MAYBE

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

Strict Trs:
  { a__primes() -> a__sieve(a__from(s(s(0()))))
  , a__primes() -> primes()
  , a__sieve(X) -> sieve(X)
  , a__sieve(cons(X, Y)) -> cons(mark(X), filter(X, sieve(Y)))
  , a__from(X) -> cons(mark(X), from(s(X)))
  , a__from(X) -> from(X)
  , mark(s(X)) -> s(mark(X))
  , mark(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(from(X)) -> a__from(mark(X))
  , mark(true()) -> true()
  , mark(false()) -> false()
  , mark(divides(X1, X2)) -> divides(mark(X1), mark(X2))
  , mark(filter(X1, X2)) -> a__filter(mark(X1), mark(X2))
  , mark(sieve(X)) -> a__sieve(mark(X))
  , mark(primes()) -> a__primes()
  , mark(head(X)) -> a__head(mark(X))
  , mark(tail(X)) -> a__tail(mark(X))
  , mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
  , a__head(X) -> head(X)
  , a__head(cons(X, Y)) -> mark(X)
  , a__tail(X) -> tail(X)
  , a__tail(cons(X, Y)) -> mark(Y)
  , a__if(X1, X2, X3) -> if(X1, X2, X3)
  , a__if(true(), X, Y) -> mark(X)
  , a__if(false(), X, Y) -> mark(Y)
  , a__filter(X1, X2) -> filter(X1, X2)
  , a__filter(s(s(X)), cons(Y, Z)) ->
    a__if(divides(s(s(mark(X))), mark(Y)),
          filter(s(s(X)), Z),
          cons(Y, filter(X, sieve(Y)))) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { a__primes^#() ->
    c_1(a__sieve^#(a__from(s(s(0())))), a__from^#(s(s(0()))))
  , a__primes^#() -> c_2()
  , a__sieve^#(X) -> c_3()
  , a__sieve^#(cons(X, Y)) -> c_4(mark^#(X))
  , a__from^#(X) -> c_5(mark^#(X))
  , a__from^#(X) -> c_6()
  , mark^#(s(X)) -> c_7(mark^#(X))
  , mark^#(0()) -> c_8()
  , mark^#(cons(X1, X2)) -> c_9(mark^#(X1))
  , mark^#(from(X)) -> c_10(a__from^#(mark(X)), mark^#(X))
  , mark^#(true()) -> c_11()
  , mark^#(false()) -> c_12()
  , mark^#(divides(X1, X2)) -> c_13(mark^#(X1), mark^#(X2))
  , mark^#(filter(X1, X2)) ->
    c_14(a__filter^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(sieve(X)) -> c_15(a__sieve^#(mark(X)), mark^#(X))
  , mark^#(primes()) -> c_16(a__primes^#())
  , mark^#(head(X)) -> c_17(a__head^#(mark(X)), mark^#(X))
  , mark^#(tail(X)) -> c_18(a__tail^#(mark(X)), mark^#(X))
  , mark^#(if(X1, X2, X3)) ->
    c_19(a__if^#(mark(X1), X2, X3), mark^#(X1))
  , a__filter^#(X1, X2) -> c_27()
  , a__filter^#(s(s(X)), cons(Y, Z)) ->
    c_28(a__if^#(divides(s(s(mark(X))), mark(Y)),
                 filter(s(s(X)), Z),
                 cons(Y, filter(X, sieve(Y)))),
         mark^#(X),
         mark^#(Y))
  , a__head^#(X) -> c_20()
  , a__head^#(cons(X, Y)) -> c_21(mark^#(X))
  , a__tail^#(X) -> c_22()
  , a__tail^#(cons(X, Y)) -> c_23(mark^#(Y))
  , a__if^#(X1, X2, X3) -> c_24()
  , a__if^#(true(), X, Y) -> c_25(mark^#(X))
  , a__if^#(false(), X, Y) -> c_26(mark^#(Y)) }

and mark the set of starting terms.

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

Strict DPs:
  { a__primes^#() ->
    c_1(a__sieve^#(a__from(s(s(0())))), a__from^#(s(s(0()))))
  , a__primes^#() -> c_2()
  , a__sieve^#(X) -> c_3()
  , a__sieve^#(cons(X, Y)) -> c_4(mark^#(X))
  , a__from^#(X) -> c_5(mark^#(X))
  , a__from^#(X) -> c_6()
  , mark^#(s(X)) -> c_7(mark^#(X))
  , mark^#(0()) -> c_8()
  , mark^#(cons(X1, X2)) -> c_9(mark^#(X1))
  , mark^#(from(X)) -> c_10(a__from^#(mark(X)), mark^#(X))
  , mark^#(true()) -> c_11()
  , mark^#(false()) -> c_12()
  , mark^#(divides(X1, X2)) -> c_13(mark^#(X1), mark^#(X2))
  , mark^#(filter(X1, X2)) ->
    c_14(a__filter^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(sieve(X)) -> c_15(a__sieve^#(mark(X)), mark^#(X))
  , mark^#(primes()) -> c_16(a__primes^#())
  , mark^#(head(X)) -> c_17(a__head^#(mark(X)), mark^#(X))
  , mark^#(tail(X)) -> c_18(a__tail^#(mark(X)), mark^#(X))
  , mark^#(if(X1, X2, X3)) ->
    c_19(a__if^#(mark(X1), X2, X3), mark^#(X1))
  , a__filter^#(X1, X2) -> c_27()
  , a__filter^#(s(s(X)), cons(Y, Z)) ->
    c_28(a__if^#(divides(s(s(mark(X))), mark(Y)),
                 filter(s(s(X)), Z),
                 cons(Y, filter(X, sieve(Y)))),
         mark^#(X),
         mark^#(Y))
  , a__head^#(X) -> c_20()
  , a__head^#(cons(X, Y)) -> c_21(mark^#(X))
  , a__tail^#(X) -> c_22()
  , a__tail^#(cons(X, Y)) -> c_23(mark^#(Y))
  , a__if^#(X1, X2, X3) -> c_24()
  , a__if^#(true(), X, Y) -> c_25(mark^#(X))
  , a__if^#(false(), X, Y) -> c_26(mark^#(Y)) }
Weak Trs:
  { a__primes() -> a__sieve(a__from(s(s(0()))))
  , a__primes() -> primes()
  , a__sieve(X) -> sieve(X)
  , a__sieve(cons(X, Y)) -> cons(mark(X), filter(X, sieve(Y)))
  , a__from(X) -> cons(mark(X), from(s(X)))
  , a__from(X) -> from(X)
  , mark(s(X)) -> s(mark(X))
  , mark(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(from(X)) -> a__from(mark(X))
  , mark(true()) -> true()
  , mark(false()) -> false()
  , mark(divides(X1, X2)) -> divides(mark(X1), mark(X2))
  , mark(filter(X1, X2)) -> a__filter(mark(X1), mark(X2))
  , mark(sieve(X)) -> a__sieve(mark(X))
  , mark(primes()) -> a__primes()
  , mark(head(X)) -> a__head(mark(X))
  , mark(tail(X)) -> a__tail(mark(X))
  , mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
  , a__head(X) -> head(X)
  , a__head(cons(X, Y)) -> mark(X)
  , a__tail(X) -> tail(X)
  , a__tail(cons(X, Y)) -> mark(Y)
  , a__if(X1, X2, X3) -> if(X1, X2, X3)
  , a__if(true(), X, Y) -> mark(X)
  , a__if(false(), X, Y) -> mark(Y)
  , a__filter(X1, X2) -> filter(X1, X2)
  , a__filter(s(s(X)), cons(Y, Z)) ->
    a__if(divides(s(s(mark(X))), mark(Y)),
          filter(s(s(X)), Z),
          cons(Y, filter(X, sieve(Y)))) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of
{2,3,6,8,11,12,20,22,24,26} by applications of
Pre({2,3,6,8,11,12,20,22,24,26}) =
{1,4,5,7,9,10,13,14,15,16,17,18,19,21,23,25,27,28}. Here rules are
labeled as follows:

  DPs:
    { 1: a__primes^#() ->
         c_1(a__sieve^#(a__from(s(s(0())))), a__from^#(s(s(0()))))
    , 2: a__primes^#() -> c_2()
    , 3: a__sieve^#(X) -> c_3()
    , 4: a__sieve^#(cons(X, Y)) -> c_4(mark^#(X))
    , 5: a__from^#(X) -> c_5(mark^#(X))
    , 6: a__from^#(X) -> c_6()
    , 7: mark^#(s(X)) -> c_7(mark^#(X))
    , 8: mark^#(0()) -> c_8()
    , 9: mark^#(cons(X1, X2)) -> c_9(mark^#(X1))
    , 10: mark^#(from(X)) -> c_10(a__from^#(mark(X)), mark^#(X))
    , 11: mark^#(true()) -> c_11()
    , 12: mark^#(false()) -> c_12()
    , 13: mark^#(divides(X1, X2)) -> c_13(mark^#(X1), mark^#(X2))
    , 14: mark^#(filter(X1, X2)) ->
          c_14(a__filter^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
    , 15: mark^#(sieve(X)) -> c_15(a__sieve^#(mark(X)), mark^#(X))
    , 16: mark^#(primes()) -> c_16(a__primes^#())
    , 17: mark^#(head(X)) -> c_17(a__head^#(mark(X)), mark^#(X))
    , 18: mark^#(tail(X)) -> c_18(a__tail^#(mark(X)), mark^#(X))
    , 19: mark^#(if(X1, X2, X3)) ->
          c_19(a__if^#(mark(X1), X2, X3), mark^#(X1))
    , 20: a__filter^#(X1, X2) -> c_27()
    , 21: a__filter^#(s(s(X)), cons(Y, Z)) ->
          c_28(a__if^#(divides(s(s(mark(X))), mark(Y)),
                       filter(s(s(X)), Z),
                       cons(Y, filter(X, sieve(Y)))),
               mark^#(X),
               mark^#(Y))
    , 22: a__head^#(X) -> c_20()
    , 23: a__head^#(cons(X, Y)) -> c_21(mark^#(X))
    , 24: a__tail^#(X) -> c_22()
    , 25: a__tail^#(cons(X, Y)) -> c_23(mark^#(Y))
    , 26: a__if^#(X1, X2, X3) -> c_24()
    , 27: a__if^#(true(), X, Y) -> c_25(mark^#(X))
    , 28: a__if^#(false(), X, Y) -> c_26(mark^#(Y)) }

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

Strict DPs:
  { a__primes^#() ->
    c_1(a__sieve^#(a__from(s(s(0())))), a__from^#(s(s(0()))))
  , a__sieve^#(cons(X, Y)) -> c_4(mark^#(X))
  , a__from^#(X) -> c_5(mark^#(X))
  , mark^#(s(X)) -> c_7(mark^#(X))
  , mark^#(cons(X1, X2)) -> c_9(mark^#(X1))
  , mark^#(from(X)) -> c_10(a__from^#(mark(X)), mark^#(X))
  , mark^#(divides(X1, X2)) -> c_13(mark^#(X1), mark^#(X2))
  , mark^#(filter(X1, X2)) ->
    c_14(a__filter^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(sieve(X)) -> c_15(a__sieve^#(mark(X)), mark^#(X))
  , mark^#(primes()) -> c_16(a__primes^#())
  , mark^#(head(X)) -> c_17(a__head^#(mark(X)), mark^#(X))
  , mark^#(tail(X)) -> c_18(a__tail^#(mark(X)), mark^#(X))
  , mark^#(if(X1, X2, X3)) ->
    c_19(a__if^#(mark(X1), X2, X3), mark^#(X1))
  , a__filter^#(s(s(X)), cons(Y, Z)) ->
    c_28(a__if^#(divides(s(s(mark(X))), mark(Y)),
                 filter(s(s(X)), Z),
                 cons(Y, filter(X, sieve(Y)))),
         mark^#(X),
         mark^#(Y))
  , a__head^#(cons(X, Y)) -> c_21(mark^#(X))
  , a__tail^#(cons(X, Y)) -> c_23(mark^#(Y))
  , a__if^#(true(), X, Y) -> c_25(mark^#(X))
  , a__if^#(false(), X, Y) -> c_26(mark^#(Y)) }
Weak DPs:
  { a__primes^#() -> c_2()
  , a__sieve^#(X) -> c_3()
  , a__from^#(X) -> c_6()
  , mark^#(0()) -> c_8()
  , mark^#(true()) -> c_11()
  , mark^#(false()) -> c_12()
  , a__filter^#(X1, X2) -> c_27()
  , a__head^#(X) -> c_20()
  , a__tail^#(X) -> c_22()
  , a__if^#(X1, X2, X3) -> c_24() }
Weak Trs:
  { a__primes() -> a__sieve(a__from(s(s(0()))))
  , a__primes() -> primes()
  , a__sieve(X) -> sieve(X)
  , a__sieve(cons(X, Y)) -> cons(mark(X), filter(X, sieve(Y)))
  , a__from(X) -> cons(mark(X), from(s(X)))
  , a__from(X) -> from(X)
  , mark(s(X)) -> s(mark(X))
  , mark(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(from(X)) -> a__from(mark(X))
  , mark(true()) -> true()
  , mark(false()) -> false()
  , mark(divides(X1, X2)) -> divides(mark(X1), mark(X2))
  , mark(filter(X1, X2)) -> a__filter(mark(X1), mark(X2))
  , mark(sieve(X)) -> a__sieve(mark(X))
  , mark(primes()) -> a__primes()
  , mark(head(X)) -> a__head(mark(X))
  , mark(tail(X)) -> a__tail(mark(X))
  , mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
  , a__head(X) -> head(X)
  , a__head(cons(X, Y)) -> mark(X)
  , a__tail(X) -> tail(X)
  , a__tail(cons(X, Y)) -> mark(Y)
  , a__if(X1, X2, X3) -> if(X1, X2, X3)
  , a__if(true(), X, Y) -> mark(X)
  , a__if(false(), X, Y) -> mark(Y)
  , a__filter(X1, X2) -> filter(X1, X2)
  , a__filter(s(s(X)), cons(Y, Z)) ->
    a__if(divides(s(s(mark(X))), mark(Y)),
          filter(s(s(X)), Z),
          cons(Y, filter(X, sieve(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.

{ a__primes^#() -> c_2()
, a__sieve^#(X) -> c_3()
, a__from^#(X) -> c_6()
, mark^#(0()) -> c_8()
, mark^#(true()) -> c_11()
, mark^#(false()) -> c_12()
, a__filter^#(X1, X2) -> c_27()
, a__head^#(X) -> c_20()
, a__tail^#(X) -> c_22()
, a__if^#(X1, X2, X3) -> c_24() }

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

Strict DPs:
  { a__primes^#() ->
    c_1(a__sieve^#(a__from(s(s(0())))), a__from^#(s(s(0()))))
  , a__sieve^#(cons(X, Y)) -> c_4(mark^#(X))
  , a__from^#(X) -> c_5(mark^#(X))
  , mark^#(s(X)) -> c_7(mark^#(X))
  , mark^#(cons(X1, X2)) -> c_9(mark^#(X1))
  , mark^#(from(X)) -> c_10(a__from^#(mark(X)), mark^#(X))
  , mark^#(divides(X1, X2)) -> c_13(mark^#(X1), mark^#(X2))
  , mark^#(filter(X1, X2)) ->
    c_14(a__filter^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(sieve(X)) -> c_15(a__sieve^#(mark(X)), mark^#(X))
  , mark^#(primes()) -> c_16(a__primes^#())
  , mark^#(head(X)) -> c_17(a__head^#(mark(X)), mark^#(X))
  , mark^#(tail(X)) -> c_18(a__tail^#(mark(X)), mark^#(X))
  , mark^#(if(X1, X2, X3)) ->
    c_19(a__if^#(mark(X1), X2, X3), mark^#(X1))
  , a__filter^#(s(s(X)), cons(Y, Z)) ->
    c_28(a__if^#(divides(s(s(mark(X))), mark(Y)),
                 filter(s(s(X)), Z),
                 cons(Y, filter(X, sieve(Y)))),
         mark^#(X),
         mark^#(Y))
  , a__head^#(cons(X, Y)) -> c_21(mark^#(X))
  , a__tail^#(cons(X, Y)) -> c_23(mark^#(Y))
  , a__if^#(true(), X, Y) -> c_25(mark^#(X))
  , a__if^#(false(), X, Y) -> c_26(mark^#(Y)) }
Weak Trs:
  { a__primes() -> a__sieve(a__from(s(s(0()))))
  , a__primes() -> primes()
  , a__sieve(X) -> sieve(X)
  , a__sieve(cons(X, Y)) -> cons(mark(X), filter(X, sieve(Y)))
  , a__from(X) -> cons(mark(X), from(s(X)))
  , a__from(X) -> from(X)
  , mark(s(X)) -> s(mark(X))
  , mark(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(from(X)) -> a__from(mark(X))
  , mark(true()) -> true()
  , mark(false()) -> false()
  , mark(divides(X1, X2)) -> divides(mark(X1), mark(X2))
  , mark(filter(X1, X2)) -> a__filter(mark(X1), mark(X2))
  , mark(sieve(X)) -> a__sieve(mark(X))
  , mark(primes()) -> a__primes()
  , mark(head(X)) -> a__head(mark(X))
  , mark(tail(X)) -> a__tail(mark(X))
  , mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
  , a__head(X) -> head(X)
  , a__head(cons(X, Y)) -> mark(X)
  , a__tail(X) -> tail(X)
  , a__tail(cons(X, Y)) -> mark(Y)
  , a__if(X1, X2, X3) -> if(X1, X2, X3)
  , a__if(true(), X, Y) -> mark(X)
  , a__if(false(), X, Y) -> mark(Y)
  , a__filter(X1, X2) -> filter(X1, X2)
  , a__filter(s(s(X)), cons(Y, Z)) ->
    a__if(divides(s(s(mark(X))), mark(Y)),
          filter(s(s(X)), Z),
          cons(Y, filter(X, sieve(Y)))) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  { a__filter^#(s(s(X)), cons(Y, Z)) ->
    c_28(a__if^#(divides(s(s(mark(X))), mark(Y)),
                 filter(s(s(X)), Z),
                 cons(Y, filter(X, sieve(Y)))),
         mark^#(X),
         mark^#(Y)) }

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

Strict DPs:
  { a__primes^#() ->
    c_1(a__sieve^#(a__from(s(s(0())))), a__from^#(s(s(0()))))
  , a__sieve^#(cons(X, Y)) -> c_2(mark^#(X))
  , a__from^#(X) -> c_3(mark^#(X))
  , mark^#(s(X)) -> c_4(mark^#(X))
  , mark^#(cons(X1, X2)) -> c_5(mark^#(X1))
  , mark^#(from(X)) -> c_6(a__from^#(mark(X)), mark^#(X))
  , mark^#(divides(X1, X2)) -> c_7(mark^#(X1), mark^#(X2))
  , mark^#(filter(X1, X2)) ->
    c_8(a__filter^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(sieve(X)) -> c_9(a__sieve^#(mark(X)), mark^#(X))
  , mark^#(primes()) -> c_10(a__primes^#())
  , mark^#(head(X)) -> c_11(a__head^#(mark(X)), mark^#(X))
  , mark^#(tail(X)) -> c_12(a__tail^#(mark(X)), mark^#(X))
  , mark^#(if(X1, X2, X3)) ->
    c_13(a__if^#(mark(X1), X2, X3), mark^#(X1))
  , a__filter^#(s(s(X)), cons(Y, Z)) -> c_14(mark^#(X), mark^#(Y))
  , a__head^#(cons(X, Y)) -> c_15(mark^#(X))
  , a__tail^#(cons(X, Y)) -> c_16(mark^#(Y))
  , a__if^#(true(), X, Y) -> c_17(mark^#(X))
  , a__if^#(false(), X, Y) -> c_18(mark^#(Y)) }
Weak Trs:
  { a__primes() -> a__sieve(a__from(s(s(0()))))
  , a__primes() -> primes()
  , a__sieve(X) -> sieve(X)
  , a__sieve(cons(X, Y)) -> cons(mark(X), filter(X, sieve(Y)))
  , a__from(X) -> cons(mark(X), from(s(X)))
  , a__from(X) -> from(X)
  , mark(s(X)) -> s(mark(X))
  , mark(0()) -> 0()
  , mark(cons(X1, X2)) -> cons(mark(X1), X2)
  , mark(from(X)) -> a__from(mark(X))
  , mark(true()) -> true()
  , mark(false()) -> false()
  , mark(divides(X1, X2)) -> divides(mark(X1), mark(X2))
  , mark(filter(X1, X2)) -> a__filter(mark(X1), mark(X2))
  , mark(sieve(X)) -> a__sieve(mark(X))
  , mark(primes()) -> a__primes()
  , mark(head(X)) -> a__head(mark(X))
  , mark(tail(X)) -> a__tail(mark(X))
  , mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
  , a__head(X) -> head(X)
  , a__head(cons(X, Y)) -> mark(X)
  , a__tail(X) -> tail(X)
  , a__tail(cons(X, Y)) -> mark(Y)
  , a__if(X1, X2, X3) -> if(X1, X2, X3)
  , a__if(true(), X, Y) -> mark(X)
  , a__if(false(), X, Y) -> mark(Y)
  , a__filter(X1, X2) -> filter(X1, X2)
  , a__filter(s(s(X)), cons(Y, Z)) ->
    a__if(divides(s(s(mark(X))), mark(Y)),
          filter(s(s(X)), Z),
          cons(Y, filter(X, sieve(Y)))) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..