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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
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Subject: =?UTF-8?Q?Re=3A_Analysis_of_Flibble=E2=80=99s_Latest=3A_Detecting_v?=
 =?UTF-8?Q?s=2E_Simulating_Infinite_Recursion_ZFC?=
Date: Wed, 21 May 2025 21:32:49 +0200
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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:
>>
>> <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.
>>
> 
> 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.
> 
> ZFC shows how a whole way of examining a problem can be
> tossed out as incorrect and replaced with a whole new way.
> 
> 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 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.
>>
>> 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.
>>
> 
> 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.
> 
>> 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.
>>
> 
> void DDD()
> {
>    HHH(DDD);
>    return;
> }
> 
> DDD correctly simulated by HHH DOES NOT HALT.
> 

Verifiable counter-factual.

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.