YES(?,O(n^1))

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

Strict Trs:
  { f(X) -> if(X, c(), n__f(n__true()))
  , f(X) -> n__f(X)
  , if(true(), X, Y) -> X
  , if(false(), X, Y) -> activate(Y)
  , true() -> n__true()
  , activate(X) -> X
  , activate(n__f(X)) -> f(activate(X))
  , activate(n__true()) -> true() }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

Arguments of following rules are not normal-forms:

{ if(true(), X, Y) -> X }

All above mentioned rules can be savely removed.

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

Strict Trs:
  { f(X) -> if(X, c(), n__f(n__true()))
  , f(X) -> n__f(X)
  , if(false(), X, Y) -> activate(Y)
  , true() -> n__true()
  , activate(X) -> X
  , activate(n__f(X)) -> f(activate(X))
  , activate(n__true()) -> true() }
Obligation:
  innermost 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_1(1) -> 1
, f_2(6) -> 1
, if_0(2, 2, 2) -> 1
, if_1(2, 3, 4) -> 1
, if_2(1, 7, 8) -> 1
, if_3(6, 10, 11) -> 1
, c_0() -> 1
, c_0() -> 2
, c_1() -> 3
, c_2() -> 7
, c_3() -> 10
, n__f_0(2) -> 1
, n__f_0(2) -> 2
, n__f_1(2) -> 1
, n__f_1(5) -> 1
, n__f_1(5) -> 4
, n__f_2(1) -> 1
, n__f_2(9) -> 1
, n__f_2(9) -> 8
, n__f_3(6) -> 1
, n__f_3(12) -> 11
, n__true_0() -> 1
, n__true_0() -> 2
, n__true_1() -> 1
, n__true_1() -> 5
, n__true_1() -> 6
, n__true_2() -> 1
, n__true_2() -> 6
, n__true_2() -> 9
, n__true_3() -> 6
, n__true_3() -> 12
, n__true_4() -> 6
, true_0() -> 1
, true_1() -> 1
, true_2() -> 6
, true_3() -> 6
, false_0() -> 1
, false_0() -> 2
, activate_0(2) -> 1
, activate_1(2) -> 1
, activate_1(4) -> 1
, activate_1(8) -> 1
, activate_2(5) -> 6
, activate_2(9) -> 6
, 2 -> 1
, 4 -> 1
, 5 -> 6
, 8 -> 1
, 9 -> 6 }

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