Re: Correcting the definition of the halting problem --- Computable functions

Am Mon, 24 Mar 2025 22:29:06 -0500 schrieb olcott:

On 3/24/2025 10:12 PM, dbush wrote:

On 3/24/2025 10:07 PM, olcott wrote:

On 3/24/2025 8:46 PM, André G. Isaak wrote:

On 2025-03-24 19:33, olcott wrote:

On 3/24/2025 7:00 PM, André G. Isaak wrote:

>

In the post you were responding to I pointed out that computable
functions are mathematical objects.

Computable functions implemented using models of computation would
seem to be more concrete than pure math functions.

Those are called computations or algorithms, not computable
functions.

https://en.wikipedia.org/wiki/Pure_function Is another way to look at
computable functions implemented by some concrete model of
computation.

And not all mathematical functions are computable, such as the halting
function.

The halting problems asks whether there *is* an algorithm which can
compute the halting function, but the halting function itself is a
purely mathematical object which exists prior to, and independent of,
any such algorithm (if one existed).

None-the-less it only has specific elements of its domain as its
entire basis. For Turing machines this always means a finite string
that (for example) encodes a specific sequence of moves.

False.  *All* turing machine are the domain of the halting function,
and the existence of UTMs show that all turning machines can be
described by a finite string.

You just aren't paying enough attention. Turing machines are never in
the domain of any computable function. <snip>

Fine, their descriptions are, and their behaviour is computable -
by running them.

The math halting function is free to "abstract away" key details that
change everything. That is why I have never been talking about the
pure math and have always been referring to its implementation in a
model of computation.

What details?

There are no details abstracted away.  The halting function is simply
uncomputable.

When the measure of the behavior specified by a finite string input DD
is when correctly emulated by HHH then DD can't reach its own final halt
state then not-halting is decidable.

Circular reasoning. You'll have to prove HHH simulates correctly first.

A halt decider cannot exist

So again, you explicitly agree with the Linz proof and all other proofs
of the halting function.
 

because the halting problem is defined incorrectly

There's nothing incorrect about wanting something that would solve the
Goldbach conjecture and make unknowable truths knowable.  Your
alternate definition won't provide that.
There's no requirement that a function be computable.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.