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, X) -> a__f(a(), b())
  , a__f(X1, X2) -> f(X1, X2)
  , a__b() -> a()
  , a__b() -> b()
  , mark(a()) -> a()
  , mark(b()) -> a__b()
  , mark(f(X1, X2)) -> a__f(mark(X1), X2) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

We add following dependency tuples:

Strict DPs:
  { a__f^#(X, X) -> c_1(a__f^#(a(), b()))
  , a__f^#(X1, X2) -> c_2()
  , a__b^#() -> c_3()
  , a__b^#() -> c_4()
  , mark^#(a()) -> c_5()
  , mark^#(b()) -> c_6(a__b^#())
  , mark^#(f(X1, X2)) -> c_7(a__f^#(mark(X1), X2), mark^#(X1)) }

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, X) -> c_1(a__f^#(a(), b()))
  , a__f^#(X1, X2) -> c_2()
  , a__b^#() -> c_3()
  , a__b^#() -> c_4()
  , mark^#(a()) -> c_5()
  , mark^#(b()) -> c_6(a__b^#())
  , mark^#(f(X1, X2)) -> c_7(a__f^#(mark(X1), X2), mark^#(X1)) }
Weak Trs:
  { a__f(X, X) -> a__f(a(), b())
  , a__f(X1, X2) -> f(X1, X2)
  , a__b() -> a()
  , a__b() -> b()
  , mark(a()) -> a()
  , mark(b()) -> a__b()
  , mark(f(X1, X2)) -> a__f(mark(X1), X2) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

We estimate the number of application of {2,3,4,5} by applications
of Pre({2,3,4,5}) = {1,6,7}. Here rules are labeled as follows:

  DPs:
    { 1: a__f^#(X, X) -> c_1(a__f^#(a(), b()))
    , 2: a__f^#(X1, X2) -> c_2()
    , 3: a__b^#() -> c_3()
    , 4: a__b^#() -> c_4()
    , 5: mark^#(a()) -> c_5()
    , 6: mark^#(b()) -> c_6(a__b^#())
    , 7: mark^#(f(X1, X2)) -> c_7(a__f^#(mark(X1), X2), mark^#(X1)) }

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

Strict DPs:
  { a__f^#(X, X) -> c_1(a__f^#(a(), b()))
  , mark^#(b()) -> c_6(a__b^#())
  , mark^#(f(X1, X2)) -> c_7(a__f^#(mark(X1), X2), mark^#(X1)) }
Weak DPs:
  { a__f^#(X1, X2) -> c_2()
  , a__b^#() -> c_3()
  , a__b^#() -> c_4()
  , mark^#(a()) -> c_5() }
Weak Trs:
  { a__f(X, X) -> a__f(a(), b())
  , a__f(X1, X2) -> f(X1, X2)
  , a__b() -> a()
  , a__b() -> b()
  , mark(a()) -> a()
  , mark(b()) -> a__b()
  , mark(f(X1, X2)) -> a__f(mark(X1), X2) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

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

  DPs:
    { 1: a__f^#(X, X) -> c_1(a__f^#(a(), b()))
    , 2: mark^#(b()) -> c_6(a__b^#())
    , 3: mark^#(f(X1, X2)) -> c_7(a__f^#(mark(X1), X2), mark^#(X1))
    , 4: a__f^#(X1, X2) -> c_2()
    , 5: a__b^#() -> c_3()
    , 6: a__b^#() -> c_4()
    , 7: mark^#(a()) -> c_5() }

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

Strict DPs:
  { mark^#(f(X1, X2)) -> c_7(a__f^#(mark(X1), X2), mark^#(X1)) }
Weak DPs:
  { a__f^#(X, X) -> c_1(a__f^#(a(), b()))
  , a__f^#(X1, X2) -> c_2()
  , a__b^#() -> c_3()
  , a__b^#() -> c_4()
  , mark^#(a()) -> c_5()
  , mark^#(b()) -> c_6(a__b^#()) }
Weak Trs:
  { a__f(X, X) -> a__f(a(), b())
  , a__f(X1, X2) -> f(X1, X2)
  , a__b() -> a()
  , a__b() -> b()
  , mark(a()) -> a()
  , mark(b()) -> a__b()
  , mark(f(X1, X2)) -> a__f(mark(X1), X2) }
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, X) -> c_1(a__f^#(a(), b()))
, a__f^#(X1, X2) -> c_2()
, a__b^#() -> c_3()
, a__b^#() -> c_4()
, mark^#(a()) -> c_5()
, mark^#(b()) -> c_6(a__b^#()) }

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

Strict DPs:
  { mark^#(f(X1, X2)) -> c_7(a__f^#(mark(X1), X2), mark^#(X1)) }
Weak Trs:
  { a__f(X, X) -> a__f(a(), b())
  , a__f(X1, X2) -> f(X1, X2)
  , a__b() -> a()
  , a__b() -> b()
  , mark(a()) -> a()
  , mark(b()) -> a__b()
  , mark(f(X1, X2)) -> a__f(mark(X1), X2) }
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(X1, X2)) -> c_7(a__f^#(mark(X1), X2), mark^#(X1)) }

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

Strict DPs: { mark^#(f(X1, X2)) -> c_1(mark^#(X1)) }
Weak Trs:
  { a__f(X, X) -> a__f(a(), b())
  , a__f(X1, X2) -> f(X1, X2)
  , a__b() -> a()
  , a__b() -> b()
  , mark(a()) -> a()
  , mark(b()) -> a__b()
  , mark(f(X1, X2)) -> a__f(mark(X1), X2) }
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(X1, X2)) -> c_1(mark^#(X1)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

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

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

   [f](x1, x2) = [1] x1 + [1] x2 + [3]
                                      
  [mark^#](x1) = [1] x1 + [3]         
                                      
     [c_1](x1) = [1] x1 + [2]         

This order satisfies following ordering constraints:

  [mark^#(f(X1, X2))] = [1] X1 + [1] X2 + [6]
                      > [1] X1 + [5]         
                      = [c_1(mark^#(X1))]    
                                             

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