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

Op 21.mei.2025 om 17:54 schreef olcott:

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.
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ZFC shows how a whole way of examining a problem can be
tossed out as incorrect and replaced with a whole new way.
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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.
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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.
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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;
}
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DDD correctly simulated by HHH DOES NOT HALT.
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 Verifiable counter-factual.
 The simulation of DDD does not reach a natural end only because HHH prevents it to halt by a premature abort.