XPost: comp.theory, sci.logic
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.
--- SoupGate-Win32 v1.05
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