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

On 5/21/25 11:54 AM, olcott wrote:

On 5/21/2025 12:56 AM, Richard Heathfield wrote:

On 21/05/2025 06:23, olcott wrote:

On 5/20/2025 9:15 PM, Richard Damon wrote:

On 5/20/25 3:10 PM, Mr Flibble wrote:

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<snip>
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Conclusion: ----------- Flibble sharpens his argument by
clarifying that SHDs are not required to simulate infinite
execution. They are expected to *detect* infinite behavior
structurally and respond in finite time. This keeps them
within the bounds of what a decider must be and
strengthens the philosophical coherence of his
redefinition of the Halting Problem.

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But you can't "redefine" the Halting Problem and then say you have answered the Halting Problem.

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Do you mean like how ZFC resolved Russell's
Paradox thus converting "set theory" into "naive set theory"?

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No, because there is no paradox in the Halting Problem. A proof by contradiction is not a paradox.
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 A self-contradictory input and a proof by contradiction
are not the same thing. A proof by contradiction would
conclude that "this sentence is not true" is true because
it cannot be proved false.

The input is NOT "self-contradictory".
The proof by contradiction is based on, If an H existed that got the right answer, the answer it gives would be wrong.

 ZFC shows how a whole way of examining a problem can be
tossed out as incorrect and replaced with a whole new way.

Nope, They didn't say we needed to "toss out" Naive Set Theory, as that was already being tossed out knowing it needed a replacement, he just provided an alternative.

 The HP proofs are based on defining a D that can
actually do the opposite of whatever value that H returns.
No such D can actually exist.

Sure they do. WHy doesn't the D in the proof do the opposite of that H does.
Remember, H is a PROGRAM, and thus H(D) has a FIXED answer, that will always be the same, and thus D can call H(D), get its answer, and the do the opposite.
What step is impossible?
If H(D) doesn't answer to D, it doesn't answer to anyone, and thus fails to be a decider.

 

A better parallel would be Cantor's proof that there are uncountably many real numbers, or Euclid's proof that there is no largest prime. Both of these proofs make a single assumption and then derive a contradiction, thus showing that the assumption must be false. No paradoxes need apply.
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In the Halting Problem's case, the assumption is that a UNIVERSAL algorithm exists for determining whether any arbitrary program halts when applied to given arbitrary input. The argument derives a contradiction showing the assumption to be false.
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 Likewise with Russell's Paradox it is assumed that there
can be a set of all sets that do not contain themselves as
members. This is "resolved" as nonsense.

No, Russell's Paradox was based on the "rules" of the old (now called Naive) set theory, that said you could have any set you could describe.
What he showed was that such a definition of what sets you could have lead to an inconsistant system, so RUSSEL is the one that made the field see the need to "throw out" that version of set theory.

 

Whatever you think your HHH determines, we know from Turing that it doesn't determine it for arbitrary programs with arbitrary input. It therefore has no bearing whatsoever on the Halting Problem.
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 void DDD()
{
   HHH(DDD);
   return;
}
 DDD correctly simulated by HHH DOES NOT HALT.
 

But the HHH that correctly simulates DDD doesn't give an answer.
And a diffferent HHH gets a different DDD, as to be an input, it needs to be a program, and thus paired with the HHH that exists, which is different for those two cases.
The HHH that doesn't correctly simulate its DDD can't use the fact that a DIFFERENT HHH, when it correctly simulated A DIFFERENT PROGRAM DDD, saw that it was non-halting.
All you are doing is proving that you are nothing but a stupid liar that can't understand the category error you are making in your faulty logic.