MAYBE

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

Strict Trs:
  { +(x, 0()) -> x
  , +(0(), x) -> x
  , +(s(x), s(y)) -> s(s(+(x, y)))
  , *(x, 0()) -> 0()
  , *(0(), x) -> 0()
  , *(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
  , sum(nil()) -> 0()
  , sum(cons(x, l)) -> +(x, sum(l))
  , prod(nil()) -> s(0())
  , prod(cons(x, l)) -> *(x, prod(l)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { +^#(x, 0()) -> c_1()
  , +^#(0(), x) -> c_2()
  , +^#(s(x), s(y)) -> c_3(+^#(x, y))
  , *^#(x, 0()) -> c_4()
  , *^#(0(), x) -> c_5()
  , *^#(s(x), s(y)) ->
    c_6(+^#(*(x, y), +(x, y)), *^#(x, y), +^#(x, y))
  , sum^#(nil()) -> c_7()
  , sum^#(cons(x, l)) -> c_8(+^#(x, sum(l)), sum^#(l))
  , prod^#(nil()) -> c_9()
  , prod^#(cons(x, l)) -> c_10(*^#(x, prod(l)), prod^#(l)) }

and mark the set of starting terms.

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

Strict DPs:
  { +^#(x, 0()) -> c_1()
  , +^#(0(), x) -> c_2()
  , +^#(s(x), s(y)) -> c_3(+^#(x, y))
  , *^#(x, 0()) -> c_4()
  , *^#(0(), x) -> c_5()
  , *^#(s(x), s(y)) ->
    c_6(+^#(*(x, y), +(x, y)), *^#(x, y), +^#(x, y))
  , sum^#(nil()) -> c_7()
  , sum^#(cons(x, l)) -> c_8(+^#(x, sum(l)), sum^#(l))
  , prod^#(nil()) -> c_9()
  , prod^#(cons(x, l)) -> c_10(*^#(x, prod(l)), prod^#(l)) }
Weak Trs:
  { +(x, 0()) -> x
  , +(0(), x) -> x
  , +(s(x), s(y)) -> s(s(+(x, y)))
  , *(x, 0()) -> 0()
  , *(0(), x) -> 0()
  , *(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
  , sum(nil()) -> 0()
  , sum(cons(x, l)) -> +(x, sum(l))
  , prod(nil()) -> s(0())
  , prod(cons(x, l)) -> *(x, prod(l)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: +^#(x, 0()) -> c_1()
    , 2: +^#(0(), x) -> c_2()
    , 3: +^#(s(x), s(y)) -> c_3(+^#(x, y))
    , 4: *^#(x, 0()) -> c_4()
    , 5: *^#(0(), x) -> c_5()
    , 6: *^#(s(x), s(y)) ->
         c_6(+^#(*(x, y), +(x, y)), *^#(x, y), +^#(x, y))
    , 7: sum^#(nil()) -> c_7()
    , 8: sum^#(cons(x, l)) -> c_8(+^#(x, sum(l)), sum^#(l))
    , 9: prod^#(nil()) -> c_9()
    , 10: prod^#(cons(x, l)) -> c_10(*^#(x, prod(l)), prod^#(l)) }

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

Strict DPs:
  { +^#(s(x), s(y)) -> c_3(+^#(x, y))
  , *^#(s(x), s(y)) ->
    c_6(+^#(*(x, y), +(x, y)), *^#(x, y), +^#(x, y))
  , sum^#(cons(x, l)) -> c_8(+^#(x, sum(l)), sum^#(l))
  , prod^#(cons(x, l)) -> c_10(*^#(x, prod(l)), prod^#(l)) }
Weak DPs:
  { +^#(x, 0()) -> c_1()
  , +^#(0(), x) -> c_2()
  , *^#(x, 0()) -> c_4()
  , *^#(0(), x) -> c_5()
  , sum^#(nil()) -> c_7()
  , prod^#(nil()) -> c_9() }
Weak Trs:
  { +(x, 0()) -> x
  , +(0(), x) -> x
  , +(s(x), s(y)) -> s(s(+(x, y)))
  , *(x, 0()) -> 0()
  , *(0(), x) -> 0()
  , *(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
  , sum(nil()) -> 0()
  , sum(cons(x, l)) -> +(x, sum(l))
  , prod(nil()) -> s(0())
  , prod(cons(x, l)) -> *(x, prod(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.

{ +^#(x, 0()) -> c_1()
, +^#(0(), x) -> c_2()
, *^#(x, 0()) -> c_4()
, *^#(0(), x) -> c_5()
, sum^#(nil()) -> c_7()
, prod^#(nil()) -> c_9() }

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

Strict DPs:
  { +^#(s(x), s(y)) -> c_3(+^#(x, y))
  , *^#(s(x), s(y)) ->
    c_6(+^#(*(x, y), +(x, y)), *^#(x, y), +^#(x, y))
  , sum^#(cons(x, l)) -> c_8(+^#(x, sum(l)), sum^#(l))
  , prod^#(cons(x, l)) -> c_10(*^#(x, prod(l)), prod^#(l)) }
Weak Trs:
  { +(x, 0()) -> x
  , +(0(), x) -> x
  , +(s(x), s(y)) -> s(s(+(x, y)))
  , *(x, 0()) -> 0()
  , *(0(), x) -> 0()
  , *(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
  , sum(nil()) -> 0()
  , sum(cons(x, l)) -> +(x, sum(l))
  , prod(nil()) -> s(0())
  , prod(cons(x, l)) -> *(x, prod(l)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..