YES(?,O(n^1))
0.00/0.25	YES(?,O(n^1))
0.00/0.25	
0.00/0.25	We are left with following problem, upon which TcT provides the
0.00/0.25	certificate YES(?,O(n^1)).
0.00/0.25	
0.00/0.25	Strict Trs:
0.00/0.25	  { first(X1, X2) -> n__first(X1, X2)
0.00/0.25	  , first(0(), X) -> nil()
0.00/0.25	  , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
0.00/0.25	  , activate(X) -> X
0.00/0.25	  , activate(n__first(X1, X2)) -> first(X1, X2)
0.00/0.25	  , activate(n__from(X)) -> from(X)
0.00/0.25	  , from(X) -> cons(X, n__from(s(X)))
0.00/0.25	  , from(X) -> n__from(X) }
0.00/0.25	Obligation:
0.00/0.25	  innermost runtime complexity
0.00/0.25	Answer:
0.00/0.25	  YES(?,O(n^1))
0.00/0.25	
0.00/0.25	The input was oriented with the instance of 'Small Polynomial Path
0.00/0.25	Order (PS)' as induced by the safe mapping
0.00/0.25	
0.00/0.25	 safe(first) = {1}, safe(0) = {}, safe(nil) = {}, safe(s) = {1},
0.00/0.25	 safe(cons) = {1, 2}, safe(n__first) = {1, 2}, safe(activate) = {},
0.00/0.25	 safe(from) = {1}, safe(n__from) = {1}
0.00/0.25	
0.00/0.25	and precedence
0.00/0.25	
0.00/0.25	 first ~ activate, first ~ from, activate ~ from .
0.00/0.25	
0.00/0.25	Following symbols are considered recursive:
0.00/0.25	
0.00/0.25	 {first, activate, from}
0.00/0.25	
0.00/0.25	The recursion depth is 1.
0.00/0.25	
0.00/0.25	For your convenience, here are the satisfied ordering constraints:
0.00/0.25	
0.00/0.25	                   first(X2; X1) > n__first(; X1,  X2)                     
0.00/0.25	                                                                           
0.00/0.25	                   first(X; 0()) > nil()                                   
0.00/0.25	                                                                           
0.00/0.25	    first(cons(; Y,  Z); s(; X)) > cons(; Y,  n__first(; X,  activate(Z;)))
0.00/0.25	                                                                           
0.00/0.25	                    activate(X;) > X                                       
0.00/0.25	                                                                           
0.00/0.25	  activate(n__first(; X1,  X2);) > first(X2; X1)                           
0.00/0.25	                                                                           
0.00/0.25	         activate(n__from(; X);) > from(; X)                               
0.00/0.25	                                                                           
0.00/0.25	                       from(; X) > cons(; X,  n__from(; s(; X)))           
0.00/0.25	                                                                           
0.00/0.25	                       from(; X) > n__from(; X)                            
0.00/0.25	                                                                           
0.00/0.25	
0.00/0.25	Hurray, we answered YES(?,O(n^1))
0.00/0.25	EOF