YES(?,O(n^1))

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

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

We add following dependency tuples:

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

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

  DPs:
    { 1: a__f^#(X) -> c_1()
    , 2: a__f^#(X) -> c_2()
    , 3: mark^#(g(X)) -> c_3()
    , 4: mark^#(h(X)) -> c_4(mark^#(X))
    , 5: mark^#(f(X)) -> c_5(a__f^#(mark(X)), mark^#(X)) }

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

Strict DPs:
  { mark^#(h(X)) -> c_4(mark^#(X))
  , mark^#(f(X)) -> c_5(a__f^#(mark(X)), mark^#(X)) }
Weak DPs:
  { a__f^#(X) -> c_1()
  , a__f^#(X) -> c_2()
  , mark^#(g(X)) -> c_3() }
Weak Trs:
  { a__f(X) -> g(h(f(X)))
  , a__f(X) -> f(X)
  , mark(g(X)) -> g(X)
  , mark(h(X)) -> h(mark(X))
  , mark(f(X)) -> a__f(mark(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.

{ a__f^#(X) -> c_1()
, a__f^#(X) -> c_2()
, mark^#(g(X)) -> c_3() }

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

Strict DPs:
  { mark^#(h(X)) -> c_4(mark^#(X))
  , mark^#(f(X)) -> c_5(a__f^#(mark(X)), mark^#(X)) }
Weak Trs:
  { a__f(X) -> g(h(f(X)))
  , a__f(X) -> f(X)
  , mark(g(X)) -> g(X)
  , mark(h(X)) -> h(mark(X))
  , mark(f(X)) -> a__f(mark(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:

  { mark^#(f(X)) -> c_5(a__f^#(mark(X)), mark^#(X)) }

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

Strict DPs:
  { mark^#(h(X)) -> c_1(mark^#(X))
  , mark^#(f(X)) -> c_2(mark^#(X)) }
Weak Trs:
  { a__f(X) -> g(h(f(X)))
  , a__f(X) -> f(X)
  , mark(g(X)) -> g(X)
  , mark(h(X)) -> h(mark(X))
  , mark(f(X)) -> a__f(mark(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:
  { mark^#(h(X)) -> c_1(mark^#(X))
  , mark^#(f(X)) -> c_2(mark^#(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(h) = {1}, safe(f) = {1}, safe(mark^#) = {}, safe(c_1) = {},
 safe(c_2) = {}

and precedence

 empty .

Following symbols are considered recursive:

 {mark^#}

The recursion depth is 1.

Further, following argument filtering is employed:

 pi(h) = [1], pi(f) = [1], pi(mark^#) = [1], pi(c_1) = [1],
 pi(c_2) = [1]

Usable defined function symbols are a subset of:

 {mark^#}

For your convenience, here are the satisfied ordering constraints:

  pi(mark^#(h(X))) = mark^#(h(; X);)   
                   > c_1(mark^#(X;);)  
                   = pi(c_1(mark^#(X)))
                                       
  pi(mark^#(f(X))) = mark^#(f(; X);)   
                   > c_2(mark^#(X;);)  
                   = pi(c_2(mark^#(X)))
                                       

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