XPost: sci.math, comp.theory   
   From: mikko.levanto@iki.fi   
      
   olcott kirjoitti 27.11.2025 klo 17.17:   
   > On 11/27/2025 1:40 AM, Mikko wrote:   
   >> olcott kirjoitti 26.11.2025 klo 17.15:   
   >>> On 11/26/2025 3:22 AM, Mikko wrote:   
   >>>> olcott kirjoitti 26.11.2025 klo 5.17:   
   >>>>> On 11/25/2025 9:09 PM, Kaz Kylheku wrote:   
   >>>>>> On 2025-11-26, olcott  wrote:   
   >>>>>>> On 11/25/2025 8:36 PM, Kaz Kylheku wrote:   
   >>>>>>>> On 2025-11-26, olcott  wrote:   
   >>>>>>>>> None of them ever had the slightest clue about Montague   
   >>>>>>>>> Grammar. Except for one they all had very severe math   
   >>>>>>>>> phobia.   
   >>>>>>>>   
   >>>>>>>> So do you; you are terribly afraid of the mathematical idea that   
   >>>>>>>> simulations that are paused still exist and have future   
   >>>>>>>> states.   
   >>>>>>>>   
   >>>>>>>   
   >>>>>>> I am not going to discuss your psychotic nonsense.   
   >>>>>>   
   >>>>>> In all honesty, you and your therapist /should/ be laser focused   
   >>>>>> on your   
   >>>>>> own psychotic nonsense.   
   >>>>>>   
   >>>>>>> You already agreed that I am correct so this subject   
   >>>>>>> is closed.   
   >>>>>>   
   >>>>>> Whaaat ...   
   >>>>>>   
   >>>>>>> news://news.eternal-september.org/20251104183329.967@kylheku.com   
   >>>>>>> On 11/4/2025 8:43 PM, Kaz Kylheku wrote:   
   >>>>>>>> On 2025-11-05, olcott  wrote:   
   >>>>>>>>>   
   >>>>>>>>> The whole point is that D simulated by H   
   >>>>>>>>> cannot possbly reach its own simulated   
   >>>>>>>>> "return" statement no matter what H does.   
   >>>>>>>>   
   >>>>>>>> Yes; this doesn't happen while H is running.   
   >>>>>>>>   
   >>>>>>>> So while H does /something/, no matter what H does,   
   >>>>>>>> that D simulation won't reach the return statement.   
   >>>>>>   
   >>>>>> But we know that. If H is nonreturning, of course D is.   
   >>>>>> Since D calls H(D), D is suspended until H(D) returns,   
   >>>>>> which means forever if H(D) is nonterminating.   
   >>>>>>   
   >>>>>> I have no idea what you are trying to milk out of this;   
   >>>>>> it is completely uncontroversial.   
   >>>>>   
   >>>>> I really did figure out how to determine the   
   >>>>> correct halt status that the halting problem's   
   >>>>> counter-example input specifies to it decider.   
   >>>>   
   >>>> The basic halting problem is about Turing machines. A Turing machine   
   >>>> specifies only one bhavour. It does not specify anything else to the   
   >>>> decider. An ambiguous program is outside of the domain of the halting   
   >>>> problem.   
   >>>   
   >>> That is inaccurate.   
   >>   
   >> No, it is not. Of course there are many ways to formulate the problem   
   >> but what I said is true about the basic formulation. All formulations   
   >> restrict the domain to unambiguous specifications.   
   >>   
   >   
   > It is a perfectly unambiguous ultimately   
   > self-contradictory specification.   
      
   If the specification of D is perfectly unambiguous there is no point   
   to say "specifies to its decider". If D specifies a halting behaviour   
   it specifies it both to its normal execution environment and to every   
   decider, otherwise it specifies non-halting behaviour both to its   
   normal execution environment and its decider. If it specifies one   
   behaviour to one and a diffetent behaviour to the other then it is   
   ambiguous. Becauyse that is what the words mean.   
      
   --   
   Mikko   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)