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(X))) -> mark(c(f(g(f(X)))))
  , active(c(X)) -> mark(d(X))
  , active(h(X)) -> mark(c(d(X)))
  , active(h(X)) -> h(active(X))
  , f(mark(X)) -> mark(f(X))
  , f(ok(X)) -> ok(f(X))
  , c(ok(X)) -> ok(c(X))
  , g(ok(X)) -> ok(g(X))
  , d(ok(X)) -> ok(d(X))
  , h(mark(X)) -> mark(h(X))
  , h(ok(X)) -> ok(h(X))
  , proper(f(X)) -> f(proper(X))
  , proper(c(X)) -> c(proper(X))
  , proper(g(X)) -> g(proper(X))
  , proper(d(X)) -> d(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(2) -> 1
, active_1(2) -> 4
, f_0(2) -> 1
, f_1(2) -> 3
, mark_0(2) -> 2
, mark_1(3) -> 1
, mark_1(3) -> 3
, c_0(2) -> 1
, c_1(2) -> 3
, g_0(2) -> 1
, g_1(2) -> 3
, d_0(2) -> 1
, d_1(2) -> 3
, h_0(2) -> 1
, h_1(2) -> 3
, proper_0(2) -> 1
, proper_1(2) -> 4
, ok_0(2) -> 2
, ok_1(3) -> 1
, ok_1(3) -> 3
, top_0(2) -> 1
, top_1(4) -> 1 }

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