YES(?,O(n^1)) 1044.97/297.05 YES(?,O(n^1)) 1044.97/297.05 1044.97/297.05 We are left with following problem, upon which TcT provides the 1044.97/297.05 certificate YES(?,O(n^1)). 1044.97/297.05 1044.97/297.05 Strict Trs: 1044.97/297.05 { active(f(X)) -> f(active(X)) 1044.97/297.05 , active(f(0())) -> mark(cons(0(), f(s(0())))) 1044.97/297.05 , active(f(s(0()))) -> mark(f(p(s(0())))) 1044.97/297.05 , active(cons(X1, X2)) -> cons(active(X1), X2) 1044.97/297.05 , active(s(X)) -> s(active(X)) 1044.97/297.05 , active(p(X)) -> p(active(X)) 1044.97/297.05 , active(p(s(X))) -> mark(X) 1044.97/297.05 , f(mark(X)) -> mark(f(X)) 1044.97/297.05 , f(ok(X)) -> ok(f(X)) 1044.97/297.05 , cons(mark(X1), X2) -> mark(cons(X1, X2)) 1044.97/297.05 , cons(ok(X1), ok(X2)) -> ok(cons(X1, X2)) 1044.97/297.05 , s(mark(X)) -> mark(s(X)) 1044.97/297.05 , s(ok(X)) -> ok(s(X)) 1044.97/297.05 , p(mark(X)) -> mark(p(X)) 1044.97/297.05 , p(ok(X)) -> ok(p(X)) 1044.97/297.05 , proper(f(X)) -> f(proper(X)) 1044.97/297.05 , proper(0()) -> ok(0()) 1044.97/297.05 , proper(cons(X1, X2)) -> cons(proper(X1), proper(X2)) 1044.97/297.05 , proper(s(X)) -> s(proper(X)) 1044.97/297.05 , proper(p(X)) -> p(proper(X)) 1044.97/297.05 , top(mark(X)) -> top(proper(X)) 1044.97/297.05 , top(ok(X)) -> top(active(X)) } 1044.97/297.05 Obligation: 1044.97/297.05 runtime complexity 1044.97/297.05 Answer: 1044.97/297.05 YES(?,O(n^1)) 1044.97/297.05 1044.97/297.05 The problem is match-bounded by 2. The enriched problem is 1044.97/297.05 compatible with the following automaton. 1044.97/297.05 { active_0(3) -> 1 1044.97/297.05 , active_0(4) -> 1 1044.97/297.05 , active_0(9) -> 1 1044.97/297.05 , active_1(3) -> 16 1044.97/297.05 , active_1(4) -> 16 1044.97/297.05 , active_1(9) -> 16 1044.97/297.05 , active_2(15) -> 17 1044.97/297.05 , f_0(3) -> 2 1044.97/297.05 , f_0(4) -> 2 1044.97/297.05 , f_0(9) -> 2 1044.97/297.05 , f_1(3) -> 11 1044.97/297.05 , f_1(4) -> 11 1044.97/297.05 , f_1(9) -> 11 1044.97/297.05 , 0_0() -> 3 1044.97/297.05 , 0_1() -> 15 1044.97/297.05 , mark_0(3) -> 4 1044.97/297.05 , mark_0(4) -> 4 1044.97/297.05 , mark_0(9) -> 4 1044.97/297.05 , mark_1(11) -> 2 1044.97/297.05 , mark_1(11) -> 11 1044.97/297.05 , mark_1(12) -> 5 1044.97/297.05 , mark_1(12) -> 12 1044.97/297.05 , mark_1(13) -> 6 1044.97/297.05 , mark_1(13) -> 13 1044.97/297.05 , mark_1(14) -> 7 1044.97/297.05 , mark_1(14) -> 14 1044.97/297.05 , cons_0(3, 3) -> 5 1044.97/297.05 , cons_0(3, 4) -> 5 1044.97/297.05 , cons_0(3, 9) -> 5 1044.97/297.05 , cons_0(4, 3) -> 5 1044.97/297.05 , cons_0(4, 4) -> 5 1044.97/297.05 , cons_0(4, 9) -> 5 1044.97/297.05 , cons_0(9, 3) -> 5 1044.97/297.05 , cons_0(9, 4) -> 5 1044.97/297.05 , cons_0(9, 9) -> 5 1044.97/297.05 , cons_1(3, 3) -> 12 1044.97/297.05 , cons_1(3, 4) -> 12 1044.97/297.05 , cons_1(3, 9) -> 12 1044.97/297.05 , cons_1(4, 3) -> 12 1044.97/297.05 , cons_1(4, 4) -> 12 1044.97/297.05 , cons_1(4, 9) -> 12 1044.97/297.05 , cons_1(9, 3) -> 12 1044.97/297.05 , cons_1(9, 4) -> 12 1044.97/297.05 , cons_1(9, 9) -> 12 1044.97/297.05 , s_0(3) -> 6 1044.97/297.05 , s_0(4) -> 6 1044.97/297.05 , s_0(9) -> 6 1044.97/297.05 , s_1(3) -> 13 1044.97/297.05 , s_1(4) -> 13 1044.97/297.05 , s_1(9) -> 13 1044.97/297.05 , p_0(3) -> 7 1044.97/297.05 , p_0(4) -> 7 1044.97/297.05 , p_0(9) -> 7 1044.97/297.05 , p_1(3) -> 14 1044.97/297.05 , p_1(4) -> 14 1044.97/297.05 , p_1(9) -> 14 1044.97/297.05 , proper_0(3) -> 8 1044.97/297.05 , proper_0(4) -> 8 1044.97/297.05 , proper_0(9) -> 8 1044.97/297.05 , proper_1(3) -> 16 1044.97/297.05 , proper_1(4) -> 16 1044.97/297.05 , proper_1(9) -> 16 1044.97/297.05 , ok_0(3) -> 9 1044.97/297.05 , ok_0(4) -> 9 1044.97/297.05 , ok_0(9) -> 9 1044.97/297.05 , ok_1(11) -> 2 1044.97/297.05 , ok_1(11) -> 11 1044.97/297.05 , ok_1(12) -> 5 1044.97/297.05 , ok_1(12) -> 12 1044.97/297.05 , ok_1(13) -> 6 1044.97/297.05 , ok_1(13) -> 13 1044.97/297.05 , ok_1(14) -> 7 1044.97/297.05 , ok_1(14) -> 14 1044.97/297.05 , ok_1(15) -> 8 1044.97/297.05 , ok_1(15) -> 16 1044.97/297.05 , top_0(3) -> 10 1044.97/297.05 , top_0(4) -> 10 1044.97/297.05 , top_0(9) -> 10 1044.97/297.05 , top_1(16) -> 10 1044.97/297.05 , top_2(17) -> 10 } 1044.97/297.05 1044.97/297.05 Hurray, we answered YES(?,O(n^1)) 1044.97/297.06 EOF