YES(?,O(n^2))
23.84/11.74	YES(?,O(n^2))
23.84/11.74	
23.84/11.74	We are left with following problem, upon which TcT provides the
23.84/11.74	certificate YES(?,O(n^2)).
23.84/11.74	
23.84/11.74	Strict Trs:
23.84/11.74	  { f(0()) -> 1()
23.84/11.74	  , f(s(x)) -> g(x, s(x))
23.84/11.74	  , g(0(), y) -> y
23.84/11.74	  , g(s(x), y) -> g(x, s(+(y, x)))
23.84/11.74	  , g(s(x), y) -> g(x, +(y, s(x)))
23.84/11.74	  , +(x, 0()) -> x
23.84/11.74	  , +(x, s(y)) -> s(+(x, y)) }
23.84/11.74	Obligation:
23.84/11.74	  innermost runtime complexity
23.84/11.74	Answer:
23.84/11.74	  YES(?,O(n^2))
23.84/11.74	
23.84/11.74	The input was oriented with the instance of 'Small Polynomial Path
23.84/11.74	Order (PS)' as induced by the safe mapping
23.84/11.74	
23.84/11.74	 safe(f) = {}, safe(0) = {}, safe(1) = {}, safe(s) = {1},
23.84/11.74	 safe(g) = {2}, safe(+) = {1}
23.84/11.74	
23.84/11.74	and precedence
23.84/11.74	
23.84/11.74	 f > +, g > +, f ~ g .
23.84/11.74	
23.84/11.74	Following symbols are considered recursive:
23.84/11.74	
23.84/11.74	 {f, g, +}
23.84/11.74	
23.84/11.74	The recursion depth is 2.
23.84/11.74	
23.84/11.74	For your convenience, here are the satisfied ordering constraints:
23.84/11.74	
23.84/11.74	       f(0();) > 1()               
23.84/11.74	                                   
23.84/11.74	    f(s(; x);) > g(x; s(; x))      
23.84/11.74	                                   
23.84/11.74	     g(0(); y) > y                 
23.84/11.74	                                   
23.84/11.74	  g(s(; x); y) > g(x; s(; +(x; y)))
23.84/11.74	                                   
23.84/11.74	  g(s(; x); y) > g(x; +(s(; x); y))
23.84/11.74	                                   
23.84/11.74	     +(0(); x) > x                 
23.84/11.74	                                   
23.84/11.74	  +(s(; y); x) > s(; +(y; x))      
23.84/11.74	                                   
23.84/11.74	
23.84/11.74	Hurray, we answered YES(?,O(n^2))
23.84/11.78	EOF