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

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

The problem is match-bounded by 6. The enriched problem is
compatible with the following automaton.
{ active_0(2) -> 1
, active_0(3) -> 1
, active_0(7) -> 1
, active_1(2) -> 14
, active_1(3) -> 14
, active_1(7) -> 14
, active_2(11) -> 15
, active_3(26) -> 21
, active_4(29) -> 30
, active_5(27) -> 33
, active_6(36) -> 37
, c_0() -> 2
, c_1() -> 11
, c_2() -> 18
, c_3() -> 25
, c_4() -> 35
, mark_0(2) -> 3
, mark_0(3) -> 3
, mark_0(7) -> 3
, mark_1(9) -> 1
, mark_1(9) -> 14
, mark_2(16) -> 15
, mark_4(28) -> 21
, mark_5(31) -> 30
, f_0(2) -> 4
, f_0(3) -> 4
, f_0(7) -> 4
, f_1(2) -> 12
, f_1(3) -> 12
, f_1(7) -> 12
, f_1(10) -> 9
, f_2(17) -> 16
, f_2(19) -> 15
, f_3(22) -> 21
, f_3(24) -> 26
, f_4(27) -> 29
, g_0(2) -> 5
, g_0(3) -> 5
, g_0(7) -> 5
, g_1(2) -> 13
, g_1(3) -> 13
, g_1(7) -> 13
, g_1(11) -> 10
, g_2(18) -> 17
, g_2(20) -> 19
, g_3(18) -> 24
, g_3(23) -> 22
, g_4(18) -> 28
, g_4(25) -> 27
, g_5(25) -> 31
, g_5(32) -> 30
, g_5(35) -> 36
, g_6(34) -> 33
, proper_0(2) -> 6
, proper_0(3) -> 6
, proper_0(7) -> 6
, proper_1(2) -> 14
, proper_1(3) -> 14
, proper_1(7) -> 14
, proper_2(9) -> 15
, proper_2(10) -> 19
, proper_2(11) -> 20
, proper_3(16) -> 21
, proper_3(17) -> 22
, proper_3(18) -> 23
, proper_4(28) -> 30
, proper_5(18) -> 32
, proper_5(31) -> 33
, proper_6(25) -> 34
, ok_0(2) -> 7
, ok_0(3) -> 7
, ok_0(7) -> 7
, ok_1(11) -> 6
, ok_1(11) -> 14
, ok_1(12) -> 4
, ok_1(12) -> 12
, ok_1(13) -> 5
, ok_1(13) -> 13
, ok_2(18) -> 20
, ok_3(24) -> 19
, ok_3(25) -> 23
, ok_3(25) -> 32
, ok_3(26) -> 15
, ok_4(27) -> 22
, ok_4(27) -> 30
, ok_4(29) -> 21
, ok_4(35) -> 34
, ok_5(36) -> 33
, top_0(2) -> 8
, top_0(3) -> 8
, top_0(7) -> 8
, top_1(14) -> 8
, top_2(15) -> 8
, top_3(21) -> 8
, top_4(30) -> 8
, top_5(33) -> 8
, top_6(37) -> 8 }

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