YES(?,O(n^1))

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

Strict Trs:
  { active(f(X)) -> f(active(X))
  , active(f(f(a()))) -> mark(c(f(g(f(a())))))
  , active(g(X)) -> g(active(X))
  , f(mark(X)) -> mark(f(X))
  , f(ok(X)) -> ok(f(X))
  , c(ok(X)) -> ok(c(X))
  , g(mark(X)) -> mark(g(X))
  , g(ok(X)) -> ok(g(X))
  , proper(f(X)) -> f(proper(X))
  , proper(a()) -> ok(a())
  , proper(c(X)) -> c(proper(X))
  , proper(g(X)) -> g(proper(X))
  , 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 2. The enriched problem is
compatible with the following automaton.
{ active_0(3) -> 1
, active_0(4) -> 1
, active_0(8) -> 1
, active_1(3) -> 14
, active_1(4) -> 14
, active_1(8) -> 14
, active_2(13) -> 15
, f_0(3) -> 2
, f_0(4) -> 2
, f_0(8) -> 2
, f_1(3) -> 10
, f_1(4) -> 10
, f_1(8) -> 10
, a_0() -> 3
, a_1() -> 13
, mark_0(3) -> 4
, mark_0(4) -> 4
, mark_0(8) -> 4
, mark_1(10) -> 2
, mark_1(10) -> 10
, mark_1(12) -> 6
, mark_1(12) -> 12
, c_0(3) -> 5
, c_0(4) -> 5
, c_0(8) -> 5
, c_1(3) -> 11
, c_1(4) -> 11
, c_1(8) -> 11
, g_0(3) -> 6
, g_0(4) -> 6
, g_0(8) -> 6
, g_1(3) -> 12
, g_1(4) -> 12
, g_1(8) -> 12
, proper_0(3) -> 7
, proper_0(4) -> 7
, proper_0(8) -> 7
, proper_1(3) -> 14
, proper_1(4) -> 14
, proper_1(8) -> 14
, ok_0(3) -> 8
, ok_0(4) -> 8
, ok_0(8) -> 8
, ok_1(10) -> 2
, ok_1(10) -> 10
, ok_1(11) -> 5
, ok_1(11) -> 11
, ok_1(12) -> 6
, ok_1(12) -> 12
, ok_1(13) -> 7
, ok_1(13) -> 14
, top_0(3) -> 9
, top_0(4) -> 9
, top_0(8) -> 9
, top_1(14) -> 9
, top_2(15) -> 9 }

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