YES(O(1),O(n^1))
0.00/0.68	YES(O(1),O(n^1))
0.00/0.68	
0.00/0.68	We are left with following problem, upon which TcT provides the
0.00/0.68	certificate YES(O(1),O(n^1)).
0.00/0.68	
0.00/0.68	Strict Trs:
0.00/0.68	  { 2nd(cons(X, n__cons(Y, Z))) -> activate(Y)
0.00/0.68	  , cons(X1, X2) -> n__cons(X1, X2)
0.00/0.68	  , activate(X) -> X
0.00/0.68	  , activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
0.00/0.68	  , activate(n__from(X)) -> from(activate(X))
0.00/0.68	  , activate(n__s(X)) -> s(activate(X))
0.00/0.68	  , from(X) -> cons(X, n__from(n__s(X)))
0.00/0.68	  , from(X) -> n__from(X)
0.00/0.68	  , s(X) -> n__s(X) }
0.00/0.68	Obligation:
0.00/0.68	  innermost runtime complexity
0.00/0.68	Answer:
0.00/0.68	  YES(O(1),O(n^1))
0.00/0.68	
0.00/0.68	Arguments of following rules are not normal-forms:
0.00/0.68	
0.00/0.68	{ 2nd(cons(X, n__cons(Y, Z))) -> activate(Y) }
0.00/0.68	
0.00/0.68	All above mentioned rules can be savely removed.
0.00/0.68	
0.00/0.68	We are left with following problem, upon which TcT provides the
0.00/0.68	certificate YES(O(1),O(n^1)).
0.00/0.68	
0.00/0.68	Strict Trs:
0.00/0.68	  { cons(X1, X2) -> n__cons(X1, X2)
0.00/0.68	  , activate(X) -> X
0.00/0.68	  , activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
0.00/0.68	  , activate(n__from(X)) -> from(activate(X))
0.00/0.68	  , activate(n__s(X)) -> s(activate(X))
0.00/0.68	  , from(X) -> cons(X, n__from(n__s(X)))
0.00/0.68	  , from(X) -> n__from(X)
0.00/0.68	  , s(X) -> n__s(X) }
0.00/0.68	Obligation:
0.00/0.68	  innermost runtime complexity
0.00/0.68	Answer:
0.00/0.68	  YES(O(1),O(n^1))
0.00/0.68	
0.00/0.68	We use the processor 'Small Polynomial Path Order (PS,1-bounded)'
0.00/0.68	to orient following rules strictly.
0.00/0.68	
0.00/0.68	Trs:
0.00/0.68	  { cons(X1, X2) -> n__cons(X1, X2)
0.00/0.68	  , activate(X) -> X
0.00/0.68	  , activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
0.00/0.68	  , activate(n__from(X)) -> from(activate(X))
0.00/0.68	  , activate(n__s(X)) -> s(activate(X))
0.00/0.68	  , from(X) -> cons(X, n__from(n__s(X)))
0.00/0.68	  , from(X) -> n__from(X)
0.00/0.68	  , s(X) -> n__s(X) }
0.00/0.68	
0.00/0.68	The induced complexity on above rules (modulo remaining rules) is
0.00/0.68	YES(?,O(n^1)) . These rules are moved into the corresponding weak
0.00/0.68	component(s).
0.00/0.68	
0.00/0.68	Sub-proof:
0.00/0.68	----------
0.00/0.68	  The input was oriented with the instance of 'Small Polynomial Path
0.00/0.68	  Order (PS,1-bounded)' as induced by the safe mapping
0.00/0.68	  
0.00/0.68	   safe(cons) = {1, 2}, safe(n__cons) = {1, 2}, safe(activate) = {},
0.00/0.68	   safe(from) = {1}, safe(n__from) = {1}, safe(n__s) = {1},
0.00/0.68	   safe(s) = {1}
0.00/0.68	  
0.00/0.68	  and precedence
0.00/0.68	  
0.00/0.68	   activate > cons, activate > from, activate > s, from > cons,
0.00/0.68	   s > cons, from ~ s .
0.00/0.68	  
0.00/0.68	  Following symbols are considered recursive:
0.00/0.68	  
0.00/0.68	   {activate}
0.00/0.68	  
0.00/0.68	  The recursion depth is 1.
0.00/0.68	  
0.00/0.68	  For your convenience, here are the satisfied ordering constraints:
0.00/0.68	  
0.00/0.68	                  cons(; X1,  X2) > n__cons(; X1,  X2)              
0.00/0.68	                                                                    
0.00/0.68	                     activate(X;) > X                               
0.00/0.68	                                                                    
0.00/0.68	    activate(n__cons(; X1,  X2);) > cons(; activate(X1;),  X2)      
0.00/0.68	                                                                    
0.00/0.68	          activate(n__from(; X);) > from(; activate(X;))            
0.00/0.68	                                                                    
0.00/0.68	             activate(n__s(; X);) > s(; activate(X;))               
0.00/0.68	                                                                    
0.00/0.68	                        from(; X) > cons(; X,  n__from(; n__s(; X)))
0.00/0.68	                                                                    
0.00/0.68	                        from(; X) > n__from(; X)                    
0.00/0.68	                                                                    
0.00/0.68	                           s(; X) > n__s(; X)                       
0.00/0.68	                                                                    
0.00/0.68	
0.00/0.68	We return to the main proof.
0.00/0.68	
0.00/0.68	We are left with following problem, upon which TcT provides the
0.00/0.68	certificate YES(O(1),O(1)).
0.00/0.68	
0.00/0.68	Weak Trs:
0.00/0.68	  { cons(X1, X2) -> n__cons(X1, X2)
0.00/0.68	  , activate(X) -> X
0.00/0.68	  , activate(n__cons(X1, X2)) -> cons(activate(X1), X2)
0.00/0.68	  , activate(n__from(X)) -> from(activate(X))
0.00/0.68	  , activate(n__s(X)) -> s(activate(X))
0.00/0.68	  , from(X) -> cons(X, n__from(n__s(X)))
0.00/0.68	  , from(X) -> n__from(X)
0.00/0.68	  , s(X) -> n__s(X) }
0.00/0.68	Obligation:
0.00/0.68	  innermost runtime complexity
0.00/0.68	Answer:
0.00/0.68	  YES(O(1),O(1))
0.00/0.68	
0.00/0.68	Empty rules are trivially bounded
0.00/0.68	
0.00/0.68	Hurray, we answered YES(O(1),O(n^1))
0.00/0.69	EOF