MAYBE

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

Strict Trs:
  { a__from(X) -> cons(mark(X), from(s(X)))
  , a__from(X) -> from(X)
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(from(X)) -> a__from(mark(X))
  , mark(s(X)) -> s(mark(X))
  , mark(nil()) -> nil()
  , mark(0()) -> 0()
  , mark(length(X)) -> a__length(X)
  , mark(length1(X)) -> a__length1(X)
  , a__length(X) -> length(X)
  , a__length(cons(X, Y)) -> s(a__length1(Y))
  , a__length(nil()) -> 0()
  , a__length1(X) -> a__length(X)
  , a__length1(X) -> length1(X) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { a__from^#(X) -> c_1(mark^#(X))
  , a__from^#(X) -> c_2()
  , mark^#(cons(X1, X2)) -> c_3(mark^#(X1))
  , mark^#(from(X)) -> c_4(a__from^#(mark(X)), mark^#(X))
  , mark^#(s(X)) -> c_5(mark^#(X))
  , mark^#(nil()) -> c_6()
  , mark^#(0()) -> c_7()
  , mark^#(length(X)) -> c_8(a__length^#(X))
  , mark^#(length1(X)) -> c_9(a__length1^#(X))
  , a__length^#(X) -> c_10()
  , a__length^#(cons(X, Y)) -> c_11(a__length1^#(Y))
  , a__length^#(nil()) -> c_12()
  , a__length1^#(X) -> c_13(a__length^#(X))
  , a__length1^#(X) -> c_14() }

and mark the set of starting terms.

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

Strict DPs:
  { a__from^#(X) -> c_1(mark^#(X))
  , a__from^#(X) -> c_2()
  , mark^#(cons(X1, X2)) -> c_3(mark^#(X1))
  , mark^#(from(X)) -> c_4(a__from^#(mark(X)), mark^#(X))
  , mark^#(s(X)) -> c_5(mark^#(X))
  , mark^#(nil()) -> c_6()
  , mark^#(0()) -> c_7()
  , mark^#(length(X)) -> c_8(a__length^#(X))
  , mark^#(length1(X)) -> c_9(a__length1^#(X))
  , a__length^#(X) -> c_10()
  , a__length^#(cons(X, Y)) -> c_11(a__length1^#(Y))
  , a__length^#(nil()) -> c_12()
  , a__length1^#(X) -> c_13(a__length^#(X))
  , a__length1^#(X) -> c_14() }
Weak Trs:
  { a__from(X) -> cons(mark(X), from(s(X)))
  , a__from(X) -> from(X)
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(from(X)) -> a__from(mark(X))
  , mark(s(X)) -> s(mark(X))
  , mark(nil()) -> nil()
  , mark(0()) -> 0()
  , mark(length(X)) -> a__length(X)
  , mark(length1(X)) -> a__length1(X)
  , a__length(X) -> length(X)
  , a__length(cons(X, Y)) -> s(a__length1(Y))
  , a__length(nil()) -> 0()
  , a__length1(X) -> a__length(X)
  , a__length1(X) -> length1(X) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of {2,6,7,10,12,14} by
applications of Pre({2,6,7,10,12,14}) = {1,3,4,5,8,9,11,13}. Here
rules are labeled as follows:

  DPs:
    { 1: a__from^#(X) -> c_1(mark^#(X))
    , 2: a__from^#(X) -> c_2()
    , 3: mark^#(cons(X1, X2)) -> c_3(mark^#(X1))
    , 4: mark^#(from(X)) -> c_4(a__from^#(mark(X)), mark^#(X))
    , 5: mark^#(s(X)) -> c_5(mark^#(X))
    , 6: mark^#(nil()) -> c_6()
    , 7: mark^#(0()) -> c_7()
    , 8: mark^#(length(X)) -> c_8(a__length^#(X))
    , 9: mark^#(length1(X)) -> c_9(a__length1^#(X))
    , 10: a__length^#(X) -> c_10()
    , 11: a__length^#(cons(X, Y)) -> c_11(a__length1^#(Y))
    , 12: a__length^#(nil()) -> c_12()
    , 13: a__length1^#(X) -> c_13(a__length^#(X))
    , 14: a__length1^#(X) -> c_14() }

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

Strict DPs:
  { a__from^#(X) -> c_1(mark^#(X))
  , mark^#(cons(X1, X2)) -> c_3(mark^#(X1))
  , mark^#(from(X)) -> c_4(a__from^#(mark(X)), mark^#(X))
  , mark^#(s(X)) -> c_5(mark^#(X))
  , mark^#(length(X)) -> c_8(a__length^#(X))
  , mark^#(length1(X)) -> c_9(a__length1^#(X))
  , a__length^#(cons(X, Y)) -> c_11(a__length1^#(Y))
  , a__length1^#(X) -> c_13(a__length^#(X)) }
Weak DPs:
  { a__from^#(X) -> c_2()
  , mark^#(nil()) -> c_6()
  , mark^#(0()) -> c_7()
  , a__length^#(X) -> c_10()
  , a__length^#(nil()) -> c_12()
  , a__length1^#(X) -> c_14() }
Weak Trs:
  { a__from(X) -> cons(mark(X), from(s(X)))
  , a__from(X) -> from(X)
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(from(X)) -> a__from(mark(X))
  , mark(s(X)) -> s(mark(X))
  , mark(nil()) -> nil()
  , mark(0()) -> 0()
  , mark(length(X)) -> a__length(X)
  , mark(length1(X)) -> a__length1(X)
  , a__length(X) -> length(X)
  , a__length(cons(X, Y)) -> s(a__length1(Y))
  , a__length(nil()) -> 0()
  , a__length1(X) -> a__length(X)
  , a__length1(X) -> length1(X) }
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.

{ a__from^#(X) -> c_2()
, mark^#(nil()) -> c_6()
, mark^#(0()) -> c_7()
, a__length^#(X) -> c_10()
, a__length^#(nil()) -> c_12()
, a__length1^#(X) -> c_14() }

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

Strict DPs:
  { a__from^#(X) -> c_1(mark^#(X))
  , mark^#(cons(X1, X2)) -> c_3(mark^#(X1))
  , mark^#(from(X)) -> c_4(a__from^#(mark(X)), mark^#(X))
  , mark^#(s(X)) -> c_5(mark^#(X))
  , mark^#(length(X)) -> c_8(a__length^#(X))
  , mark^#(length1(X)) -> c_9(a__length1^#(X))
  , a__length^#(cons(X, Y)) -> c_11(a__length1^#(Y))
  , a__length1^#(X) -> c_13(a__length^#(X)) }
Weak Trs:
  { a__from(X) -> cons(mark(X), from(s(X)))
  , a__from(X) -> from(X)
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(from(X)) -> a__from(mark(X))
  , mark(s(X)) -> s(mark(X))
  , mark(nil()) -> nil()
  , mark(0()) -> 0()
  , mark(length(X)) -> a__length(X)
  , mark(length1(X)) -> a__length1(X)
  , a__length(X) -> length(X)
  , a__length(cons(X, Y)) -> s(a__length1(Y))
  , a__length(nil()) -> 0()
  , a__length1(X) -> a__length(X)
  , a__length1(X) -> length1(X) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..