XPost: comp.theory, sci.logic, sci.math   
   From: 643-408-1753@kylheku.com   
      
   On 2025-10-23, olcott  wrote:   
   > On 10/22/2025 6:01 PM, Kaz Kylheku wrote:   
   >> On 2025-10-22, olcott  wrote:   
   >>> On 10/22/2025 3:20 PM, Kaz Kylheku wrote:   
   >>>> On 2025-10-22, olcott  wrote:   
   >>>>> On 10/22/2025 2:52 PM, Kaz Kylheku wrote:   
   >>>>>> On 2025-10-22, André G  Isaak  wrote:   
   >>>>>>> On 2025-10-22 12:40, Kaz Kylheku wrote:   
   >>>>>>>> But that entire bundle is one fixed case DD, with a single behavior,   
   >>>>>>>> which is a property of DD, which is a finite string.   
   >>>>>>>   
   >>>>>>> I think part of the problem here is that Olcott doesn't grasp that the   
   >>>>>>> "finite string input" DD *must* include as a substring the entire   
   >>>>>>> description of HHH.   
   >>>>>>   
   >>>>>> Furthermore, he doesn't get that it doesn't literally have to be HHH,   
   >>>>>> but the same algorithm: a workalike.   
   >>>>>>   
   >>>>>> The HHH analyzing DD's halting could be in C, while the HHH   
   >>>>>> called by DD could be in Python.   
   >>>>>   
   >>>>> DD does call HHH(DD) in recursive simulation   
   >>>>> and you try to get away with lying about it.   
   >>>>   
   >>>> I'm saying that's not a requirement in the halting problem.   
   >>>>   
   >>>> DD does not have to use that implementation of HHH; it can have   
   >>>> its own clean-room implementation and it can be in any language.   
   >>>>   
   >>>> But nonetheless, yes, there will still be a nested simulation tower.   
   >>>>   
   >>>   
   >>> I made sure to read what you said all the way through   
   >>> this time. DD correctly simulated by HHH cannot possibly   
   >>> reach its own final halt state no matter what HHH does.   
   >>   
   >> The /simulation/ of DD by HHH will not /reproduce/ the halt   
   >> state of DD, which DD undeniably /has/.   
   >>   
   > *Hence the halting problem is wrong*   
      
   The much simpler explanation is that the decider is wrong.  By being   
   wrong, it contributes one point of evidence that confirms the theorem.   
      
   > Turing machine deciders only compute the mapping   
   > from their finite string inputs to an accept state   
   > or reject state on the basis that this input finite   
   > string specifies a semantic or syntactic property.   
      
   I don't understand what you think you're achieving by repeating   
   this, but its inclusion does mean that your poting has four   
   correct lines, improving its correctness average.   
      
   > The halting problem requires that halt deciders do what   
   > no Turing machine decider can do report on the semantic   
   > property of non-inputs.   
      
   It positively doesn't.  Even the diagonal cases that defeat deciders are   
   all valid inputs: self-contained finite strings denoting machines, which   
   perpetrate their trick without referencing anything outside of their own   
   description.   
      
   You are just making up nonsense and presenting without a shred of   
   rational evidence (which, of course, doesn't exist for a falsehood).   
      
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