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(n__g(n__f(X))))
  , c(X) -> d(activate(X))
  , d(X) -> n__d(X)
  , activate(X) -> X
  , activate(n__f(X)) -> f(activate(X))
  , activate(n__g(X)) -> g(X)
  , activate(n__d(X)) -> d(X)
  , h(X) -> c(n__d(X))
  , g(X) -> n__g(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(3) -> 1
, f_0(4) -> 1
, f_0(8) -> 1
, f_1(10) -> 6
, f_1(10) -> 10
, f_3(15) -> 14
, c_0(3) -> 2
, c_0(4) -> 2
, c_0(8) -> 2
, c_1(5) -> 7
, c_2(11) -> 6
, c_2(11) -> 10
, n__f_0(3) -> 3
, n__f_0(3) -> 6
, n__f_0(3) -> 10
, n__f_0(4) -> 3
, n__f_0(4) -> 6
, n__f_0(4) -> 10
, n__f_0(8) -> 3
, n__f_0(8) -> 6
, n__f_0(8) -> 10
, n__f_1(3) -> 1
, n__f_1(4) -> 1
, n__f_1(8) -> 1
, n__f_2(10) -> 6
, n__f_2(10) -> 10
, n__f_2(12) -> 11
, n__f_2(12) -> 14
, n__f_4(15) -> 14
, n__g_0(3) -> 4
, n__g_0(3) -> 6
, n__g_0(3) -> 10
, n__g_0(4) -> 4
, n__g_0(4) -> 6
, n__g_0(4) -> 10
, n__g_0(8) -> 4
, n__g_0(8) -> 6
, n__g_0(8) -> 10
, n__g_1(3) -> 9
, n__g_1(4) -> 9
, n__g_1(8) -> 9
, n__g_2(3) -> 6
, n__g_2(3) -> 10
, n__g_2(4) -> 6
, n__g_2(4) -> 10
, n__g_2(8) -> 6
, n__g_2(8) -> 10
, n__g_2(10) -> 12
, n__g_2(10) -> 15
, n__g_4(10) -> 15
, d_0(3) -> 5
, d_0(3) -> 13
, d_0(4) -> 5
, d_0(4) -> 13
, d_0(8) -> 5
, d_0(8) -> 13
, d_1(3) -> 6
, d_1(3) -> 10
, d_1(4) -> 6
, d_1(4) -> 10
, d_1(8) -> 6
, d_1(8) -> 10
, d_1(10) -> 2
, d_2(3) -> 13
, d_2(4) -> 13
, d_2(8) -> 13
, d_2(13) -> 7
, d_3(14) -> 6
, d_3(14) -> 10
, activate_0(3) -> 6
, activate_0(4) -> 6
, activate_0(8) -> 6
, activate_1(3) -> 10
, activate_1(4) -> 10
, activate_1(8) -> 10
, activate_2(5) -> 13
, activate_3(11) -> 14
, activate_3(12) -> 15
, h_0(3) -> 7
, h_0(4) -> 7
, h_0(8) -> 7
, n__d_0(3) -> 6
, n__d_0(3) -> 8
, n__d_0(3) -> 10
, n__d_0(4) -> 6
, n__d_0(4) -> 8
, n__d_0(4) -> 10
, n__d_0(8) -> 6
, n__d_0(8) -> 8
, n__d_0(8) -> 10
, n__d_1(3) -> 5
, n__d_1(3) -> 13
, n__d_1(4) -> 5
, n__d_1(4) -> 13
, n__d_1(8) -> 5
, n__d_1(8) -> 13
, n__d_2(3) -> 6
, n__d_2(3) -> 10
, n__d_2(4) -> 6
, n__d_2(4) -> 10
, n__d_2(8) -> 6
, n__d_2(8) -> 10
, n__d_2(10) -> 2
, n__d_3(3) -> 13
, n__d_3(4) -> 13
, n__d_3(8) -> 13
, n__d_3(13) -> 7
, n__d_4(14) -> 6
, n__d_4(14) -> 10
, g_0(3) -> 9
, g_0(4) -> 9
, g_0(8) -> 9
, g_1(3) -> 6
, g_1(3) -> 10
, g_1(4) -> 6
, g_1(4) -> 10
, g_1(8) -> 6
, g_1(8) -> 10
, g_3(10) -> 15 }

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