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

Strict Trs:
  { f(X) -> n__f(X)
  , f(0()) -> cons(0(), n__f(n__s(n__0())))
  , f(s(0())) -> f(p(s(0())))
  , 0() -> n__0()
  , s(X) -> n__s(X)
  , p(s(0())) -> 0()
  , activate(X) -> X
  , activate(n__f(X)) -> f(activate(X))
  , activate(n__s(X)) -> s(activate(X))
  , activate(n__0()) -> 0() }
Obligation:
  runtime complexity
Answer:
  YES(?,O(n^1))

The problem is match-bounded by 4. The enriched problem is
compatible with the following automaton.
{ f_0(3) -> 1
, f_0(4) -> 1
, f_0(5) -> 1
, f_0(6) -> 1
, f_1(10) -> 9
, f_1(10) -> 10
, f_2(15) -> 9
, f_2(15) -> 10
, 0_0() -> 2
, 0_1() -> 9
, 0_1() -> 10
, 0_2() -> 11
, 0_3() -> 15
, cons_0(3, 3) -> 3
, cons_0(3, 3) -> 9
, cons_0(3, 3) -> 10
, cons_0(3, 4) -> 3
, cons_0(3, 4) -> 9
, cons_0(3, 4) -> 10
, cons_0(3, 5) -> 3
, cons_0(3, 5) -> 9
, cons_0(3, 5) -> 10
, cons_0(3, 6) -> 3
, cons_0(3, 6) -> 9
, cons_0(3, 6) -> 10
, cons_0(4, 3) -> 3
, cons_0(4, 3) -> 9
, cons_0(4, 3) -> 10
, cons_0(4, 4) -> 3
, cons_0(4, 4) -> 9
, cons_0(4, 4) -> 10
, cons_0(4, 5) -> 3
, cons_0(4, 5) -> 9
, cons_0(4, 5) -> 10
, cons_0(4, 6) -> 3
, cons_0(4, 6) -> 9
, cons_0(4, 6) -> 10
, cons_0(5, 3) -> 3
, cons_0(5, 3) -> 9
, cons_0(5, 3) -> 10
, cons_0(5, 4) -> 3
, cons_0(5, 4) -> 9
, cons_0(5, 4) -> 10
, cons_0(5, 5) -> 3
, cons_0(5, 5) -> 9
, cons_0(5, 5) -> 10
, cons_0(5, 6) -> 3
, cons_0(5, 6) -> 9
, cons_0(5, 6) -> 10
, cons_0(6, 3) -> 3
, cons_0(6, 3) -> 9
, cons_0(6, 3) -> 10
, cons_0(6, 4) -> 3
, cons_0(6, 4) -> 9
, cons_0(6, 4) -> 10
, cons_0(6, 5) -> 3
, cons_0(6, 5) -> 9
, cons_0(6, 5) -> 10
, cons_0(6, 6) -> 3
, cons_0(6, 6) -> 9
, cons_0(6, 6) -> 10
, cons_2(11, 12) -> 9
, cons_2(11, 12) -> 10
, cons_3(15, 17) -> 9
, cons_3(15, 17) -> 10
, n__f_0(3) -> 4
, n__f_0(3) -> 9
, n__f_0(3) -> 10
, n__f_0(4) -> 4
, n__f_0(4) -> 9
, n__f_0(4) -> 10
, n__f_0(5) -> 4
, n__f_0(5) -> 9
, n__f_0(5) -> 10
, n__f_0(6) -> 4
, n__f_0(6) -> 9
, n__f_0(6) -> 10
, n__f_1(3) -> 1
, n__f_1(4) -> 1
, n__f_1(5) -> 1
, n__f_1(6) -> 1
, n__f_2(10) -> 9
, n__f_2(10) -> 10
, n__f_2(13) -> 12
, n__f_3(15) -> 9
, n__f_3(15) -> 10
, n__f_3(16) -> 17
, n__s_0(3) -> 5
, n__s_0(3) -> 9
, n__s_0(3) -> 10
, n__s_0(4) -> 5
, n__s_0(4) -> 9
, n__s_0(4) -> 10
, n__s_0(5) -> 5
, n__s_0(5) -> 9
, n__s_0(5) -> 10
, n__s_0(6) -> 5
, n__s_0(6) -> 9
, n__s_0(6) -> 10
, n__s_1(3) -> 7
, n__s_1(4) -> 7
, n__s_1(5) -> 7
, n__s_1(6) -> 7
, n__s_2(10) -> 9
, n__s_2(10) -> 10
, n__s_2(14) -> 13
, n__s_3(11) -> 16
, n__0_0() -> 6
, n__0_0() -> 9
, n__0_0() -> 10
, n__0_1() -> 2
, n__0_2() -> 9
, n__0_2() -> 10
, n__0_2() -> 14
, n__0_3() -> 11
, n__0_4() -> 15
, s_0(3) -> 7
, s_0(4) -> 7
, s_0(5) -> 7
, s_0(6) -> 7
, s_1(10) -> 9
, s_1(10) -> 10
, s_2(11) -> 16
, p_0(3) -> 8
, p_0(4) -> 8
, p_0(5) -> 8
, p_0(6) -> 8
, p_2(16) -> 15
, activate_0(3) -> 9
, activate_0(4) -> 9
, activate_0(5) -> 9
, activate_0(6) -> 9
, activate_1(3) -> 10
, activate_1(4) -> 10
, activate_1(5) -> 10
, activate_1(6) -> 10 }

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