MAYBE

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

Strict Trs:
  { a__nats() -> a__adx(a__zeros())
  , a__nats() -> nats()
  , a__adx(X) -> adx(X)
  , a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y)))
  , a__zeros() -> cons(0(), zeros())
  , a__zeros() -> zeros()
  , a__incr(X) -> incr(X)
  , a__incr(cons(X, Y)) -> cons(s(X), incr(Y))
  , a__hd(X) -> hd(X)
  , a__hd(cons(X, Y)) -> mark(X)
  , mark(cons(X1, X2)) -> cons(X1, X2)
  , mark(0()) -> 0()
  , mark(zeros()) -> a__zeros()
  , mark(s(X)) -> s(X)
  , mark(incr(X)) -> a__incr(mark(X))
  , mark(adx(X)) -> a__adx(mark(X))
  , mark(nats()) -> a__nats()
  , mark(hd(X)) -> a__hd(mark(X))
  , mark(tl(X)) -> a__tl(mark(X))
  , a__tl(X) -> tl(X)
  , a__tl(cons(X, Y)) -> mark(Y) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { a__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#())
  , a__nats^#() -> c_2()
  , a__adx^#(X) -> c_3()
  , a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y))))
  , a__zeros^#() -> c_5()
  , a__zeros^#() -> c_6()
  , a__incr^#(X) -> c_7()
  , a__incr^#(cons(X, Y)) -> c_8()
  , a__hd^#(X) -> c_9()
  , a__hd^#(cons(X, Y)) -> c_10(mark^#(X))
  , mark^#(cons(X1, X2)) -> c_11()
  , mark^#(0()) -> c_12()
  , mark^#(zeros()) -> c_13(a__zeros^#())
  , mark^#(s(X)) -> c_14()
  , mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X))
  , mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X))
  , mark^#(nats()) -> c_17(a__nats^#())
  , mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X))
  , mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X))
  , a__tl^#(X) -> c_20()
  , a__tl^#(cons(X, Y)) -> c_21(mark^#(Y)) }

and mark the set of starting terms.

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

Strict DPs:
  { a__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#())
  , a__nats^#() -> c_2()
  , a__adx^#(X) -> c_3()
  , a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y))))
  , a__zeros^#() -> c_5()
  , a__zeros^#() -> c_6()
  , a__incr^#(X) -> c_7()
  , a__incr^#(cons(X, Y)) -> c_8()
  , a__hd^#(X) -> c_9()
  , a__hd^#(cons(X, Y)) -> c_10(mark^#(X))
  , mark^#(cons(X1, X2)) -> c_11()
  , mark^#(0()) -> c_12()
  , mark^#(zeros()) -> c_13(a__zeros^#())
  , mark^#(s(X)) -> c_14()
  , mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X))
  , mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X))
  , mark^#(nats()) -> c_17(a__nats^#())
  , mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X))
  , mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X))
  , a__tl^#(X) -> c_20()
  , a__tl^#(cons(X, Y)) -> c_21(mark^#(Y)) }
Weak Trs:
  { a__nats() -> a__adx(a__zeros())
  , a__nats() -> nats()
  , a__adx(X) -> adx(X)
  , a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y)))
  , a__zeros() -> cons(0(), zeros())
  , a__zeros() -> zeros()
  , a__incr(X) -> incr(X)
  , a__incr(cons(X, Y)) -> cons(s(X), incr(Y))
  , a__hd(X) -> hd(X)
  , a__hd(cons(X, Y)) -> mark(X)
  , mark(cons(X1, X2)) -> cons(X1, X2)
  , mark(0()) -> 0()
  , mark(zeros()) -> a__zeros()
  , mark(s(X)) -> s(X)
  , mark(incr(X)) -> a__incr(mark(X))
  , mark(adx(X)) -> a__adx(mark(X))
  , mark(nats()) -> a__nats()
  , mark(hd(X)) -> a__hd(mark(X))
  , mark(tl(X)) -> a__tl(mark(X))
  , a__tl(X) -> tl(X)
  , a__tl(cons(X, Y)) -> mark(Y) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: a__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#())
    , 2: a__nats^#() -> c_2()
    , 3: a__adx^#(X) -> c_3()
    , 4: a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y))))
    , 5: a__zeros^#() -> c_5()
    , 6: a__zeros^#() -> c_6()
    , 7: a__incr^#(X) -> c_7()
    , 8: a__incr^#(cons(X, Y)) -> c_8()
    , 9: a__hd^#(X) -> c_9()
    , 10: a__hd^#(cons(X, Y)) -> c_10(mark^#(X))
    , 11: mark^#(cons(X1, X2)) -> c_11()
    , 12: mark^#(0()) -> c_12()
    , 13: mark^#(zeros()) -> c_13(a__zeros^#())
    , 14: mark^#(s(X)) -> c_14()
    , 15: mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X))
    , 16: mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X))
    , 17: mark^#(nats()) -> c_17(a__nats^#())
    , 18: mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X))
    , 19: mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X))
    , 20: a__tl^#(X) -> c_20()
    , 21: a__tl^#(cons(X, Y)) -> c_21(mark^#(Y)) }

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

Strict DPs:
  { a__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#())
  , a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y))))
  , a__hd^#(cons(X, Y)) -> c_10(mark^#(X))
  , mark^#(zeros()) -> c_13(a__zeros^#())
  , mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X))
  , mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X))
  , mark^#(nats()) -> c_17(a__nats^#())
  , mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X))
  , mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X))
  , a__tl^#(cons(X, Y)) -> c_21(mark^#(Y)) }
Weak DPs:
  { a__nats^#() -> c_2()
  , a__adx^#(X) -> c_3()
  , a__zeros^#() -> c_5()
  , a__zeros^#() -> c_6()
  , a__incr^#(X) -> c_7()
  , a__incr^#(cons(X, Y)) -> c_8()
  , a__hd^#(X) -> c_9()
  , mark^#(cons(X1, X2)) -> c_11()
  , mark^#(0()) -> c_12()
  , mark^#(s(X)) -> c_14()
  , a__tl^#(X) -> c_20() }
Weak Trs:
  { a__nats() -> a__adx(a__zeros())
  , a__nats() -> nats()
  , a__adx(X) -> adx(X)
  , a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y)))
  , a__zeros() -> cons(0(), zeros())
  , a__zeros() -> zeros()
  , a__incr(X) -> incr(X)
  , a__incr(cons(X, Y)) -> cons(s(X), incr(Y))
  , a__hd(X) -> hd(X)
  , a__hd(cons(X, Y)) -> mark(X)
  , mark(cons(X1, X2)) -> cons(X1, X2)
  , mark(0()) -> 0()
  , mark(zeros()) -> a__zeros()
  , mark(s(X)) -> s(X)
  , mark(incr(X)) -> a__incr(mark(X))
  , mark(adx(X)) -> a__adx(mark(X))
  , mark(nats()) -> a__nats()
  , mark(hd(X)) -> a__hd(mark(X))
  , mark(tl(X)) -> a__tl(mark(X))
  , a__tl(X) -> tl(X)
  , a__tl(cons(X, Y)) -> mark(Y) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: a__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#())
    , 2: a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y))))
    , 3: a__hd^#(cons(X, Y)) -> c_10(mark^#(X))
    , 4: mark^#(zeros()) -> c_13(a__zeros^#())
    , 5: mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X))
    , 6: mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X))
    , 7: mark^#(nats()) -> c_17(a__nats^#())
    , 8: mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X))
    , 9: mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X))
    , 10: a__tl^#(cons(X, Y)) -> c_21(mark^#(Y))
    , 11: a__nats^#() -> c_2()
    , 12: a__adx^#(X) -> c_3()
    , 13: a__zeros^#() -> c_5()
    , 14: a__zeros^#() -> c_6()
    , 15: a__incr^#(X) -> c_7()
    , 16: a__incr^#(cons(X, Y)) -> c_8()
    , 17: a__hd^#(X) -> c_9()
    , 18: mark^#(cons(X1, X2)) -> c_11()
    , 19: mark^#(0()) -> c_12()
    , 20: mark^#(s(X)) -> c_14()
    , 21: a__tl^#(X) -> c_20() }

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

Strict DPs:
  { a__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#())
  , a__hd^#(cons(X, Y)) -> c_10(mark^#(X))
  , mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X))
  , mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X))
  , mark^#(nats()) -> c_17(a__nats^#())
  , mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X))
  , mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X))
  , a__tl^#(cons(X, Y)) -> c_21(mark^#(Y)) }
Weak DPs:
  { a__nats^#() -> c_2()
  , a__adx^#(X) -> c_3()
  , a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y))))
  , a__zeros^#() -> c_5()
  , a__zeros^#() -> c_6()
  , a__incr^#(X) -> c_7()
  , a__incr^#(cons(X, Y)) -> c_8()
  , a__hd^#(X) -> c_9()
  , mark^#(cons(X1, X2)) -> c_11()
  , mark^#(0()) -> c_12()
  , mark^#(zeros()) -> c_13(a__zeros^#())
  , mark^#(s(X)) -> c_14()
  , a__tl^#(X) -> c_20() }
Weak Trs:
  { a__nats() -> a__adx(a__zeros())
  , a__nats() -> nats()
  , a__adx(X) -> adx(X)
  , a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y)))
  , a__zeros() -> cons(0(), zeros())
  , a__zeros() -> zeros()
  , a__incr(X) -> incr(X)
  , a__incr(cons(X, Y)) -> cons(s(X), incr(Y))
  , a__hd(X) -> hd(X)
  , a__hd(cons(X, Y)) -> mark(X)
  , mark(cons(X1, X2)) -> cons(X1, X2)
  , mark(0()) -> 0()
  , mark(zeros()) -> a__zeros()
  , mark(s(X)) -> s(X)
  , mark(incr(X)) -> a__incr(mark(X))
  , mark(adx(X)) -> a__adx(mark(X))
  , mark(nats()) -> a__nats()
  , mark(hd(X)) -> a__hd(mark(X))
  , mark(tl(X)) -> a__tl(mark(X))
  , a__tl(X) -> tl(X)
  , a__tl(cons(X, Y)) -> mark(Y) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: a__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#())
    , 2: a__hd^#(cons(X, Y)) -> c_10(mark^#(X))
    , 3: mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X))
    , 4: mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X))
    , 5: mark^#(nats()) -> c_17(a__nats^#())
    , 6: mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X))
    , 7: mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X))
    , 8: a__tl^#(cons(X, Y)) -> c_21(mark^#(Y))
    , 9: a__nats^#() -> c_2()
    , 10: a__adx^#(X) -> c_3()
    , 11: a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y))))
    , 12: a__zeros^#() -> c_5()
    , 13: a__zeros^#() -> c_6()
    , 14: a__incr^#(X) -> c_7()
    , 15: a__incr^#(cons(X, Y)) -> c_8()
    , 16: a__hd^#(X) -> c_9()
    , 17: mark^#(cons(X1, X2)) -> c_11()
    , 18: mark^#(0()) -> c_12()
    , 19: mark^#(zeros()) -> c_13(a__zeros^#())
    , 20: mark^#(s(X)) -> c_14()
    , 21: a__tl^#(X) -> c_20() }

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

Strict DPs:
  { a__hd^#(cons(X, Y)) -> c_10(mark^#(X))
  , mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X))
  , mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X))
  , mark^#(nats()) -> c_17(a__nats^#())
  , mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X))
  , mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X))
  , a__tl^#(cons(X, Y)) -> c_21(mark^#(Y)) }
Weak DPs:
  { a__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#())
  , a__nats^#() -> c_2()
  , a__adx^#(X) -> c_3()
  , a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y))))
  , a__zeros^#() -> c_5()
  , a__zeros^#() -> c_6()
  , a__incr^#(X) -> c_7()
  , a__incr^#(cons(X, Y)) -> c_8()
  , a__hd^#(X) -> c_9()
  , mark^#(cons(X1, X2)) -> c_11()
  , mark^#(0()) -> c_12()
  , mark^#(zeros()) -> c_13(a__zeros^#())
  , mark^#(s(X)) -> c_14()
  , a__tl^#(X) -> c_20() }
Weak Trs:
  { a__nats() -> a__adx(a__zeros())
  , a__nats() -> nats()
  , a__adx(X) -> adx(X)
  , a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y)))
  , a__zeros() -> cons(0(), zeros())
  , a__zeros() -> zeros()
  , a__incr(X) -> incr(X)
  , a__incr(cons(X, Y)) -> cons(s(X), incr(Y))
  , a__hd(X) -> hd(X)
  , a__hd(cons(X, Y)) -> mark(X)
  , mark(cons(X1, X2)) -> cons(X1, X2)
  , mark(0()) -> 0()
  , mark(zeros()) -> a__zeros()
  , mark(s(X)) -> s(X)
  , mark(incr(X)) -> a__incr(mark(X))
  , mark(adx(X)) -> a__adx(mark(X))
  , mark(nats()) -> a__nats()
  , mark(hd(X)) -> a__hd(mark(X))
  , mark(tl(X)) -> a__tl(mark(X))
  , a__tl(X) -> tl(X)
  , a__tl(cons(X, Y)) -> mark(Y) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: a__hd^#(cons(X, Y)) -> c_10(mark^#(X))
    , 2: mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X))
    , 3: mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X))
    , 4: mark^#(nats()) -> c_17(a__nats^#())
    , 5: mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X))
    , 6: mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X))
    , 7: a__tl^#(cons(X, Y)) -> c_21(mark^#(Y))
    , 8: a__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#())
    , 9: a__nats^#() -> c_2()
    , 10: a__adx^#(X) -> c_3()
    , 11: a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y))))
    , 12: a__zeros^#() -> c_5()
    , 13: a__zeros^#() -> c_6()
    , 14: a__incr^#(X) -> c_7()
    , 15: a__incr^#(cons(X, Y)) -> c_8()
    , 16: a__hd^#(X) -> c_9()
    , 17: mark^#(cons(X1, X2)) -> c_11()
    , 18: mark^#(0()) -> c_12()
    , 19: mark^#(zeros()) -> c_13(a__zeros^#())
    , 20: mark^#(s(X)) -> c_14()
    , 21: a__tl^#(X) -> c_20() }

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

Strict DPs:
  { a__hd^#(cons(X, Y)) -> c_10(mark^#(X))
  , mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X))
  , mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X))
  , mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X))
  , mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X))
  , a__tl^#(cons(X, Y)) -> c_21(mark^#(Y)) }
Weak DPs:
  { a__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#())
  , a__nats^#() -> c_2()
  , a__adx^#(X) -> c_3()
  , a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y))))
  , a__zeros^#() -> c_5()
  , a__zeros^#() -> c_6()
  , a__incr^#(X) -> c_7()
  , a__incr^#(cons(X, Y)) -> c_8()
  , a__hd^#(X) -> c_9()
  , mark^#(cons(X1, X2)) -> c_11()
  , mark^#(0()) -> c_12()
  , mark^#(zeros()) -> c_13(a__zeros^#())
  , mark^#(s(X)) -> c_14()
  , mark^#(nats()) -> c_17(a__nats^#())
  , a__tl^#(X) -> c_20() }
Weak Trs:
  { a__nats() -> a__adx(a__zeros())
  , a__nats() -> nats()
  , a__adx(X) -> adx(X)
  , a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y)))
  , a__zeros() -> cons(0(), zeros())
  , a__zeros() -> zeros()
  , a__incr(X) -> incr(X)
  , a__incr(cons(X, Y)) -> cons(s(X), incr(Y))
  , a__hd(X) -> hd(X)
  , a__hd(cons(X, Y)) -> mark(X)
  , mark(cons(X1, X2)) -> cons(X1, X2)
  , mark(0()) -> 0()
  , mark(zeros()) -> a__zeros()
  , mark(s(X)) -> s(X)
  , mark(incr(X)) -> a__incr(mark(X))
  , mark(adx(X)) -> a__adx(mark(X))
  , mark(nats()) -> a__nats()
  , mark(hd(X)) -> a__hd(mark(X))
  , mark(tl(X)) -> a__tl(mark(X))
  , a__tl(X) -> tl(X)
  , a__tl(cons(X, Y)) -> mark(Y) }
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__nats^#() -> c_1(a__adx^#(a__zeros()), a__zeros^#())
, a__nats^#() -> c_2()
, a__adx^#(X) -> c_3()
, a__adx^#(cons(X, Y)) -> c_4(a__incr^#(cons(X, adx(Y))))
, a__zeros^#() -> c_5()
, a__zeros^#() -> c_6()
, a__incr^#(X) -> c_7()
, a__incr^#(cons(X, Y)) -> c_8()
, a__hd^#(X) -> c_9()
, mark^#(cons(X1, X2)) -> c_11()
, mark^#(0()) -> c_12()
, mark^#(zeros()) -> c_13(a__zeros^#())
, mark^#(s(X)) -> c_14()
, mark^#(nats()) -> c_17(a__nats^#())
, a__tl^#(X) -> c_20() }

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

Strict DPs:
  { a__hd^#(cons(X, Y)) -> c_10(mark^#(X))
  , mark^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X))
  , mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X))
  , mark^#(hd(X)) -> c_18(a__hd^#(mark(X)), mark^#(X))
  , mark^#(tl(X)) -> c_19(a__tl^#(mark(X)), mark^#(X))
  , a__tl^#(cons(X, Y)) -> c_21(mark^#(Y)) }
Weak Trs:
  { a__nats() -> a__adx(a__zeros())
  , a__nats() -> nats()
  , a__adx(X) -> adx(X)
  , a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y)))
  , a__zeros() -> cons(0(), zeros())
  , a__zeros() -> zeros()
  , a__incr(X) -> incr(X)
  , a__incr(cons(X, Y)) -> cons(s(X), incr(Y))
  , a__hd(X) -> hd(X)
  , a__hd(cons(X, Y)) -> mark(X)
  , mark(cons(X1, X2)) -> cons(X1, X2)
  , mark(0()) -> 0()
  , mark(zeros()) -> a__zeros()
  , mark(s(X)) -> s(X)
  , mark(incr(X)) -> a__incr(mark(X))
  , mark(adx(X)) -> a__adx(mark(X))
  , mark(nats()) -> a__nats()
  , mark(hd(X)) -> a__hd(mark(X))
  , mark(tl(X)) -> a__tl(mark(X))
  , a__tl(X) -> tl(X)
  , a__tl(cons(X, Y)) -> mark(Y) }
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^#(incr(X)) -> c_15(a__incr^#(mark(X)), mark^#(X))
  , mark^#(adx(X)) -> c_16(a__adx^#(mark(X)), mark^#(X)) }

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

Strict DPs:
  { a__hd^#(cons(X, Y)) -> c_1(mark^#(X))
  , mark^#(incr(X)) -> c_2(mark^#(X))
  , mark^#(adx(X)) -> c_3(mark^#(X))
  , mark^#(hd(X)) -> c_4(a__hd^#(mark(X)), mark^#(X))
  , mark^#(tl(X)) -> c_5(a__tl^#(mark(X)), mark^#(X))
  , a__tl^#(cons(X, Y)) -> c_6(mark^#(Y)) }
Weak Trs:
  { a__nats() -> a__adx(a__zeros())
  , a__nats() -> nats()
  , a__adx(X) -> adx(X)
  , a__adx(cons(X, Y)) -> a__incr(cons(X, adx(Y)))
  , a__zeros() -> cons(0(), zeros())
  , a__zeros() -> zeros()
  , a__incr(X) -> incr(X)
  , a__incr(cons(X, Y)) -> cons(s(X), incr(Y))
  , a__hd(X) -> hd(X)
  , a__hd(cons(X, Y)) -> mark(X)
  , mark(cons(X1, X2)) -> cons(X1, X2)
  , mark(0()) -> 0()
  , mark(zeros()) -> a__zeros()
  , mark(s(X)) -> s(X)
  , mark(incr(X)) -> a__incr(mark(X))
  , mark(adx(X)) -> a__adx(mark(X))
  , mark(nats()) -> a__nats()
  , mark(hd(X)) -> a__hd(mark(X))
  , mark(tl(X)) -> a__tl(mark(X))
  , a__tl(X) -> tl(X)
  , a__tl(cons(X, Y)) -> mark(Y) }
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..