YES(?,O(n^1))

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

Strict Trs:
  { active(zeros()) -> mark(cons(0(), zeros()))
  , active(cons(X1, X2)) -> cons(active(X1), X2)
  , active(tail(X)) -> tail(active(X))
  , active(tail(cons(X, XS))) -> mark(XS)
  , cons(mark(X1), X2) -> mark(cons(X1, X2))
  , cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
  , tail(mark(X)) -> mark(tail(X))
  , tail(ok(X)) -> ok(tail(X))
  , proper(zeros()) -> ok(zeros())
  , proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
  , proper(0()) -> ok(0())
  , proper(tail(X)) -> tail(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 5. The enriched problem is
compatible with the following automaton.
{ active_0(2) -> 1
, active_0(3) -> 1
, active_0(5) -> 1
, active_0(8) -> 1
, active_1(2) -> 15
, active_1(3) -> 15
, active_1(5) -> 15
, active_1(8) -> 15
, active_2(11) -> 16
, active_2(12) -> 16
, active_3(26) -> 22
, active_4(18) -> 28
, active_4(29) -> 30
, active_5(25) -> 31
, zeros_0() -> 2
, zeros_1() -> 12
, zeros_2() -> 19
, zeros_3() -> 27
, mark_0(2) -> 3
, mark_0(3) -> 3
, mark_0(5) -> 3
, mark_0(8) -> 3
, mark_1(10) -> 1
, mark_1(10) -> 15
, mark_1(13) -> 4
, mark_1(13) -> 13
, mark_1(14) -> 6
, mark_1(14) -> 14
, mark_2(17) -> 16
, cons_0(2, 2) -> 4
, cons_0(2, 3) -> 4
, cons_0(2, 5) -> 4
, cons_0(2, 8) -> 4
, cons_0(3, 2) -> 4
, cons_0(3, 3) -> 4
, cons_0(3, 5) -> 4
, cons_0(3, 8) -> 4
, cons_0(5, 2) -> 4
, cons_0(5, 3) -> 4
, cons_0(5, 5) -> 4
, cons_0(5, 8) -> 4
, cons_0(8, 2) -> 4
, cons_0(8, 3) -> 4
, cons_0(8, 5) -> 4
, cons_0(8, 8) -> 4
, cons_1(2, 2) -> 13
, cons_1(2, 3) -> 13
, cons_1(2, 5) -> 13
, cons_1(2, 8) -> 13
, cons_1(3, 2) -> 13
, cons_1(3, 3) -> 13
, cons_1(3, 5) -> 13
, cons_1(3, 8) -> 13
, cons_1(5, 2) -> 13
, cons_1(5, 3) -> 13
, cons_1(5, 5) -> 13
, cons_1(5, 8) -> 13
, cons_1(8, 2) -> 13
, cons_1(8, 3) -> 13
, cons_1(8, 5) -> 13
, cons_1(8, 8) -> 13
, cons_1(11, 12) -> 10
, cons_2(18, 19) -> 17
, cons_2(20, 21) -> 16
, cons_3(18, 19) -> 26
, cons_3(23, 24) -> 22
, cons_4(25, 27) -> 29
, cons_4(28, 19) -> 22
, cons_5(31, 27) -> 30
, 0_0() -> 5
, 0_1() -> 11
, 0_2() -> 18
, 0_3() -> 25
, tail_0(2) -> 6
, tail_0(3) -> 6
, tail_0(5) -> 6
, tail_0(8) -> 6
, tail_1(2) -> 14
, tail_1(3) -> 14
, tail_1(5) -> 14
, tail_1(8) -> 14
, proper_0(2) -> 7
, proper_0(3) -> 7
, proper_0(5) -> 7
, proper_0(8) -> 7
, proper_1(2) -> 15
, proper_1(3) -> 15
, proper_1(5) -> 15
, proper_1(8) -> 15
, proper_2(10) -> 16
, proper_2(11) -> 20
, proper_2(12) -> 21
, proper_3(17) -> 22
, proper_3(18) -> 23
, proper_3(19) -> 24
, ok_0(2) -> 8
, ok_0(3) -> 8
, ok_0(5) -> 8
, ok_0(8) -> 8
, ok_1(11) -> 7
, ok_1(11) -> 15
, ok_1(12) -> 7
, ok_1(12) -> 15
, ok_1(13) -> 4
, ok_1(13) -> 13
, ok_1(14) -> 6
, ok_1(14) -> 14
, ok_2(18) -> 20
, ok_2(19) -> 21
, ok_3(25) -> 23
, ok_3(26) -> 16
, ok_3(27) -> 24
, ok_4(29) -> 22
, top_0(2) -> 9
, top_0(3) -> 9
, top_0(5) -> 9
, top_0(8) -> 9
, top_1(15) -> 9
, top_2(16) -> 9
, top_3(22) -> 9
, top_4(30) -> 9 }

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