Re: Flibble’s Leap: Why Behavioral Divergence Implies a Type Distinction in the Halting Problem

On 5/11/25 11:56 AM, olcott wrote:

On 5/11/2025 10:49 AM, Mr Flibble wrote:

On Sun, 11 May 2025 16:47:09 +0100, Richard Heathfield wrote:
>

On 11/05/2025 16:34, Mr Flibble wrote:

On Sun, 11 May 2025 16:25:14 +0100, Richard Heathfield wrote:
>

For a question to be semantically incorrect, it takes more than just
you and your allies to be unhappy with it.

>
For a question to be semantically correct, it takes more than just you
and your allies to be happy with it.

>
Indeed. It has to have meaning. It does. That meaning has to be
understood by sufficiently intelligent people. It is.
>
You don't like the question. I get that. I don't know /why/ you don't
like it, because all your explanations to date have been complete
expletive deleted. For a Usenet article to be semantically correct, it
helps if your readers can understand what the <exp. del.> you're talking
about.
>
What I get from your stand is that you agree with olcott that a
'pathological' input halts... no, never halts... well, you can't decide
between you, but you're agreed that it's definitely decidable, right?

>
Re-read the OP for my answer:
>
Flibble’s Leap: Why Behavioral Divergence Implies a Type Distinction in
the Halting Problem
===========================================================================================
>
Summary
-------
Flibble argues that the Halting Problem's undecidability proof is built on
a category (type) error: it assumes a program and its own representation
(as a finite string) are interchangeable. This assumption fails under
simulating deciders, revealing a type distinction through behavioral
divergence. As such, all deciders must respect this boundary, and
diagonalization becomes ill-formed. This reframing dissolves the paradox
by making the Halting Problem itself an ill-posed question.
>
1. Operational Evidence of Type Distinction
-------------------------------------------
- When a program (e.g., `DD()`) is passed to a simulating halt decider
(`HHH`), it leads to infinite recursion.
- This behavior differs from direct execution (e.g., a crash due to a
stack overflow).

 The directly executed DD() simply halts because
HHH has stopped the infinite recursion that it
specifies on its second recursive call.

Right, HHH ALWAYS breaks that loop by aborting, and thus DD is non-halting, and *ANY* correct (and thus complete) emulation will reach a final state.

 DD emulated by HHH according to the rules of the
x86 language cannot possibly halt. Because all deciders
are required to report on what their finite string input
specifies HHH must reject DD as non-halting.
 

But HHH doesn't do that. You just admitted in the previous paragraph that HHH stops its simulation, and thus breaks the rules of the x86 processor, so your criteria is just based on lies. The fact that a partial emulation doesn't reach the final state that the complete emulation does isn't actually evidence of non-halting.
You don't get to assume that the impossible happens.