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On Wed, 21 May 2025 21:32:49 +0200, Fred. Zwarts wrote:Its even worse than that.
Op 21.mei.2025 om 17:54 schreef olcott:You are making the classic mistake of conflating the decider halting withOn 5/21/2025 12:56 AM, Richard Heathfield wrote:Verifiable counter-factual.On 21/05/2025 06:23, olcott wrote:A self-contradictory input and a proof by contradiction are not theOn 5/20/2025 9:15 PM, Richard Damon wrote:>On 5/20/25 3:10 PM, Mr Flibble wrote:
<snip>
>>>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.
But you can't "redefine" the Halting Problem and then say you have
answered the Halting Problem.
Do you mean like how ZFC resolved Russell's Paradox thus converting
"set theory" into "naive set theory"?
No, because there is no paradox in the Halting Problem. A proof by
contradiction is not a paradox.
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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.
>A better parallel would be Cantor's proof that there are uncountablyLikewise with Russell's Paradox it is assumed that there can be a set
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.
>
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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of all sets that do not contain themselves as members. This is
"resolved" as nonsense.
>Whatever you think your HHH determines, we know from Turing that itvoid DDD()
doesn't determine it for arbitrary programs with arbitrary input. It
therefore has no bearing whatsoever on the Halting Problem.
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{
HHH(DDD);
return;
}
>
DDD correctly simulated by HHH DOES NOT HALT.
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The simulation of DDD does not reach a natural end only because HHH
prevents it to halt by a premature abort.
Due this premature abort HHH misses the part of the input that specifies
the conditional abort in Halt7.c, which specifies that the program
halts. If a simulator would not abort this input, it would reach the
natural end of the program. Proven by direct execution and world-class
simulators. But HHH has a bug, which makes that it aborts before it can
see that the input halts, only because its programmer dreamed of an
infinite recursion, where there is only a finite recursion.
Come out of rebuttal mode, which makes that you do not pay enough
attention to this logic, but reject it immediately when it disturbs your
dreams, after which you only repeat the clueless claims.
the program being analysed halting.
/Flibble
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