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(take(X1, X2)) -> a__take(mark(X1), mark(X2))
  , mark(head(X)) -> a__head(mark(X))
  , mark(2nd(X)) -> a__2nd(mark(X))
  , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
  , a__head(X) -> head(X)
  , a__head(cons(X, XS)) -> mark(X)
  , a__2nd(X) -> 2nd(X)
  , a__2nd(cons(X, XS)) -> a__head(mark(XS))
  , a__take(X1, X2) -> take(X1, X2)
  , a__take(s(N), cons(X, XS)) -> cons(mark(X), take(N, XS))
  , a__take(0(), XS) -> nil()
  , a__sel(X1, X2) -> sel(X1, X2)
  , a__sel(s(N), cons(X, XS)) -> a__sel(mark(N), mark(XS))
  , a__sel(0(), cons(X, XS)) -> 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^#(take(X1, X2)) ->
    c_8(a__take^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(head(X)) -> c_9(a__head^#(mark(X)), mark^#(X))
  , mark^#(2nd(X)) -> c_10(a__2nd^#(mark(X)), mark^#(X))
  , mark^#(sel(X1, X2)) ->
    c_11(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__take^#(X1, X2) -> c_16()
  , a__take^#(s(N), cons(X, XS)) -> c_17(mark^#(X))
  , a__take^#(0(), XS) -> c_18()
  , a__head^#(X) -> c_12()
  , a__head^#(cons(X, XS)) -> c_13(mark^#(X))
  , a__2nd^#(X) -> c_14()
  , a__2nd^#(cons(X, XS)) -> c_15(a__head^#(mark(XS)), mark^#(XS))
  , a__sel^#(X1, X2) -> c_19()
  , a__sel^#(s(N), cons(X, XS)) ->
    c_20(a__sel^#(mark(N), mark(XS)), mark^#(N), mark^#(XS))
  , a__sel^#(0(), cons(X, XS)) -> c_21(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^#(take(X1, X2)) ->
    c_8(a__take^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(head(X)) -> c_9(a__head^#(mark(X)), mark^#(X))
  , mark^#(2nd(X)) -> c_10(a__2nd^#(mark(X)), mark^#(X))
  , mark^#(sel(X1, X2)) ->
    c_11(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__take^#(X1, X2) -> c_16()
  , a__take^#(s(N), cons(X, XS)) -> c_17(mark^#(X))
  , a__take^#(0(), XS) -> c_18()
  , a__head^#(X) -> c_12()
  , a__head^#(cons(X, XS)) -> c_13(mark^#(X))
  , a__2nd^#(X) -> c_14()
  , a__2nd^#(cons(X, XS)) -> c_15(a__head^#(mark(XS)), mark^#(XS))
  , a__sel^#(X1, X2) -> c_19()
  , a__sel^#(s(N), cons(X, XS)) ->
    c_20(a__sel^#(mark(N), mark(XS)), mark^#(N), mark^#(XS))
  , a__sel^#(0(), cons(X, XS)) -> c_21(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(take(X1, X2)) -> a__take(mark(X1), mark(X2))
  , mark(head(X)) -> a__head(mark(X))
  , mark(2nd(X)) -> a__2nd(mark(X))
  , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
  , a__head(X) -> head(X)
  , a__head(cons(X, XS)) -> mark(X)
  , a__2nd(X) -> 2nd(X)
  , a__2nd(cons(X, XS)) -> a__head(mark(XS))
  , a__take(X1, X2) -> take(X1, X2)
  , a__take(s(N), cons(X, XS)) -> cons(mark(X), take(N, XS))
  , a__take(0(), XS) -> nil()
  , a__sel(X1, X2) -> sel(X1, X2)
  , a__sel(s(N), cons(X, XS)) -> a__sel(mark(N), mark(XS))
  , a__sel(0(), cons(X, XS)) -> mark(X) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of {2,6,7,12,14,15,17,19} by
applications of Pre({2,6,7,12,14,15,17,19}) =
{1,3,4,5,8,9,10,11,13,16,18,20,21}. 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^#(take(X1, X2)) ->
         c_8(a__take^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
    , 9: mark^#(head(X)) -> c_9(a__head^#(mark(X)), mark^#(X))
    , 10: mark^#(2nd(X)) -> c_10(a__2nd^#(mark(X)), mark^#(X))
    , 11: mark^#(sel(X1, X2)) ->
          c_11(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
    , 12: a__take^#(X1, X2) -> c_16()
    , 13: a__take^#(s(N), cons(X, XS)) -> c_17(mark^#(X))
    , 14: a__take^#(0(), XS) -> c_18()
    , 15: a__head^#(X) -> c_12()
    , 16: a__head^#(cons(X, XS)) -> c_13(mark^#(X))
    , 17: a__2nd^#(X) -> c_14()
    , 18: a__2nd^#(cons(X, XS)) ->
          c_15(a__head^#(mark(XS)), mark^#(XS))
    , 19: a__sel^#(X1, X2) -> c_19()
    , 20: a__sel^#(s(N), cons(X, XS)) ->
          c_20(a__sel^#(mark(N), mark(XS)), mark^#(N), mark^#(XS))
    , 21: a__sel^#(0(), cons(X, XS)) -> c_21(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^#(take(X1, X2)) ->
    c_8(a__take^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(head(X)) -> c_9(a__head^#(mark(X)), mark^#(X))
  , mark^#(2nd(X)) -> c_10(a__2nd^#(mark(X)), mark^#(X))
  , mark^#(sel(X1, X2)) ->
    c_11(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__take^#(s(N), cons(X, XS)) -> c_17(mark^#(X))
  , a__head^#(cons(X, XS)) -> c_13(mark^#(X))
  , a__2nd^#(cons(X, XS)) -> c_15(a__head^#(mark(XS)), mark^#(XS))
  , a__sel^#(s(N), cons(X, XS)) ->
    c_20(a__sel^#(mark(N), mark(XS)), mark^#(N), mark^#(XS))
  , a__sel^#(0(), cons(X, XS)) -> c_21(mark^#(X)) }
Weak DPs:
  { a__from^#(X) -> c_2()
  , mark^#(0()) -> c_6()
  , mark^#(nil()) -> c_7()
  , a__take^#(X1, X2) -> c_16()
  , a__take^#(0(), XS) -> c_18()
  , a__head^#(X) -> c_12()
  , a__2nd^#(X) -> c_14()
  , a__sel^#(X1, X2) -> c_19() }
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(take(X1, X2)) -> a__take(mark(X1), mark(X2))
  , mark(head(X)) -> a__head(mark(X))
  , mark(2nd(X)) -> a__2nd(mark(X))
  , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
  , a__head(X) -> head(X)
  , a__head(cons(X, XS)) -> mark(X)
  , a__2nd(X) -> 2nd(X)
  , a__2nd(cons(X, XS)) -> a__head(mark(XS))
  , a__take(X1, X2) -> take(X1, X2)
  , a__take(s(N), cons(X, XS)) -> cons(mark(X), take(N, XS))
  , a__take(0(), XS) -> nil()
  , a__sel(X1, X2) -> sel(X1, X2)
  , a__sel(s(N), cons(X, XS)) -> a__sel(mark(N), mark(XS))
  , a__sel(0(), cons(X, XS)) -> 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__take^#(X1, X2) -> c_16()
, a__take^#(0(), XS) -> c_18()
, a__head^#(X) -> c_12()
, a__2nd^#(X) -> c_14()
, a__sel^#(X1, X2) -> c_19() }

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^#(take(X1, X2)) ->
    c_8(a__take^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(head(X)) -> c_9(a__head^#(mark(X)), mark^#(X))
  , mark^#(2nd(X)) -> c_10(a__2nd^#(mark(X)), mark^#(X))
  , mark^#(sel(X1, X2)) ->
    c_11(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , a__take^#(s(N), cons(X, XS)) -> c_17(mark^#(X))
  , a__head^#(cons(X, XS)) -> c_13(mark^#(X))
  , a__2nd^#(cons(X, XS)) -> c_15(a__head^#(mark(XS)), mark^#(XS))
  , a__sel^#(s(N), cons(X, XS)) ->
    c_20(a__sel^#(mark(N), mark(XS)), mark^#(N), mark^#(XS))
  , a__sel^#(0(), cons(X, XS)) -> c_21(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(take(X1, X2)) -> a__take(mark(X1), mark(X2))
  , mark(head(X)) -> a__head(mark(X))
  , mark(2nd(X)) -> a__2nd(mark(X))
  , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
  , a__head(X) -> head(X)
  , a__head(cons(X, XS)) -> mark(X)
  , a__2nd(X) -> 2nd(X)
  , a__2nd(cons(X, XS)) -> a__head(mark(XS))
  , a__take(X1, X2) -> take(X1, X2)
  , a__take(s(N), cons(X, XS)) -> cons(mark(X), take(N, XS))
  , a__take(0(), XS) -> nil()
  , a__sel(X1, X2) -> sel(X1, X2)
  , a__sel(s(N), cons(X, XS)) -> a__sel(mark(N), mark(XS))
  , a__sel(0(), cons(X, XS)) -> mark(X) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..