MAYBE

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

Strict Trs:
  { a__incr(X) -> incr(X)
  , a__incr(nil()) -> nil()
  , a__incr(cons(X, L)) -> cons(s(mark(X)), incr(L))
  , mark(nil()) -> nil()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(s(X)) -> s(mark(X))
  , mark(incr(X)) -> a__incr(mark(X))
  , mark(adx(X)) -> a__adx(mark(X))
  , mark(0()) -> 0()
  , mark(zeros()) -> a__zeros()
  , mark(nats()) -> a__nats()
  , mark(head(X)) -> a__head(mark(X))
  , mark(tail(X)) -> a__tail(mark(X))
  , a__adx(X) -> adx(X)
  , a__adx(nil()) -> nil()
  , a__adx(cons(X, L)) -> a__incr(cons(mark(X), adx(L)))
  , a__nats() -> a__adx(a__zeros())
  , a__nats() -> nats()
  , a__zeros() -> cons(0(), zeros())
  , a__zeros() -> zeros()
  , a__head(X) -> head(X)
  , a__head(cons(X, L)) -> mark(X)
  , a__tail(X) -> tail(X)
  , a__tail(cons(X, L)) -> mark(L) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { a__incr^#(X) -> c_1()
  , a__incr^#(nil()) -> c_2()
  , a__incr^#(cons(X, L)) -> c_3(mark^#(X))
  , mark^#(nil()) -> c_4()
  , mark^#(cons(X1, X2)) -> c_5(mark^#(X1))
  , mark^#(s(X)) -> c_6(mark^#(X))
  , mark^#(incr(X)) -> c_7(a__incr^#(mark(X)), mark^#(X))
  , mark^#(adx(X)) -> c_8(a__adx^#(mark(X)), mark^#(X))
  , mark^#(0()) -> c_9()
  , mark^#(zeros()) -> c_10(a__zeros^#())
  , mark^#(nats()) -> c_11(a__nats^#())
  , mark^#(head(X)) -> c_12(a__head^#(mark(X)), mark^#(X))
  , mark^#(tail(X)) -> c_13(a__tail^#(mark(X)), mark^#(X))
  , a__adx^#(X) -> c_14()
  , a__adx^#(nil()) -> c_15()
  , a__adx^#(cons(X, L)) ->
    c_16(a__incr^#(cons(mark(X), adx(L))), mark^#(X))
  , a__zeros^#() -> c_19()
  , a__zeros^#() -> c_20()
  , a__nats^#() -> c_17(a__adx^#(a__zeros()), a__zeros^#())
  , a__nats^#() -> c_18()
  , a__head^#(X) -> c_21()
  , a__head^#(cons(X, L)) -> c_22(mark^#(X))
  , a__tail^#(X) -> c_23()
  , a__tail^#(cons(X, L)) -> c_24(mark^#(L)) }

and mark the set of starting terms.

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

Strict DPs:
  { a__incr^#(X) -> c_1()
  , a__incr^#(nil()) -> c_2()
  , a__incr^#(cons(X, L)) -> c_3(mark^#(X))
  , mark^#(nil()) -> c_4()
  , mark^#(cons(X1, X2)) -> c_5(mark^#(X1))
  , mark^#(s(X)) -> c_6(mark^#(X))
  , mark^#(incr(X)) -> c_7(a__incr^#(mark(X)), mark^#(X))
  , mark^#(adx(X)) -> c_8(a__adx^#(mark(X)), mark^#(X))
  , mark^#(0()) -> c_9()
  , mark^#(zeros()) -> c_10(a__zeros^#())
  , mark^#(nats()) -> c_11(a__nats^#())
  , mark^#(head(X)) -> c_12(a__head^#(mark(X)), mark^#(X))
  , mark^#(tail(X)) -> c_13(a__tail^#(mark(X)), mark^#(X))
  , a__adx^#(X) -> c_14()
  , a__adx^#(nil()) -> c_15()
  , a__adx^#(cons(X, L)) ->
    c_16(a__incr^#(cons(mark(X), adx(L))), mark^#(X))
  , a__zeros^#() -> c_19()
  , a__zeros^#() -> c_20()
  , a__nats^#() -> c_17(a__adx^#(a__zeros()), a__zeros^#())
  , a__nats^#() -> c_18()
  , a__head^#(X) -> c_21()
  , a__head^#(cons(X, L)) -> c_22(mark^#(X))
  , a__tail^#(X) -> c_23()
  , a__tail^#(cons(X, L)) -> c_24(mark^#(L)) }
Weak Trs:
  { a__incr(X) -> incr(X)
  , a__incr(nil()) -> nil()
  , a__incr(cons(X, L)) -> cons(s(mark(X)), incr(L))
  , mark(nil()) -> nil()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(s(X)) -> s(mark(X))
  , mark(incr(X)) -> a__incr(mark(X))
  , mark(adx(X)) -> a__adx(mark(X))
  , mark(0()) -> 0()
  , mark(zeros()) -> a__zeros()
  , mark(nats()) -> a__nats()
  , mark(head(X)) -> a__head(mark(X))
  , mark(tail(X)) -> a__tail(mark(X))
  , a__adx(X) -> adx(X)
  , a__adx(nil()) -> nil()
  , a__adx(cons(X, L)) -> a__incr(cons(mark(X), adx(L)))
  , a__nats() -> a__adx(a__zeros())
  , a__nats() -> nats()
  , a__zeros() -> cons(0(), zeros())
  , a__zeros() -> zeros()
  , a__head(X) -> head(X)
  , a__head(cons(X, L)) -> mark(X)
  , a__tail(X) -> tail(X)
  , a__tail(cons(X, L)) -> mark(L) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of
{1,2,4,9,14,15,17,18,20,21,23} by applications of
Pre({1,2,4,9,14,15,17,18,20,21,23}) =
{3,5,6,7,8,10,11,12,13,16,19,22,24}. Here rules are labeled as
follows:

  DPs:
    { 1: a__incr^#(X) -> c_1()
    , 2: a__incr^#(nil()) -> c_2()
    , 3: a__incr^#(cons(X, L)) -> c_3(mark^#(X))
    , 4: mark^#(nil()) -> c_4()
    , 5: mark^#(cons(X1, X2)) -> c_5(mark^#(X1))
    , 6: mark^#(s(X)) -> c_6(mark^#(X))
    , 7: mark^#(incr(X)) -> c_7(a__incr^#(mark(X)), mark^#(X))
    , 8: mark^#(adx(X)) -> c_8(a__adx^#(mark(X)), mark^#(X))
    , 9: mark^#(0()) -> c_9()
    , 10: mark^#(zeros()) -> c_10(a__zeros^#())
    , 11: mark^#(nats()) -> c_11(a__nats^#())
    , 12: mark^#(head(X)) -> c_12(a__head^#(mark(X)), mark^#(X))
    , 13: mark^#(tail(X)) -> c_13(a__tail^#(mark(X)), mark^#(X))
    , 14: a__adx^#(X) -> c_14()
    , 15: a__adx^#(nil()) -> c_15()
    , 16: a__adx^#(cons(X, L)) ->
          c_16(a__incr^#(cons(mark(X), adx(L))), mark^#(X))
    , 17: a__zeros^#() -> c_19()
    , 18: a__zeros^#() -> c_20()
    , 19: a__nats^#() -> c_17(a__adx^#(a__zeros()), a__zeros^#())
    , 20: a__nats^#() -> c_18()
    , 21: a__head^#(X) -> c_21()
    , 22: a__head^#(cons(X, L)) -> c_22(mark^#(X))
    , 23: a__tail^#(X) -> c_23()
    , 24: a__tail^#(cons(X, L)) -> c_24(mark^#(L)) }

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

Strict DPs:
  { a__incr^#(cons(X, L)) -> c_3(mark^#(X))
  , mark^#(cons(X1, X2)) -> c_5(mark^#(X1))
  , mark^#(s(X)) -> c_6(mark^#(X))
  , mark^#(incr(X)) -> c_7(a__incr^#(mark(X)), mark^#(X))
  , mark^#(adx(X)) -> c_8(a__adx^#(mark(X)), mark^#(X))
  , mark^#(zeros()) -> c_10(a__zeros^#())
  , mark^#(nats()) -> c_11(a__nats^#())
  , mark^#(head(X)) -> c_12(a__head^#(mark(X)), mark^#(X))
  , mark^#(tail(X)) -> c_13(a__tail^#(mark(X)), mark^#(X))
  , a__adx^#(cons(X, L)) ->
    c_16(a__incr^#(cons(mark(X), adx(L))), mark^#(X))
  , a__nats^#() -> c_17(a__adx^#(a__zeros()), a__zeros^#())
  , a__head^#(cons(X, L)) -> c_22(mark^#(X))
  , a__tail^#(cons(X, L)) -> c_24(mark^#(L)) }
Weak DPs:
  { a__incr^#(X) -> c_1()
  , a__incr^#(nil()) -> c_2()
  , mark^#(nil()) -> c_4()
  , mark^#(0()) -> c_9()
  , a__adx^#(X) -> c_14()
  , a__adx^#(nil()) -> c_15()
  , a__zeros^#() -> c_19()
  , a__zeros^#() -> c_20()
  , a__nats^#() -> c_18()
  , a__head^#(X) -> c_21()
  , a__tail^#(X) -> c_23() }
Weak Trs:
  { a__incr(X) -> incr(X)
  , a__incr(nil()) -> nil()
  , a__incr(cons(X, L)) -> cons(s(mark(X)), incr(L))
  , mark(nil()) -> nil()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(s(X)) -> s(mark(X))
  , mark(incr(X)) -> a__incr(mark(X))
  , mark(adx(X)) -> a__adx(mark(X))
  , mark(0()) -> 0()
  , mark(zeros()) -> a__zeros()
  , mark(nats()) -> a__nats()
  , mark(head(X)) -> a__head(mark(X))
  , mark(tail(X)) -> a__tail(mark(X))
  , a__adx(X) -> adx(X)
  , a__adx(nil()) -> nil()
  , a__adx(cons(X, L)) -> a__incr(cons(mark(X), adx(L)))
  , a__nats() -> a__adx(a__zeros())
  , a__nats() -> nats()
  , a__zeros() -> cons(0(), zeros())
  , a__zeros() -> zeros()
  , a__head(X) -> head(X)
  , a__head(cons(X, L)) -> mark(X)
  , a__tail(X) -> tail(X)
  , a__tail(cons(X, L)) -> mark(L) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: a__incr^#(cons(X, L)) -> c_3(mark^#(X))
    , 2: mark^#(cons(X1, X2)) -> c_5(mark^#(X1))
    , 3: mark^#(s(X)) -> c_6(mark^#(X))
    , 4: mark^#(incr(X)) -> c_7(a__incr^#(mark(X)), mark^#(X))
    , 5: mark^#(adx(X)) -> c_8(a__adx^#(mark(X)), mark^#(X))
    , 6: mark^#(zeros()) -> c_10(a__zeros^#())
    , 7: mark^#(nats()) -> c_11(a__nats^#())
    , 8: mark^#(head(X)) -> c_12(a__head^#(mark(X)), mark^#(X))
    , 9: mark^#(tail(X)) -> c_13(a__tail^#(mark(X)), mark^#(X))
    , 10: a__adx^#(cons(X, L)) ->
          c_16(a__incr^#(cons(mark(X), adx(L))), mark^#(X))
    , 11: a__nats^#() -> c_17(a__adx^#(a__zeros()), a__zeros^#())
    , 12: a__head^#(cons(X, L)) -> c_22(mark^#(X))
    , 13: a__tail^#(cons(X, L)) -> c_24(mark^#(L))
    , 14: a__incr^#(X) -> c_1()
    , 15: a__incr^#(nil()) -> c_2()
    , 16: mark^#(nil()) -> c_4()
    , 17: mark^#(0()) -> c_9()
    , 18: a__adx^#(X) -> c_14()
    , 19: a__adx^#(nil()) -> c_15()
    , 20: a__zeros^#() -> c_19()
    , 21: a__zeros^#() -> c_20()
    , 22: a__nats^#() -> c_18()
    , 23: a__head^#(X) -> c_21()
    , 24: a__tail^#(X) -> c_23() }

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

Strict DPs:
  { a__incr^#(cons(X, L)) -> c_3(mark^#(X))
  , mark^#(cons(X1, X2)) -> c_5(mark^#(X1))
  , mark^#(s(X)) -> c_6(mark^#(X))
  , mark^#(incr(X)) -> c_7(a__incr^#(mark(X)), mark^#(X))
  , mark^#(adx(X)) -> c_8(a__adx^#(mark(X)), mark^#(X))
  , mark^#(nats()) -> c_11(a__nats^#())
  , mark^#(head(X)) -> c_12(a__head^#(mark(X)), mark^#(X))
  , mark^#(tail(X)) -> c_13(a__tail^#(mark(X)), mark^#(X))
  , a__adx^#(cons(X, L)) ->
    c_16(a__incr^#(cons(mark(X), adx(L))), mark^#(X))
  , a__nats^#() -> c_17(a__adx^#(a__zeros()), a__zeros^#())
  , a__head^#(cons(X, L)) -> c_22(mark^#(X))
  , a__tail^#(cons(X, L)) -> c_24(mark^#(L)) }
Weak DPs:
  { a__incr^#(X) -> c_1()
  , a__incr^#(nil()) -> c_2()
  , mark^#(nil()) -> c_4()
  , mark^#(0()) -> c_9()
  , mark^#(zeros()) -> c_10(a__zeros^#())
  , a__adx^#(X) -> c_14()
  , a__adx^#(nil()) -> c_15()
  , a__zeros^#() -> c_19()
  , a__zeros^#() -> c_20()
  , a__nats^#() -> c_18()
  , a__head^#(X) -> c_21()
  , a__tail^#(X) -> c_23() }
Weak Trs:
  { a__incr(X) -> incr(X)
  , a__incr(nil()) -> nil()
  , a__incr(cons(X, L)) -> cons(s(mark(X)), incr(L))
  , mark(nil()) -> nil()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(s(X)) -> s(mark(X))
  , mark(incr(X)) -> a__incr(mark(X))
  , mark(adx(X)) -> a__adx(mark(X))
  , mark(0()) -> 0()
  , mark(zeros()) -> a__zeros()
  , mark(nats()) -> a__nats()
  , mark(head(X)) -> a__head(mark(X))
  , mark(tail(X)) -> a__tail(mark(X))
  , a__adx(X) -> adx(X)
  , a__adx(nil()) -> nil()
  , a__adx(cons(X, L)) -> a__incr(cons(mark(X), adx(L)))
  , a__nats() -> a__adx(a__zeros())
  , a__nats() -> nats()
  , a__zeros() -> cons(0(), zeros())
  , a__zeros() -> zeros()
  , a__head(X) -> head(X)
  , a__head(cons(X, L)) -> mark(X)
  , a__tail(X) -> tail(X)
  , a__tail(cons(X, L)) -> mark(L) }
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__incr^#(X) -> c_1()
, a__incr^#(nil()) -> c_2()
, mark^#(nil()) -> c_4()
, mark^#(0()) -> c_9()
, mark^#(zeros()) -> c_10(a__zeros^#())
, a__adx^#(X) -> c_14()
, a__adx^#(nil()) -> c_15()
, a__zeros^#() -> c_19()
, a__zeros^#() -> c_20()
, a__nats^#() -> c_18()
, a__head^#(X) -> c_21()
, a__tail^#(X) -> c_23() }

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

Strict DPs:
  { a__incr^#(cons(X, L)) -> c_3(mark^#(X))
  , mark^#(cons(X1, X2)) -> c_5(mark^#(X1))
  , mark^#(s(X)) -> c_6(mark^#(X))
  , mark^#(incr(X)) -> c_7(a__incr^#(mark(X)), mark^#(X))
  , mark^#(adx(X)) -> c_8(a__adx^#(mark(X)), mark^#(X))
  , mark^#(nats()) -> c_11(a__nats^#())
  , mark^#(head(X)) -> c_12(a__head^#(mark(X)), mark^#(X))
  , mark^#(tail(X)) -> c_13(a__tail^#(mark(X)), mark^#(X))
  , a__adx^#(cons(X, L)) ->
    c_16(a__incr^#(cons(mark(X), adx(L))), mark^#(X))
  , a__nats^#() -> c_17(a__adx^#(a__zeros()), a__zeros^#())
  , a__head^#(cons(X, L)) -> c_22(mark^#(X))
  , a__tail^#(cons(X, L)) -> c_24(mark^#(L)) }
Weak Trs:
  { a__incr(X) -> incr(X)
  , a__incr(nil()) -> nil()
  , a__incr(cons(X, L)) -> cons(s(mark(X)), incr(L))
  , mark(nil()) -> nil()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(s(X)) -> s(mark(X))
  , mark(incr(X)) -> a__incr(mark(X))
  , mark(adx(X)) -> a__adx(mark(X))
  , mark(0()) -> 0()
  , mark(zeros()) -> a__zeros()
  , mark(nats()) -> a__nats()
  , mark(head(X)) -> a__head(mark(X))
  , mark(tail(X)) -> a__tail(mark(X))
  , a__adx(X) -> adx(X)
  , a__adx(nil()) -> nil()
  , a__adx(cons(X, L)) -> a__incr(cons(mark(X), adx(L)))
  , a__nats() -> a__adx(a__zeros())
  , a__nats() -> nats()
  , a__zeros() -> cons(0(), zeros())
  , a__zeros() -> zeros()
  , a__head(X) -> head(X)
  , a__head(cons(X, L)) -> mark(X)
  , a__tail(X) -> tail(X)
  , a__tail(cons(X, L)) -> mark(L) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  { a__nats^#() -> c_17(a__adx^#(a__zeros()), a__zeros^#()) }

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

Strict DPs:
  { a__incr^#(cons(X, L)) -> c_1(mark^#(X))
  , mark^#(cons(X1, X2)) -> c_2(mark^#(X1))
  , mark^#(s(X)) -> c_3(mark^#(X))
  , mark^#(incr(X)) -> c_4(a__incr^#(mark(X)), mark^#(X))
  , mark^#(adx(X)) -> c_5(a__adx^#(mark(X)), mark^#(X))
  , mark^#(nats()) -> c_6(a__nats^#())
  , mark^#(head(X)) -> c_7(a__head^#(mark(X)), mark^#(X))
  , mark^#(tail(X)) -> c_8(a__tail^#(mark(X)), mark^#(X))
  , a__adx^#(cons(X, L)) ->
    c_9(a__incr^#(cons(mark(X), adx(L))), mark^#(X))
  , a__nats^#() -> c_10(a__adx^#(a__zeros()))
  , a__head^#(cons(X, L)) -> c_11(mark^#(X))
  , a__tail^#(cons(X, L)) -> c_12(mark^#(L)) }
Weak Trs:
  { a__incr(X) -> incr(X)
  , a__incr(nil()) -> nil()
  , a__incr(cons(X, L)) -> cons(s(mark(X)), incr(L))
  , mark(nil()) -> nil()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(s(X)) -> s(mark(X))
  , mark(incr(X)) -> a__incr(mark(X))
  , mark(adx(X)) -> a__adx(mark(X))
  , mark(0()) -> 0()
  , mark(zeros()) -> a__zeros()
  , mark(nats()) -> a__nats()
  , mark(head(X)) -> a__head(mark(X))
  , mark(tail(X)) -> a__tail(mark(X))
  , a__adx(X) -> adx(X)
  , a__adx(nil()) -> nil()
  , a__adx(cons(X, L)) -> a__incr(cons(mark(X), adx(L)))
  , a__nats() -> a__adx(a__zeros())
  , a__nats() -> nats()
  , a__zeros() -> cons(0(), zeros())
  , a__zeros() -> zeros()
  , a__head(X) -> head(X)
  , a__head(cons(X, L)) -> mark(X)
  , a__tail(X) -> tail(X)
  , a__tail(cons(X, L)) -> mark(L) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..