MAYBE

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

Strict Trs:
  { a__dbl(X) -> dbl(X)
  , a__dbl(0()) -> 0()
  , a__dbl(s(X)) -> s(s(dbl(X)))
  , a__dbls(X) -> dbls(X)
  , a__dbls(nil()) -> nil()
  , a__dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y))
  , a__sel(X1, X2) -> sel(X1, X2)
  , a__sel(0(), cons(X, Y)) -> mark(X)
  , a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z))
  , mark(0()) -> 0()
  , mark(s(X)) -> s(X)
  , mark(dbl(X)) -> a__dbl(mark(X))
  , mark(nil()) -> nil()
  , mark(cons(X1, X2)) -> cons(X1, X2)
  , mark(dbls(X)) -> a__dbls(mark(X))
  , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
  , mark(indx(X1, X2)) -> a__indx(mark(X1), X2)
  , mark(from(X)) -> a__from(X)
  , a__indx(X1, X2) -> indx(X1, X2)
  , a__indx(nil(), X) -> nil()
  , a__indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
  , a__from(X) -> cons(X, from(s(X)))
  , a__from(X) -> from(X) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { a__dbl^#(X) -> c_1()
  , a__dbl^#(0()) -> c_2()
  , a__dbl^#(s(X)) -> c_3()
  , a__dbls^#(X) -> c_4()
  , a__dbls^#(nil()) -> c_5()
  , a__dbls^#(cons(X, Y)) -> c_6()
  , a__sel^#(X1, X2) -> c_7()
  , a__sel^#(0(), cons(X, Y)) -> c_8(mark^#(X))
  , a__sel^#(s(X), cons(Y, Z)) ->
    c_9(a__sel^#(mark(X), mark(Z)), mark^#(X), mark^#(Z))
  , mark^#(0()) -> c_10()
  , mark^#(s(X)) -> c_11()
  , mark^#(dbl(X)) -> c_12(a__dbl^#(mark(X)), mark^#(X))
  , mark^#(nil()) -> c_13()
  , mark^#(cons(X1, X2)) -> c_14()
  , mark^#(dbls(X)) -> c_15(a__dbls^#(mark(X)), mark^#(X))
  , mark^#(sel(X1, X2)) ->
    c_16(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(indx(X1, X2)) -> c_17(a__indx^#(mark(X1), X2), mark^#(X1))
  , mark^#(from(X)) -> c_18(a__from^#(X))
  , a__indx^#(X1, X2) -> c_19()
  , a__indx^#(nil(), X) -> c_20()
  , a__indx^#(cons(X, Y), Z) -> c_21()
  , a__from^#(X) -> c_22()
  , a__from^#(X) -> c_23() }

and mark the set of starting terms.

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

Strict DPs:
  { a__dbl^#(X) -> c_1()
  , a__dbl^#(0()) -> c_2()
  , a__dbl^#(s(X)) -> c_3()
  , a__dbls^#(X) -> c_4()
  , a__dbls^#(nil()) -> c_5()
  , a__dbls^#(cons(X, Y)) -> c_6()
  , a__sel^#(X1, X2) -> c_7()
  , a__sel^#(0(), cons(X, Y)) -> c_8(mark^#(X))
  , a__sel^#(s(X), cons(Y, Z)) ->
    c_9(a__sel^#(mark(X), mark(Z)), mark^#(X), mark^#(Z))
  , mark^#(0()) -> c_10()
  , mark^#(s(X)) -> c_11()
  , mark^#(dbl(X)) -> c_12(a__dbl^#(mark(X)), mark^#(X))
  , mark^#(nil()) -> c_13()
  , mark^#(cons(X1, X2)) -> c_14()
  , mark^#(dbls(X)) -> c_15(a__dbls^#(mark(X)), mark^#(X))
  , mark^#(sel(X1, X2)) ->
    c_16(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(indx(X1, X2)) -> c_17(a__indx^#(mark(X1), X2), mark^#(X1))
  , mark^#(from(X)) -> c_18(a__from^#(X))
  , a__indx^#(X1, X2) -> c_19()
  , a__indx^#(nil(), X) -> c_20()
  , a__indx^#(cons(X, Y), Z) -> c_21()
  , a__from^#(X) -> c_22()
  , a__from^#(X) -> c_23() }
Weak Trs:
  { a__dbl(X) -> dbl(X)
  , a__dbl(0()) -> 0()
  , a__dbl(s(X)) -> s(s(dbl(X)))
  , a__dbls(X) -> dbls(X)
  , a__dbls(nil()) -> nil()
  , a__dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y))
  , a__sel(X1, X2) -> sel(X1, X2)
  , a__sel(0(), cons(X, Y)) -> mark(X)
  , a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z))
  , mark(0()) -> 0()
  , mark(s(X)) -> s(X)
  , mark(dbl(X)) -> a__dbl(mark(X))
  , mark(nil()) -> nil()
  , mark(cons(X1, X2)) -> cons(X1, X2)
  , mark(dbls(X)) -> a__dbls(mark(X))
  , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
  , mark(indx(X1, X2)) -> a__indx(mark(X1), X2)
  , mark(from(X)) -> a__from(X)
  , a__indx(X1, X2) -> indx(X1, X2)
  , a__indx(nil(), X) -> nil()
  , a__indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
  , a__from(X) -> cons(X, from(s(X)))
  , a__from(X) -> from(X) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of
{1,2,3,4,5,6,7,10,11,13,14,19,20,21,22,23} by applications of
Pre({1,2,3,4,5,6,7,10,11,13,14,19,20,21,22,23}) =
{8,9,12,15,16,17,18}. Here rules are labeled as follows:

  DPs:
    { 1: a__dbl^#(X) -> c_1()
    , 2: a__dbl^#(0()) -> c_2()
    , 3: a__dbl^#(s(X)) -> c_3()
    , 4: a__dbls^#(X) -> c_4()
    , 5: a__dbls^#(nil()) -> c_5()
    , 6: a__dbls^#(cons(X, Y)) -> c_6()
    , 7: a__sel^#(X1, X2) -> c_7()
    , 8: a__sel^#(0(), cons(X, Y)) -> c_8(mark^#(X))
    , 9: a__sel^#(s(X), cons(Y, Z)) ->
         c_9(a__sel^#(mark(X), mark(Z)), mark^#(X), mark^#(Z))
    , 10: mark^#(0()) -> c_10()
    , 11: mark^#(s(X)) -> c_11()
    , 12: mark^#(dbl(X)) -> c_12(a__dbl^#(mark(X)), mark^#(X))
    , 13: mark^#(nil()) -> c_13()
    , 14: mark^#(cons(X1, X2)) -> c_14()
    , 15: mark^#(dbls(X)) -> c_15(a__dbls^#(mark(X)), mark^#(X))
    , 16: mark^#(sel(X1, X2)) ->
          c_16(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
    , 17: mark^#(indx(X1, X2)) ->
          c_17(a__indx^#(mark(X1), X2), mark^#(X1))
    , 18: mark^#(from(X)) -> c_18(a__from^#(X))
    , 19: a__indx^#(X1, X2) -> c_19()
    , 20: a__indx^#(nil(), X) -> c_20()
    , 21: a__indx^#(cons(X, Y), Z) -> c_21()
    , 22: a__from^#(X) -> c_22()
    , 23: a__from^#(X) -> c_23() }

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

Strict DPs:
  { a__sel^#(0(), cons(X, Y)) -> c_8(mark^#(X))
  , a__sel^#(s(X), cons(Y, Z)) ->
    c_9(a__sel^#(mark(X), mark(Z)), mark^#(X), mark^#(Z))
  , mark^#(dbl(X)) -> c_12(a__dbl^#(mark(X)), mark^#(X))
  , mark^#(dbls(X)) -> c_15(a__dbls^#(mark(X)), mark^#(X))
  , mark^#(sel(X1, X2)) ->
    c_16(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(indx(X1, X2)) -> c_17(a__indx^#(mark(X1), X2), mark^#(X1))
  , mark^#(from(X)) -> c_18(a__from^#(X)) }
Weak DPs:
  { a__dbl^#(X) -> c_1()
  , a__dbl^#(0()) -> c_2()
  , a__dbl^#(s(X)) -> c_3()
  , a__dbls^#(X) -> c_4()
  , a__dbls^#(nil()) -> c_5()
  , a__dbls^#(cons(X, Y)) -> c_6()
  , a__sel^#(X1, X2) -> c_7()
  , mark^#(0()) -> c_10()
  , mark^#(s(X)) -> c_11()
  , mark^#(nil()) -> c_13()
  , mark^#(cons(X1, X2)) -> c_14()
  , a__indx^#(X1, X2) -> c_19()
  , a__indx^#(nil(), X) -> c_20()
  , a__indx^#(cons(X, Y), Z) -> c_21()
  , a__from^#(X) -> c_22()
  , a__from^#(X) -> c_23() }
Weak Trs:
  { a__dbl(X) -> dbl(X)
  , a__dbl(0()) -> 0()
  , a__dbl(s(X)) -> s(s(dbl(X)))
  , a__dbls(X) -> dbls(X)
  , a__dbls(nil()) -> nil()
  , a__dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y))
  , a__sel(X1, X2) -> sel(X1, X2)
  , a__sel(0(), cons(X, Y)) -> mark(X)
  , a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z))
  , mark(0()) -> 0()
  , mark(s(X)) -> s(X)
  , mark(dbl(X)) -> a__dbl(mark(X))
  , mark(nil()) -> nil()
  , mark(cons(X1, X2)) -> cons(X1, X2)
  , mark(dbls(X)) -> a__dbls(mark(X))
  , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
  , mark(indx(X1, X2)) -> a__indx(mark(X1), X2)
  , mark(from(X)) -> a__from(X)
  , a__indx(X1, X2) -> indx(X1, X2)
  , a__indx(nil(), X) -> nil()
  , a__indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
  , a__from(X) -> cons(X, from(s(X)))
  , a__from(X) -> from(X) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: a__sel^#(0(), cons(X, Y)) -> c_8(mark^#(X))
    , 2: a__sel^#(s(X), cons(Y, Z)) ->
         c_9(a__sel^#(mark(X), mark(Z)), mark^#(X), mark^#(Z))
    , 3: mark^#(dbl(X)) -> c_12(a__dbl^#(mark(X)), mark^#(X))
    , 4: mark^#(dbls(X)) -> c_15(a__dbls^#(mark(X)), mark^#(X))
    , 5: mark^#(sel(X1, X2)) ->
         c_16(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
    , 6: mark^#(indx(X1, X2)) ->
         c_17(a__indx^#(mark(X1), X2), mark^#(X1))
    , 7: mark^#(from(X)) -> c_18(a__from^#(X))
    , 8: a__dbl^#(X) -> c_1()
    , 9: a__dbl^#(0()) -> c_2()
    , 10: a__dbl^#(s(X)) -> c_3()
    , 11: a__dbls^#(X) -> c_4()
    , 12: a__dbls^#(nil()) -> c_5()
    , 13: a__dbls^#(cons(X, Y)) -> c_6()
    , 14: a__sel^#(X1, X2) -> c_7()
    , 15: mark^#(0()) -> c_10()
    , 16: mark^#(s(X)) -> c_11()
    , 17: mark^#(nil()) -> c_13()
    , 18: mark^#(cons(X1, X2)) -> c_14()
    , 19: a__indx^#(X1, X2) -> c_19()
    , 20: a__indx^#(nil(), X) -> c_20()
    , 21: a__indx^#(cons(X, Y), Z) -> c_21()
    , 22: a__from^#(X) -> c_22()
    , 23: a__from^#(X) -> c_23() }

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

Strict DPs:
  { a__sel^#(0(), cons(X, Y)) -> c_8(mark^#(X))
  , a__sel^#(s(X), cons(Y, Z)) ->
    c_9(a__sel^#(mark(X), mark(Z)), mark^#(X), mark^#(Z))
  , mark^#(dbl(X)) -> c_12(a__dbl^#(mark(X)), mark^#(X))
  , mark^#(dbls(X)) -> c_15(a__dbls^#(mark(X)), mark^#(X))
  , mark^#(sel(X1, X2)) ->
    c_16(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(indx(X1, X2)) ->
    c_17(a__indx^#(mark(X1), X2), mark^#(X1)) }
Weak DPs:
  { a__dbl^#(X) -> c_1()
  , a__dbl^#(0()) -> c_2()
  , a__dbl^#(s(X)) -> c_3()
  , a__dbls^#(X) -> c_4()
  , a__dbls^#(nil()) -> c_5()
  , a__dbls^#(cons(X, Y)) -> c_6()
  , a__sel^#(X1, X2) -> c_7()
  , mark^#(0()) -> c_10()
  , mark^#(s(X)) -> c_11()
  , mark^#(nil()) -> c_13()
  , mark^#(cons(X1, X2)) -> c_14()
  , mark^#(from(X)) -> c_18(a__from^#(X))
  , a__indx^#(X1, X2) -> c_19()
  , a__indx^#(nil(), X) -> c_20()
  , a__indx^#(cons(X, Y), Z) -> c_21()
  , a__from^#(X) -> c_22()
  , a__from^#(X) -> c_23() }
Weak Trs:
  { a__dbl(X) -> dbl(X)
  , a__dbl(0()) -> 0()
  , a__dbl(s(X)) -> s(s(dbl(X)))
  , a__dbls(X) -> dbls(X)
  , a__dbls(nil()) -> nil()
  , a__dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y))
  , a__sel(X1, X2) -> sel(X1, X2)
  , a__sel(0(), cons(X, Y)) -> mark(X)
  , a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z))
  , mark(0()) -> 0()
  , mark(s(X)) -> s(X)
  , mark(dbl(X)) -> a__dbl(mark(X))
  , mark(nil()) -> nil()
  , mark(cons(X1, X2)) -> cons(X1, X2)
  , mark(dbls(X)) -> a__dbls(mark(X))
  , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
  , mark(indx(X1, X2)) -> a__indx(mark(X1), X2)
  , mark(from(X)) -> a__from(X)
  , a__indx(X1, X2) -> indx(X1, X2)
  , a__indx(nil(), X) -> nil()
  , a__indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
  , a__from(X) -> cons(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__dbl^#(X) -> c_1()
, a__dbl^#(0()) -> c_2()
, a__dbl^#(s(X)) -> c_3()
, a__dbls^#(X) -> c_4()
, a__dbls^#(nil()) -> c_5()
, a__dbls^#(cons(X, Y)) -> c_6()
, a__sel^#(X1, X2) -> c_7()
, mark^#(0()) -> c_10()
, mark^#(s(X)) -> c_11()
, mark^#(nil()) -> c_13()
, mark^#(cons(X1, X2)) -> c_14()
, mark^#(from(X)) -> c_18(a__from^#(X))
, a__indx^#(X1, X2) -> c_19()
, a__indx^#(nil(), X) -> c_20()
, a__indx^#(cons(X, Y), Z) -> c_21()
, a__from^#(X) -> c_22()
, a__from^#(X) -> c_23() }

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

Strict DPs:
  { a__sel^#(0(), cons(X, Y)) -> c_8(mark^#(X))
  , a__sel^#(s(X), cons(Y, Z)) ->
    c_9(a__sel^#(mark(X), mark(Z)), mark^#(X), mark^#(Z))
  , mark^#(dbl(X)) -> c_12(a__dbl^#(mark(X)), mark^#(X))
  , mark^#(dbls(X)) -> c_15(a__dbls^#(mark(X)), mark^#(X))
  , mark^#(sel(X1, X2)) ->
    c_16(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(indx(X1, X2)) ->
    c_17(a__indx^#(mark(X1), X2), mark^#(X1)) }
Weak Trs:
  { a__dbl(X) -> dbl(X)
  , a__dbl(0()) -> 0()
  , a__dbl(s(X)) -> s(s(dbl(X)))
  , a__dbls(X) -> dbls(X)
  , a__dbls(nil()) -> nil()
  , a__dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y))
  , a__sel(X1, X2) -> sel(X1, X2)
  , a__sel(0(), cons(X, Y)) -> mark(X)
  , a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z))
  , mark(0()) -> 0()
  , mark(s(X)) -> s(X)
  , mark(dbl(X)) -> a__dbl(mark(X))
  , mark(nil()) -> nil()
  , mark(cons(X1, X2)) -> cons(X1, X2)
  , mark(dbls(X)) -> a__dbls(mark(X))
  , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
  , mark(indx(X1, X2)) -> a__indx(mark(X1), X2)
  , mark(from(X)) -> a__from(X)
  , a__indx(X1, X2) -> indx(X1, X2)
  , a__indx(nil(), X) -> nil()
  , a__indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
  , a__from(X) -> cons(X, from(s(X)))
  , a__from(X) -> from(X) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  { mark^#(dbl(X)) -> c_12(a__dbl^#(mark(X)), mark^#(X))
  , mark^#(dbls(X)) -> c_15(a__dbls^#(mark(X)), mark^#(X))
  , mark^#(indx(X1, X2)) ->
    c_17(a__indx^#(mark(X1), X2), mark^#(X1)) }

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

Strict DPs:
  { a__sel^#(0(), cons(X, Y)) -> c_1(mark^#(X))
  , a__sel^#(s(X), cons(Y, Z)) ->
    c_2(a__sel^#(mark(X), mark(Z)), mark^#(X), mark^#(Z))
  , mark^#(dbl(X)) -> c_3(mark^#(X))
  , mark^#(dbls(X)) -> c_4(mark^#(X))
  , mark^#(sel(X1, X2)) ->
    c_5(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(indx(X1, X2)) -> c_6(mark^#(X1)) }
Weak Trs:
  { a__dbl(X) -> dbl(X)
  , a__dbl(0()) -> 0()
  , a__dbl(s(X)) -> s(s(dbl(X)))
  , a__dbls(X) -> dbls(X)
  , a__dbls(nil()) -> nil()
  , a__dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y))
  , a__sel(X1, X2) -> sel(X1, X2)
  , a__sel(0(), cons(X, Y)) -> mark(X)
  , a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z))
  , mark(0()) -> 0()
  , mark(s(X)) -> s(X)
  , mark(dbl(X)) -> a__dbl(mark(X))
  , mark(nil()) -> nil()
  , mark(cons(X1, X2)) -> cons(X1, X2)
  , mark(dbls(X)) -> a__dbls(mark(X))
  , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2))
  , mark(indx(X1, X2)) -> a__indx(mark(X1), X2)
  , mark(from(X)) -> a__from(X)
  , a__indx(X1, X2) -> indx(X1, X2)
  , a__indx(nil(), X) -> nil()
  , a__indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
  , a__from(X) -> cons(X, from(s(X)))
  , a__from(X) -> from(X) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

None of the processors succeeded.

Details of failed attempt(s):
-----------------------------
1) 'matrices' failed due to the following reason:
   
   None of the processors succeeded.
   
   Details of failed attempt(s):
   -----------------------------
   1) 'matrix interpretation of dimension 4' failed due to the
      following reason:
      
      Following exception was raised:
        stack overflow
   
   2) 'matrix interpretation of dimension 3' failed due to the
      following reason:
      
      The input cannot be shown compatible
   
   3) 'matrix interpretation of dimension 3' failed due to the
      following reason:
      
      The input cannot be shown compatible
   
   4) 'matrix interpretation of dimension 2' failed due to the
      following reason:
      
      The input cannot be shown compatible
   
   5) 'matrix interpretation of dimension 2' failed due to the
      following reason:
      
      The input cannot be shown compatible
   
   6) 'matrix interpretation of dimension 1' failed due to the
      following reason:
      
      The input cannot be shown compatible
   

2) 'empty' failed due to the following reason:
   
   Empty strict component of the problem is NOT empty.


Arrrr..