MAYBE

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

Strict Trs:
  { a__first(X1, X2) -> first(X1, X2)
  , a__first(0(), X) -> nil()
  , a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
  , mark(0()) -> 0()
  , mark(nil()) -> nil()
  , mark(s(X)) -> s(mark(X))
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
  , mark(from(X)) -> a__from(mark(X))
  , a__from(X) -> cons(mark(X), from(s(X)))
  , a__from(X) -> from(X) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { a__first^#(X1, X2) -> c_1()
  , a__first^#(0(), X) -> c_2()
  , a__first^#(s(X), cons(Y, Z)) -> c_3(mark^#(Y))
  , mark^#(0()) -> c_4()
  , mark^#(nil()) -> c_5()
  , mark^#(s(X)) -> c_6(mark^#(X))
  , mark^#(cons(X1, X2)) -> c_7(mark^#(X1))
  , mark^#(first(X1, X2)) ->
    c_8(a__first^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(from(X)) -> c_9(a__from^#(mark(X)), mark^#(X))
  , a__from^#(X) -> c_10(mark^#(X))
  , a__from^#(X) -> c_11() }

and mark the set of starting terms.

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

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

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

  DPs:
    { 1: a__first^#(X1, X2) -> c_1()
    , 2: a__first^#(0(), X) -> c_2()
    , 3: a__first^#(s(X), cons(Y, Z)) -> c_3(mark^#(Y))
    , 4: mark^#(0()) -> c_4()
    , 5: mark^#(nil()) -> c_5()
    , 6: mark^#(s(X)) -> c_6(mark^#(X))
    , 7: mark^#(cons(X1, X2)) -> c_7(mark^#(X1))
    , 8: mark^#(first(X1, X2)) ->
         c_8(a__first^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
    , 9: mark^#(from(X)) -> c_9(a__from^#(mark(X)), mark^#(X))
    , 10: a__from^#(X) -> c_10(mark^#(X))
    , 11: a__from^#(X) -> c_11() }

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

Strict DPs:
  { a__first^#(s(X), cons(Y, Z)) -> c_3(mark^#(Y))
  , mark^#(s(X)) -> c_6(mark^#(X))
  , mark^#(cons(X1, X2)) -> c_7(mark^#(X1))
  , mark^#(first(X1, X2)) ->
    c_8(a__first^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(from(X)) -> c_9(a__from^#(mark(X)), mark^#(X))
  , a__from^#(X) -> c_10(mark^#(X)) }
Weak DPs:
  { a__first^#(X1, X2) -> c_1()
  , a__first^#(0(), X) -> c_2()
  , mark^#(0()) -> c_4()
  , mark^#(nil()) -> c_5()
  , a__from^#(X) -> c_11() }
Weak Trs:
  { a__first(X1, X2) -> first(X1, X2)
  , a__first(0(), X) -> nil()
  , a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
  , mark(0()) -> 0()
  , mark(nil()) -> nil()
  , mark(s(X)) -> s(mark(X))
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
  , mark(from(X)) -> a__from(mark(X))
  , a__from(X) -> cons(mark(X), from(s(X)))
  , a__from(X) -> from(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__first^#(X1, X2) -> c_1()
, a__first^#(0(), X) -> c_2()
, mark^#(0()) -> c_4()
, mark^#(nil()) -> c_5()
, a__from^#(X) -> c_11() }

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

Strict DPs:
  { a__first^#(s(X), cons(Y, Z)) -> c_3(mark^#(Y))
  , mark^#(s(X)) -> c_6(mark^#(X))
  , mark^#(cons(X1, X2)) -> c_7(mark^#(X1))
  , mark^#(first(X1, X2)) ->
    c_8(a__first^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(from(X)) -> c_9(a__from^#(mark(X)), mark^#(X))
  , a__from^#(X) -> c_10(mark^#(X)) }
Weak Trs:
  { a__first(X1, X2) -> first(X1, X2)
  , a__first(0(), X) -> nil()
  , a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
  , mark(0()) -> 0()
  , mark(nil()) -> nil()
  , mark(s(X)) -> s(mark(X))
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
  , mark(from(X)) -> a__from(mark(X))
  , a__from(X) -> cons(mark(X), from(s(X)))
  , a__from(X) -> from(X) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..