XPost: comp.theory, sci.logic   
   From: polcott333@gmail.com   
      
   On 7/19/2025 2:59 PM, wij wrote:   
   > On Sat, 2025-07-19 at 14:47 -0500, olcott wrote:   
   >> On 7/19/2025 2:29 PM, wij wrote:   
   >>> On Sat, 2025-07-19 at 14:19 -0500, olcott wrote:   
   >>>> On 7/19/2025 12:02 PM, Richard Damon wrote:   
   >>>>> On 7/19/25 10:42 AM, olcott wrote:   
   >>>>>> On 7/18/2025 3:49 AM, joes wrote:   
   >>>>>>   
   >>>>>>> That is wrong. It is, as you say, very obvious that HHH cannot simulate   
   >>>>>>> DDD past the call to HHH. You just draw the wrong conclusion from it.   
   >>>>>>> (Aside: what "seems" to you will convince no one. You can just call   
   >>>>>>> everybody dishonest. Also, they are not "your reviewers".)   
   >>>>>>>   
   >>>>>>   
   >>>>>> For the purposes of this discussion this is the   
   >>>>>> 100% complete definition of HHH. It is the exact   
   >>>>>> same one that I give to all the chat bots.   
   >>>>>>   
   >>>>>> Termination Analyzer HHH simulates its input until   
   >>>>>> it detects a non-terminating behavior pattern. When   
   >>>>>> HHH detects such a pattern it aborts its simulation   
   >>>>>> and returns 0.   
   >>>>>   
   >>>>> So, the only HHH that meets your definition is the HHH that never   
   >>>>> detects the pattern and aborts, and thus never returns.   
   >>>>>   
   >>>>   
   >>>> All of the Chat bots conclude that HHH(DDD) is correct   
   >>>> to reject its input as non-halting because this input   
   >>>> specified recursive simulation. They figure this out   
   >>>> on their own without any prompting.   
   >>>>   
   >>>> https://chatgpt.com/share/687aa4c2-b814-8011-9e7d-b85c03b291eb   
   >>>   
   >>> It is still nothing to do with the Halting Problem proof (Because it is   
   POOH)   
   >>>   
   >>   
   >> It is a key element of my refutation of this proof   
   >> because HHH also correctly determines that HHH(DD)==0.   
   >>   
   >> DD correctly simulated by HHH cannot possibly ever   
   >> reach past its first statement because it specifies   
   >> recursive simulation.   
   >>   
   >> int DD()   
   >> {   
   >>     int Halt_Status = HHH(DD);   
   >>     if (Halt_Status)   
   >>       HERE: goto HERE;   
   >>     return Halt_Status;   
   >> }   
   >>   
   >   
   > Boring. HHH cannot do what the HP says.   
   >   
      
   Turing machine (at least partial) halt deciders only compute   
   the mapping from their finite string inputs to the actual   
   behavior that this input finite string actually specifies.   
      
   Conventional notation of a Turing Machine: Ĥ   
   Conventional notation of a TM description: ⟨Ĥ⟩   
      
   Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞,   
      if Ĥ applied to ⟨Ĥ⟩ halts, and   
   Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn   
      if Ĥ applied to ⟨Ĥ⟩ does not halt.   
      
   *Is corrected to this*   
      
   Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞   
        ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H reaches   
        its simulated final halt state of ⟨Ĥ.qn⟩, and   
   Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn   
        ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H cannot possibly   
        reach its simulated final halt state of ⟨Ĥ.qn⟩.   
      
   *Original proof*   
   https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf   
      
   --   
   Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius   
   hits a target no one else can see." Arthur Schopenhauer   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)