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(X)) -> mark(g(h(f(X))))
  , active(h(X)) -> h(active(X))
  , f(mark(X)) -> mark(f(X))
  , f(ok(X)) -> ok(f(X))
  , g(ok(X)) -> ok(g(X))
  , h(mark(X)) -> mark(h(X))
  , h(ok(X)) -> ok(h(X))
  , proper(f(X)) -> f(proper(X))
  , proper(g(X)) -> g(proper(X))
  , proper(h(X)) -> h(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 1. The enriched problem is
compatible with the following automaton.
{ active_0(3) -> 1
, active_0(7) -> 1
, active_1(3) -> 12
, active_1(7) -> 12
, f_0(3) -> 2
, f_0(7) -> 2
, f_1(3) -> 9
, f_1(7) -> 9
, mark_0(3) -> 3
, mark_0(7) -> 3
, mark_1(9) -> 2
, mark_1(9) -> 9
, mark_1(11) -> 5
, mark_1(11) -> 11
, g_0(3) -> 4
, g_0(7) -> 4
, g_1(3) -> 10
, g_1(7) -> 10
, h_0(3) -> 5
, h_0(7) -> 5
, h_1(3) -> 11
, h_1(7) -> 11
, proper_0(3) -> 6
, proper_0(7) -> 6
, proper_1(3) -> 12
, proper_1(7) -> 12
, ok_0(3) -> 7
, ok_0(7) -> 7
, ok_1(9) -> 2
, ok_1(9) -> 9
, ok_1(10) -> 4
, ok_1(10) -> 10
, ok_1(11) -> 5
, ok_1(11) -> 11
, top_0(3) -> 8
, top_0(7) -> 8
, top_1(12) -> 8 }

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