Damon wrote:

On 8/6/25 7:15 AM, olcott wrote:

On 8/6/2025 6:01 AM, Richard Damon wrote:

On 8/5/25 10:56 PM, olcott wrote:

It is a requirement of the aspect of Ĥ named Ĥ.embedded_H.

Depends on which system you are working in.

IF you are talking about Linz system, where H *IS* a Halt Decider,

If H *is* a halt decider then the proof that no halt
decider exists is done.

Nope, because while in that system, H *IS* a Halt Decider, but might not actually exist.

A lack of sufficient precision in your words again.
H *is* a halt decider that may not exist is a contradiction.

It seems you don't undertstand that concept,

You are not being careful enough with the precision of your words.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.

*Lines 2 and 4 above*
*Does insist that Ĥ.embedded_H report on its own behavior*

Which is perfectly valid, if done by giving the decider an input representing a program that uses that decider.

Note, the input doesn't "refer" to the decider, it just has a
description of that decider within the description of the program.

    "The contradiction in Linz's (or Turing's) self-referential
     halting construction only appears if one insists that the
     machine can and must decide on its own behavior, which is
     neither possible nor required."

https://chatgpt.com/share/6890ee5a-52bc-8011-852e-3d9f97bcfbd8

Which since you LIED and said deciders aren't responsible for all input
in the domain (the descriptions of all programs) that answer is just meaningless.

Turing machines are not in the domain of any other Turing machine.

Sorry, you are just proving you are just a stupid liar.

It is you not being precise enough with the meaning of words.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: comp.ai.philosophy

On Wed, 06 Aug 2025 07:01:18 -0500, olcott wrote:

On 8/6/2025 6:29 AM, Richard Damon wrote:

On 8/6/25 7:15 AM, olcott wrote:

On 8/6/2025 6:01 AM, Richard Damon wrote:

On 8/5/25 10:56 PM, olcott wrote:

It is a requirement of the aspect of Ĥ named Ĥ.embedded_H.

Depends on which system you are working in.

IF you are talking about Linz system, where H *IS* a Halt Decider,

If H *is* a halt decider then the proof that no halt decider exists is
done.

Nope, because while in that system, H *IS* a Halt Decider, but might
not actually exist.

A lack of sufficient precision in your words again.
H *is* a halt decider that may not exist is a contradiction.

No, the proofs are asking if such a halt decider exists so it is a
*premise* (a statement that is provisionally accepted as true for the
purpose of the argument) that a *hypothetical* H exists and the proofs
then go on to show that such an H cannot exist.

/Flibble
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: comp.ai.philosophy

On 8/6/2025 12:38 PM, Mr Flibble wrote:

On Wed, 06 Aug 2025 07:01:18 -0500, olcott wrote:

On 8/6/2025 6:29 AM, Richard Damon wrote:

On 8/6/25 7:15 AM, olcott wrote:

On 8/6/2025 6:01 AM, Richard Damon wrote:

On 8/5/25 10:56 PM, olcott wrote:

It is a requirement of the aspect of Ĥ named Ĥ.embedded_H.

Depends on which system you are working in.

IF you are talking about Linz system, where H *IS* a Halt Decider,

If H *is* a halt decider then the proof that no halt decider exists is >>>> done.

Nope, because while in that system, H *IS* a Halt Decider, but might
not actually exist.

A lack of sufficient precision in your words again.
H *is* a halt decider that may not exist is a contradiction.

No, the proofs are asking if such a halt decider exists so it is a
*premise* (a statement that is provisionally accepted as true for the
purpose of the argument) that a *hypothetical* H exists and the proofs

Yes you have that exactly correct.

then go on to show that such an H cannot exist.

/Flibble

--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: comp.ai.philosophy

On Wed, 06 Aug 2025 12:44:56 -0500, olcott wrote:

On 8/6/2025 12:38 PM, Mr Flibble wrote:

On Wed, 06 Aug 2025 07:01:18 -0500, olcott wrote:

On 8/6/2025 6:29 AM, Richard Damon wrote:

On 8/6/25 7:15 AM, olcott wrote:

On 8/6/2025 6:01 AM, Richard Damon wrote:

On 8/5/25 10:56 PM, olcott wrote:

It is a requirement of the aspect of Ĥ named Ĥ.embedded_H.

Depends on which system you are working in.

IF you are talking about Linz system, where H *IS* a Halt Decider,

If H *is* a halt decider then the proof that no halt decider exists
is done.

Nope, because while in that system, H *IS* a Halt Decider, but might
not actually exist.

A lack of sufficient precision in your words again.
H *is* a halt decider that may not exist is a contradiction.

No, the proofs are asking if such a halt decider exists so it is a
*premise* (a statement that is provisionally accepted as true for the
purpose of the argument) that a *hypothetical* H exists and the proofs

Yes you have that exactly correct.

The next bit is exactly correct too:

then go on to show that such an H cannot exist.

/Flibble

/Flibble
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: comp.ai.philosophy

On 8/6/2025 1:42 PM, Mr Flibble wrote:

On Wed, 06 Aug 2025 12:44:56 -0500, olcott wrote:

On 8/6/2025 12:38 PM, Mr Flibble wrote:

On Wed, 06 Aug 2025 07:01:18 -0500, olcott wrote:

On 8/6/2025 6:29 AM, Richard Damon wrote:

On 8/6/25 7:15 AM, olcott wrote:

On 8/6/2025 6:01 AM, Richard Damon wrote:

On 8/5/25 10:56 PM, olcott wrote:

It is a requirement of the aspect of Ĥ named Ĥ.embedded_H.

Depends on which system you are working in.

IF you are talking about Linz system, where H *IS* a Halt Decider, >>>>>> If H *is* a halt decider then the proof that no halt decider exists >>>>>> is done.

Nope, because while in that system, H *IS* a Halt Decider, but might >>>>> not actually exist.

A lack of sufficient precision in your words again.
H *is* a halt decider that may not exist is a contradiction.

No, the proofs are asking if such a halt decider exists so it is a
*premise* (a statement that is provisionally accepted as true for the
purpose of the argument) that a *hypothetical* H exists and the proofs

Yes you have that exactly correct.

The next bit is exactly correct too:

then go on to show that such an H cannot exist.

/Flibble

/Flibble

Until you spend 22 years on this and examine it
from angles that did not occur to anyone else.

This same self-referential structure that is based
on the Liar Paradox: "This sentence is not true"
is at the heart of several other issues in logic
and mathematics.

After 2000 years the best experts in the field
of the philosophy of logic still don't understand
that all of these expressions of language are
semantically ill-formed.

The philosophical term is "not a bearer of truth".
Logicians tend to have never heard of this term
thus dismiss it as nonsense.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: comp.ai.philosophy

On 8/6/25 8:01 AM, olcott wrote:

On 8/6/2025 6:29 AM, Richard Damon wrote:

On 8/6/25 7:15 AM, olcott wrote:

On 8/6/2025 6:01 AM, Richard Damon wrote:

On 8/5/25 10:56 PM, olcott wrote:

It is a requirement of the aspect of Ĥ named Ĥ.embedded_H.

Depends on which system you are working in.

IF you are talking about Linz system, where H *IS* a Halt Decider,

If H *is* a halt decider then the proof that no halt
decider exists is done.

Nope, because while in that system, H *IS* a Halt Decider, but might
not actually exist.

A lack of sufficient precision in your words again.
H *is* a halt decider that may not exist is a contradiction.

Not at all, your logic just can't handle that reality.

H is a Halt Decider means that H is an element of the set of Halt Deciders.

From that, we can estabish a number of properties that H must posees.

This does NOT assert that H exists, just the properties that H must have
it it does.

The fact that the set of Halt Deciders is an empty set, means that it
doesn't exist, but the properties we proved for it, still stand.

It seems you don't undertstand that concept,

You are not being careful enough with the precision of your words.

No, you just don't understand the logic.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,
if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if Ĥ applied to ⟨Ĥ⟩ does not halt.

*Lines 2 and 4 above*
*Does insist that Ĥ.embedded_H report on its own behavior*

Which is perfectly valid, if done by giving the decider an input
representing a program that uses that decider.

Note, the input doesn't "refer" to the decider, it just has a
description of that decider within the description of the program.

    "The contradiction in Linz's (or Turing's) self-referential
     halting construction only appears if one insists that the
     machine can and must decide on its own behavior, which is
     neither possible nor required."

https://chatgpt.com/share/6890ee5a-52bc-8011-852e-3d9f97bcfbd8

Which since you LIED and said deciders aren't responsible for all
input in the domain (the descriptions of all programs) that answer is
just meaningless.

Turing machines are not in the domain of any other Turing machine.

Sure they are, via representations.

I guess you think arithmatic isn't computedable, as numbers aren't in
the domain of Turing Machines by your definiton, and thus no Turing
machine can compute the sum of One and Two.

Sorry, you are just proving you are just a stupid liar.

It is you not being precise enough with the meaning of words.

Nope, Your stupidity doesn't invalidate the concepts.
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: comp.ai.philosophy

On 8/6/25 3:02 PM, olcott wrote:

On 8/6/2025 1:42 PM, Mr Flibble wrote:

On Wed, 06 Aug 2025 12:44:56 -0500, olcott wrote:

On 8/6/2025 12:38 PM, Mr Flibble wrote:

On Wed, 06 Aug 2025 07:01:18 -0500, olcott wrote:

On 8/6/2025 6:29 AM, Richard Damon wrote:

On 8/6/25 7:15 AM, olcott wrote:

On 8/6/2025 6:01 AM, Richard Damon wrote:

On 8/5/25 10:56 PM, olcott wrote:

It is a requirement of the aspect of Ĥ named Ĥ.embedded_H. >>>>>>>>>

Depends on which system you are working in.

IF you are talking about Linz system, where H *IS* a Halt Decider, >>>>>>> If H *is* a halt decider then the proof that no halt decider exists >>>>>>> is done.

Nope, because while in that system, H *IS* a Halt Decider, but might >>>>>> not actually exist.

A lack of sufficient precision in your words again.
H *is* a halt decider that may not exist is a contradiction.

No, the proofs are asking if such a halt decider exists so it is a
*premise* (a statement that is provisionally accepted as true for the
purpose of the argument) that a *hypothetical* H exists and the proofs

Yes you have that exactly correct.

The next bit is exactly correct too:

then go on to show that such an H cannot exist.

/Flibble

/Flibble

Until you spend 22 years on this and examine it
from angles that did not occur to anyone else.

This same self-referential structure that is based
on the Liar Paradox: "This sentence is not true"
is at the heart of several other issues in logic
and mathematics.

No, your problem is you have lies on the brain, and think everything is
just a lie.

After 2000 years the best experts in the field
of the philosophy of logic still don't understand
that all of these expressions of language are
semantically ill-formed.

The philosophical term is "not a bearer of truth".
Logicians tend to have never heard of this term
thus dismiss it as nonsense.

But since you don't understand what truth is, especially in formal
systems, nothing you say actually makes sense.

When someone claims that NOBODY else understands something, that is a
good clue that THEY don't understand what they are talking about.

YOU have proved this about yourself, but of course, since you have
brainwashed yourself to prevent yourself form understanding what you are talking about, you can't understand the truth that others say, as it
doesn't match the lies you have convinced yourself of.

Sorry, you are just showing that you are mentally incompetent and don't uderstand what you are saying.

Yes, there HAVE been brilliant men who found misconceptions and errors
in what has been commonly taken as truth, but they have been able to
break that idea down into the true fundamentals that allow others to see
what they saw.

You can't do that, as you hit the point of your lie, and to break it
down smaller just reveals the lie clearly, so you can't let yourself do
that, as it might break you out of your brainwashing.

Sorry, all you are doing is proving that you are just a pathological
liar, that just won't understand the truth of what he is talking about.

When you claim the definitiona are incorrect, and try to change them in system, rather than creating a new system, you show you don't understand
what you are doing.

If Classical Computation Theory has a wrong definition that causes it to
be disagreeable to you, fully define POOPS just like Zermelo did with
ZFC. The problem is that task seems to be beyound you ability, so
instead you just live a lie.
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