XPost: sci.math, comp.theory   
   From: mikko.levanto@iki.fi   
      
   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.   
      
   --   
   Mikko   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)