Re: Analysis of Flibble’s Latest: Detecting vs. Simulating Infinite Recursion ZFC

On 5/23/25 12:00 PM, olcott wrote:

On 5/23/2025 8:00 AM, Ben Bacarisse wrote:

Mike Terry <news.dead.person.stones@darjeeling.plus.com> writes:
>

On 22/05/2025 06:41, Richard Heathfield wrote:

On 22/05/2025 06:23, Keith Thompson wrote:

Richard Heathfield <rjh@cpax.org.uk> writes:

On 22/05/2025 00:14, olcott wrote:

On 5/21/2025 6:11 PM, Richard Heathfield wrote:

[...]

Turing proved that what you're asking is impossible.
>

That is not what he proved.

>
Then you'll be able to write a universal termination analyser that can
correctly report for any program and any input whether it halts. Good
luck with that.

>
Not necessarily.

Of course not. But I'm just reflecting. He seemed to think that my
inability to write the kind of program Turing envisaged (an inability
that I readily concede) is evidence for his argument. Well, what's sauce
for the goose is sauce for the gander.
>

Even if olcott had refuted the proofs of the
insolvability of the Halting Problem -- or even if he had proved
that a universal halt decider is possible

And we both know what we both think of that idea.
>

-- that doesn't imply
that he or anyone else would be able to write one.

Indeed.
>

I've never been entirely clear on what olcott is claiming.

Nor I. Mike Terry seems to have a pretty good handle on it, but no matter
how clearly he explains it to me my eyes glaze over and I start to snore.

>
Hey, it's the way I tell 'em!
>
Here's what the tabloids might have said about it, if it had made the
front pages when the story broke:
>
   COMPUTER BOFFIN IS TURING IN HIS GRAVE!
>
   An Internet crank claims to have refuted Linz HP proof by creating a
   Halt Decider that CORRECTLY decides its own "impossible input"!
   The computing world is underwhelmed.
>
Better?  (Appologies for the headline, it's the best I could come up
with.)

>
And the big picture is that this can be done because false is the
correct halting decision for some halting computations.  He has said
this explicitly (as I have posted before) but he has also explained it
in words:
>
| When-so-ever a halt decider correctly determines that its input would
| never halt unless forced to halt by this halt decider this halt
| decider has made a correct not-halting determination.
>

 <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     If simulating halt decider H correctly simulates its
     input D until H correctly determines that its simulated D
     would never stop running unless aborted then

Right, and since The specified PROGRAM D (built on the specified program H which returns 0 as you claim to be correct), when correctly simulated will reach an end in finite time, it is IMPOSSIBLE for H to CORRECTLY determine that said simulation will never stop running unless aborted.
The Hypothetical H simulating that DIFFERENT hypothetical D is irrelevent, as it isn't THAT D but something else. Your Hypothetical H when simulating THIS D (which includes the code of the actual H) will find the end. The fact that you can't put that hypothetical H in the same memory location as the original H was without changing the full input (which needed to include that code) just show the ERROR in your computation setup.

 

Or perhaps you prefer this explanation from 2023:
>
| (1) A return value of 1 from H(D,D) means the input to H(D,D) has halted
|
| (2) A return value of 0 from H(D,D) has been redefined to mean
|     (a) D does not halt
|     (b) D has been defined to do the opposite of whatever Boolean value
|         that H returns.
>
All very clear.  Of course (2)(b) is undeciable in general.
>

 (2)(b) is not sufficiently precise.
 INPUT D to simulating termination analyzer H
has been defined to do the opposite of whatever
Boolean value that H returns.
 *Cannot possibly be defined*
 int main()
{
   DD(); // The HHH that DD calls cannot possibly report
}       // on the behavior of its caller.
         // That is just NOT the way that computations work.