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”.
--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
On 10/26/23 8:45 AM, olcott wrote:
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.
It must give to correct answer to the problem given to it.
Where is the "rule" that says it can't consider that case? It *MUST* consider that case or it violates the "All Inputs" requirement.
How can that possibly affect the behavior of the code that already
exists in H / embedded_H.
It only CAN use what it can compute, but the key is that it turns out
this is insufficient to actually be able to answer the question.
If embedded_H needs to rely on its correct simulation of the input, then
it needs to correctly simulate that input, which means its simulation
needs to reflect what the direct execution of the program the input specifies will do. Since that program halts, since you claim H(D,D)
returns 0, that simulation must indicate the input is Halting, or the simulation is incorrect, and embedded_H needs to blaim its own failing
to correctly simulate.
You want to ban the correct answer, because it shows that the
computation is impossible to do, but that it part of the question, *IS*
the computatoin possible.
You are just showing you don't understand that basics of computability,
or Truth, in your unfounded assumption that all quetions need to be computable, or Truths provable.
Your illogical adherence to this false premise is just making you whole logic system unsound and inconsistant.
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