Re: Final Statement on the Halting Problem

On 6/16/25 2:33 PM, Mr Flibble wrote:

On Mon, 16 Jun 2025 13:11:23 +0300, Mikko wrote:
 

On 2025-06-15 19:21:09 +0000, Mr Flibble said:
>

On Sun, 15 Jun 2025 14:23:36 -0400, Richard Damon wrote:
>

On 6/15/25 9:55 AM, Mr Flibble wrote:

The halting problem as defined ignores recursive self reference
focusing on the paradox instead, I would argue the recursive self
reference leads to infinite regress in the definition of the problem
thus creating a category error making the problem definition itself
ill-formed.
>
/Flibble

>
But there is no recursive self-reference in the halting problem.
>
You only get that recursion when you assume that there exists a
program that can solve it, which is what shows that there is not
computation that can solve the halting problem.
>
You have just fallen for Peter Olcotts deceptive strawman definition
of the halting problem, because you don't really understand what you
are talking about.

>
Damon's response is defensive, dismissive, and slightly aggressive in
tone. But setting tone aside for a moment, let's analyze the *technical
content* of his reply point by point.
>
---
>
### 🔍 **Claim-by-Claim Analysis**
>
#### **1. "But there is no recursive self-reference in the halting
problem."**
>
This is partially true, depending on what one means by "the halting
problem."

>
Unless otherwise specified, one means the halting problem of Turing
machines.
>

* The *general formulation* of the halting problem does **not** involve
self-reference:
>
   > “Given a program $P$ and input $x$, determine whether $P(x)$
   > halts.”

>
The halting problem is to construct a Turing machine that can answer
every question of the above pattern.
>

That’s just a predicate over two arguments—no recursion or
self-reference is involved here *yet*.
>
* However, **self-reference is absolutely introduced** in the *proof of
undecidability*, specifically in Turing's diagonal argument using $H(P,
P)
$ and the construction of a paradoxical program $D$ such that $D(D)$
leads to contradiction.

>
There is no self-reference in Turing's construction of $D$. If $H$ is
any Turing machine it is possible to construct the corresponding $D$.
Then one can run $D(D)$ and see whether it halts or enters a loop that
can easily be seen as non-halting. Turing's $D$ does not do anything
else. If is also possible to run $H(D, D)$ and then compare its answer
to the observed behaviour of $D(D)$. There is nothing paradoxcal there,
the answer is simply wrong.

 BULLSHIT
 /Flibble

Just showing your stupidity.
There is no "self-reference" in the construction of Turing's machine, that takes the input (which becomes the description of itself, but not generated by itself, so not a "self-reference").
The no input version also doesn't have a "self-reference" either, but is just carefully constructed like the C program that prints its own source code where a large part of the code is just a "text string" that just matches other code of the program that processes a prints it twice to give the full output. (This is why Turing didn't use that version, it actually is very hard to build in a Turing Machine, it was used in Professor Sipser's version, which was using a higher level language)
Absolutely NO "self-reference"
In fact, it turns out that Turing Machines don't use "references" at all, so can't have "self-references".
You are just showing that you have fallen for PO's lies.