YES(?,POLY)
68.06/21.68	YES(?,POLY)
68.06/21.68	
68.06/21.68	We are left with following problem, upon which TcT provides the
68.06/21.68	certificate YES(?,POLY).
68.06/21.68	
68.06/21.68	Strict Trs:
68.06/21.68	  { f(s(x1), x2, x3, x4, x5) -> f(x1, x2, x3, x4, x5)
68.06/21.68	  , f(0(), s(x2), x3, x4, x5) -> f(x2, x2, x3, x4, x5)
68.06/21.68	  , f(0(), 0(), s(x3), x4, x5) -> f(x3, x3, x3, x4, x5)
68.06/21.68	  , f(0(), 0(), 0(), s(x4), x5) -> f(x4, x4, x4, x4, x5)
68.06/21.68	  , f(0(), 0(), 0(), 0(), s(x5)) -> f(x5, x5, x5, x5, x5)
68.06/21.68	  , f(0(), 0(), 0(), 0(), 0()) -> 0() }
68.06/21.68	Obligation:
68.06/21.68	  innermost runtime complexity
68.06/21.68	Answer:
68.06/21.68	  YES(?,POLY)
68.06/21.68	
68.06/21.68	The input was oriented with the instance of 'Polynomial Path Order
68.06/21.68	(PS)' as induced by the safe mapping
68.06/21.68	
68.06/21.68	 safe(f) = {}, safe(s) = {1}, safe(0) = {}
68.06/21.68	
68.06/21.68	and precedence
68.06/21.68	
68.06/21.68	 empty .
68.06/21.68	
68.06/21.68	Following symbols are considered recursive:
68.06/21.68	
68.06/21.68	 {f}
68.06/21.68	
68.06/21.68	The recursion depth is 1.
68.06/21.68	
68.06/21.68	For your convenience, here are the satisfied ordering constraints:
68.06/21.68	
68.06/21.68	      f(s(; x1),  x2,  x3,  x4,  x5;) > f(x1,  x2,  x3,  x4,  x5;)
68.06/21.68	                                                                  
68.06/21.68	     f(0(),  s(; x2),  x3,  x4,  x5;) > f(x2,  x2,  x3,  x4,  x5;)
68.06/21.68	                                                                  
68.06/21.68	    f(0(),  0(),  s(; x3),  x4,  x5;) > f(x3,  x3,  x3,  x4,  x5;)
68.06/21.68	                                                                  
68.06/21.68	   f(0(),  0(),  0(),  s(; x4),  x5;) > f(x4,  x4,  x4,  x4,  x5;)
68.06/21.68	                                                                  
68.06/21.68	  f(0(),  0(),  0(),  0(),  s(; x5);) > f(x5,  x5,  x5,  x5,  x5;)
68.06/21.68	                                                                  
68.06/21.68	      f(0(),  0(),  0(),  0(),  0();) > 0()                       
68.06/21.68	                                                                  
68.06/21.68	
68.06/21.68	Hurray, we answered YES(?,POLY)
68.43/21.70	EOF