YES(?,O(n^1))
0.00/0.08	YES(?,O(n^1))
0.00/0.08	
0.00/0.08	We are left with following problem, upon which TcT provides the
0.00/0.08	certificate YES(?,O(n^1)).
0.00/0.08	
0.00/0.08	Strict Trs:
0.00/0.08	  { f(g(x)) -> g(g(f(x)))
0.00/0.08	  , f(g(x)) -> g(g(g(x))) }
0.00/0.08	Obligation:
0.00/0.08	  runtime complexity
0.00/0.08	Answer:
0.00/0.08	  YES(?,O(n^1))
0.00/0.08	
0.00/0.08	The input is overlay and right-linear. Switching to innermost
0.00/0.08	rewriting.
0.00/0.08	
0.00/0.08	We are left with following problem, upon which TcT provides the
0.00/0.08	certificate YES(?,O(n^1)).
0.00/0.08	
0.00/0.08	Strict Trs:
0.00/0.08	  { f(g(x)) -> g(g(f(x)))
0.00/0.08	  , f(g(x)) -> g(g(g(x))) }
0.00/0.08	Obligation:
0.00/0.08	  innermost runtime complexity
0.00/0.08	Answer:
0.00/0.08	  YES(?,O(n^1))
0.00/0.08	
0.00/0.08	The input was oriented with the instance of 'Small Polynomial Path
0.00/0.08	Order (PS)' as induced by the safe mapping
0.00/0.08	
0.00/0.08	 safe(f) = {}, safe(g) = {1}
0.00/0.08	
0.00/0.08	and precedence
0.00/0.08	
0.00/0.08	 empty .
0.00/0.08	
0.00/0.08	Following symbols are considered recursive:
0.00/0.08	
0.00/0.08	 {f}
0.00/0.08	
0.00/0.08	The recursion depth is 1.
0.00/0.08	
0.00/0.08	For your convenience, here are the satisfied ordering constraints:
0.00/0.08	
0.00/0.08	  f(g(; x);) > g(; g(; f(x;))) 
0.00/0.08	                               
0.00/0.08	  f(g(; x);) > g(; g(; g(; x)))
0.00/0.08	                               
0.00/0.08	
0.00/0.08	Hurray, we answered YES(?,O(n^1))
0.00/0.08	EOF