YES(?,O(n^1))

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

The input is overlay and right-linear. Switching to innermost
rewriting.

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(s(0())))
  , f(s(0())) -> f(p(s(0())))
  , p(s(X)) -> X
  , activate(X) -> X
  , activate(n__f(X)) -> f(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.
{ f_0(2) -> 1
, f_0(3) -> 1
, f_0(4) -> 1
, f_0(5) -> 1
, f_1(2) -> 7
, f_1(3) -> 7
, f_1(4) -> 7
, f_1(5) -> 7
, f_1(12) -> 1
, f_1(12) -> 7
, 0_0() -> 2
, 0_0() -> 6
, 0_0() -> 7
, 0_1() -> 8
, 0_1() -> 11
, 0_1() -> 12
, 0_2() -> 13
, 0_2() -> 16
, cons_0(2, 2) -> 3
, cons_0(2, 2) -> 6
, cons_0(2, 2) -> 7
, cons_0(2, 3) -> 3
, cons_0(2, 3) -> 6
, cons_0(2, 3) -> 7
, cons_0(2, 4) -> 3
, cons_0(2, 4) -> 6
, cons_0(2, 4) -> 7
, cons_0(2, 5) -> 3
, cons_0(2, 5) -> 6
, cons_0(2, 5) -> 7
, cons_0(3, 2) -> 3
, cons_0(3, 2) -> 6
, cons_0(3, 2) -> 7
, cons_0(3, 3) -> 3
, cons_0(3, 3) -> 6
, cons_0(3, 3) -> 7
, cons_0(3, 4) -> 3
, cons_0(3, 4) -> 6
, cons_0(3, 4) -> 7
, cons_0(3, 5) -> 3
, cons_0(3, 5) -> 6
, cons_0(3, 5) -> 7
, cons_0(4, 2) -> 3
, cons_0(4, 2) -> 6
, cons_0(4, 2) -> 7
, cons_0(4, 3) -> 3
, cons_0(4, 3) -> 6
, cons_0(4, 3) -> 7
, cons_0(4, 4) -> 3
, cons_0(4, 4) -> 6
, cons_0(4, 4) -> 7
, cons_0(4, 5) -> 3
, cons_0(4, 5) -> 6
, cons_0(4, 5) -> 7
, cons_0(5, 2) -> 3
, cons_0(5, 2) -> 6
, cons_0(5, 2) -> 7
, cons_0(5, 3) -> 3
, cons_0(5, 3) -> 6
, cons_0(5, 3) -> 7
, cons_0(5, 4) -> 3
, cons_0(5, 4) -> 6
, cons_0(5, 4) -> 7
, cons_0(5, 5) -> 3
, cons_0(5, 5) -> 6
, cons_0(5, 5) -> 7
, cons_1(8, 9) -> 1
, cons_1(12, 9) -> 7
, cons_2(13, 14) -> 1
, cons_2(16, 14) -> 7
, n__f_0(2) -> 4
, n__f_0(2) -> 6
, n__f_0(2) -> 7
, n__f_0(3) -> 4
, n__f_0(3) -> 6
, n__f_0(3) -> 7
, n__f_0(4) -> 4
, n__f_0(4) -> 6
, n__f_0(4) -> 7
, n__f_0(5) -> 4
, n__f_0(5) -> 6
, n__f_0(5) -> 7
, n__f_1(2) -> 1
, n__f_1(3) -> 1
, n__f_1(4) -> 1
, n__f_1(5) -> 1
, n__f_1(10) -> 9
, n__f_2(2) -> 7
, n__f_2(3) -> 7
, n__f_2(4) -> 7
, n__f_2(5) -> 7
, n__f_2(12) -> 1
, n__f_2(12) -> 7
, n__f_2(15) -> 14
, s_0(2) -> 5
, s_0(2) -> 6
, s_0(2) -> 7
, s_0(3) -> 5
, s_0(3) -> 6
, s_0(3) -> 7
, s_0(4) -> 5
, s_0(4) -> 6
, s_0(4) -> 7
, s_0(5) -> 5
, s_0(5) -> 6
, s_0(5) -> 7
, s_1(11) -> 10
, s_2(16) -> 15
, p_0(2) -> 6
, p_0(3) -> 6
, p_0(4) -> 6
, p_0(5) -> 6
, p_1(10) -> 12
, activate_0(2) -> 7
, activate_0(3) -> 7
, activate_0(4) -> 7
, activate_0(5) -> 7
, 2 -> 6
, 2 -> 7
, 3 -> 6
, 3 -> 7
, 4 -> 6
, 4 -> 7
, 5 -> 6
, 5 -> 7
, 11 -> 12 }

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