MAYBE

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

Strict Trs:
  { cond1(true(), x) -> cond2(even(x), x)
  , cond2(true(), x) -> cond1(neq(x, 0()), div2(x))
  , cond2(false(), x) -> cond1(neq(x, 0()), p(x))
  , even(0()) -> true()
  , even(s(0())) -> false()
  , even(s(s(x))) -> even(x)
  , neq(0(), 0()) -> false()
  , neq(0(), s(x)) -> true()
  , neq(s(x), 0()) -> true()
  , neq(s(x), s(y())) -> neq(x, y())
  , div2(0()) -> 0()
  , div2(s(0())) -> 0()
  , div2(s(s(x))) -> s(div2(x))
  , p(0()) -> 0()
  , p(s(x)) -> x }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { cond1^#(true(), x) -> c_1(cond2^#(even(x), x), even^#(x))
  , cond2^#(true(), x) ->
    c_2(cond1^#(neq(x, 0()), div2(x)), neq^#(x, 0()), div2^#(x))
  , cond2^#(false(), x) ->
    c_3(cond1^#(neq(x, 0()), p(x)), neq^#(x, 0()), p^#(x))
  , even^#(0()) -> c_4()
  , even^#(s(0())) -> c_5()
  , even^#(s(s(x))) -> c_6(even^#(x))
  , neq^#(0(), 0()) -> c_7()
  , neq^#(0(), s(x)) -> c_8()
  , neq^#(s(x), 0()) -> c_9()
  , neq^#(s(x), s(y())) -> c_10(neq^#(x, y()))
  , div2^#(0()) -> c_11()
  , div2^#(s(0())) -> c_12()
  , div2^#(s(s(x))) -> c_13(div2^#(x))
  , p^#(0()) -> c_14()
  , p^#(s(x)) -> c_15() }

and mark the set of starting terms.

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

Strict DPs:
  { cond1^#(true(), x) -> c_1(cond2^#(even(x), x), even^#(x))
  , cond2^#(true(), x) ->
    c_2(cond1^#(neq(x, 0()), div2(x)), neq^#(x, 0()), div2^#(x))
  , cond2^#(false(), x) ->
    c_3(cond1^#(neq(x, 0()), p(x)), neq^#(x, 0()), p^#(x))
  , even^#(0()) -> c_4()
  , even^#(s(0())) -> c_5()
  , even^#(s(s(x))) -> c_6(even^#(x))
  , neq^#(0(), 0()) -> c_7()
  , neq^#(0(), s(x)) -> c_8()
  , neq^#(s(x), 0()) -> c_9()
  , neq^#(s(x), s(y())) -> c_10(neq^#(x, y()))
  , div2^#(0()) -> c_11()
  , div2^#(s(0())) -> c_12()
  , div2^#(s(s(x))) -> c_13(div2^#(x))
  , p^#(0()) -> c_14()
  , p^#(s(x)) -> c_15() }
Weak Trs:
  { cond1(true(), x) -> cond2(even(x), x)
  , cond2(true(), x) -> cond1(neq(x, 0()), div2(x))
  , cond2(false(), x) -> cond1(neq(x, 0()), p(x))
  , even(0()) -> true()
  , even(s(0())) -> false()
  , even(s(s(x))) -> even(x)
  , neq(0(), 0()) -> false()
  , neq(0(), s(x)) -> true()
  , neq(s(x), 0()) -> true()
  , neq(s(x), s(y())) -> neq(x, y())
  , div2(0()) -> 0()
  , div2(s(0())) -> 0()
  , div2(s(s(x))) -> s(div2(x))
  , p(0()) -> 0()
  , p(s(x)) -> x }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: cond1^#(true(), x) -> c_1(cond2^#(even(x), x), even^#(x))
    , 2: cond2^#(true(), x) ->
         c_2(cond1^#(neq(x, 0()), div2(x)), neq^#(x, 0()), div2^#(x))
    , 3: cond2^#(false(), x) ->
         c_3(cond1^#(neq(x, 0()), p(x)), neq^#(x, 0()), p^#(x))
    , 4: even^#(0()) -> c_4()
    , 5: even^#(s(0())) -> c_5()
    , 6: even^#(s(s(x))) -> c_6(even^#(x))
    , 7: neq^#(0(), 0()) -> c_7()
    , 8: neq^#(0(), s(x)) -> c_8()
    , 9: neq^#(s(x), 0()) -> c_9()
    , 10: neq^#(s(x), s(y())) -> c_10(neq^#(x, y()))
    , 11: div2^#(0()) -> c_11()
    , 12: div2^#(s(0())) -> c_12()
    , 13: div2^#(s(s(x))) -> c_13(div2^#(x))
    , 14: p^#(0()) -> c_14()
    , 15: p^#(s(x)) -> c_15() }

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

Strict DPs:
  { cond1^#(true(), x) -> c_1(cond2^#(even(x), x), even^#(x))
  , cond2^#(true(), x) ->
    c_2(cond1^#(neq(x, 0()), div2(x)), neq^#(x, 0()), div2^#(x))
  , cond2^#(false(), x) ->
    c_3(cond1^#(neq(x, 0()), p(x)), neq^#(x, 0()), p^#(x))
  , even^#(s(s(x))) -> c_6(even^#(x))
  , div2^#(s(s(x))) -> c_13(div2^#(x)) }
Weak DPs:
  { even^#(0()) -> c_4()
  , even^#(s(0())) -> c_5()
  , neq^#(0(), 0()) -> c_7()
  , neq^#(0(), s(x)) -> c_8()
  , neq^#(s(x), 0()) -> c_9()
  , neq^#(s(x), s(y())) -> c_10(neq^#(x, y()))
  , div2^#(0()) -> c_11()
  , div2^#(s(0())) -> c_12()
  , p^#(0()) -> c_14()
  , p^#(s(x)) -> c_15() }
Weak Trs:
  { cond1(true(), x) -> cond2(even(x), x)
  , cond2(true(), x) -> cond1(neq(x, 0()), div2(x))
  , cond2(false(), x) -> cond1(neq(x, 0()), p(x))
  , even(0()) -> true()
  , even(s(0())) -> false()
  , even(s(s(x))) -> even(x)
  , neq(0(), 0()) -> false()
  , neq(0(), s(x)) -> true()
  , neq(s(x), 0()) -> true()
  , neq(s(x), s(y())) -> neq(x, y())
  , div2(0()) -> 0()
  , div2(s(0())) -> 0()
  , div2(s(s(x))) -> s(div2(x))
  , p(0()) -> 0()
  , p(s(x)) -> 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.

{ even^#(0()) -> c_4()
, even^#(s(0())) -> c_5()
, neq^#(0(), 0()) -> c_7()
, neq^#(0(), s(x)) -> c_8()
, neq^#(s(x), 0()) -> c_9()
, neq^#(s(x), s(y())) -> c_10(neq^#(x, y()))
, div2^#(0()) -> c_11()
, div2^#(s(0())) -> c_12()
, p^#(0()) -> c_14()
, p^#(s(x)) -> c_15() }

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

Strict DPs:
  { cond1^#(true(), x) -> c_1(cond2^#(even(x), x), even^#(x))
  , cond2^#(true(), x) ->
    c_2(cond1^#(neq(x, 0()), div2(x)), neq^#(x, 0()), div2^#(x))
  , cond2^#(false(), x) ->
    c_3(cond1^#(neq(x, 0()), p(x)), neq^#(x, 0()), p^#(x))
  , even^#(s(s(x))) -> c_6(even^#(x))
  , div2^#(s(s(x))) -> c_13(div2^#(x)) }
Weak Trs:
  { cond1(true(), x) -> cond2(even(x), x)
  , cond2(true(), x) -> cond1(neq(x, 0()), div2(x))
  , cond2(false(), x) -> cond1(neq(x, 0()), p(x))
  , even(0()) -> true()
  , even(s(0())) -> false()
  , even(s(s(x))) -> even(x)
  , neq(0(), 0()) -> false()
  , neq(0(), s(x)) -> true()
  , neq(s(x), 0()) -> true()
  , neq(s(x), s(y())) -> neq(x, y())
  , div2(0()) -> 0()
  , div2(s(0())) -> 0()
  , div2(s(s(x))) -> s(div2(x))
  , p(0()) -> 0()
  , p(s(x)) -> 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:

  { cond2^#(true(), x) ->
    c_2(cond1^#(neq(x, 0()), div2(x)), neq^#(x, 0()), div2^#(x))
  , cond2^#(false(), x) ->
    c_3(cond1^#(neq(x, 0()), p(x)), neq^#(x, 0()), p^#(x)) }

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

Strict DPs:
  { cond1^#(true(), x) -> c_1(cond2^#(even(x), x), even^#(x))
  , cond2^#(true(), x) ->
    c_2(cond1^#(neq(x, 0()), div2(x)), div2^#(x))
  , cond2^#(false(), x) -> c_3(cond1^#(neq(x, 0()), p(x)))
  , even^#(s(s(x))) -> c_4(even^#(x))
  , div2^#(s(s(x))) -> c_5(div2^#(x)) }
Weak Trs:
  { cond1(true(), x) -> cond2(even(x), x)
  , cond2(true(), x) -> cond1(neq(x, 0()), div2(x))
  , cond2(false(), x) -> cond1(neq(x, 0()), p(x))
  , even(0()) -> true()
  , even(s(0())) -> false()
  , even(s(s(x))) -> even(x)
  , neq(0(), 0()) -> false()
  , neq(0(), s(x)) -> true()
  , neq(s(x), 0()) -> true()
  , neq(s(x), s(y())) -> neq(x, y())
  , div2(0()) -> 0()
  , div2(s(0())) -> 0()
  , div2(s(s(x))) -> s(div2(x))
  , p(0()) -> 0()
  , p(s(x)) -> x }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { even(0()) -> true()
    , even(s(0())) -> false()
    , even(s(s(x))) -> even(x)
    , neq(0(), 0()) -> false()
    , neq(0(), s(x)) -> true()
    , neq(s(x), 0()) -> true()
    , neq(s(x), s(y())) -> neq(x, y())
    , div2(0()) -> 0()
    , div2(s(0())) -> 0()
    , div2(s(s(x))) -> s(div2(x))
    , p(0()) -> 0()
    , p(s(x)) -> x }

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

Strict DPs:
  { cond1^#(true(), x) -> c_1(cond2^#(even(x), x), even^#(x))
  , cond2^#(true(), x) ->
    c_2(cond1^#(neq(x, 0()), div2(x)), div2^#(x))
  , cond2^#(false(), x) -> c_3(cond1^#(neq(x, 0()), p(x)))
  , even^#(s(s(x))) -> c_4(even^#(x))
  , div2^#(s(s(x))) -> c_5(div2^#(x)) }
Weak Trs:
  { even(0()) -> true()
  , even(s(0())) -> false()
  , even(s(s(x))) -> even(x)
  , neq(0(), 0()) -> false()
  , neq(0(), s(x)) -> true()
  , neq(s(x), 0()) -> true()
  , neq(s(x), s(y())) -> neq(x, y())
  , div2(0()) -> 0()
  , div2(s(0())) -> 0()
  , div2(s(s(x))) -> s(div2(x))
  , p(0()) -> 0()
  , p(s(x)) -> x }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..