XPost: comp.theory, comp.lang.c++, comp.lang.c   
   From: 643-408-1753@kylheku.com   
      
   On 2025-10-02, olcott  wrote:   
   > void P()   
   > {   
   >    if H(P)  // returns 1 for halts 0 for loops   
   >      HERE: goto HERE;   
   > }   
   >   
   > For *any* decider H and input P   
   > If H says halts then P loops   
   > If H says loops then P halts   
   > Making this HP diagonal case unsatisfiable.   
      
   You stil don't get it: there isn't one diagonal case.   
   The diagonal is an ifninite number of cells running through   
   a table.   
      
   Each case is different. For each i from 0 to infinity, Pi is   
   a different case which calls Hi.   
      
     // Diagonal case 1   
     void P1()   
     {   
        if H1(P1)   
          HERE: goto HERE;   
     }   
      
     // Diagonal case 2   
     void P2()   
     {   
        if H2(P2) // different case, different code!   
          HERE: goto HERE;   
     }   
      
     // Diagonal case 3   
     void P3()   
     {   
        if H3(P3) // different case again, different code!   
          HERE: goto HERE;   
     }   
      
     // Diagonal case 4   
     void P4() { .... H4(P4) ... }   
     // Diagonal case 4   
     void P5() { .... }   
     // ...   
      
   There isn't a single P that is not satisfiable. There are   
   multiple Pi's each of which is shown not to be solved   
   correctly by its /respective/ Hi.   
      
   Everyone of these Pi's has a definite halting status,   
      
   It's possible for Hj(Pi) to give a correct answer,   
   as long as j != i. When j == i, we are on the diagonal:   
   we have H7(P7), or H129(P129) ...   
      
   Do you finally understand it?   
      
   --   
   TXR Programming Language: http://nongnu.org/txr   
   Cygnal: Cygwin Native Application Library: http://kylheku.com/cygnal   
   Mastodon: @Kazinator@mstdn.ca   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)