YES(?,O(1))
0.00/0.13	YES(?,O(1))
0.00/0.13	
0.00/0.13	We are left with following problem, upon which TcT provides the
0.00/0.13	certificate YES(?,O(1)).
0.00/0.13	
0.00/0.13	Strict Trs:
0.00/0.13	  { fst(0(), Z) -> nil()
0.00/0.13	  , fst(s(), cons(Y)) -> cons(Y)
0.00/0.13	  , from(X) -> cons(X)
0.00/0.13	  , add(0(), X) -> X
0.00/0.13	  , add(s(), Y) -> s()
0.00/0.13	  , len(nil()) -> 0()
0.00/0.13	  , len(cons(X)) -> s() }
0.00/0.13	Obligation:
0.00/0.13	  innermost runtime complexity
0.00/0.13	Answer:
0.00/0.13	  YES(?,O(1))
0.00/0.13	
0.00/0.13	The input was oriented with the instance of 'Small Polynomial Path
0.00/0.13	Order (PS)' as induced by the safe mapping
0.00/0.13	
0.00/0.13	 safe(fst) = {1, 2}, safe(0) = {}, safe(nil) = {}, safe(s) = {},
0.00/0.13	 safe(cons) = {1}, safe(from) = {}, safe(add) = {}, safe(len) = {}
0.00/0.13	
0.00/0.13	and precedence
0.00/0.13	
0.00/0.13	 fst > from, add > from, len > from, fst ~ add, fst ~ len,
0.00/0.13	 add ~ len .
0.00/0.13	
0.00/0.13	Following symbols are considered recursive:
0.00/0.13	
0.00/0.13	 {}
0.00/0.13	
0.00/0.13	The recursion depth is 0.
0.00/0.13	
0.00/0.13	For your convenience, here are the satisfied ordering constraints:
0.00/0.13	
0.00/0.13	          fst(; 0(),  Z) > nil()    
0.00/0.13	                                    
0.00/0.13	  fst(; s(),  cons(; Y)) > cons(; Y)
0.00/0.13	                                    
0.00/0.13	                from(X;) > cons(; X)
0.00/0.13	                                    
0.00/0.13	           add(0(),  X;) > X        
0.00/0.13	                                    
0.00/0.13	           add(s(),  Y;) > s()      
0.00/0.13	                                    
0.00/0.13	             len(nil();) > 0()      
0.00/0.13	                                    
0.00/0.13	         len(cons(; X);) > s()      
0.00/0.13	                                    
0.00/0.13	
0.00/0.13	Hurray, we answered YES(?,O(1))
0.00/0.13	EOF