YES(?,O(n^1))
0.00/0.35	YES(?,O(n^1))
0.00/0.35	
0.00/0.35	We are left with following problem, upon which TcT provides the
0.00/0.35	certificate YES(?,O(n^1)).
0.00/0.35	
0.00/0.35	Strict Trs:
0.00/0.35	  { f(X) -> n__f(X)
0.00/0.35	  , f(0()) -> cons(0(), n__f(n__s(n__0())))
0.00/0.35	  , f(s(0())) -> f(p(s(0())))
0.00/0.35	  , 0() -> n__0()
0.00/0.35	  , s(X) -> n__s(X)
0.00/0.35	  , p(s(0())) -> 0()
0.00/0.35	  , activate(X) -> X
0.00/0.35	  , activate(n__f(X)) -> f(activate(X))
0.00/0.35	  , activate(n__s(X)) -> s(activate(X))
0.00/0.35	  , activate(n__0()) -> 0() }
0.00/0.35	Obligation:
0.00/0.35	  innermost runtime complexity
0.00/0.35	Answer:
0.00/0.35	  YES(?,O(n^1))
0.00/0.35	
0.00/0.35	Arguments of following rules are not normal-forms:
0.00/0.35	
0.00/0.35	{ f(0()) -> cons(0(), n__f(n__s(n__0())))
0.00/0.35	, f(s(0())) -> f(p(s(0())))
0.00/0.35	, p(s(0())) -> 0() }
0.00/0.35	
0.00/0.35	All above mentioned rules can be savely removed.
0.00/0.35	
0.00/0.35	We are left with following problem, upon which TcT provides the
0.00/0.35	certificate YES(?,O(n^1)).
0.00/0.35	
0.00/0.35	Strict Trs:
0.00/0.35	  { f(X) -> n__f(X)
0.00/0.35	  , 0() -> n__0()
0.00/0.35	  , s(X) -> n__s(X)
0.00/0.35	  , activate(X) -> X
0.00/0.35	  , activate(n__f(X)) -> f(activate(X))
0.00/0.35	  , activate(n__s(X)) -> s(activate(X))
0.00/0.35	  , activate(n__0()) -> 0() }
0.00/0.35	Obligation:
0.00/0.35	  innermost runtime complexity
0.00/0.35	Answer:
0.00/0.35	  YES(?,O(n^1))
0.00/0.35	
0.00/0.35	The input was oriented with the instance of 'Small Polynomial Path
0.00/0.35	Order (PS,1-bounded)' as induced by the safe mapping
0.00/0.35	
0.00/0.35	 safe(f) = {1}, safe(0) = {}, safe(n__f) = {1}, safe(n__s) = {1},
0.00/0.35	 safe(n__0) = {}, safe(s) = {1}, safe(activate) = {}
0.00/0.35	
0.00/0.35	and precedence
0.00/0.35	
0.00/0.35	 activate > f, activate > 0, activate > s, f ~ 0, f ~ s, 0 ~ s .
0.00/0.35	
0.00/0.35	Following symbols are considered recursive:
0.00/0.35	
0.00/0.35	 {activate}
0.00/0.35	
0.00/0.35	The recursion depth is 1.
0.00/0.35	
0.00/0.35	For your convenience, here are the satisfied ordering constraints:
0.00/0.35	
0.00/0.35	                f(; X) > n__f(; X)        
0.00/0.35	                                          
0.00/0.35	                   0() > n__0()           
0.00/0.35	                                          
0.00/0.35	                s(; X) > n__s(; X)        
0.00/0.35	                                          
0.00/0.35	          activate(X;) > X                
0.00/0.35	                                          
0.00/0.35	  activate(n__f(; X);) > f(; activate(X;))
0.00/0.35	                                          
0.00/0.35	  activate(n__s(; X);) > s(; activate(X;))
0.00/0.35	                                          
0.00/0.35	     activate(n__0();) > 0()              
0.00/0.35	                                          
0.00/0.35	
0.00/0.35	Hurray, we answered YES(?,O(n^1))
0.00/0.35	EOF