XPost: sci.physics.relativity, comp.lang.prolog   
   From: janburse@fastmail.fm   
      
   Hi,   
      
   If you know BB(N), you have a halting decision procedure   
   for N-turing machines. Since if BB(N) is maximum number   
   S(N) of steps before halting,   
      
   you can just run an arbitrary turing machine, and when   
   its steps exceeds S(N), you know its not a halting   
   turing machine.   
      
   So knowing BB(N) makes the halting problem decidable.   
   But the halting problem is not decidable. So there   
   must be some M maybe where BB(M) has no S(N) , no   
      
   maximum. Idea is to construct turing machines that   
   relate to consistency problems, consistency problems   
   can be even harder than halting problems, we might   
      
   ask for the opposite, does a program never halt.   
   Since never halt could be interpreted that no   
   inconsistency is derived. Again knowing BB(N) would   
      
   help, since dedidability via S(N) is established both   
   ways, saying "Yes" to halt, and saying "No" to not halt.   
   So we can show a reducibility from consistency   
      
   to busy beaver, I guess.   
      
   Bye   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)