XPost: comp.theory, sci.logic   
   From: polcott333@gmail.com   
      
   On 8/6/2025 9:13 PM, Richard Damon wrote:   
   > On 8/6/25 8:18 AM, olcott wrote:   
   >> On 8/6/2025 6:41 AM, Richard Damon wrote:   
   >>> On 8/6/25 7:39 AM, olcott wrote:   
   >>>> On 8/6/2025 4:00 AM, Fred. Zwarts wrote:   
   >>>>> Op 06.aug.2025 om 05:47 schreef olcott:   
   >>>>>>   
   >>>>>> It corrects the error of the requirement that   
   >>>>>> a halt decider reports on its own behavior.   
   >>>>>>   
   >>>>>>     "The contradiction in Linz's (or Turing's) self-referential   
   >>>>>>      halting construction only appears if one insists that the   
   >>>>>>      machine can and must decide on its own behavior, which is   
   >>>>>>      neither possible nor required."   
   >>>>>>   
   >>>>>> https://chatgpt.com/share/6890ee5a-52bc-8011-852e-3d9f97bcfbd8   
   >>>>>>   
   >>>>>   
   >>>>> That seems to be one of your misunderstanding. A decider must   
   >>>>> report on the behaviour of its input, even if this behaviour   
   >>>>> resembles its own behaviour. That is very different from 'deciding   
   >>>>> on its own behaviour'.   
   >>>> *This is my correction*   
   >>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞   
   >>>>     if ⟨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   
   >>>>     if ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by Ĥ.embedded_H cannot possibly   
   >>>>     reach its simulated final halt state of ⟨Ĥ.qn⟩.   
   >>>>   
   >>>> *This is the original erroneous one*   
   >>>> 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.   
   >>>>   
   >>>> *Lines two and four requires Ĥ.embedded_H to report on its own   
   >>>> behavior*   
   >>>>   
   >>>   
   >>>   
   >>> Which is valid.   
   >>>   
   >>> And you are just stupid to think otherwise.   
   >>   
   >> It is only valid when Turing machine deciders can   
   >> take themselves as inputs.   
   >>   
   >   
   > They only need to take representation of themselves as input.   
   >   
      
   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.   
      
   The Linz proof has Ĥ take its own machine description   
   as input (this is fine) yet requires Ĥ.embedded_H to   
   directly report on its own behavior. No TM can ever   
   do that.   
      
      
   --   
   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   
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