XPost: comp.theory, sci.logic   
   From: F.Zwarts@HetNet.nl   
      
   Op 07.aug.2025 om 05:15 schreef olcott:   
   > 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.   
   That is your mistake. It does not require to report on its own   
   behaviour, but on the behaviour of Ĥ. It is irrelevant whether the   
   behaviour of Ĥ resembles that of embedded_H.   
      
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