YES(?,O(n^1)) 0.00/0.17 YES(?,O(n^1)) 0.00/0.17 0.00/0.17 We are left with following problem, upon which TcT provides the 0.00/0.17 certificate YES(?,O(n^1)). 0.00/0.17 0.00/0.17 Strict Trs: 0.00/0.17 { f(x, g(x)) -> x 0.00/0.17 , f(x, h(y)) -> f(h(x), y) } 0.00/0.17 Obligation: 0.00/0.17 runtime complexity 0.00/0.17 Answer: 0.00/0.17 YES(?,O(n^1)) 0.00/0.17 0.00/0.17 The input is overlay and right-linear. Switching to innermost 0.00/0.17 rewriting. 0.00/0.17 0.00/0.17 We are left with following problem, upon which TcT provides the 0.00/0.17 certificate YES(?,O(n^1)). 0.00/0.17 0.00/0.17 Strict Trs: 0.00/0.17 { f(x, g(x)) -> x 0.00/0.17 , f(x, h(y)) -> f(h(x), y) } 0.00/0.17 Obligation: 0.00/0.17 innermost runtime complexity 0.00/0.17 Answer: 0.00/0.17 YES(?,O(n^1)) 0.00/0.17 0.00/0.17 The input was oriented with the instance of 'Small Polynomial Path 0.00/0.17 Order (PS)' as induced by the safe mapping 0.00/0.17 0.00/0.17 safe(f) = {1}, safe(g) = {1}, safe(h) = {1} 0.00/0.17 0.00/0.17 and precedence 0.00/0.17 0.00/0.17 empty . 0.00/0.17 0.00/0.17 Following symbols are considered recursive: 0.00/0.17 0.00/0.17 {f} 0.00/0.17 0.00/0.17 The recursion depth is 1. 0.00/0.17 0.00/0.17 For your convenience, here are the satisfied ordering constraints: 0.00/0.17 0.00/0.17 f(g(; x); x) > x 0.00/0.17 0.00/0.17 f(h(; y); x) > f(y; h(; x)) 0.00/0.17 0.00/0.17 0.00/0.17 Hurray, we answered YES(?,O(n^1)) 0.00/0.17 EOF