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(n__f(n__a())) -> f(n__g(n__f(n__a())))
  , a() -> n__a()
  , g(X) -> n__g(X)
  , activate(X) -> X
  , activate(n__f(X)) -> f(X)
  , activate(n__a()) -> a()
  , activate(n__g(X)) -> g(activate(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(n__f(n__a())) -> f(n__g(n__f(n__a())))
  , a() -> n__a()
  , g(X) -> n__g(X)
  , activate(X) -> X
  , activate(n__f(X)) -> f(X)
  , activate(n__a()) -> a()
  , activate(n__g(X)) -> g(activate(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_1(2) -> 1
, f_1(2) -> 6
, f_1(3) -> 1
, f_1(3) -> 6
, n__f_0(2) -> 1
, n__f_0(2) -> 2
, n__f_0(2) -> 6
, n__f_1(2) -> 1
, n__f_1(5) -> 4
, n__f_2(2) -> 1
, n__f_2(2) -> 6
, n__f_2(3) -> 1
, n__f_2(3) -> 6
, n__a_0() -> 1
, n__a_0() -> 2
, n__a_0() -> 6
, n__a_1() -> 1
, n__a_1() -> 5
, n__a_2() -> 1
, n__a_2() -> 6
, n__g_0(2) -> 1
, n__g_0(2) -> 2
, n__g_0(2) -> 6
, n__g_1(2) -> 1
, n__g_1(4) -> 3
, n__g_2(6) -> 1
, n__g_2(6) -> 6
, a_0() -> 1
, a_1() -> 1
, a_1() -> 6
, g_0(2) -> 1
, g_1(6) -> 1
, g_1(6) -> 6
, activate_0(2) -> 1
, activate_1(2) -> 6
, 2 -> 1
, 2 -> 6 }

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