YES(?,O(n^1))
0.00/0.21	YES(?,O(n^1))
0.00/0.21	
0.00/0.21	We are left with following problem, upon which TcT provides the
0.00/0.21	certificate YES(?,O(n^1)).
0.00/0.21	
0.00/0.21	Strict Trs:
0.00/0.21	  { +(0(), y) -> y
0.00/0.21	  , +(s(x), y) -> +(x, s(y))
0.00/0.21	  , +(s(x), y) -> s(+(x, y)) }
0.00/0.21	Obligation:
0.00/0.21	  runtime complexity
0.00/0.21	Answer:
0.00/0.21	  YES(?,O(n^1))
0.00/0.21	
0.00/0.21	The input is overlay and right-linear. Switching to innermost
0.00/0.21	rewriting.
0.00/0.21	
0.00/0.21	We are left with following problem, upon which TcT provides the
0.00/0.21	certificate YES(?,O(n^1)).
0.00/0.21	
0.00/0.21	Strict Trs:
0.00/0.21	  { +(0(), y) -> y
0.00/0.21	  , +(s(x), y) -> +(x, s(y))
0.00/0.21	  , +(s(x), y) -> s(+(x, y)) }
0.00/0.21	Obligation:
0.00/0.21	  innermost runtime complexity
0.00/0.21	Answer:
0.00/0.21	  YES(?,O(n^1))
0.00/0.21	
0.00/0.21	The input was oriented with the instance of 'Small Polynomial Path
0.00/0.21	Order (PS)' as induced by the safe mapping
0.00/0.21	
0.00/0.21	 safe(+) = {2}, safe(0) = {}, safe(s) = {1}
0.00/0.21	
0.00/0.21	and precedence
0.00/0.21	
0.00/0.21	 empty .
0.00/0.21	
0.00/0.21	Following symbols are considered recursive:
0.00/0.21	
0.00/0.21	 {+}
0.00/0.21	
0.00/0.21	The recursion depth is 1.
0.00/0.21	
0.00/0.21	For your convenience, here are the satisfied ordering constraints:
0.00/0.21	
0.00/0.21	     +(0(); y) > y           
0.00/0.21	                             
0.00/0.21	  +(s(; x); y) > +(x; s(; y))
0.00/0.21	                             
0.00/0.21	  +(s(; x); y) > s(; +(x; y))
0.00/0.21	                             
0.00/0.21	
0.00/0.21	Hurray, we answered YES(?,O(n^1))
0.00/0.21	EOF