YES(?,O(n^1))
0.00/0.39	YES(?,O(n^1))
0.00/0.39	
0.00/0.39	We are left with following problem, upon which TcT provides the
0.00/0.39	certificate YES(?,O(n^1)).
0.00/0.39	
0.00/0.39	Strict Trs:
0.00/0.39	  { from(X) -> cons(X, n__from(s(X)))
0.00/0.39	  , from(X) -> n__from(X)
0.00/0.39	  , after(s(N), cons(X, XS)) -> after(N, activate(XS))
0.00/0.39	  , after(0(), XS) -> XS
0.00/0.39	  , activate(X) -> X
0.00/0.39	  , activate(n__from(X)) -> from(X) }
0.00/0.39	Obligation:
0.00/0.39	  innermost runtime complexity
0.00/0.39	Answer:
0.00/0.39	  YES(?,O(n^1))
0.00/0.39	
0.00/0.39	The input was oriented with the instance of 'Small Polynomial Path
0.00/0.39	Order (PS,1-bounded)' as induced by the safe mapping
0.00/0.39	
0.00/0.39	 safe(from) = {1}, safe(cons) = {1, 2}, safe(n__from) = {1},
0.00/0.39	 safe(s) = {1}, safe(after) = {2}, safe(0) = {},
0.00/0.39	 safe(activate) = {1}
0.00/0.39	
0.00/0.39	and precedence
0.00/0.39	
0.00/0.39	 after > from, after > activate, activate > from .
0.00/0.39	
0.00/0.39	Following symbols are considered recursive:
0.00/0.39	
0.00/0.39	 {after}
0.00/0.39	
0.00/0.39	The recursion depth is 1.
0.00/0.39	
0.00/0.39	For your convenience, here are the satisfied ordering constraints:
0.00/0.39	
0.00/0.39	                      from(; X) > cons(; X,  n__from(; s(; X)))
0.00/0.39	                                                               
0.00/0.39	                      from(; X) > n__from(; X)                 
0.00/0.39	                                                               
0.00/0.39	  after(s(; N); cons(; X,  XS)) > after(N; activate(; XS))     
0.00/0.39	                                                               
0.00/0.39	                 after(0(); XS) > XS                           
0.00/0.39	                                                               
0.00/0.39	                  activate(; X) > X                            
0.00/0.39	                                                               
0.00/0.39	       activate(; n__from(; X)) > from(; X)                    
0.00/0.39	                                                               
0.00/0.39	
0.00/0.39	Hurray, we answered YES(?,O(n^1))
0.00/0.39	EOF