MAYBE

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

Strict Trs:
  { cond1(true(), x, y, z) -> cond2(gr(x, 0()), x, y, z)
  , cond2(true(), x, y, z) -> cond1(gr(add(x, y), z), p(x), y, z)
  , cond2(false(), x, y, z) -> cond3(gr(y, 0()), x, y, z)
  , gr(0(), x) -> false()
  , gr(s(x), 0()) -> true()
  , gr(s(x), s(y)) -> gr(x, y)
  , add(0(), x) -> x
  , add(s(x), y) -> s(add(x, y))
  , p(0()) -> 0()
  , p(s(x)) -> x
  , cond3(true(), x, y, z) -> cond1(gr(add(x, y), z), x, p(y), z)
  , cond3(false(), x, y, z) -> cond1(gr(add(x, y), z), x, y, z) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { cond1^#(true(), x, y, z) ->
    c_1(cond2^#(gr(x, 0()), x, y, z), gr^#(x, 0()))
  , cond2^#(true(), x, y, z) ->
    c_2(cond1^#(gr(add(x, y), z), p(x), y, z),
        gr^#(add(x, y), z),
        add^#(x, y),
        p^#(x))
  , cond2^#(false(), x, y, z) ->
    c_3(cond3^#(gr(y, 0()), x, y, z), gr^#(y, 0()))
  , gr^#(0(), x) -> c_4()
  , gr^#(s(x), 0()) -> c_5()
  , gr^#(s(x), s(y)) -> c_6(gr^#(x, y))
  , add^#(0(), x) -> c_7()
  , add^#(s(x), y) -> c_8(add^#(x, y))
  , p^#(0()) -> c_9()
  , p^#(s(x)) -> c_10()
  , cond3^#(true(), x, y, z) ->
    c_11(cond1^#(gr(add(x, y), z), x, p(y), z),
         gr^#(add(x, y), z),
         add^#(x, y),
         p^#(y))
  , cond3^#(false(), x, y, z) ->
    c_12(cond1^#(gr(add(x, y), z), x, y, z),
         gr^#(add(x, y), z),
         add^#(x, y)) }

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, y, z) ->
    c_1(cond2^#(gr(x, 0()), x, y, z), gr^#(x, 0()))
  , cond2^#(true(), x, y, z) ->
    c_2(cond1^#(gr(add(x, y), z), p(x), y, z),
        gr^#(add(x, y), z),
        add^#(x, y),
        p^#(x))
  , cond2^#(false(), x, y, z) ->
    c_3(cond3^#(gr(y, 0()), x, y, z), gr^#(y, 0()))
  , gr^#(0(), x) -> c_4()
  , gr^#(s(x), 0()) -> c_5()
  , gr^#(s(x), s(y)) -> c_6(gr^#(x, y))
  , add^#(0(), x) -> c_7()
  , add^#(s(x), y) -> c_8(add^#(x, y))
  , p^#(0()) -> c_9()
  , p^#(s(x)) -> c_10()
  , cond3^#(true(), x, y, z) ->
    c_11(cond1^#(gr(add(x, y), z), x, p(y), z),
         gr^#(add(x, y), z),
         add^#(x, y),
         p^#(y))
  , cond3^#(false(), x, y, z) ->
    c_12(cond1^#(gr(add(x, y), z), x, y, z),
         gr^#(add(x, y), z),
         add^#(x, y)) }
Weak Trs:
  { cond1(true(), x, y, z) -> cond2(gr(x, 0()), x, y, z)
  , cond2(true(), x, y, z) -> cond1(gr(add(x, y), z), p(x), y, z)
  , cond2(false(), x, y, z) -> cond3(gr(y, 0()), x, y, z)
  , gr(0(), x) -> false()
  , gr(s(x), 0()) -> true()
  , gr(s(x), s(y)) -> gr(x, y)
  , add(0(), x) -> x
  , add(s(x), y) -> s(add(x, y))
  , p(0()) -> 0()
  , p(s(x)) -> x
  , cond3(true(), x, y, z) -> cond1(gr(add(x, y), z), x, p(y), z)
  , cond3(false(), x, y, z) -> cond1(gr(add(x, y), z), x, y, z) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: cond1^#(true(), x, y, z) ->
         c_1(cond2^#(gr(x, 0()), x, y, z), gr^#(x, 0()))
    , 2: cond2^#(true(), x, y, z) ->
         c_2(cond1^#(gr(add(x, y), z), p(x), y, z),
             gr^#(add(x, y), z),
             add^#(x, y),
             p^#(x))
    , 3: cond2^#(false(), x, y, z) ->
         c_3(cond3^#(gr(y, 0()), x, y, z), gr^#(y, 0()))
    , 4: gr^#(0(), x) -> c_4()
    , 5: gr^#(s(x), 0()) -> c_5()
    , 6: gr^#(s(x), s(y)) -> c_6(gr^#(x, y))
    , 7: add^#(0(), x) -> c_7()
    , 8: add^#(s(x), y) -> c_8(add^#(x, y))
    , 9: p^#(0()) -> c_9()
    , 10: p^#(s(x)) -> c_10()
    , 11: cond3^#(true(), x, y, z) ->
          c_11(cond1^#(gr(add(x, y), z), x, p(y), z),
               gr^#(add(x, y), z),
               add^#(x, y),
               p^#(y))
    , 12: cond3^#(false(), x, y, z) ->
          c_12(cond1^#(gr(add(x, y), z), x, y, z),
               gr^#(add(x, y), z),
               add^#(x, y)) }

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

Strict DPs:
  { cond1^#(true(), x, y, z) ->
    c_1(cond2^#(gr(x, 0()), x, y, z), gr^#(x, 0()))
  , cond2^#(true(), x, y, z) ->
    c_2(cond1^#(gr(add(x, y), z), p(x), y, z),
        gr^#(add(x, y), z),
        add^#(x, y),
        p^#(x))
  , cond2^#(false(), x, y, z) ->
    c_3(cond3^#(gr(y, 0()), x, y, z), gr^#(y, 0()))
  , gr^#(s(x), s(y)) -> c_6(gr^#(x, y))
  , add^#(s(x), y) -> c_8(add^#(x, y))
  , cond3^#(true(), x, y, z) ->
    c_11(cond1^#(gr(add(x, y), z), x, p(y), z),
         gr^#(add(x, y), z),
         add^#(x, y),
         p^#(y))
  , cond3^#(false(), x, y, z) ->
    c_12(cond1^#(gr(add(x, y), z), x, y, z),
         gr^#(add(x, y), z),
         add^#(x, y)) }
Weak DPs:
  { gr^#(0(), x) -> c_4()
  , gr^#(s(x), 0()) -> c_5()
  , add^#(0(), x) -> c_7()
  , p^#(0()) -> c_9()
  , p^#(s(x)) -> c_10() }
Weak Trs:
  { cond1(true(), x, y, z) -> cond2(gr(x, 0()), x, y, z)
  , cond2(true(), x, y, z) -> cond1(gr(add(x, y), z), p(x), y, z)
  , cond2(false(), x, y, z) -> cond3(gr(y, 0()), x, y, z)
  , gr(0(), x) -> false()
  , gr(s(x), 0()) -> true()
  , gr(s(x), s(y)) -> gr(x, y)
  , add(0(), x) -> x
  , add(s(x), y) -> s(add(x, y))
  , p(0()) -> 0()
  , p(s(x)) -> x
  , cond3(true(), x, y, z) -> cond1(gr(add(x, y), z), x, p(y), z)
  , cond3(false(), x, y, z) -> cond1(gr(add(x, y), z), x, y, z) }
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.

{ gr^#(0(), x) -> c_4()
, gr^#(s(x), 0()) -> c_5()
, add^#(0(), x) -> c_7()
, p^#(0()) -> c_9()
, p^#(s(x)) -> c_10() }

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

Strict DPs:
  { cond1^#(true(), x, y, z) ->
    c_1(cond2^#(gr(x, 0()), x, y, z), gr^#(x, 0()))
  , cond2^#(true(), x, y, z) ->
    c_2(cond1^#(gr(add(x, y), z), p(x), y, z),
        gr^#(add(x, y), z),
        add^#(x, y),
        p^#(x))
  , cond2^#(false(), x, y, z) ->
    c_3(cond3^#(gr(y, 0()), x, y, z), gr^#(y, 0()))
  , gr^#(s(x), s(y)) -> c_6(gr^#(x, y))
  , add^#(s(x), y) -> c_8(add^#(x, y))
  , cond3^#(true(), x, y, z) ->
    c_11(cond1^#(gr(add(x, y), z), x, p(y), z),
         gr^#(add(x, y), z),
         add^#(x, y),
         p^#(y))
  , cond3^#(false(), x, y, z) ->
    c_12(cond1^#(gr(add(x, y), z), x, y, z),
         gr^#(add(x, y), z),
         add^#(x, y)) }
Weak Trs:
  { cond1(true(), x, y, z) -> cond2(gr(x, 0()), x, y, z)
  , cond2(true(), x, y, z) -> cond1(gr(add(x, y), z), p(x), y, z)
  , cond2(false(), x, y, z) -> cond3(gr(y, 0()), x, y, z)
  , gr(0(), x) -> false()
  , gr(s(x), 0()) -> true()
  , gr(s(x), s(y)) -> gr(x, y)
  , add(0(), x) -> x
  , add(s(x), y) -> s(add(x, y))
  , p(0()) -> 0()
  , p(s(x)) -> x
  , cond3(true(), x, y, z) -> cond1(gr(add(x, y), z), x, p(y), z)
  , cond3(false(), x, y, z) -> cond1(gr(add(x, y), z), x, y, z) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  { cond1^#(true(), x, y, z) ->
    c_1(cond2^#(gr(x, 0()), x, y, z), gr^#(x, 0()))
  , cond2^#(true(), x, y, z) ->
    c_2(cond1^#(gr(add(x, y), z), p(x), y, z),
        gr^#(add(x, y), z),
        add^#(x, y),
        p^#(x))
  , cond2^#(false(), x, y, z) ->
    c_3(cond3^#(gr(y, 0()), x, y, z), gr^#(y, 0()))
  , cond3^#(true(), x, y, z) ->
    c_11(cond1^#(gr(add(x, y), z), x, p(y), z),
         gr^#(add(x, y), z),
         add^#(x, y),
         p^#(y)) }

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

Strict DPs:
  { cond1^#(true(), x, y, z) -> c_1(cond2^#(gr(x, 0()), x, y, z))
  , cond2^#(true(), x, y, z) ->
    c_2(cond1^#(gr(add(x, y), z), p(x), y, z),
        gr^#(add(x, y), z),
        add^#(x, y))
  , cond2^#(false(), x, y, z) -> c_3(cond3^#(gr(y, 0()), x, y, z))
  , gr^#(s(x), s(y)) -> c_4(gr^#(x, y))
  , add^#(s(x), y) -> c_5(add^#(x, y))
  , cond3^#(true(), x, y, z) ->
    c_6(cond1^#(gr(add(x, y), z), x, p(y), z),
        gr^#(add(x, y), z),
        add^#(x, y))
  , cond3^#(false(), x, y, z) ->
    c_7(cond1^#(gr(add(x, y), z), x, y, z),
        gr^#(add(x, y), z),
        add^#(x, y)) }
Weak Trs:
  { cond1(true(), x, y, z) -> cond2(gr(x, 0()), x, y, z)
  , cond2(true(), x, y, z) -> cond1(gr(add(x, y), z), p(x), y, z)
  , cond2(false(), x, y, z) -> cond3(gr(y, 0()), x, y, z)
  , gr(0(), x) -> false()
  , gr(s(x), 0()) -> true()
  , gr(s(x), s(y)) -> gr(x, y)
  , add(0(), x) -> x
  , add(s(x), y) -> s(add(x, y))
  , p(0()) -> 0()
  , p(s(x)) -> x
  , cond3(true(), x, y, z) -> cond1(gr(add(x, y), z), x, p(y), z)
  , cond3(false(), x, y, z) -> cond1(gr(add(x, y), z), x, y, z) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { gr(0(), x) -> false()
    , gr(s(x), 0()) -> true()
    , gr(s(x), s(y)) -> gr(x, y)
    , add(0(), x) -> x
    , add(s(x), y) -> s(add(x, y))
    , 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, y, z) -> c_1(cond2^#(gr(x, 0()), x, y, z))
  , cond2^#(true(), x, y, z) ->
    c_2(cond1^#(gr(add(x, y), z), p(x), y, z),
        gr^#(add(x, y), z),
        add^#(x, y))
  , cond2^#(false(), x, y, z) -> c_3(cond3^#(gr(y, 0()), x, y, z))
  , gr^#(s(x), s(y)) -> c_4(gr^#(x, y))
  , add^#(s(x), y) -> c_5(add^#(x, y))
  , cond3^#(true(), x, y, z) ->
    c_6(cond1^#(gr(add(x, y), z), x, p(y), z),
        gr^#(add(x, y), z),
        add^#(x, y))
  , cond3^#(false(), x, y, z) ->
    c_7(cond1^#(gr(add(x, y), z), x, y, z),
        gr^#(add(x, y), z),
        add^#(x, y)) }
Weak Trs:
  { gr(0(), x) -> false()
  , gr(s(x), 0()) -> true()
  , gr(s(x), s(y)) -> gr(x, y)
  , add(0(), x) -> x
  , add(s(x), y) -> s(add(x, y))
  , p(0()) -> 0()
  , p(s(x)) -> x }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..