XPost: comp.theory, sci.logic   
   From: polcott333@gmail.com   
      
   On 7/19/2025 3:57 PM, wij wrote:   
   > On Sat, 2025-07-19 at 15:41 -0500, olcott wrote:   
   >> On 7/19/2025 3:14 PM, wij wrote:   
   >>>   
   >>> HP is very simple: H(D)=1 if D halts, H(D)=0 if D does not halt.   
   >>>   
   >>   
   >> The standard proof assumes a decider   
   >> H(M,x) that determines whether machine   
   >> M halts on input x.   
   >>   
   >> But this formulation is flawed, because:   
   >   
   > Whatever the 'formulation' is, the HP result is a fact that no H can decide   
   > the halting status of any given D.   
   >   
      
   And that is wrong because H(⟨D⟩) is correctly determined.   
   It has always been a type mismatch error when H(D) was   
   assumed.   
      
   >> Turing machines can only process finite encodings   
   >> (e.g. ⟨M⟩), not executable entities like M.   
   >>   
   >> So the valid formulation must be   
   >> H(⟨M⟩,x), where ⟨M⟩ is a string.   
   >   
   > Halting Problem::= H(D)=1 if D halts, H(D)=0 if D does not halt.   
   > The conclusion is, no such H exists.   
   >   
      
   And that is wrong because H(⟨D⟩) is correctly determined.   
   It has always been a type mismatch error when H(D) was   
   assumed.   
      
   int DD()   
   {   
      int Halt_Status = HHH(DD);   
      if (Halt_Status)   
        HERE: goto HERE;   
      return Halt_Status;   
   }   
      
   DD correctly simulated by HHH cannot reach past   
   the "if" statement thus cannot reach the "return"   
   statement. This makes HHH(DD)==0 correct.   
      
   > 'formulation' does not really matter.   
   > If 'formulation' matters, it is another problem.   
   >   
      
      
      
   --   
   Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius   
   hits a target no one else can see." Arthur Schopenhauer   
      
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