YES(?,O(n^1))

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

Strict Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , a__f(a(), X, X) -> a__f(X, a__b(), b())
  , a__b() -> a()
  , a__b() -> b()
  , mark(a()) -> a()
  , mark(b()) -> a__b()
  , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

We add following dependency tuples:

Strict DPs:
  { a__f^#(X1, X2, X3) -> c_1()
  , a__f^#(a(), X, X) -> c_2(a__f^#(X, a__b(), b()), a__b^#())
  , a__b^#() -> c_3()
  , a__b^#() -> c_4()
  , mark^#(a()) -> c_5()
  , mark^#(b()) -> c_6(a__b^#())
  , mark^#(f(X1, X2, X3)) ->
    c_7(a__f^#(X1, mark(X2), X3), mark^#(X2)) }

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^#(X1, X2, X3) -> c_1()
  , a__f^#(a(), X, X) -> c_2(a__f^#(X, a__b(), b()), a__b^#())
  , a__b^#() -> c_3()
  , a__b^#() -> c_4()
  , mark^#(a()) -> c_5()
  , mark^#(b()) -> c_6(a__b^#())
  , mark^#(f(X1, X2, X3)) ->
    c_7(a__f^#(X1, mark(X2), X3), mark^#(X2)) }
Weak Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , a__f(a(), X, X) -> a__f(X, a__b(), b())
  , a__b() -> a()
  , a__b() -> b()
  , mark(a()) -> a()
  , mark(b()) -> a__b()
  , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

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

  DPs:
    { 1: a__f^#(X1, X2, X3) -> c_1()
    , 2: a__f^#(a(), X, X) -> c_2(a__f^#(X, a__b(), b()), a__b^#())
    , 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, X3)) ->
         c_7(a__f^#(X1, mark(X2), X3), mark^#(X2)) }

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

Strict DPs:
  { a__f^#(a(), X, X) -> c_2(a__f^#(X, a__b(), b()), a__b^#())
  , mark^#(b()) -> c_6(a__b^#())
  , mark^#(f(X1, X2, X3)) ->
    c_7(a__f^#(X1, mark(X2), X3), mark^#(X2)) }
Weak DPs:
  { a__f^#(X1, X2, X3) -> c_1()
  , a__b^#() -> c_3()
  , a__b^#() -> c_4()
  , mark^#(a()) -> c_5() }
Weak Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , a__f(a(), X, X) -> a__f(X, a__b(), b())
  , a__b() -> a()
  , a__b() -> b()
  , mark(a()) -> a()
  , mark(b()) -> a__b()
  , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

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

  DPs:
    { 1: a__f^#(a(), X, X) -> c_2(a__f^#(X, a__b(), b()), a__b^#())
    , 2: mark^#(b()) -> c_6(a__b^#())
    , 3: mark^#(f(X1, X2, X3)) ->
         c_7(a__f^#(X1, mark(X2), X3), mark^#(X2))
    , 4: a__f^#(X1, X2, X3) -> c_1()
    , 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:
  { a__f^#(a(), X, X) -> c_2(a__f^#(X, a__b(), b()), a__b^#())
  , mark^#(f(X1, X2, X3)) ->
    c_7(a__f^#(X1, mark(X2), X3), mark^#(X2)) }
Weak DPs:
  { a__f^#(X1, X2, X3) -> c_1()
  , a__b^#() -> c_3()
  , a__b^#() -> c_4()
  , mark^#(a()) -> c_5()
  , mark^#(b()) -> c_6(a__b^#()) }
Weak Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , a__f(a(), X, X) -> a__f(X, a__b(), b())
  , a__b() -> a()
  , a__b() -> b()
  , mark(a()) -> a()
  , mark(b()) -> a__b()
  , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3) }
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^#(X1, X2, X3) -> c_1()
, 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:
  { a__f^#(a(), X, X) -> c_2(a__f^#(X, a__b(), b()), a__b^#())
  , mark^#(f(X1, X2, X3)) ->
    c_7(a__f^#(X1, mark(X2), X3), mark^#(X2)) }
Weak Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , a__f(a(), X, X) -> a__f(X, a__b(), b())
  , a__b() -> a()
  , a__b() -> b()
  , mark(a()) -> a()
  , mark(b()) -> a__b()
  , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3) }
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:

  { a__f^#(a(), X, X) -> c_2(a__f^#(X, a__b(), b()), a__b^#()) }

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

Strict DPs:
  { a__f^#(a(), X, X) -> c_1(a__f^#(X, a__b(), b()))
  , mark^#(f(X1, X2, X3)) ->
    c_2(a__f^#(X1, mark(X2), X3), mark^#(X2)) }
Weak Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , a__f(a(), X, X) -> a__f(X, a__b(), b())
  , a__b() -> a()
  , a__b() -> b()
  , mark(a()) -> a()
  , mark(b()) -> a__b()
  , mark(f(X1, X2, X3)) -> a__f(X1, mark(X2), X3) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The following argument positions are usable:
  Uargs(c_1) = {1}, Uargs(c_2) = {1, 2}

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

    [a__f](x1, x2, x3) = [0]                           
                                                       
                   [a] = [2]                           
                                                       
                [a__b] = [0]                           
                                                       
                   [b] = [0]                           
                                                       
            [mark](x1) = [0]                           
                                                       
       [f](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [1]
                                                       
  [a__f^#](x1, x2, x3) = [1] x1 + [1] x3 + [0]         
                                                       
          [mark^#](x1) = [1] x1 + [0]                  
                                                       
             [c_1](x1) = [1] x1 + [1]                  
                                                       
         [c_2](x1, x2) = [1] x1 + [1] x2 + [0]         

This order satisfies following ordering constraints:

      [a__f^#(a(), X, X)] = [1] X + [2]                                
                          > [1] X + [1]                                
                          = [c_1(a__f^#(X, a__b(), b()))]              
                                                                       
  [mark^#(f(X1, X2, X3))] = [1] X1 + [1] X2 + [1] X3 + [1]             
                          > [1] X1 + [1] X2 + [1] X3 + [0]             
                          = [c_2(a__f^#(X1, mark(X2), X3), mark^#(X2))]
                                                                       

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