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

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

We add the following dependency tuples:

Strict DPs:
  { a__f^#(X1, X2, X3) -> c_1()
  , a__f^#(X, g(X), Y) -> c_2(a__f^#(Y, Y, Y))
  , a__g^#(X) -> c_3()
  , a__g^#(b()) -> c_4()
  , a__b^#() -> c_5()
  , a__b^#() -> c_6()
  , mark^#(g(X)) -> c_7(a__g^#(mark(X)), mark^#(X))
  , mark^#(b()) -> c_8(a__b^#())
  , mark^#(c()) -> c_9()
  , mark^#(f(X1, X2, X3)) -> c_10(a__f^#(X1, X2, X3)) }

and mark the set of starting terms.

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

Strict DPs:
  { a__f^#(X1, X2, X3) -> c_1()
  , a__f^#(X, g(X), Y) -> c_2(a__f^#(Y, Y, Y))
  , a__g^#(X) -> c_3()
  , a__g^#(b()) -> c_4()
  , a__b^#() -> c_5()
  , a__b^#() -> c_6()
  , mark^#(g(X)) -> c_7(a__g^#(mark(X)), mark^#(X))
  , mark^#(b()) -> c_8(a__b^#())
  , mark^#(c()) -> c_9()
  , mark^#(f(X1, X2, X3)) -> c_10(a__f^#(X1, X2, X3)) }
Weak Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , a__f(X, g(X), Y) -> a__f(Y, Y, Y)
  , a__g(X) -> g(X)
  , a__g(b()) -> c()
  , a__b() -> b()
  , a__b() -> c()
  , mark(g(X)) -> a__g(mark(X))
  , mark(b()) -> a__b()
  , mark(c()) -> c()
  , mark(f(X1, X2, X3)) -> a__f(X1, X2, X3) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We estimate the number of application of {1,3,4,5,6,9} by
applications of Pre({1,3,4,5,6,9}) = {2,7,8,10}. Here rules are
labeled as follows:

  DPs:
    { 1: a__f^#(X1, X2, X3) -> c_1()
    , 2: a__f^#(X, g(X), Y) -> c_2(a__f^#(Y, Y, Y))
    , 3: a__g^#(X) -> c_3()
    , 4: a__g^#(b()) -> c_4()
    , 5: a__b^#() -> c_5()
    , 6: a__b^#() -> c_6()
    , 7: mark^#(g(X)) -> c_7(a__g^#(mark(X)), mark^#(X))
    , 8: mark^#(b()) -> c_8(a__b^#())
    , 9: mark^#(c()) -> c_9()
    , 10: mark^#(f(X1, X2, X3)) -> c_10(a__f^#(X1, X2, X3)) }

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

Strict DPs:
  { a__f^#(X, g(X), Y) -> c_2(a__f^#(Y, Y, Y))
  , mark^#(g(X)) -> c_7(a__g^#(mark(X)), mark^#(X))
  , mark^#(b()) -> c_8(a__b^#())
  , mark^#(f(X1, X2, X3)) -> c_10(a__f^#(X1, X2, X3)) }
Weak DPs:
  { a__f^#(X1, X2, X3) -> c_1()
  , a__g^#(X) -> c_3()
  , a__g^#(b()) -> c_4()
  , a__b^#() -> c_5()
  , a__b^#() -> c_6()
  , mark^#(c()) -> c_9() }
Weak Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , a__f(X, g(X), Y) -> a__f(Y, Y, Y)
  , a__g(X) -> g(X)
  , a__g(b()) -> c()
  , a__b() -> b()
  , a__b() -> c()
  , mark(g(X)) -> a__g(mark(X))
  , mark(b()) -> a__b()
  , mark(c()) -> c()
  , mark(f(X1, X2, X3)) -> a__f(X1, X2, X3) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

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

  DPs:
    { 1: a__f^#(X, g(X), Y) -> c_2(a__f^#(Y, Y, Y))
    , 2: mark^#(g(X)) -> c_7(a__g^#(mark(X)), mark^#(X))
    , 3: mark^#(b()) -> c_8(a__b^#())
    , 4: mark^#(f(X1, X2, X3)) -> c_10(a__f^#(X1, X2, X3))
    , 5: a__f^#(X1, X2, X3) -> c_1()
    , 6: a__g^#(X) -> c_3()
    , 7: a__g^#(b()) -> c_4()
    , 8: a__b^#() -> c_5()
    , 9: a__b^#() -> c_6()
    , 10: mark^#(c()) -> c_9() }

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

Strict DPs:
  { mark^#(g(X)) -> c_7(a__g^#(mark(X)), mark^#(X))
  , mark^#(f(X1, X2, X3)) -> c_10(a__f^#(X1, X2, X3)) }
Weak DPs:
  { a__f^#(X1, X2, X3) -> c_1()
  , a__f^#(X, g(X), Y) -> c_2(a__f^#(Y, Y, Y))
  , a__g^#(X) -> c_3()
  , a__g^#(b()) -> c_4()
  , a__b^#() -> c_5()
  , a__b^#() -> c_6()
  , mark^#(b()) -> c_8(a__b^#())
  , mark^#(c()) -> c_9() }
Weak Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , a__f(X, g(X), Y) -> a__f(Y, Y, Y)
  , a__g(X) -> g(X)
  , a__g(b()) -> c()
  , a__b() -> b()
  , a__b() -> c()
  , mark(g(X)) -> a__g(mark(X))
  , mark(b()) -> a__b()
  , mark(c()) -> c()
  , mark(f(X1, X2, X3)) -> a__f(X1, X2, X3) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

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

  DPs:
    { 1: mark^#(g(X)) -> c_7(a__g^#(mark(X)), mark^#(X))
    , 2: mark^#(f(X1, X2, X3)) -> c_10(a__f^#(X1, X2, X3))
    , 3: a__f^#(X1, X2, X3) -> c_1()
    , 4: a__f^#(X, g(X), Y) -> c_2(a__f^#(Y, Y, Y))
    , 5: a__g^#(X) -> c_3()
    , 6: a__g^#(b()) -> c_4()
    , 7: a__b^#() -> c_5()
    , 8: a__b^#() -> c_6()
    , 9: mark^#(b()) -> c_8(a__b^#())
    , 10: mark^#(c()) -> c_9() }

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

Strict DPs: { mark^#(g(X)) -> c_7(a__g^#(mark(X)), mark^#(X)) }
Weak DPs:
  { a__f^#(X1, X2, X3) -> c_1()
  , a__f^#(X, g(X), Y) -> c_2(a__f^#(Y, Y, Y))
  , a__g^#(X) -> c_3()
  , a__g^#(b()) -> c_4()
  , a__b^#() -> c_5()
  , a__b^#() -> c_6()
  , mark^#(b()) -> c_8(a__b^#())
  , mark^#(c()) -> c_9()
  , mark^#(f(X1, X2, X3)) -> c_10(a__f^#(X1, X2, X3)) }
Weak Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , a__f(X, g(X), Y) -> a__f(Y, Y, Y)
  , a__g(X) -> g(X)
  , a__g(b()) -> c()
  , a__b() -> b()
  , a__b() -> c()
  , mark(g(X)) -> a__g(mark(X))
  , mark(b()) -> a__b()
  , mark(c()) -> c()
  , mark(f(X1, X2, X3)) -> a__f(X1, X2, X3) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),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__f^#(X, g(X), Y) -> c_2(a__f^#(Y, Y, Y))
, a__g^#(X) -> c_3()
, a__g^#(b()) -> c_4()
, a__b^#() -> c_5()
, a__b^#() -> c_6()
, mark^#(b()) -> c_8(a__b^#())
, mark^#(c()) -> c_9()
, mark^#(f(X1, X2, X3)) -> c_10(a__f^#(X1, X2, X3)) }

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

Strict DPs: { mark^#(g(X)) -> c_7(a__g^#(mark(X)), mark^#(X)) }
Weak Trs:
  { a__f(X1, X2, X3) -> f(X1, X2, X3)
  , a__f(X, g(X), Y) -> a__f(Y, Y, Y)
  , a__g(X) -> g(X)
  , a__g(b()) -> c()
  , a__b() -> b()
  , a__b() -> c()
  , mark(g(X)) -> a__g(mark(X))
  , mark(b()) -> a__b()
  , mark(c()) -> c()
  , mark(f(X1, X2, X3)) -> a__f(X1, X2, X3) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

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

  { mark^#(g(X)) -> c_7(a__g^#(mark(X)), mark^#(X)) }

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

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

Strict DPs: { mark^#(g(X)) -> c_1(mark^#(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'Small Polynomial Path Order (PS,1-bounded)'
to orient following rules strictly.

DPs:
  { 1: mark^#(g(X)) -> c_1(mark^#(X)) }

Sub-proof:
----------
  The input was oriented with the instance of 'Small Polynomial Path
  Order (PS,1-bounded)' as induced by the safe mapping
  
   safe(g) = {1}, safe(mark^#) = {}, safe(c_1) = {}
  
  and precedence
  
   empty .
  
  Following symbols are considered recursive:
  
   {mark^#}
  
  The recursion depth is 1.
  
  Further, following argument filtering is employed:
  
   pi(g) = [1], pi(mark^#) = [1], pi(c_1) = [1]
  
  Usable defined function symbols are a subset of:
  
   {mark^#}
  
  For your convenience, here are the satisfied ordering constraints:
  
    pi(mark^#(g(X))) = mark^#(g(; X);)   
                     > c_1(mark^#(X;);)  
                     = pi(c_1(mark^#(X)))
                                         

The strictly oriented rules are moved into the weak component.

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

Weak DPs: { mark^#(g(X)) -> c_1(mark^#(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

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

{ mark^#(g(X)) -> c_1(mark^#(X)) }

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

Rules: Empty
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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