Re: The philosophy of computation reformulates existing ideas on a new basis ---

On 10/28/24 8:57 PM, olcott wrote:

On 10/28/2024 6:56 PM, Richard Damon wrote:

On 10/28/24 11:04 AM, olcott wrote:

On 10/28/2024 6:16 AM, Richard Damon wrote:

The machine being used to compute the Halting Function has taken a finite string description, the Halting Function itself always took a Turing Machine,
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That is incorrect. It has always been the finite string Turing Machine
description of a Turing machine is the input to the halt decider.
There are always been a distinction between the abstraction and the
encoding.

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Nope, read the problem you have quoted in the past.
>

 Ultimately I trust Linz the most on this:
 the problem is: given the description of a Turing machine
M and an input w, does M, when started in the initial
configuration qow, perform a computation that eventually halts?
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf

So, he begins by asking if we can built that Turing Machine to compute the Halting Function.
Note, the criteria:
Does M, when started in the initial configuration qow, perform a computation that eventually halts?
That is the Halting Function. What is its "inputs" the Machine M and the input tape.
NOT the description. That is just what was given to the decider.

 Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
 Linz also makes sure to ignore that the behavior of ⟨Ĥ⟩ ⟨Ĥ⟩
correctly simulated by embedded_H cannot possibly reach
either ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩ because like everyone else he rejects
simulation out of hand:

Because the PARTIAL emulation that you describe doesn't matter, and you seem to forget the BY DEFINITION embedded_H does exactly the same thing as H.
He doesn't "reject" simulation, he is just using the fact that the only "correct simuation" is a simulation that exactly reproduces the behavior of that which it is simulating.

 We cannot find the answer by simulating the action of M on w,
say by performing it on a universal Turing machine, because
there is no limit on the length of the computation.

So?
Just because it is impossible to do doesn't meaan we can't be asked to do it. Especially when the ultimate question is CAN we do it.

 DDD emulated by HHH according to the semantics of the x86
language cannot possibly reach its own "return" instruction
whether or not HHH ever aborts its emulation of DDD.

because the emulation of DDD by HHH can never be complete and correct and have HHH give an answer.

 IS FREAKING ISOMORPHIC TO
 ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H cannot possibly reach
either ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.
 

Nope. Since partial emulation is not "isomorphic" to complete behavior.
Now, perhaps you can use the ismorphism of both being impossible to do to try to demonstart that the halting problem is uncomputable.
But, trying to claim that the partial emulation by some copy of H not reaching the final state says that H^ <H^> is non-halting is just bad logic, and a LIE.