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(f(X)) -> c(n__f(g(n__f(X))))
  , c(X) -> d(activate(X))
  , d(X) -> n__d(X)
  , activate(X) -> X
  , activate(n__f(X)) -> f(X)
  , activate(n__d(X)) -> d(X)
  , h(X) -> c(n__d(X)) }
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(2) -> 1
, f_0(2) -> 4
, f_1(2) -> 1
, f_1(2) -> 3
, f_1(2) -> 4
, f_2(2) -> 4
, f_3(2) -> 4
, c_0(2) -> 1
, c_0(2) -> 4
, c_1(1) -> 1
, c_1(1) -> 4
, n__f_0(2) -> 1
, n__f_0(2) -> 2
, n__f_0(2) -> 3
, n__f_0(2) -> 4
, n__f_1(2) -> 1
, n__f_1(2) -> 4
, n__f_2(2) -> 1
, n__f_2(2) -> 3
, n__f_2(2) -> 4
, n__f_3(2) -> 4
, n__f_4(2) -> 4
, g_0(2) -> 1
, g_0(2) -> 2
, g_0(2) -> 3
, g_0(2) -> 4
, d_0(2) -> 1
, d_0(2) -> 4
, d_1(2) -> 1
, d_1(2) -> 3
, d_1(2) -> 4
, d_1(3) -> 1
, d_1(3) -> 4
, d_2(2) -> 4
, d_2(4) -> 1
, d_2(4) -> 4
, d_3(2) -> 4
, d_3(3) -> 4
, d_3(4) -> 4
, activate_0(2) -> 1
, activate_0(2) -> 4
, activate_1(2) -> 3
, activate_2(1) -> 4
, h_0(2) -> 1
, h_0(2) -> 4
, n__d_0(2) -> 1
, n__d_0(2) -> 2
, n__d_0(2) -> 3
, n__d_0(2) -> 4
, n__d_1(2) -> 1
, n__d_1(2) -> 4
, n__d_2(2) -> 1
, n__d_2(2) -> 3
, n__d_2(2) -> 4
, n__d_2(3) -> 1
, n__d_2(3) -> 4
, n__d_3(2) -> 4
, n__d_3(4) -> 1
, n__d_3(4) -> 4
, n__d_4(2) -> 4
, n__d_4(3) -> 4
, n__d_4(4) -> 4
, 1 -> 4
, 2 -> 1
, 2 -> 3
, 2 -> 4 }

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