YES(?,O(n^1))

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

Strict Trs:
  { f(X) -> g(n__h(n__f(X)))
  , f(X) -> n__f(X)
  , h(X) -> n__h(X)
  , activate(X) -> X
  , activate(n__h(X)) -> h(activate(X))
  , activate(n__f(X)) -> f(activate(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

We add following dependency tuples:

Strict DPs:
  { f^#(X) -> c_1()
  , f^#(X) -> c_2()
  , h^#(X) -> c_3()
  , activate^#(X) -> c_4()
  , activate^#(n__h(X)) -> c_5(h^#(activate(X)), activate^#(X))
  , activate^#(n__f(X)) -> c_6(f^#(activate(X)), activate^#(X)) }

and mark the set of starting terms.

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

Strict DPs:
  { f^#(X) -> c_1()
  , f^#(X) -> c_2()
  , h^#(X) -> c_3()
  , activate^#(X) -> c_4()
  , activate^#(n__h(X)) -> c_5(h^#(activate(X)), activate^#(X))
  , activate^#(n__f(X)) -> c_6(f^#(activate(X)), activate^#(X)) }
Weak Trs:
  { f(X) -> g(n__h(n__f(X)))
  , f(X) -> n__f(X)
  , h(X) -> n__h(X)
  , activate(X) -> X
  , activate(n__h(X)) -> h(activate(X))
  , activate(n__f(X)) -> f(activate(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

We estimate the number of application of {1,2,3,4} by applications
of Pre({1,2,3,4}) = {5,6}. Here rules are labeled as follows:

  DPs:
    { 1: f^#(X) -> c_1()
    , 2: f^#(X) -> c_2()
    , 3: h^#(X) -> c_3()
    , 4: activate^#(X) -> c_4()
    , 5: activate^#(n__h(X)) -> c_5(h^#(activate(X)), activate^#(X))
    , 6: activate^#(n__f(X)) -> c_6(f^#(activate(X)), activate^#(X)) }

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

Strict DPs:
  { activate^#(n__h(X)) -> c_5(h^#(activate(X)), activate^#(X))
  , activate^#(n__f(X)) -> c_6(f^#(activate(X)), activate^#(X)) }
Weak DPs:
  { f^#(X) -> c_1()
  , f^#(X) -> c_2()
  , h^#(X) -> c_3()
  , activate^#(X) -> c_4() }
Weak Trs:
  { f(X) -> g(n__h(n__f(X)))
  , f(X) -> n__f(X)
  , h(X) -> n__h(X)
  , activate(X) -> X
  , activate(n__h(X)) -> h(activate(X))
  , activate(n__f(X)) -> f(activate(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ f^#(X) -> c_1()
, f^#(X) -> c_2()
, h^#(X) -> c_3()
, activate^#(X) -> c_4() }

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

Strict DPs:
  { activate^#(n__h(X)) -> c_5(h^#(activate(X)), activate^#(X))
  , activate^#(n__f(X)) -> c_6(f^#(activate(X)), activate^#(X)) }
Weak Trs:
  { f(X) -> g(n__h(n__f(X)))
  , f(X) -> n__f(X)
  , h(X) -> n__h(X)
  , activate(X) -> X
  , activate(n__h(X)) -> h(activate(X))
  , activate(n__f(X)) -> f(activate(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:

  { activate^#(n__h(X)) -> c_5(h^#(activate(X)), activate^#(X))
  , activate^#(n__f(X)) -> c_6(f^#(activate(X)), activate^#(X)) }

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

Strict DPs:
  { activate^#(n__h(X)) -> c_1(activate^#(X))
  , activate^#(n__f(X)) -> c_2(activate^#(X)) }
Weak Trs:
  { f(X) -> g(n__h(n__f(X)))
  , f(X) -> n__f(X)
  , h(X) -> n__h(X)
  , activate(X) -> X
  , activate(n__h(X)) -> h(activate(X))
  , activate(n__f(X)) -> f(activate(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

No rule is usable, rules are removed from the input problem.

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

Strict DPs:
  { activate^#(n__h(X)) -> c_1(activate^#(X))
  , activate^#(n__f(X)) -> c_2(activate^#(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The input was oriented with the instance of 'Small Polynomial Path
Order (PS)' as induced by the safe mapping

 safe(n__h) = {1}, safe(n__f) = {1}, safe(activate^#) = {},
 safe(c_1) = {}, safe(c_2) = {}

and precedence

 empty .

Following symbols are considered recursive:

 {activate^#}

The recursion depth is 1.

Further, following argument filtering is employed:

 pi(n__h) = [1], pi(n__f) = [1], pi(activate^#) = [1],
 pi(c_1) = [1], pi(c_2) = [1]

Usable defined function symbols are a subset of:

 {activate^#}

For your convenience, here are the satisfied ordering constraints:

  pi(activate^#(n__h(X))) = activate^#(n__h(; X);)
                          > c_1(activate^#(X;);)  
                          = pi(c_1(activate^#(X)))
                                                  
  pi(activate^#(n__f(X))) = activate^#(n__f(; X);)
                          > c_2(activate^#(X;);)  
                          = pi(c_2(activate^#(X)))
                                                  

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