YES(?,O(n^1))
0.00/0.73	YES(?,O(n^1))
0.00/0.73	
0.00/0.73	We are left with following problem, upon which TcT provides the
0.00/0.73	certificate YES(?,O(n^1)).
0.00/0.73	
0.00/0.73	Strict Trs:
0.00/0.73	  { from(X) -> cons(X, n__from(n__s(X)))
0.00/0.73	  , from(X) -> n__from(X)
0.00/0.73	  , 2ndspos(0(), Z) -> rnil()
0.00/0.73	  , 2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
0.00/0.73	  , 2ndspos(s(N), cons2(X, cons(Y, Z))) ->
0.00/0.73	    rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
0.00/0.73	  , s(X) -> n__s(X)
0.00/0.73	  , activate(X) -> X
0.00/0.73	  , activate(n__from(X)) -> from(activate(X))
0.00/0.73	  , activate(n__s(X)) -> s(activate(X))
0.00/0.73	  , 2ndsneg(0(), Z) -> rnil()
0.00/0.73	  , 2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
0.00/0.73	  , 2ndsneg(s(N), cons2(X, cons(Y, Z))) ->
0.00/0.73	    rcons(negrecip(Y), 2ndspos(N, activate(Z)))
0.00/0.73	  , pi(X) -> 2ndspos(X, from(0()))
0.00/0.73	  , plus(0(), Y) -> Y
0.00/0.73	  , plus(s(X), Y) -> s(plus(X, Y))
0.00/0.73	  , times(0(), Y) -> 0()
0.00/0.73	  , times(s(X), Y) -> plus(Y, times(X, Y))
0.00/0.73	  , square(X) -> times(X, X) }
0.00/0.73	Obligation:
0.00/0.73	  innermost runtime complexity
0.00/0.73	Answer:
0.00/0.73	  YES(?,O(n^1))
0.00/0.73	
0.00/0.73	Arguments of following rules are not normal-forms:
0.00/0.73	
0.00/0.73	{ 2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
0.00/0.73	, 2ndspos(s(N), cons2(X, cons(Y, Z))) ->
0.00/0.73	  rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
0.00/0.73	, 2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
0.00/0.73	, 2ndsneg(s(N), cons2(X, cons(Y, Z))) ->
0.00/0.73	  rcons(negrecip(Y), 2ndspos(N, activate(Z)))
0.00/0.73	, plus(s(X), Y) -> s(plus(X, Y))
0.00/0.73	, times(s(X), Y) -> plus(Y, times(X, Y)) }
0.00/0.73	
0.00/0.73	All above mentioned rules can be savely removed.
0.00/0.73	
0.00/0.73	We are left with following problem, upon which TcT provides the
0.00/0.73	certificate YES(?,O(n^1)).
0.00/0.73	
0.00/0.73	Strict Trs:
0.00/0.73	  { from(X) -> cons(X, n__from(n__s(X)))
0.00/0.73	  , from(X) -> n__from(X)
0.00/0.73	  , 2ndspos(0(), Z) -> rnil()
0.00/0.73	  , s(X) -> n__s(X)
0.00/0.73	  , activate(X) -> X
0.00/0.73	  , activate(n__from(X)) -> from(activate(X))
0.00/0.73	  , activate(n__s(X)) -> s(activate(X))
0.00/0.73	  , 2ndsneg(0(), Z) -> rnil()
0.00/0.73	  , pi(X) -> 2ndspos(X, from(0()))
0.00/0.73	  , plus(0(), Y) -> Y
0.00/0.73	  , times(0(), Y) -> 0()
0.00/0.73	  , square(X) -> times(X, X) }
0.00/0.73	Obligation:
0.00/0.73	  innermost runtime complexity
0.00/0.73	Answer:
0.00/0.73	  YES(?,O(n^1))
0.00/0.73	
0.00/0.73	The input was oriented with the instance of 'Small Polynomial Path
0.00/0.73	Order (PS,1-bounded)' as induced by the safe mapping
0.00/0.73	
0.00/0.73	 safe(from) = {1}, safe(cons) = {1, 2}, safe(n__from) = {1},
0.00/0.73	 safe(n__s) = {1}, safe(2ndspos) = {1, 2}, safe(0) = {},
0.00/0.73	 safe(rnil) = {}, safe(s) = {1}, safe(activate) = {},
0.00/0.73	 safe(2ndsneg) = {}, safe(pi) = {1}, safe(plus) = {},
0.00/0.73	 safe(times) = {1, 2}, safe(square) = {1}
0.00/0.73	
0.00/0.73	and precedence
0.00/0.73	
0.00/0.73	 from > s, from > plus, 2ndspos > s, 2ndspos > plus,
0.00/0.73	 activate > from, activate > s, activate > plus, pi > from,
0.00/0.73	 pi > 2ndspos, pi > s, pi > 2ndsneg, pi > plus, pi > times,
0.00/0.73	 times > s, times > plus, square > from, square > 2ndspos,
0.00/0.73	 square > s, square > 2ndsneg, square > plus, square > times,
0.00/0.73	 2ndspos ~ times, s ~ plus, activate ~ pi, activate ~ square,
0.00/0.73	 pi ~ square .
0.00/0.73	
0.00/0.73	Following symbols are considered recursive:
0.00/0.73	
0.00/0.73	 {2ndspos, activate, times}
0.00/0.73	
0.00/0.73	The recursion depth is 1.
0.00/0.73	
0.00/0.73	For your convenience, here are the satisfied ordering constraints:
0.00/0.73	
0.00/0.73	                from(; X) > cons(; X,  n__from(; n__s(; X)))
0.00/0.73	                                                            
0.00/0.73	                from(; X) > n__from(; X)                    
0.00/0.73	                                                            
0.00/0.73	       2ndspos(; 0(),  Z) > rnil()                          
0.00/0.73	                                                            
0.00/0.73	                   s(; X) > n__s(; X)                       
0.00/0.73	                                                            
0.00/0.73	             activate(X;) > X                               
0.00/0.73	                                                            
0.00/0.73	  activate(n__from(; X);) > from(; activate(X;))            
0.00/0.73	                                                            
0.00/0.73	     activate(n__s(; X);) > s(; activate(X;))               
0.00/0.73	                                                            
0.00/0.73	        2ndsneg(0(),  Z;) > rnil()                          
0.00/0.73	                                                            
0.00/0.73	                  pi(; X) > 2ndspos(; X,  from(; 0()))      
0.00/0.73	                                                            
0.00/0.73	           plus(0(),  Y;) > Y                               
0.00/0.73	                                                            
0.00/0.73	         times(; 0(),  Y) > 0()                             
0.00/0.73	                                                            
0.00/0.73	              square(; X) > times(; X,  X)                  
0.00/0.73	                                                            
0.00/0.73	
0.00/0.73	Hurray, we answered YES(?,O(n^1))
0.00/0.73	EOF