YES(?,O(n^1))

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

Strict Trs:
  { active(__(X1, X2)) -> __(X1, active(X2))
  , active(__(X1, X2)) -> __(active(X1), X2)
  , active(__(X, nil())) -> mark(X)
  , active(__(__(X, Y), Z)) -> mark(__(X, __(Y, Z)))
  , active(__(nil(), X)) -> mark(X)
  , active(and(X1, X2)) -> and(active(X1), X2)
  , active(and(tt(), X)) -> mark(X)
  , active(isNePal(X)) -> isNePal(active(X))
  , active(isNePal(__(I, __(P, I)))) -> mark(tt())
  , __(X1, mark(X2)) -> mark(__(X1, X2))
  , __(mark(X1), X2) -> mark(__(X1, X2))
  , __(ok(X1), ok(X2)) -> ok(__(X1, X2))
  , and(mark(X1), X2) -> mark(and(X1, X2))
  , and(ok(X1), ok(X2)) -> ok(and(X1, X2))
  , isNePal(mark(X)) -> mark(isNePal(X))
  , isNePal(ok(X)) -> ok(isNePal(X))
  , proper(__(X1, X2)) -> __(proper(X1), proper(X2))
  , proper(nil()) -> ok(nil())
  , proper(and(X1, X2)) -> and(proper(X1), proper(X2))
  , proper(tt()) -> ok(tt())
  , proper(isNePal(X)) -> isNePal(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(6) -> 1
, active_0(9) -> 1
, active_1(3) -> 15
, active_1(4) -> 15
, active_1(6) -> 15
, active_1(9) -> 15
, active_2(14) -> 16
, ___0(3, 3) -> 2
, ___0(3, 4) -> 2
, ___0(3, 6) -> 2
, ___0(3, 9) -> 2
, ___0(4, 3) -> 2
, ___0(4, 4) -> 2
, ___0(4, 6) -> 2
, ___0(4, 9) -> 2
, ___0(6, 3) -> 2
, ___0(6, 4) -> 2
, ___0(6, 6) -> 2
, ___0(6, 9) -> 2
, ___0(9, 3) -> 2
, ___0(9, 4) -> 2
, ___0(9, 6) -> 2
, ___0(9, 9) -> 2
, ___1(3, 3) -> 11
, ___1(3, 4) -> 11
, ___1(3, 6) -> 11
, ___1(3, 9) -> 11
, ___1(4, 3) -> 11
, ___1(4, 4) -> 11
, ___1(4, 6) -> 11
, ___1(4, 9) -> 11
, ___1(6, 3) -> 11
, ___1(6, 4) -> 11
, ___1(6, 6) -> 11
, ___1(6, 9) -> 11
, ___1(9, 3) -> 11
, ___1(9, 4) -> 11
, ___1(9, 6) -> 11
, ___1(9, 9) -> 11
, mark_0(3) -> 3
, mark_0(4) -> 3
, mark_0(6) -> 3
, mark_0(9) -> 3
, mark_1(11) -> 2
, mark_1(11) -> 11
, mark_1(12) -> 5
, mark_1(12) -> 12
, mark_1(13) -> 7
, mark_1(13) -> 13
, nil_0() -> 4
, nil_1() -> 14
, and_0(3, 3) -> 5
, and_0(3, 4) -> 5
, and_0(3, 6) -> 5
, and_0(3, 9) -> 5
, and_0(4, 3) -> 5
, and_0(4, 4) -> 5
, and_0(4, 6) -> 5
, and_0(4, 9) -> 5
, and_0(6, 3) -> 5
, and_0(6, 4) -> 5
, and_0(6, 6) -> 5
, and_0(6, 9) -> 5
, and_0(9, 3) -> 5
, and_0(9, 4) -> 5
, and_0(9, 6) -> 5
, and_0(9, 9) -> 5
, and_1(3, 3) -> 12
, and_1(3, 4) -> 12
, and_1(3, 6) -> 12
, and_1(3, 9) -> 12
, and_1(4, 3) -> 12
, and_1(4, 4) -> 12
, and_1(4, 6) -> 12
, and_1(4, 9) -> 12
, and_1(6, 3) -> 12
, and_1(6, 4) -> 12
, and_1(6, 6) -> 12
, and_1(6, 9) -> 12
, and_1(9, 3) -> 12
, and_1(9, 4) -> 12
, and_1(9, 6) -> 12
, and_1(9, 9) -> 12
, tt_0() -> 6
, tt_1() -> 14
, isNePal_0(3) -> 7
, isNePal_0(4) -> 7
, isNePal_0(6) -> 7
, isNePal_0(9) -> 7
, isNePal_1(3) -> 13
, isNePal_1(4) -> 13
, isNePal_1(6) -> 13
, isNePal_1(9) -> 13
, proper_0(3) -> 8
, proper_0(4) -> 8
, proper_0(6) -> 8
, proper_0(9) -> 8
, proper_1(3) -> 15
, proper_1(4) -> 15
, proper_1(6) -> 15
, proper_1(9) -> 15
, ok_0(3) -> 9
, ok_0(4) -> 9
, ok_0(6) -> 9
, ok_0(9) -> 9
, ok_1(11) -> 2
, ok_1(11) -> 11
, ok_1(12) -> 5
, ok_1(12) -> 12
, ok_1(13) -> 7
, ok_1(13) -> 13
, ok_1(14) -> 8
, ok_1(14) -> 15
, top_0(3) -> 10
, top_0(4) -> 10
, top_0(6) -> 10
, top_0(9) -> 10
, top_1(15) -> 10
, top_2(16) -> 10 }

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