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|    comp.ai.philosophy    |    Perhaps we should ask SkyNet about this    |    59,235 messages    |
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|    Message 57,780 of 59,235    |
|    olcott to Richard Damon    |
|    Re: Proof that DDD is correctly emulated    |
|    02 Aug 25 10:16:45    |
   
   XPost: comp.theory, sci.logic   
   From: polcott333@gmail.com   
      
   On 8/2/2025 8:07 AM, Richard Damon wrote:   
   > On 8/1/25 9:52 PM, olcott wrote:   
   >> On 8/1/2025 7:32 PM, Richard Damon wrote:   
   >>> On 8/1/25 8:25 PM, olcott wrote:   
   >>>>   
   >>>> My sole purpose is to show that the conventional HP proofs   
   >>>> do not prove undecidability. I have done that.   
   >>>>   
   >>>   
   >>> And you ignore that it has been shown that any finite pattern you try   
   >>> to detect becomes wrong,   
   >> No this is counter-factual.   
   >> A pattern cannot possibly become wrong.   
   >   
   > It was always wrong.   
   >   
   > Just because a pattern is found in a non-halting program doesn't mean   
   > that the pattern proves non-halting behavior.   
   >   
   >>   
   >> void DDD()   
   >> {   
   >> HHH(DDD);   
   >> return;   
   >> }   
   >>   
   >> HHH simulates DDD that calls HHH(DDD)   
   >> that simulates DDD that calls HHH(DDD)   
   >> that simulates DDD that calls HHH(DDD)   
   >> that simulates DDD that calls HHH(DDD)   
   >> that simulates DDD that calls HHH(DDD)   
   >> that simulates DDD that calls HHH(DDD)   
   >> that simulates DDD that calls HHH(DDD)   
   >> that simulates DDD that calls HHH(DDD)   
   >> that simulates DDD that calls HHH(DDD)   
   >>   
   >> This is stipulated by the basic structure   
   >> of the relationship between DDD and each HHH.   
   >   
   > Only if HHH is stipulated to always do a correct simulation, which means   
   > it can not abort.   
   >   
      
   A correct simulation of 2 statements of DDD proves   
   a repeating pattern that cannot possibly reach any   
   simulated final halt state.   
      
   I showed a correct simulation of 10 statements of DDD   
   so that you would look foolish denying that any repeating   
   pattern exists.   
      
   int Simulate(ptr x)   
   {   
    x();   
    return 1;   
   }   
      
   void Infinite_Recursion()   
   {   
    Simulate(Infinite_Recursion);   
    return;   
   }   
      
   A correct simulation of 2 statements of Infinite_Recursion   
   proves a repeating pattern that cannot possibly reach any   
   simulated final halt state.   
      
   --   
   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)   
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