MAYBE

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

Strict Trs:
  { a__zeros() -> cons(0(), zeros())
  , a__zeros() -> zeros()
  , a__U11(X1, X2) -> U11(X1, X2)
  , a__U11(tt(), L) -> a__U12(tt(), L)
  , a__U12(X1, X2) -> U12(X1, X2)
  , a__U12(tt(), L) -> s(a__length(mark(L)))
  , a__length(X) -> length(X)
  , a__length(cons(N, L)) -> a__U11(tt(), L)
  , a__length(nil()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(0()) -> 0()
  , mark(zeros()) -> a__zeros()
  , mark(tt()) -> tt()
  , mark(s(X)) -> s(mark(X))
  , mark(nil()) -> nil()
  , mark(U11(X1, X2)) -> a__U11(mark(X1), X2)
  , mark(U12(X1, X2)) -> a__U12(mark(X1), X2)
  , mark(length(X)) -> a__length(mark(X)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { a__zeros^#() -> c_1()
  , a__zeros^#() -> c_2()
  , a__U11^#(X1, X2) -> c_3()
  , a__U11^#(tt(), L) -> c_4(a__U12^#(tt(), L))
  , a__U12^#(X1, X2) -> c_5()
  , a__U12^#(tt(), L) -> c_6(a__length^#(mark(L)), mark^#(L))
  , a__length^#(X) -> c_7()
  , a__length^#(cons(N, L)) -> c_8(a__U11^#(tt(), L))
  , a__length^#(nil()) -> c_9()
  , mark^#(cons(X1, X2)) -> c_10(mark^#(X1))
  , mark^#(0()) -> c_11()
  , mark^#(zeros()) -> c_12(a__zeros^#())
  , mark^#(tt()) -> c_13()
  , mark^#(s(X)) -> c_14(mark^#(X))
  , mark^#(nil()) -> c_15()
  , mark^#(U11(X1, X2)) -> c_16(a__U11^#(mark(X1), X2), mark^#(X1))
  , mark^#(U12(X1, X2)) -> c_17(a__U12^#(mark(X1), X2), mark^#(X1))
  , mark^#(length(X)) -> c_18(a__length^#(mark(X)), 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__zeros^#() -> c_1()
  , a__zeros^#() -> c_2()
  , a__U11^#(X1, X2) -> c_3()
  , a__U11^#(tt(), L) -> c_4(a__U12^#(tt(), L))
  , a__U12^#(X1, X2) -> c_5()
  , a__U12^#(tt(), L) -> c_6(a__length^#(mark(L)), mark^#(L))
  , a__length^#(X) -> c_7()
  , a__length^#(cons(N, L)) -> c_8(a__U11^#(tt(), L))
  , a__length^#(nil()) -> c_9()
  , mark^#(cons(X1, X2)) -> c_10(mark^#(X1))
  , mark^#(0()) -> c_11()
  , mark^#(zeros()) -> c_12(a__zeros^#())
  , mark^#(tt()) -> c_13()
  , mark^#(s(X)) -> c_14(mark^#(X))
  , mark^#(nil()) -> c_15()
  , mark^#(U11(X1, X2)) -> c_16(a__U11^#(mark(X1), X2), mark^#(X1))
  , mark^#(U12(X1, X2)) -> c_17(a__U12^#(mark(X1), X2), mark^#(X1))
  , mark^#(length(X)) -> c_18(a__length^#(mark(X)), mark^#(X)) }
Weak Trs:
  { a__zeros() -> cons(0(), zeros())
  , a__zeros() -> zeros()
  , a__U11(X1, X2) -> U11(X1, X2)
  , a__U11(tt(), L) -> a__U12(tt(), L)
  , a__U12(X1, X2) -> U12(X1, X2)
  , a__U12(tt(), L) -> s(a__length(mark(L)))
  , a__length(X) -> length(X)
  , a__length(cons(N, L)) -> a__U11(tt(), L)
  , a__length(nil()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(0()) -> 0()
  , mark(zeros()) -> a__zeros()
  , mark(tt()) -> tt()
  , mark(s(X)) -> s(mark(X))
  , mark(nil()) -> nil()
  , mark(U11(X1, X2)) -> a__U11(mark(X1), X2)
  , mark(U12(X1, X2)) -> a__U12(mark(X1), X2)
  , mark(length(X)) -> a__length(mark(X)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: a__zeros^#() -> c_1()
    , 2: a__zeros^#() -> c_2()
    , 3: a__U11^#(X1, X2) -> c_3()
    , 4: a__U11^#(tt(), L) -> c_4(a__U12^#(tt(), L))
    , 5: a__U12^#(X1, X2) -> c_5()
    , 6: a__U12^#(tt(), L) -> c_6(a__length^#(mark(L)), mark^#(L))
    , 7: a__length^#(X) -> c_7()
    , 8: a__length^#(cons(N, L)) -> c_8(a__U11^#(tt(), L))
    , 9: a__length^#(nil()) -> c_9()
    , 10: mark^#(cons(X1, X2)) -> c_10(mark^#(X1))
    , 11: mark^#(0()) -> c_11()
    , 12: mark^#(zeros()) -> c_12(a__zeros^#())
    , 13: mark^#(tt()) -> c_13()
    , 14: mark^#(s(X)) -> c_14(mark^#(X))
    , 15: mark^#(nil()) -> c_15()
    , 16: mark^#(U11(X1, X2)) ->
          c_16(a__U11^#(mark(X1), X2), mark^#(X1))
    , 17: mark^#(U12(X1, X2)) ->
          c_17(a__U12^#(mark(X1), X2), mark^#(X1))
    , 18: mark^#(length(X)) -> c_18(a__length^#(mark(X)), mark^#(X)) }

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

Strict DPs:
  { a__U11^#(tt(), L) -> c_4(a__U12^#(tt(), L))
  , a__U12^#(tt(), L) -> c_6(a__length^#(mark(L)), mark^#(L))
  , a__length^#(cons(N, L)) -> c_8(a__U11^#(tt(), L))
  , mark^#(cons(X1, X2)) -> c_10(mark^#(X1))
  , mark^#(zeros()) -> c_12(a__zeros^#())
  , mark^#(s(X)) -> c_14(mark^#(X))
  , mark^#(U11(X1, X2)) -> c_16(a__U11^#(mark(X1), X2), mark^#(X1))
  , mark^#(U12(X1, X2)) -> c_17(a__U12^#(mark(X1), X2), mark^#(X1))
  , mark^#(length(X)) -> c_18(a__length^#(mark(X)), mark^#(X)) }
Weak DPs:
  { a__zeros^#() -> c_1()
  , a__zeros^#() -> c_2()
  , a__U11^#(X1, X2) -> c_3()
  , a__U12^#(X1, X2) -> c_5()
  , a__length^#(X) -> c_7()
  , a__length^#(nil()) -> c_9()
  , mark^#(0()) -> c_11()
  , mark^#(tt()) -> c_13()
  , mark^#(nil()) -> c_15() }
Weak Trs:
  { a__zeros() -> cons(0(), zeros())
  , a__zeros() -> zeros()
  , a__U11(X1, X2) -> U11(X1, X2)
  , a__U11(tt(), L) -> a__U12(tt(), L)
  , a__U12(X1, X2) -> U12(X1, X2)
  , a__U12(tt(), L) -> s(a__length(mark(L)))
  , a__length(X) -> length(X)
  , a__length(cons(N, L)) -> a__U11(tt(), L)
  , a__length(nil()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(0()) -> 0()
  , mark(zeros()) -> a__zeros()
  , mark(tt()) -> tt()
  , mark(s(X)) -> s(mark(X))
  , mark(nil()) -> nil()
  , mark(U11(X1, X2)) -> a__U11(mark(X1), X2)
  , mark(U12(X1, X2)) -> a__U12(mark(X1), X2)
  , mark(length(X)) -> a__length(mark(X)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: a__U11^#(tt(), L) -> c_4(a__U12^#(tt(), L))
    , 2: a__U12^#(tt(), L) -> c_6(a__length^#(mark(L)), mark^#(L))
    , 3: a__length^#(cons(N, L)) -> c_8(a__U11^#(tt(), L))
    , 4: mark^#(cons(X1, X2)) -> c_10(mark^#(X1))
    , 5: mark^#(zeros()) -> c_12(a__zeros^#())
    , 6: mark^#(s(X)) -> c_14(mark^#(X))
    , 7: mark^#(U11(X1, X2)) ->
         c_16(a__U11^#(mark(X1), X2), mark^#(X1))
    , 8: mark^#(U12(X1, X2)) ->
         c_17(a__U12^#(mark(X1), X2), mark^#(X1))
    , 9: mark^#(length(X)) -> c_18(a__length^#(mark(X)), mark^#(X))
    , 10: a__zeros^#() -> c_1()
    , 11: a__zeros^#() -> c_2()
    , 12: a__U11^#(X1, X2) -> c_3()
    , 13: a__U12^#(X1, X2) -> c_5()
    , 14: a__length^#(X) -> c_7()
    , 15: a__length^#(nil()) -> c_9()
    , 16: mark^#(0()) -> c_11()
    , 17: mark^#(tt()) -> c_13()
    , 18: mark^#(nil()) -> c_15() }

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

Strict DPs:
  { a__U11^#(tt(), L) -> c_4(a__U12^#(tt(), L))
  , a__U12^#(tt(), L) -> c_6(a__length^#(mark(L)), mark^#(L))
  , a__length^#(cons(N, L)) -> c_8(a__U11^#(tt(), L))
  , mark^#(cons(X1, X2)) -> c_10(mark^#(X1))
  , mark^#(s(X)) -> c_14(mark^#(X))
  , mark^#(U11(X1, X2)) -> c_16(a__U11^#(mark(X1), X2), mark^#(X1))
  , mark^#(U12(X1, X2)) -> c_17(a__U12^#(mark(X1), X2), mark^#(X1))
  , mark^#(length(X)) -> c_18(a__length^#(mark(X)), mark^#(X)) }
Weak DPs:
  { a__zeros^#() -> c_1()
  , a__zeros^#() -> c_2()
  , a__U11^#(X1, X2) -> c_3()
  , a__U12^#(X1, X2) -> c_5()
  , a__length^#(X) -> c_7()
  , a__length^#(nil()) -> c_9()
  , mark^#(0()) -> c_11()
  , mark^#(zeros()) -> c_12(a__zeros^#())
  , mark^#(tt()) -> c_13()
  , mark^#(nil()) -> c_15() }
Weak Trs:
  { a__zeros() -> cons(0(), zeros())
  , a__zeros() -> zeros()
  , a__U11(X1, X2) -> U11(X1, X2)
  , a__U11(tt(), L) -> a__U12(tt(), L)
  , a__U12(X1, X2) -> U12(X1, X2)
  , a__U12(tt(), L) -> s(a__length(mark(L)))
  , a__length(X) -> length(X)
  , a__length(cons(N, L)) -> a__U11(tt(), L)
  , a__length(nil()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(0()) -> 0()
  , mark(zeros()) -> a__zeros()
  , mark(tt()) -> tt()
  , mark(s(X)) -> s(mark(X))
  , mark(nil()) -> nil()
  , mark(U11(X1, X2)) -> a__U11(mark(X1), X2)
  , mark(U12(X1, X2)) -> a__U12(mark(X1), X2)
  , mark(length(X)) -> a__length(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__zeros^#() -> c_1()
, a__zeros^#() -> c_2()
, a__U11^#(X1, X2) -> c_3()
, a__U12^#(X1, X2) -> c_5()
, a__length^#(X) -> c_7()
, a__length^#(nil()) -> c_9()
, mark^#(0()) -> c_11()
, mark^#(zeros()) -> c_12(a__zeros^#())
, mark^#(tt()) -> c_13()
, mark^#(nil()) -> c_15() }

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

Strict DPs:
  { a__U11^#(tt(), L) -> c_4(a__U12^#(tt(), L))
  , a__U12^#(tt(), L) -> c_6(a__length^#(mark(L)), mark^#(L))
  , a__length^#(cons(N, L)) -> c_8(a__U11^#(tt(), L))
  , mark^#(cons(X1, X2)) -> c_10(mark^#(X1))
  , mark^#(s(X)) -> c_14(mark^#(X))
  , mark^#(U11(X1, X2)) -> c_16(a__U11^#(mark(X1), X2), mark^#(X1))
  , mark^#(U12(X1, X2)) -> c_17(a__U12^#(mark(X1), X2), mark^#(X1))
  , mark^#(length(X)) -> c_18(a__length^#(mark(X)), mark^#(X)) }
Weak Trs:
  { a__zeros() -> cons(0(), zeros())
  , a__zeros() -> zeros()
  , a__U11(X1, X2) -> U11(X1, X2)
  , a__U11(tt(), L) -> a__U12(tt(), L)
  , a__U12(X1, X2) -> U12(X1, X2)
  , a__U12(tt(), L) -> s(a__length(mark(L)))
  , a__length(X) -> length(X)
  , a__length(cons(N, L)) -> a__U11(tt(), L)
  , a__length(nil()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(0()) -> 0()
  , mark(zeros()) -> a__zeros()
  , mark(tt()) -> tt()
  , mark(s(X)) -> s(mark(X))
  , mark(nil()) -> nil()
  , mark(U11(X1, X2)) -> a__U11(mark(X1), X2)
  , mark(U12(X1, X2)) -> a__U12(mark(X1), X2)
  , mark(length(X)) -> a__length(mark(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:
      
      The input cannot be shown compatible
   
   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..