YES(?,O(1))
0.00/0.11	YES(?,O(1))
0.00/0.11	
0.00/0.11	We are left with following problem, upon which TcT provides the
0.00/0.11	certificate YES(?,O(1)).
0.00/0.11	
0.00/0.11	Strict Trs:
0.00/0.11	  { f(c(a(), z, x)) -> b(a(), z)
0.00/0.11	  , b(y, z) -> z
0.00/0.11	  , b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) }
0.00/0.11	Obligation:
0.00/0.11	  innermost runtime complexity
0.00/0.11	Answer:
0.00/0.11	  YES(?,O(1))
0.00/0.11	
0.00/0.11	Arguments of following rules are not normal-forms:
0.00/0.11	
0.00/0.11	{ b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y))) }
0.00/0.11	
0.00/0.11	All above mentioned rules can be savely removed.
0.00/0.11	
0.00/0.11	We are left with following problem, upon which TcT provides the
0.00/0.11	certificate YES(?,O(1)).
0.00/0.11	
0.00/0.11	Strict Trs:
0.00/0.11	  { f(c(a(), z, x)) -> b(a(), z)
0.00/0.11	  , b(y, z) -> z }
0.00/0.11	Obligation:
0.00/0.11	  innermost runtime complexity
0.00/0.11	Answer:
0.00/0.11	  YES(?,O(1))
0.00/0.11	
0.00/0.11	The input was oriented with the instance of 'Small Polynomial Path
0.00/0.11	Order (PS,0-bounded)' as induced by the safe mapping
0.00/0.11	
0.00/0.11	 safe(f) = {}, safe(c) = {1, 2, 3}, safe(a) = {}, safe(b) = {1, 2}
0.00/0.11	
0.00/0.11	and precedence
0.00/0.11	
0.00/0.11	 f > b .
0.00/0.11	
0.00/0.11	Following symbols are considered recursive:
0.00/0.11	
0.00/0.11	 {}
0.00/0.11	
0.00/0.11	The recursion depth is 0.
0.00/0.11	
0.00/0.11	For your convenience, here are the satisfied ordering constraints:
0.00/0.11	
0.00/0.11	  f(c(; a(),  z,  x);) > b(; a(),  z)
0.00/0.11	                                     
0.00/0.11	            b(; y,  z) > z           
0.00/0.11	                                     
0.00/0.11	
0.00/0.11	Hurray, we answered YES(?,O(1))
0.00/0.11	EOF