YES

Problem:
 a__filter(cons(X,Y),0(),M) -> cons(0(),filter(Y,M,M))
 a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
 a__sieve(cons(0(),Y)) -> cons(0(),sieve(Y))
 a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
 a__nats(N) -> cons(mark(N),nats(s(N)))
 a__zprimes() -> a__sieve(a__nats(s(s(0()))))
 mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
 mark(sieve(X)) -> a__sieve(mark(X))
 mark(nats(X)) -> a__nats(mark(X))
 mark(zprimes()) -> a__zprimes()
 mark(cons(X1,X2)) -> cons(mark(X1),X2)
 mark(0()) -> 0()
 mark(s(X)) -> s(mark(X))
 a__filter(X1,X2,X3) -> filter(X1,X2,X3)
 a__sieve(X) -> sieve(X)
 a__nats(X) -> nats(X)
 a__zprimes() -> zprimes()

Proof:
 WPO Processor:
  algebra: Max
  weight function:
   w0 = 0
   w(zprimes) = w(a__zprimes) = 5
   w(nats) = w(a__nats) = w(sieve) = w(a__sieve) = w(mark) = w(s) = w(
   filter) = w(a__filter) = w(0) = w(cons) = 0
  status function:
   st(filter) = [2, 0, 1]
   st(zprimes) = st(a__zprimes) = []
   st(nats) = st(a__nats) = st(sieve) = st(a__sieve) = st(mark) = st(
   s) = [0]
   st(a__filter) = [0, 1, 2]
   st(0) = []
   st(cons) = [0, 1]
  subterm penalty function:
   sp(nats, 0) = sp(a__nats, 0) = 4
   sp(cons, 0) = 2
   sp(sieve, 0) = sp(a__sieve, 0) = sp(mark, 0) = sp(s, 0) = sp(filter, 2) = sp(
   filter, 1) = sp(filter, 0) = sp(a__filter, 2) = sp(a__filter, 1) = sp(
   a__filter, 0) = sp(cons, 1) = 0
  precedence:
   mark > a__nats ~ a__sieve > a__zprimes ~ a__filter > zprimes ~ nats ~ 
   sieve ~ s ~ filter ~ 0 ~ cons
  problem:
   
  Qed