YES(?,O(n^1)) 485.15/125.34 YES(?,O(n^1)) 485.15/125.34 485.15/125.34 We are left with following problem, upon which TcT provides the 485.15/125.34 certificate YES(?,O(n^1)). 485.15/125.34 485.15/125.34 Strict Trs: 485.15/125.34 { active(f(X)) -> f(active(X)) 485.15/125.34 , active(f(f(a()))) -> mark(f(g(f(a())))) 485.15/125.34 , f(mark(X)) -> mark(f(X)) 485.15/125.34 , f(ok(X)) -> ok(f(X)) 485.15/125.34 , g(ok(X)) -> ok(g(X)) 485.15/125.34 , proper(f(X)) -> f(proper(X)) 485.15/125.34 , proper(a()) -> ok(a()) 485.15/125.34 , proper(g(X)) -> g(proper(X)) 485.15/125.34 , top(mark(X)) -> top(proper(X)) 485.15/125.34 , top(ok(X)) -> top(active(X)) } 485.15/125.34 Obligation: 485.15/125.34 innermost runtime complexity 485.15/125.34 Answer: 485.15/125.34 YES(?,O(n^1)) 485.15/125.34 485.15/125.34 The problem is match-bounded by 2. The enriched problem is 485.15/125.34 compatible with the following automaton. 485.15/125.34 { active_0(3) -> 1 485.15/125.34 , active_0(4) -> 1 485.15/125.34 , active_0(7) -> 1 485.15/125.34 , active_1(3) -> 12 485.15/125.34 , active_1(4) -> 12 485.15/125.34 , active_1(7) -> 12 485.15/125.34 , active_2(11) -> 13 485.15/125.34 , f_0(3) -> 2 485.15/125.34 , f_0(4) -> 2 485.15/125.34 , f_0(7) -> 2 485.15/125.34 , f_1(3) -> 9 485.15/125.34 , f_1(4) -> 9 485.15/125.34 , f_1(7) -> 9 485.15/125.34 , a_0() -> 3 485.15/125.34 , a_1() -> 11 485.15/125.34 , mark_0(3) -> 4 485.15/125.34 , mark_0(4) -> 4 485.15/125.34 , mark_0(7) -> 4 485.15/125.34 , mark_1(9) -> 2 485.15/125.34 , mark_1(9) -> 9 485.15/125.34 , g_0(3) -> 5 485.15/125.34 , g_0(4) -> 5 485.15/125.34 , g_0(7) -> 5 485.15/125.34 , g_1(3) -> 10 485.15/125.34 , g_1(4) -> 10 485.15/125.34 , g_1(7) -> 10 485.15/125.34 , proper_0(3) -> 6 485.15/125.34 , proper_0(4) -> 6 485.15/125.34 , proper_0(7) -> 6 485.15/125.34 , proper_1(3) -> 12 485.15/125.34 , proper_1(4) -> 12 485.15/125.34 , proper_1(7) -> 12 485.15/125.34 , ok_0(3) -> 7 485.15/125.34 , ok_0(4) -> 7 485.15/125.34 , ok_0(7) -> 7 485.15/125.34 , ok_1(9) -> 2 485.15/125.34 , ok_1(9) -> 9 485.15/125.34 , ok_1(10) -> 5 485.15/125.34 , ok_1(10) -> 10 485.15/125.34 , ok_1(11) -> 6 485.15/125.34 , ok_1(11) -> 12 485.15/125.34 , top_0(3) -> 8 485.15/125.34 , top_0(4) -> 8 485.15/125.34 , top_0(7) -> 8 485.15/125.34 , top_1(12) -> 8 485.15/125.34 , top_2(13) -> 8 } 485.15/125.34 485.15/125.34 Hurray, we answered YES(?,O(n^1)) 485.15/125.35 EOF