MAYBE

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict Trs:
  { g(A()) -> A()
  , g(B()) -> A()
  , g(B()) -> B()
  , g(C()) -> A()
  , g(C()) -> B()
  , g(C()) -> C()
  , foldB(t, 0()) -> t
  , foldB(t, s(n)) -> f(foldB(t, n), B())
  , f(t, x) -> f'(t, g(x))
  , foldC(t, 0()) -> t
  , foldC(t, s(n)) -> f(foldC(t, n), C())
  , f'(triple(a, b, c), A()) -> f''(foldB(triple(s(a), 0(), c), b))
  , f'(triple(a, b, c), B()) -> f(triple(a, b, c), A())
  , f'(triple(a, b, c), C()) -> triple(a, b, s(c))
  , f''(triple(a, b, c)) -> foldC(triple(a, b, 0()), c) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { g^#(A()) -> c_1()
  , g^#(B()) -> c_2()
  , g^#(B()) -> c_3()
  , g^#(C()) -> c_4()
  , g^#(C()) -> c_5()
  , g^#(C()) -> c_6()
  , foldB^#(t, 0()) -> c_7()
  , foldB^#(t, s(n)) -> c_8(f^#(foldB(t, n), B()), foldB^#(t, n))
  , f^#(t, x) -> c_9(f'^#(t, g(x)), g^#(x))
  , f'^#(triple(a, b, c), A()) ->
    c_12(f''^#(foldB(triple(s(a), 0(), c), b)),
         foldB^#(triple(s(a), 0(), c), b))
  , f'^#(triple(a, b, c), B()) -> c_13(f^#(triple(a, b, c), A()))
  , f'^#(triple(a, b, c), C()) -> c_14()
  , foldC^#(t, 0()) -> c_10()
  , foldC^#(t, s(n)) -> c_11(f^#(foldC(t, n), C()), foldC^#(t, n))
  , f''^#(triple(a, b, c)) -> c_15(foldC^#(triple(a, b, 0()), c)) }

and mark the set of starting terms.

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { g^#(A()) -> c_1()
  , g^#(B()) -> c_2()
  , g^#(B()) -> c_3()
  , g^#(C()) -> c_4()
  , g^#(C()) -> c_5()
  , g^#(C()) -> c_6()
  , foldB^#(t, 0()) -> c_7()
  , foldB^#(t, s(n)) -> c_8(f^#(foldB(t, n), B()), foldB^#(t, n))
  , f^#(t, x) -> c_9(f'^#(t, g(x)), g^#(x))
  , f'^#(triple(a, b, c), A()) ->
    c_12(f''^#(foldB(triple(s(a), 0(), c), b)),
         foldB^#(triple(s(a), 0(), c), b))
  , f'^#(triple(a, b, c), B()) -> c_13(f^#(triple(a, b, c), A()))
  , f'^#(triple(a, b, c), C()) -> c_14()
  , foldC^#(t, 0()) -> c_10()
  , foldC^#(t, s(n)) -> c_11(f^#(foldC(t, n), C()), foldC^#(t, n))
  , f''^#(triple(a, b, c)) -> c_15(foldC^#(triple(a, b, 0()), c)) }
Weak Trs:
  { g(A()) -> A()
  , g(B()) -> A()
  , g(B()) -> B()
  , g(C()) -> A()
  , g(C()) -> B()
  , g(C()) -> C()
  , foldB(t, 0()) -> t
  , foldB(t, s(n)) -> f(foldB(t, n), B())
  , f(t, x) -> f'(t, g(x))
  , foldC(t, 0()) -> t
  , foldC(t, s(n)) -> f(foldC(t, n), C())
  , f'(triple(a, b, c), A()) -> f''(foldB(triple(s(a), 0(), c), b))
  , f'(triple(a, b, c), B()) -> f(triple(a, b, c), A())
  , f'(triple(a, b, c), C()) -> triple(a, b, s(c))
  , f''(triple(a, b, c)) -> foldC(triple(a, b, 0()), c) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of {1,2,3,4,5,6,7,12,13} by
applications of Pre({1,2,3,4,5,6,7,12,13}) = {8,9,10,14,15}. Here
rules are labeled as follows:

  DPs:
    { 1: g^#(A()) -> c_1()
    , 2: g^#(B()) -> c_2()
    , 3: g^#(B()) -> c_3()
    , 4: g^#(C()) -> c_4()
    , 5: g^#(C()) -> c_5()
    , 6: g^#(C()) -> c_6()
    , 7: foldB^#(t, 0()) -> c_7()
    , 8: foldB^#(t, s(n)) -> c_8(f^#(foldB(t, n), B()), foldB^#(t, n))
    , 9: f^#(t, x) -> c_9(f'^#(t, g(x)), g^#(x))
    , 10: f'^#(triple(a, b, c), A()) ->
          c_12(f''^#(foldB(triple(s(a), 0(), c), b)),
               foldB^#(triple(s(a), 0(), c), b))
    , 11: f'^#(triple(a, b, c), B()) -> c_13(f^#(triple(a, b, c), A()))
    , 12: f'^#(triple(a, b, c), C()) -> c_14()
    , 13: foldC^#(t, 0()) -> c_10()
    , 14: foldC^#(t, s(n)) ->
          c_11(f^#(foldC(t, n), C()), foldC^#(t, n))
    , 15: f''^#(triple(a, b, c)) ->
          c_15(foldC^#(triple(a, b, 0()), c)) }

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { foldB^#(t, s(n)) -> c_8(f^#(foldB(t, n), B()), foldB^#(t, n))
  , f^#(t, x) -> c_9(f'^#(t, g(x)), g^#(x))
  , f'^#(triple(a, b, c), A()) ->
    c_12(f''^#(foldB(triple(s(a), 0(), c), b)),
         foldB^#(triple(s(a), 0(), c), b))
  , f'^#(triple(a, b, c), B()) -> c_13(f^#(triple(a, b, c), A()))
  , foldC^#(t, s(n)) -> c_11(f^#(foldC(t, n), C()), foldC^#(t, n))
  , f''^#(triple(a, b, c)) -> c_15(foldC^#(triple(a, b, 0()), c)) }
Weak DPs:
  { g^#(A()) -> c_1()
  , g^#(B()) -> c_2()
  , g^#(B()) -> c_3()
  , g^#(C()) -> c_4()
  , g^#(C()) -> c_5()
  , g^#(C()) -> c_6()
  , foldB^#(t, 0()) -> c_7()
  , f'^#(triple(a, b, c), C()) -> c_14()
  , foldC^#(t, 0()) -> c_10() }
Weak Trs:
  { g(A()) -> A()
  , g(B()) -> A()
  , g(B()) -> B()
  , g(C()) -> A()
  , g(C()) -> B()
  , g(C()) -> C()
  , foldB(t, 0()) -> t
  , foldB(t, s(n)) -> f(foldB(t, n), B())
  , f(t, x) -> f'(t, g(x))
  , foldC(t, 0()) -> t
  , foldC(t, s(n)) -> f(foldC(t, n), C())
  , f'(triple(a, b, c), A()) -> f''(foldB(triple(s(a), 0(), c), b))
  , f'(triple(a, b, c), B()) -> f(triple(a, b, c), A())
  , f'(triple(a, b, c), C()) -> triple(a, b, s(c))
  , f''(triple(a, b, c)) -> foldC(triple(a, b, 0()), c) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

{ g^#(A()) -> c_1()
, g^#(B()) -> c_2()
, g^#(B()) -> c_3()
, g^#(C()) -> c_4()
, g^#(C()) -> c_5()
, g^#(C()) -> c_6()
, foldB^#(t, 0()) -> c_7()
, f'^#(triple(a, b, c), C()) -> c_14()
, foldC^#(t, 0()) -> c_10() }

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { foldB^#(t, s(n)) -> c_8(f^#(foldB(t, n), B()), foldB^#(t, n))
  , f^#(t, x) -> c_9(f'^#(t, g(x)), g^#(x))
  , f'^#(triple(a, b, c), A()) ->
    c_12(f''^#(foldB(triple(s(a), 0(), c), b)),
         foldB^#(triple(s(a), 0(), c), b))
  , f'^#(triple(a, b, c), B()) -> c_13(f^#(triple(a, b, c), A()))
  , foldC^#(t, s(n)) -> c_11(f^#(foldC(t, n), C()), foldC^#(t, n))
  , f''^#(triple(a, b, c)) -> c_15(foldC^#(triple(a, b, 0()), c)) }
Weak Trs:
  { g(A()) -> A()
  , g(B()) -> A()
  , g(B()) -> B()
  , g(C()) -> A()
  , g(C()) -> B()
  , g(C()) -> C()
  , foldB(t, 0()) -> t
  , foldB(t, s(n)) -> f(foldB(t, n), B())
  , f(t, x) -> f'(t, g(x))
  , foldC(t, 0()) -> t
  , foldC(t, s(n)) -> f(foldC(t, n), C())
  , f'(triple(a, b, c), A()) -> f''(foldB(triple(s(a), 0(), c), b))
  , f'(triple(a, b, c), B()) -> f(triple(a, b, c), A())
  , f'(triple(a, b, c), C()) -> triple(a, b, s(c))
  , f''(triple(a, b, c)) -> foldC(triple(a, b, 0()), c) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  { f^#(t, x) -> c_9(f'^#(t, g(x)), g^#(x)) }

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { foldB^#(t, s(n)) -> c_1(f^#(foldB(t, n), B()), foldB^#(t, n))
  , f^#(t, x) -> c_2(f'^#(t, g(x)))
  , f'^#(triple(a, b, c), A()) ->
    c_3(f''^#(foldB(triple(s(a), 0(), c), b)),
        foldB^#(triple(s(a), 0(), c), b))
  , f'^#(triple(a, b, c), B()) -> c_4(f^#(triple(a, b, c), A()))
  , foldC^#(t, s(n)) -> c_5(f^#(foldC(t, n), C()), foldC^#(t, n))
  , f''^#(triple(a, b, c)) -> c_6(foldC^#(triple(a, b, 0()), c)) }
Weak Trs:
  { g(A()) -> A()
  , g(B()) -> A()
  , g(B()) -> B()
  , g(C()) -> A()
  , g(C()) -> B()
  , g(C()) -> C()
  , foldB(t, 0()) -> t
  , foldB(t, s(n)) -> f(foldB(t, n), B())
  , f(t, x) -> f'(t, g(x))
  , foldC(t, 0()) -> t
  , foldC(t, s(n)) -> f(foldC(t, n), C())
  , f'(triple(a, b, c), A()) -> f''(foldB(triple(s(a), 0(), c), b))
  , f'(triple(a, b, c), B()) -> f(triple(a, b, c), A())
  , f'(triple(a, b, c), C()) -> triple(a, b, s(c))
  , f''(triple(a, b, c)) -> foldC(triple(a, b, 0()), c) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..