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) -> f(X)
  , a__f(f(a())) -> c(f(g(f(a()))))
  , mark(f(X)) -> a__f(mark(X))
  , mark(a()) -> a()
  , mark(c(X)) -> c(X)
  , mark(g(X)) -> g(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^#(f(a())) -> c_2()
  , mark^#(f(X)) -> c_3(a__f^#(mark(X)), mark^#(X))
  , mark^#(a()) -> c_4()
  , mark^#(c(X)) -> c_5()
  , mark^#(g(X)) -> c_6(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^#(f(a())) -> c_2()
  , mark^#(f(X)) -> c_3(a__f^#(mark(X)), mark^#(X))
  , mark^#(a()) -> c_4()
  , mark^#(c(X)) -> c_5()
  , mark^#(g(X)) -> c_6(mark^#(X)) }
Weak Trs:
  { a__f(X) -> f(X)
  , a__f(f(a())) -> c(f(g(f(a()))))
  , mark(f(X)) -> a__f(mark(X))
  , mark(a()) -> a()
  , mark(c(X)) -> c(X)
  , mark(g(X)) -> g(mark(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

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

  DPs:
    { 1: a__f^#(X) -> c_1()
    , 2: a__f^#(f(a())) -> c_2()
    , 3: mark^#(f(X)) -> c_3(a__f^#(mark(X)), mark^#(X))
    , 4: mark^#(a()) -> c_4()
    , 5: mark^#(c(X)) -> c_5()
    , 6: mark^#(g(X)) -> c_6(mark^#(X)) }

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

Strict DPs:
  { mark^#(f(X)) -> c_3(a__f^#(mark(X)), mark^#(X))
  , mark^#(g(X)) -> c_6(mark^#(X)) }
Weak DPs:
  { a__f^#(X) -> c_1()
  , a__f^#(f(a())) -> c_2()
  , mark^#(a()) -> c_4()
  , mark^#(c(X)) -> c_5() }
Weak Trs:
  { a__f(X) -> f(X)
  , a__f(f(a())) -> c(f(g(f(a()))))
  , mark(f(X)) -> a__f(mark(X))
  , mark(a()) -> a()
  , mark(c(X)) -> c(X)
  , mark(g(X)) -> g(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^#(f(a())) -> c_2()
, mark^#(a()) -> c_4()
, mark^#(c(X)) -> c_5() }

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

Strict DPs:
  { mark^#(f(X)) -> c_3(a__f^#(mark(X)), mark^#(X))
  , mark^#(g(X)) -> c_6(mark^#(X)) }
Weak Trs:
  { a__f(X) -> f(X)
  , a__f(f(a())) -> c(f(g(f(a()))))
  , mark(f(X)) -> a__f(mark(X))
  , mark(a()) -> a()
  , mark(c(X)) -> c(X)
  , mark(g(X)) -> g(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_3(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^#(f(X)) -> c_1(mark^#(X))
  , mark^#(g(X)) -> c_2(mark^#(X)) }
Weak Trs:
  { a__f(X) -> f(X)
  , a__f(f(a())) -> c(f(g(f(a()))))
  , mark(f(X)) -> a__f(mark(X))
  , mark(a()) -> a()
  , mark(c(X)) -> c(X)
  , mark(g(X)) -> g(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^#(f(X)) -> c_1(mark^#(X))
  , mark^#(g(X)) -> c_2(mark^#(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The following argument positions are usable:
  Uargs(c_1) = {1}, Uargs(c_2) = {1}

TcT has computed following constructor-based matrix interpretation
satisfying not(EDA).

       [f](x1) = [1] x1 + [2]
                             
       [g](x1) = [1] x1 + [2]
                             
  [mark^#](x1) = [2] x1 + [0]
                             
     [c_1](x1) = [1] x1 + [1]
                             
     [c_2](x1) = [1] x1 + [1]

This order satisfies following ordering constraints:

  [mark^#(f(X))] = [2] X + [4]     
                 > [2] X + [1]     
                 = [c_1(mark^#(X))]
                                   
  [mark^#(g(X))] = [2] X + [4]     
                 > [2] X + [1]     
                 = [c_2(mark^#(X))]
                                   

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