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(0()) -> 0()
  , mark(nil()) -> nil()
  , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
  , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
  , a__first(X1, X2) -> first(X1, X2)
  , a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
  , a__first(0(), Z) -> nil()
  , a__sel(X1, X2) -> sel(X1, X2)
  , a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z))
  , a__sel(0(), cons(X, Z)) -> mark(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^#(0()) -> c_6()
  , mark^#(nil()) -> c_7()
  , mark^#(first(X1, X2)) ->
    c_8(a__first^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(sel(X1, X2)) ->
    c_9(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__first^#(X1, X2) -> c_10()
  , a__first^#(s(X), cons(Y, Z)) -> c_11(mark^#(Y))
  , a__first^#(0(), Z) -> c_12()
  , a__sel^#(X1, X2) -> c_13()
  , a__sel^#(s(X), cons(Y, Z)) ->
    c_14(a__sel^#(mark(X), mark(Z)), mark^#(X), mark^#(Z))
  , a__sel^#(0(), cons(X, Z)) -> c_15(mark^#(X)) }

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^#(0()) -> c_6()
  , mark^#(nil()) -> c_7()
  , mark^#(first(X1, X2)) ->
    c_8(a__first^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(sel(X1, X2)) ->
    c_9(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__first^#(X1, X2) -> c_10()
  , a__first^#(s(X), cons(Y, Z)) -> c_11(mark^#(Y))
  , a__first^#(0(), Z) -> c_12()
  , a__sel^#(X1, X2) -> c_13()
  , a__sel^#(s(X), cons(Y, Z)) ->
    c_14(a__sel^#(mark(X), mark(Z)), mark^#(X), mark^#(Z))
  , a__sel^#(0(), cons(X, Z)) -> c_15(mark^#(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(0()) -> 0()
  , mark(nil()) -> nil()
  , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
  , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
  , a__first(X1, X2) -> first(X1, X2)
  , a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
  , a__first(0(), Z) -> nil()
  , a__sel(X1, X2) -> sel(X1, X2)
  , a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z))
  , a__sel(0(), cons(X, Z)) -> mark(X) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of {2,6,7,10,12,13} by
applications of Pre({2,6,7,10,12,13}) = {1,3,4,5,8,9,11,14,15}.
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^#(0()) -> c_6()
    , 7: mark^#(nil()) -> c_7()
    , 8: mark^#(first(X1, X2)) ->
         c_8(a__first^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
    , 9: mark^#(sel(X1, X2)) ->
         c_9(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
    , 10: a__first^#(X1, X2) -> c_10()
    , 11: a__first^#(s(X), cons(Y, Z)) -> c_11(mark^#(Y))
    , 12: a__first^#(0(), Z) -> c_12()
    , 13: a__sel^#(X1, X2) -> c_13()
    , 14: a__sel^#(s(X), cons(Y, Z)) ->
          c_14(a__sel^#(mark(X), mark(Z)), mark^#(X), mark^#(Z))
    , 15: a__sel^#(0(), cons(X, Z)) -> c_15(mark^#(X)) }

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^#(first(X1, X2)) ->
    c_8(a__first^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(sel(X1, X2)) ->
    c_9(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__first^#(s(X), cons(Y, Z)) -> c_11(mark^#(Y))
  , a__sel^#(s(X), cons(Y, Z)) ->
    c_14(a__sel^#(mark(X), mark(Z)), mark^#(X), mark^#(Z))
  , a__sel^#(0(), cons(X, Z)) -> c_15(mark^#(X)) }
Weak DPs:
  { a__from^#(X) -> c_2()
  , mark^#(0()) -> c_6()
  , mark^#(nil()) -> c_7()
  , a__first^#(X1, X2) -> c_10()
  , a__first^#(0(), Z) -> c_12()
  , a__sel^#(X1, X2) -> c_13() }
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(0()) -> 0()
  , mark(nil()) -> nil()
  , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
  , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
  , a__first(X1, X2) -> first(X1, X2)
  , a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
  , a__first(0(), Z) -> nil()
  , a__sel(X1, X2) -> sel(X1, X2)
  , a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z))
  , a__sel(0(), cons(X, Z)) -> mark(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^#(0()) -> c_6()
, mark^#(nil()) -> c_7()
, a__first^#(X1, X2) -> c_10()
, a__first^#(0(), Z) -> c_12()
, a__sel^#(X1, X2) -> c_13() }

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^#(first(X1, X2)) ->
    c_8(a__first^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(sel(X1, X2)) ->
    c_9(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__first^#(s(X), cons(Y, Z)) -> c_11(mark^#(Y))
  , a__sel^#(s(X), cons(Y, Z)) ->
    c_14(a__sel^#(mark(X), mark(Z)), mark^#(X), mark^#(Z))
  , a__sel^#(0(), cons(X, Z)) -> c_15(mark^#(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(0()) -> 0()
  , mark(nil()) -> nil()
  , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
  , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
  , a__first(X1, X2) -> first(X1, X2)
  , a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
  , a__first(0(), Z) -> nil()
  , a__sel(X1, X2) -> sel(X1, X2)
  , a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z))
  , a__sel(0(), cons(X, Z)) -> mark(X) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..