MAYBE

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

Strict Trs:
  { numbers() -> d(0())
  , d(x) -> if(le(x, nr()), x)
  , if(true(), x) -> cons(x, d(s(x)))
  , if(false(), x) -> nil()
  , le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , nr() -> ack(s(s(s(s(s(s(0())))))), 0())
  , ack(0(), x) -> s(x)
  , ack(s(x), 0()) -> ack(x, s(0()))
  , ack(s(x), s(y)) -> ack(x, ack(s(x), y)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { numbers^#() -> c_1(d^#(0()))
  , d^#(x) -> c_2(if^#(le(x, nr()), x), le^#(x, nr()), nr^#())
  , if^#(true(), x) -> c_3(d^#(s(x)))
  , if^#(false(), x) -> c_4()
  , le^#(0(), y) -> c_5()
  , le^#(s(x), 0()) -> c_6()
  , le^#(s(x), s(y)) -> c_7(le^#(x, y))
  , nr^#() -> c_8(ack^#(s(s(s(s(s(s(0())))))), 0()))
  , ack^#(0(), x) -> c_9()
  , ack^#(s(x), 0()) -> c_10(ack^#(x, s(0())))
  , ack^#(s(x), s(y)) ->
    c_11(ack^#(x, ack(s(x), y)), ack^#(s(x), y)) }

and mark the set of starting terms.

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

Strict DPs:
  { numbers^#() -> c_1(d^#(0()))
  , d^#(x) -> c_2(if^#(le(x, nr()), x), le^#(x, nr()), nr^#())
  , if^#(true(), x) -> c_3(d^#(s(x)))
  , if^#(false(), x) -> c_4()
  , le^#(0(), y) -> c_5()
  , le^#(s(x), 0()) -> c_6()
  , le^#(s(x), s(y)) -> c_7(le^#(x, y))
  , nr^#() -> c_8(ack^#(s(s(s(s(s(s(0())))))), 0()))
  , ack^#(0(), x) -> c_9()
  , ack^#(s(x), 0()) -> c_10(ack^#(x, s(0())))
  , ack^#(s(x), s(y)) ->
    c_11(ack^#(x, ack(s(x), y)), ack^#(s(x), y)) }
Weak Trs:
  { numbers() -> d(0())
  , d(x) -> if(le(x, nr()), x)
  , if(true(), x) -> cons(x, d(s(x)))
  , if(false(), x) -> nil()
  , le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , nr() -> ack(s(s(s(s(s(s(0())))))), 0())
  , ack(0(), x) -> s(x)
  , ack(s(x), 0()) -> ack(x, s(0()))
  , ack(s(x), s(y)) -> ack(x, ack(s(x), y)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of {4,5,6,9} by applications
of Pre({4,5,6,9}) = {2,7,10,11}. Here rules are labeled as follows:

  DPs:
    { 1: numbers^#() -> c_1(d^#(0()))
    , 2: d^#(x) -> c_2(if^#(le(x, nr()), x), le^#(x, nr()), nr^#())
    , 3: if^#(true(), x) -> c_3(d^#(s(x)))
    , 4: if^#(false(), x) -> c_4()
    , 5: le^#(0(), y) -> c_5()
    , 6: le^#(s(x), 0()) -> c_6()
    , 7: le^#(s(x), s(y)) -> c_7(le^#(x, y))
    , 8: nr^#() -> c_8(ack^#(s(s(s(s(s(s(0())))))), 0()))
    , 9: ack^#(0(), x) -> c_9()
    , 10: ack^#(s(x), 0()) -> c_10(ack^#(x, s(0())))
    , 11: ack^#(s(x), s(y)) ->
          c_11(ack^#(x, ack(s(x), y)), ack^#(s(x), y)) }

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

Strict DPs:
  { numbers^#() -> c_1(d^#(0()))
  , d^#(x) -> c_2(if^#(le(x, nr()), x), le^#(x, nr()), nr^#())
  , if^#(true(), x) -> c_3(d^#(s(x)))
  , le^#(s(x), s(y)) -> c_7(le^#(x, y))
  , nr^#() -> c_8(ack^#(s(s(s(s(s(s(0())))))), 0()))
  , ack^#(s(x), 0()) -> c_10(ack^#(x, s(0())))
  , ack^#(s(x), s(y)) ->
    c_11(ack^#(x, ack(s(x), y)), ack^#(s(x), y)) }
Weak DPs:
  { if^#(false(), x) -> c_4()
  , le^#(0(), y) -> c_5()
  , le^#(s(x), 0()) -> c_6()
  , ack^#(0(), x) -> c_9() }
Weak Trs:
  { numbers() -> d(0())
  , d(x) -> if(le(x, nr()), x)
  , if(true(), x) -> cons(x, d(s(x)))
  , if(false(), x) -> nil()
  , le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , nr() -> ack(s(s(s(s(s(s(0())))))), 0())
  , ack(0(), x) -> s(x)
  , ack(s(x), 0()) -> ack(x, s(0()))
  , ack(s(x), s(y)) -> ack(x, ack(s(x), 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.

{ if^#(false(), x) -> c_4()
, le^#(0(), y) -> c_5()
, le^#(s(x), 0()) -> c_6()
, ack^#(0(), x) -> c_9() }

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

Strict DPs:
  { numbers^#() -> c_1(d^#(0()))
  , d^#(x) -> c_2(if^#(le(x, nr()), x), le^#(x, nr()), nr^#())
  , if^#(true(), x) -> c_3(d^#(s(x)))
  , le^#(s(x), s(y)) -> c_7(le^#(x, y))
  , nr^#() -> c_8(ack^#(s(s(s(s(s(s(0())))))), 0()))
  , ack^#(s(x), 0()) -> c_10(ack^#(x, s(0())))
  , ack^#(s(x), s(y)) ->
    c_11(ack^#(x, ack(s(x), y)), ack^#(s(x), y)) }
Weak Trs:
  { numbers() -> d(0())
  , d(x) -> if(le(x, nr()), x)
  , if(true(), x) -> cons(x, d(s(x)))
  , if(false(), x) -> nil()
  , le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , nr() -> ack(s(s(s(s(s(s(0())))))), 0())
  , ack(0(), x) -> s(x)
  , ack(s(x), 0()) -> ack(x, s(0()))
  , ack(s(x), s(y)) -> ack(x, ack(s(x), y)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Consider the dependency graph

  1: numbers^#() -> c_1(d^#(0()))
     -->_1 d^#(x) -> c_2(if^#(le(x, nr()), x), le^#(x, nr()), nr^#()) :2
  
  2: d^#(x) -> c_2(if^#(le(x, nr()), x), le^#(x, nr()), nr^#())
     -->_3 nr^#() -> c_8(ack^#(s(s(s(s(s(s(0())))))), 0())) :5
     -->_2 le^#(s(x), s(y)) -> c_7(le^#(x, y)) :4
     -->_1 if^#(true(), x) -> c_3(d^#(s(x))) :3
  
  3: if^#(true(), x) -> c_3(d^#(s(x)))
     -->_1 d^#(x) -> c_2(if^#(le(x, nr()), x), le^#(x, nr()), nr^#()) :2
  
  4: le^#(s(x), s(y)) -> c_7(le^#(x, y))
     -->_1 le^#(s(x), s(y)) -> c_7(le^#(x, y)) :4
  
  5: nr^#() -> c_8(ack^#(s(s(s(s(s(s(0())))))), 0()))
     -->_1 ack^#(s(x), 0()) -> c_10(ack^#(x, s(0()))) :6
  
  6: ack^#(s(x), 0()) -> c_10(ack^#(x, s(0())))
     -->_1 ack^#(s(x), s(y)) ->
           c_11(ack^#(x, ack(s(x), y)), ack^#(s(x), y)) :7
  
  7: ack^#(s(x), s(y)) ->
     c_11(ack^#(x, ack(s(x), y)), ack^#(s(x), y))
     -->_2 ack^#(s(x), s(y)) ->
           c_11(ack^#(x, ack(s(x), y)), ack^#(s(x), y)) :7
     -->_1 ack^#(s(x), s(y)) ->
           c_11(ack^#(x, ack(s(x), y)), ack^#(s(x), y)) :7
     -->_2 ack^#(s(x), 0()) -> c_10(ack^#(x, s(0()))) :6
  

Following roots of the dependency graph are removed, as the
considered set of starting terms is closed under reduction with
respect to these rules (modulo compound contexts).

  { numbers^#() -> c_1(d^#(0())) }


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

Strict DPs:
  { d^#(x) -> c_2(if^#(le(x, nr()), x), le^#(x, nr()), nr^#())
  , if^#(true(), x) -> c_3(d^#(s(x)))
  , le^#(s(x), s(y)) -> c_7(le^#(x, y))
  , nr^#() -> c_8(ack^#(s(s(s(s(s(s(0())))))), 0()))
  , ack^#(s(x), 0()) -> c_10(ack^#(x, s(0())))
  , ack^#(s(x), s(y)) ->
    c_11(ack^#(x, ack(s(x), y)), ack^#(s(x), y)) }
Weak Trs:
  { numbers() -> d(0())
  , d(x) -> if(le(x, nr()), x)
  , if(true(), x) -> cons(x, d(s(x)))
  , if(false(), x) -> nil()
  , le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , nr() -> ack(s(s(s(s(s(s(0())))))), 0())
  , ack(0(), x) -> s(x)
  , ack(s(x), 0()) -> ack(x, s(0()))
  , ack(s(x), s(y)) -> ack(x, ack(s(x), y)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { le(0(), y) -> true()
    , le(s(x), 0()) -> false()
    , le(s(x), s(y)) -> le(x, y)
    , nr() -> ack(s(s(s(s(s(s(0())))))), 0())
    , ack(0(), x) -> s(x)
    , ack(s(x), 0()) -> ack(x, s(0()))
    , ack(s(x), s(y)) -> ack(x, ack(s(x), y)) }

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

Strict DPs:
  { d^#(x) -> c_2(if^#(le(x, nr()), x), le^#(x, nr()), nr^#())
  , if^#(true(), x) -> c_3(d^#(s(x)))
  , le^#(s(x), s(y)) -> c_7(le^#(x, y))
  , nr^#() -> c_8(ack^#(s(s(s(s(s(s(0())))))), 0()))
  , ack^#(s(x), 0()) -> c_10(ack^#(x, s(0())))
  , ack^#(s(x), s(y)) ->
    c_11(ack^#(x, ack(s(x), y)), ack^#(s(x), y)) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , nr() -> ack(s(s(s(s(s(s(0())))))), 0())
  , ack(0(), x) -> s(x)
  , ack(s(x), 0()) -> ack(x, s(0()))
  , ack(s(x), s(y)) -> ack(x, ack(s(x), y)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..