XPost: comp.theory, comp.ai.philosophy, sci.math   
   From: tristan.wibberley+netnews2@alumni.manchester.ac.uk   
      
   On 20/11/2025 20:14, Kaz Kylheku wrote:   
   > On 2025-11-20, dart200  wrote:   
   >> On 11/20/25 11:55 AM, Kaz Kylheku wrote:   
   >>> On 2025-11-20, dart200  wrote:   
   >>>> On 11/19/25 6:58 PM, Kaz Kylheku wrote:   
   >>>>> On 2025-11-20, dart200  wrote:   
   >>>>>> On 11/19/25 6:29 PM, Kaz Kylheku wrote:   
   >>>>>>> On 2025-11-20, dart200  wrote:   
   >>>>>>>> a) you can construct halting paradoxes that contradicts multiple and   
   >>>>>>>> possibly even infinite deciders. certainly any finite set, after which   
   >>>>>>>   
   >>>>>>> This is not possible in general. The diagonal test case must make   
   >>>>>>> exactly one decision and then behave in a contradictory way: halt or   
   >>>>>>> not. If it interrogates as few as two deciders, it becomes intractable   
   >>>>>>> if their decisions differ: to contradict one is to agree with the   
   other.   
   >>>>>>>   
   >>>>>>> If the deciders are H0(P) { return 0; } and H1(P) { return 1; } you can   
   >>>>>>> see that between the two of them, they cover the entire space: there   
   >>>>>>> cannot be a signal case whch both of these don't get right. One   
   >>>>>>> correctly decides all nonterminating cases; the other correctly decies   
   >>>>>>> all terminating cases, and every case is one or the other.   
   >>>>>>   
   >>>>>> common man those deciders do not provide an /effectively computable/   
   >>>>>> interface and you know it   
   >>>>>>   
   >>>>>> try again, it's quite simple to produce a paradox that confounds two   
   >>>>>> legitimate deciders that genuinely never give a wrong answer   
   >>>>>   
   >>>>> But we have a proof that deciders which never give a wrong answer do not   
   >>>>> exist.   
   >>>>>   
   >>>>> If halting algorithms existed   
   >>>>>   
   >>>>> - they would all agree with each other and thus look the same from the   
   >>>>>     ouside and so wouldn't constitute a multi-decider aggregate.   
   >>>>>   
   >>>>> - it would not be /possible/ to contradict them: they never give   
   >>>>>     a wrong answer!   
   >>>>>   
   >>>>> So if we want to develop diagonal cases whch contradict deciders,   
   >>>>> we have to accept that we are targeting imperfect, partial deciders   
   >>>>> (by doing so, showing them to be that way).   
   >>>>   
   >>>> for the sake of proof/example assume they are honest until you produce   
   >>>> the paradox   
   >>>   
   >>> By "until", are you referring to some temporal concept? There is a time   
   >>> variable in the system such that a decider can be introduced, and then   
   >>> for at time, there exists no diagonal (or other) case until someone   
   >>> writes it?   
   >>>   
   >>   
   >> assume the premise exist and show a contradiction for both deciders   
   >> within one machine   
   >   
   > If one decider says true, and the other false, for the contradicting   
   > case, then no can do. Infinitely looping, or terminating, contradicts   
   > only one of the decider, agreeing with the other.  There isn't a third   
   > choice of behavior.   
   >   
      
   oh I see, not that the thwarted deciders go different ways, but the raw   
   deciders go different ways before thwarting!   
      
   well, the results of the deciders are combined to form a single decider,   
   if that is (const . first) it's just equal to the first decider, if it's   
   (const . second) it's just equal to the second decider, if it's 'or' or   
   and then it's a bit complicated. There's a whole thing about computing   
   the conjunction of all deciders that approaches the universal decision   
   as you approach the universal inclusion but how long does the approach   
   take... well, we can make an intuitive leap based on what we know about   
   halting.   
      
      
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   Tristan Wibberley   
      
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