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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Every HHH(DDD) is correct to reject its input
Date: Sat, 31 May 2025 15:40:15 -0500
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On 5/31/2025 3:30 PM, Mr Flibble wrote:
> On Sat, 31 May 2025 15:10:39 -0500, olcott wrote:
> 
>> On 5/31/2025 2:44 PM, Mr Flibble wrote:
>>> Flibble's Argument: Execution vs Simulation in SHDs
>>> ====================================================
>>>
>>> In the context of Simulating Halt Deciders (SHDs), the distinction
>>> between execution and simulation is fundamental.
>>>
>>> Correct: External Simulation ----------------------------
>>> int main() {
>>>       HHH(DDD); // SHD simulates/analyzes DDD from the outside.
>>> }
>>>
>>> - In this model, DDD is not being executed — it's being passed as data
>>> to HHH, which is analyzing it.
>>> - Even if DDD() (the function definition) contains a recursive call to
>>> HHH(DDD), this is just part of the code being simulated, not something
>>> that is actively executing.
>>> - Thus, the simulation can detect infinite recursion structurally,
>>> without running DDD.
>>>
>>> Incorrect: Active Execution ---------------------------
>>> int main() {
>>>       DDD(); // Directly executes DDD, which calls HHH(DDD) during
>>>       runtime.
>>> }
>>>
>>>
>> We can't simply reject this as incorrect since it is the basis of every
>> rebuttal of my work.
>>
>> It *is* incorrect to assume that the HHH that DDD calls is supposed to
>> report on the behavior of its caller.
>>
>>> - In this scenario, you’re actually running DDD, not simulating it.
>>> - If DDD() calls HHH(DDD) at runtime, you're now mixing execution and
>>> analysis in the same layer, violating the stratified model.
>>> - This results in self-referential execution that undermines
>>> decidability — a category error akin to the original halting paradox.
>>>
>>> Key Insight -----------
>>> As long as DDD is not executing and is only being simulated by HHH, it
>>> doesn’t matter that DDD would call HHH(DDD) — because that call is
>>> never actually made. It exists in the simulated model, not in the
>>> runtime environment. Thus, structural recursion can be detected safely
>>> and treated as non-halting without triggering a paradox.
>>>
>>> This stratification (meta → base) is what keeps the model coherent.
>>
>> A PhD computer scientist Eric Hehner has this same view. He explains
>> this view as the analyzer and the analyzed are in different programming
>> languages where the input cannot directly call its analyzer.
>>
>> I only very recently discovered that it is 100% impossible to actually
>> define *an input* that does the opposite of whatever value its analyzer
>> returns.
>>
>> In every conventional proof of the halting problem it has always been
>> that the decider cannot correctly report on the behavior of its caller.
>>
>> You will find thousands of messages posted in this forum where everyone
>> says that I am wrong because HHH does not report on the behavior of the
>> direct execution of DDD() (AKA its caller).
> 
> You cannot both execute and simulate DDD as part of the same analysis, if
> you do that then you are WRONG.
> 
> /Flibble

int main()
{
   DDD(); // calls HHH(DDD) that simulates its own separate
}        // instance of DDD. The analysis does not begin
          // until after HHH(DDD) is called.
-- 
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer