XPost: comp.theory   
   From: polcott333@gmail.com   
      
   On 7/20/2024 3:10 PM, Richard Damon wrote:   
   > On 7/20/24 4:02 PM, olcott wrote:   
   >> On 7/20/2024 2:36 PM, Fred. Zwarts wrote:   
   >>> Op 20.jul.2024 om 21:09 schreef olcott:   
   >>>> On 7/20/2024 2:00 PM, Fred. Zwarts wrote:   
   >>>>> Op 20.jul.2024 om 17:28 schreef olcott:   
   >>>>>> void DDD()   
   >>>>>> {   
   >>>>>>    HHH(DDD);   
   >>>>>> }   
   >>>>>>   
   >>>>>> int main()   
   >>>>>> {   
   >>>>>>    DDD();   
   >>>>>> }   
   >>>>>>   
   >>>>>> (a) Termination Analyzers / Partial Halt Deciders must halt   
   >>>>>> this is a design requirement.   
   >>>>>>   
   >>>>>> (b) Every simulating termination analyzer HHH either   
   >>>>>> aborts the simulation of its input or not.   
   >>>>>>   
   >>>>>> (c) Within the hypothetical case where HHH does not abort   
   >>>>>> the simulation of its input {HHH, emulated DDD and executed DDD}   
   >>>>>> never stop running.   
   >>>>>>   
   >>>>>> This violates the design requirement of (a) therefore HHH must   
   >>>>>> abort the simulation of its input.   
   >>>>>   
   >>>>> And when it aborts, the simulation is incorrect. When HHH aborts   
   >>>>> and halts, it is not needed to abort its simulation, because it   
   >>>>> will halt of its own.   
   >>>>   
   >>>> So you are trying to get away with saying that no HHH   
   >>>> ever needs to abort the simulation of its input and HHH   
   >>>> will stop running?   
   >>>>   
   >>>   
   >>> No, you try to get away with saying that a HHH that is coded to abort   
   >>> and halt, will never stop running, only because you are dreaming of   
   >>> *another* HHH that does not abort.   
   >>>   
   >>   
   >> *You know that I didn't say anything like that*   
   >>   
   >> Unless I refer to the infinite set of every possible   
   >> HHH my reviewers try to get away with saying that I am   
   >> referring to the wrong HHH.   
   >>   
   >> void DDD()   
   >> {   
   >>    HHH(DDD);   
   >>    return;   
   >> }   
   >>   
   >> DDD correctly simulated by pure function HHH cannot   
   >> possibly reach its own return instruction.   
   >>   
   >>   
   >   
   > And the problem you ignore is that each HHH is given the DDD that calls   
   > itself, and not some other HHH, and thus you can't look at the other   
      
   Yet again trying to get away with saying that when every element   
   of an infinite set cannot reach its return instruction that some   
   of these elements still reach their return instruction.   
      
   My God have mercy on your soul (if you have a soul).   
      
   --   
   Copyright 2024 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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