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(0())) -> mark(cons(0(), f(s(0()))))
  , active(f(s(0()))) -> mark(f(p(s(0()))))
  , active(cons(X1, X2)) -> cons(active(X1), X2)
  , active(s(X)) -> s(active(X))
  , active(p(X)) -> p(active(X))
  , active(p(s(0()))) -> mark(0())
  , f(mark(X)) -> mark(f(X))
  , f(ok(X)) -> ok(f(X))
  , cons(mark(X1), X2) -> mark(cons(X1, X2))
  , cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
  , s(mark(X)) -> mark(s(X))
  , s(ok(X)) -> ok(s(X))
  , p(mark(X)) -> mark(p(X))
  , p(ok(X)) -> ok(p(X))
  , proper(f(X)) -> f(proper(X))
  , proper(0()) -> ok(0())
  , proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
  , proper(s(X)) -> s(proper(X))
  , proper(p(X)) -> p(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(9) -> 1
, active_1(3) -> 16
, active_1(4) -> 16
, active_1(9) -> 16
, active_2(15) -> 17
, f_0(3) -> 2
, f_0(4) -> 2
, f_0(9) -> 2
, f_1(3) -> 11
, f_1(4) -> 11
, f_1(9) -> 11
, 0_0() -> 3
, 0_1() -> 15
, mark_0(3) -> 4
, mark_0(4) -> 4
, mark_0(9) -> 4
, mark_1(11) -> 2
, mark_1(11) -> 11
, mark_1(12) -> 5
, mark_1(12) -> 12
, mark_1(13) -> 6
, mark_1(13) -> 13
, mark_1(14) -> 7
, mark_1(14) -> 14
, cons_0(3, 3) -> 5
, cons_0(3, 4) -> 5
, cons_0(3, 9) -> 5
, cons_0(4, 3) -> 5
, cons_0(4, 4) -> 5
, cons_0(4, 9) -> 5
, cons_0(9, 3) -> 5
, cons_0(9, 4) -> 5
, cons_0(9, 9) -> 5
, cons_1(3, 3) -> 12
, cons_1(3, 4) -> 12
, cons_1(3, 9) -> 12
, cons_1(4, 3) -> 12
, cons_1(4, 4) -> 12
, cons_1(4, 9) -> 12
, cons_1(9, 3) -> 12
, cons_1(9, 4) -> 12
, cons_1(9, 9) -> 12
, s_0(3) -> 6
, s_0(4) -> 6
, s_0(9) -> 6
, s_1(3) -> 13
, s_1(4) -> 13
, s_1(9) -> 13
, p_0(3) -> 7
, p_0(4) -> 7
, p_0(9) -> 7
, p_1(3) -> 14
, p_1(4) -> 14
, p_1(9) -> 14
, proper_0(3) -> 8
, proper_0(4) -> 8
, proper_0(9) -> 8
, proper_1(3) -> 16
, proper_1(4) -> 16
, proper_1(9) -> 16
, ok_0(3) -> 9
, ok_0(4) -> 9
, ok_0(9) -> 9
, ok_1(11) -> 2
, ok_1(11) -> 11
, ok_1(12) -> 5
, ok_1(12) -> 12
, ok_1(13) -> 6
, ok_1(13) -> 13
, ok_1(14) -> 7
, ok_1(14) -> 14
, ok_1(15) -> 8
, ok_1(15) -> 16
, top_0(3) -> 10
, top_0(4) -> 10
, top_0(9) -> 10
, top_1(16) -> 10
, top_2(17) -> 10 }

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