YES(?,O(1))
0.00/0.53	YES(?,O(1))
0.00/0.53	
0.00/0.53	We are left with following problem, upon which TcT provides the
0.00/0.53	certificate YES(?,O(1)).
0.00/0.53	
0.00/0.53	Strict Trs:
0.00/0.53	  { filter(cons(X), 0(), M) -> cons(0())
0.00/0.53	  , filter(cons(X), s(N), M) -> cons(X)
0.00/0.53	  , sieve(cons(0())) -> cons(0())
0.00/0.53	  , sieve(cons(s(N))) -> cons(s(N))
0.00/0.53	  , nats(N) -> cons(N)
0.00/0.53	  , zprimes() -> sieve(nats(s(s(0())))) }
0.00/0.53	Obligation:
0.00/0.53	  innermost runtime complexity
0.00/0.53	Answer:
0.00/0.53	  YES(?,O(1))
0.00/0.53	
0.00/0.53	The input was oriented with the instance of 'Small Polynomial Path
0.00/0.53	Order (PS)' as induced by the safe mapping
0.00/0.53	
0.00/0.53	 safe(filter) = {3}, safe(cons) = {1}, safe(0) = {}, safe(s) = {1},
0.00/0.53	 safe(sieve) = {1}, safe(nats) = {1}, safe(zprimes) = {}
0.00/0.53	
0.00/0.53	and precedence
0.00/0.53	
0.00/0.53	 filter > sieve, filter > nats, zprimes > sieve, zprimes > nats,
0.00/0.53	 filter ~ zprimes, sieve ~ nats .
0.00/0.53	
0.00/0.53	Following symbols are considered recursive:
0.00/0.53	
0.00/0.53	 {}
0.00/0.53	
0.00/0.53	The recursion depth is 0.
0.00/0.53	
0.00/0.53	For your convenience, here are the satisfied ordering constraints:
0.00/0.53	
0.00/0.53	     filter(cons(; X),  0(); M) > cons(; 0())                   
0.00/0.53	                                                                
0.00/0.53	  filter(cons(; X),  s(; N); M) > cons(; X)                     
0.00/0.53	                                                                
0.00/0.53	           sieve(; cons(; 0())) > cons(; 0())                   
0.00/0.53	                                                                
0.00/0.53	        sieve(; cons(; s(; N))) > cons(; s(; N))                
0.00/0.53	                                                                
0.00/0.53	                      nats(; N) > cons(; N)                     
0.00/0.53	                                                                
0.00/0.53	                      zprimes() > sieve(; nats(; s(; s(; 0()))))
0.00/0.53	                                                                
0.00/0.53	
0.00/0.53	Hurray, we answered YES(?,O(1))
0.00/0.53	EOF