XPost: comp.theory, sci.logic   
   From: flibble@red-dwarf.jmc.corp   
      
   On Mon, 04 Aug 2025 17:57:30 -0500, olcott wrote:   
      
   > On 8/4/2025 5:44 PM, Mr Flibble wrote:   
   >> On Mon, 04 Aug 2025 17:42:24 -0500, olcott wrote:   
   >>   
   >>> On 8/4/2025 5:34 PM, Mr Flibble wrote:   
   >>>> On Mon, 04 Aug 2025 13:29:04 -0500, olcott wrote:   
   >>>>   
   >>>>> Diagonalization only arises when one assumes that a Turing machine   
   >>>>> decider must report on its own behavior instead of the behavior   
   >>>>> specified by its machine description.   
   >>>>>   
   >>>>> Everyone assumes that these must always be the same.   
   >>>>> That assumption is proven to be incorrect.   
   >>>>>   
   >>>>> When one assumes a halt decider based on a UTM then the simulated   
   >>>>> input remains stuck in recursive simulation never reaching simulated   
   >>>>> states ⟨Ĥ.∞⟩ or ⟨Ĥ.qn⟩.   
   >>>>>   
   >>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞ Ĥ.q0   
   ⟨Ĥ⟩ ⊢*   
   >>>> Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩   
   >>>>> ⊢* Ĥ.qn   
   >>>>>   
   >>>>> (a) Ĥ copies its input ⟨Ĥ⟩   
   >>>>> (b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩   
   >>>>> (c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩   
   >>>>> (d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩   
   >>>>> (e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩   
   >>>>> (f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩   
   >>>>> on and on never reaching any simulated final state of ⟨Ĥ.qn⟩   
   >>>>>   
   >>>>> When embedded_H aborts its simulation and transitions to Ĥ.qn on the   
   >>>>> basis that its simulated input cannot possibly reach its own   
   >>>>> simulated final halt state of ⟨Ĥ.qn⟩ embedded_H is correct.   
   >>>>>   
   >>>>> This causes embedded_H itself to halt, thus contradicting its result   
   >>>>> *only if a Turing machine decider can be applied to its actual self*   
   >>>>> and not merely its own machine description.   
   >>>>   
   >>>> Your Ĥ is not a halt decider as defined by the Halting Problem so has   
   >>>> nothing to do with the Halting Problem.   
   >>>>   
   >>>> /Flibble   
   >>>   
   >>> You have this part incorrectly. Ask Richard because of what he   
   >>> explained to you the other night he may correct you on this.   
   >>   
   >> No, your halt decider is a partial decider, Halting Problem deciders   
   >> are total not partial.   
   >>   
   >> /Flibble   
   >   
   > Not exactly. The HP proofs attempt to prove that no total halt decider   
   > exists on the basis of one self-referential input cannot be decided by   
   > any decider including partial deciders.   
      
   Wrong. Partial deciders have nothing to do with the Halting Problem.   
      
   >   
   > The technical term "decider" does not mean its conventional meaning of   
   > one who decides. It means an infallible Turing machine that always   
   > decides correctly. Since this is too misleading for most people I used   
   > "termination analyzer".   
      
   Halting deciders and termination analyzers are different things and you do   
   not get to redefine terms to suit your bogus argument.   
      
   /Flibble   
      
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