YES(?,O(n^1))

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^1)).

Strict Trs:
  { active(g(X)) -> mark(h(X))
  , active(h(d())) -> mark(g(c()))
  , active(c()) -> mark(d())
  , g(ok(X)) -> ok(g(X))
  , h(ok(X)) -> ok(h(X))
  , proper(g(X)) -> g(proper(X))
  , proper(h(X)) -> h(proper(X))
  , proper(c()) -> ok(c())
  , proper(d()) -> ok(d())
  , top(mark(X)) -> top(proper(X))
  , top(ok(X)) -> top(active(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The problem is match-bounded by 4. The enriched problem is
compatible with the following automaton.
{ active_0(3) -> 1
, active_0(5) -> 1
, active_0(6) -> 1
, active_0(8) -> 1
, active_1(3) -> 14
, active_1(5) -> 14
, active_1(6) -> 14
, active_1(8) -> 14
, active_2(10) -> 15
, active_2(13) -> 15
, active_3(16) -> 17
, active_4(18) -> 19
, g_0(3) -> 2
, g_0(5) -> 2
, g_0(6) -> 2
, g_0(8) -> 2
, g_1(3) -> 11
, g_1(5) -> 11
, g_1(6) -> 11
, g_1(8) -> 11
, mark_0(3) -> 3
, mark_0(5) -> 3
, mark_0(6) -> 3
, mark_0(8) -> 3
, mark_1(10) -> 1
, mark_1(10) -> 14
, mark_2(16) -> 15
, h_0(3) -> 4
, h_0(5) -> 4
, h_0(6) -> 4
, h_0(8) -> 4
, h_1(3) -> 12
, h_1(5) -> 12
, h_1(6) -> 12
, h_1(8) -> 12
, c_0() -> 5
, c_1() -> 13
, d_0() -> 6
, d_1() -> 10
, d_2() -> 16
, d_3() -> 18
, proper_0(3) -> 7
, proper_0(5) -> 7
, proper_0(6) -> 7
, proper_0(8) -> 7
, proper_1(3) -> 14
, proper_1(5) -> 14
, proper_1(6) -> 14
, proper_1(8) -> 14
, proper_2(10) -> 15
, proper_3(16) -> 17
, ok_0(3) -> 8
, ok_0(5) -> 8
, ok_0(6) -> 8
, ok_0(8) -> 8
, ok_1(10) -> 7
, ok_1(10) -> 14
, ok_1(11) -> 2
, ok_1(11) -> 11
, ok_1(12) -> 4
, ok_1(12) -> 12
, ok_1(13) -> 7
, ok_1(13) -> 14
, ok_2(16) -> 15
, ok_3(18) -> 17
, top_0(3) -> 9
, top_0(5) -> 9
, top_0(6) -> 9
, top_0(8) -> 9
, top_1(14) -> 9
, top_2(15) -> 9
, top_3(17) -> 9
, top_4(19) -> 9 }

Hurray, we answered YES(?,O(n^1))