MAYBE

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

Strict Trs:
  { sort(nil()) -> nil()
  , sort(cons(x, y)) -> insert(x, sort(y))
  , insert(x, nil()) -> cons(x, nil())
  , insert(x, cons(v, w)) -> choose(x, cons(v, w), x, v)
  , choose(x, cons(v, w), y, 0()) -> cons(x, cons(v, w))
  , choose(x, cons(v, w), 0(), s(z)) -> cons(v, insert(x, w))
  , choose(x, cons(v, w), s(y), s(z)) ->
    choose(x, cons(v, w), y, z) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { sort^#(nil()) -> c_1()
  , sort^#(cons(x, y)) -> c_2(insert^#(x, sort(y)), sort^#(y))
  , insert^#(x, nil()) -> c_3()
  , insert^#(x, cons(v, w)) -> c_4(choose^#(x, cons(v, w), x, v))
  , choose^#(x, cons(v, w), y, 0()) -> c_5()
  , choose^#(x, cons(v, w), 0(), s(z)) -> c_6(insert^#(x, w))
  , choose^#(x, cons(v, w), s(y), s(z)) ->
    c_7(choose^#(x, cons(v, w), y, z)) }

and mark the set of starting terms.

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

Strict DPs:
  { sort^#(nil()) -> c_1()
  , sort^#(cons(x, y)) -> c_2(insert^#(x, sort(y)), sort^#(y))
  , insert^#(x, nil()) -> c_3()
  , insert^#(x, cons(v, w)) -> c_4(choose^#(x, cons(v, w), x, v))
  , choose^#(x, cons(v, w), y, 0()) -> c_5()
  , choose^#(x, cons(v, w), 0(), s(z)) -> c_6(insert^#(x, w))
  , choose^#(x, cons(v, w), s(y), s(z)) ->
    c_7(choose^#(x, cons(v, w), y, z)) }
Weak Trs:
  { sort(nil()) -> nil()
  , sort(cons(x, y)) -> insert(x, sort(y))
  , insert(x, nil()) -> cons(x, nil())
  , insert(x, cons(v, w)) -> choose(x, cons(v, w), x, v)
  , choose(x, cons(v, w), y, 0()) -> cons(x, cons(v, w))
  , choose(x, cons(v, w), 0(), s(z)) -> cons(v, insert(x, w))
  , choose(x, cons(v, w), s(y), s(z)) ->
    choose(x, cons(v, w), y, z) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: sort^#(nil()) -> c_1()
    , 2: sort^#(cons(x, y)) -> c_2(insert^#(x, sort(y)), sort^#(y))
    , 3: insert^#(x, nil()) -> c_3()
    , 4: insert^#(x, cons(v, w)) -> c_4(choose^#(x, cons(v, w), x, v))
    , 5: choose^#(x, cons(v, w), y, 0()) -> c_5()
    , 6: choose^#(x, cons(v, w), 0(), s(z)) -> c_6(insert^#(x, w))
    , 7: choose^#(x, cons(v, w), s(y), s(z)) ->
         c_7(choose^#(x, cons(v, w), y, z)) }

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

Strict DPs:
  { sort^#(cons(x, y)) -> c_2(insert^#(x, sort(y)), sort^#(y))
  , insert^#(x, cons(v, w)) -> c_4(choose^#(x, cons(v, w), x, v))
  , choose^#(x, cons(v, w), 0(), s(z)) -> c_6(insert^#(x, w))
  , choose^#(x, cons(v, w), s(y), s(z)) ->
    c_7(choose^#(x, cons(v, w), y, z)) }
Weak DPs:
  { sort^#(nil()) -> c_1()
  , insert^#(x, nil()) -> c_3()
  , choose^#(x, cons(v, w), y, 0()) -> c_5() }
Weak Trs:
  { sort(nil()) -> nil()
  , sort(cons(x, y)) -> insert(x, sort(y))
  , insert(x, nil()) -> cons(x, nil())
  , insert(x, cons(v, w)) -> choose(x, cons(v, w), x, v)
  , choose(x, cons(v, w), y, 0()) -> cons(x, cons(v, w))
  , choose(x, cons(v, w), 0(), s(z)) -> cons(v, insert(x, w))
  , choose(x, cons(v, w), s(y), s(z)) ->
    choose(x, cons(v, w), 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.

{ sort^#(nil()) -> c_1()
, insert^#(x, nil()) -> c_3()
, choose^#(x, cons(v, w), y, 0()) -> c_5() }

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

Strict DPs:
  { sort^#(cons(x, y)) -> c_2(insert^#(x, sort(y)), sort^#(y))
  , insert^#(x, cons(v, w)) -> c_4(choose^#(x, cons(v, w), x, v))
  , choose^#(x, cons(v, w), 0(), s(z)) -> c_6(insert^#(x, w))
  , choose^#(x, cons(v, w), s(y), s(z)) ->
    c_7(choose^#(x, cons(v, w), y, z)) }
Weak Trs:
  { sort(nil()) -> nil()
  , sort(cons(x, y)) -> insert(x, sort(y))
  , insert(x, nil()) -> cons(x, nil())
  , insert(x, cons(v, w)) -> choose(x, cons(v, w), x, v)
  , choose(x, cons(v, w), y, 0()) -> cons(x, cons(v, w))
  , choose(x, cons(v, w), 0(), s(z)) -> cons(v, insert(x, w))
  , choose(x, cons(v, w), s(y), s(z)) ->
    choose(x, cons(v, w), y, z) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..