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(n__s(X)))
0.00/0.39	  , from(X) -> n__from(X)
0.00/0.39	  , sel(0(), cons(X, Y)) -> X
0.00/0.39	  , sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
0.00/0.39	  , s(X) -> n__s(X)
0.00/0.39	  , activate(X) -> X
0.00/0.39	  , activate(n__from(X)) -> from(activate(X))
0.00/0.39	  , activate(n__s(X)) -> s(activate(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	Arguments of following rules are not normal-forms:
0.00/0.39	
0.00/0.39	{ sel(s(X), cons(Y, Z)) -> sel(X, activate(Z)) }
0.00/0.39	
0.00/0.39	All above mentioned rules can be savely removed.
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(n__s(X)))
0.00/0.39	  , from(X) -> n__from(X)
0.00/0.39	  , sel(0(), cons(X, Y)) -> X
0.00/0.39	  , s(X) -> n__s(X)
0.00/0.39	  , activate(X) -> X
0.00/0.39	  , activate(n__from(X)) -> from(activate(X))
0.00/0.39	  , activate(n__s(X)) -> s(activate(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(n__s) = {1}, safe(sel) = {}, safe(0) = {}, safe(s) = {1},
0.00/0.39	 safe(activate) = {}
0.00/0.39	
0.00/0.39	and precedence
0.00/0.39	
0.00/0.39	 sel > from, sel > s, activate > from, activate > s, from ~ s,
0.00/0.39	 sel ~ activate .
0.00/0.39	
0.00/0.39	Following symbols are considered recursive:
0.00/0.39	
0.00/0.39	 {activate}
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(; n__s(; X)))
0.00/0.39	                                                              
0.00/0.39	                  from(; X) > n__from(; X)                    
0.00/0.39	                                                              
0.00/0.39	  sel(0(),  cons(; X,  Y);) > X                               
0.00/0.39	                                                              
0.00/0.39	                     s(; X) > n__s(; X)                       
0.00/0.39	                                                              
0.00/0.39	               activate(X;) > X                               
0.00/0.39	                                                              
0.00/0.39	    activate(n__from(; X);) > from(; activate(X;))            
0.00/0.39	                                                              
0.00/0.39	       activate(n__s(; X);) > s(; activate(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