YES

TRS:

  filter(cons(X,Y),0(),M) -> cons(0(),n__filter(activate(Y),M,M))
  filter(cons(X,Y),s(N),M) -> cons(X,n__filter(activate(Y),N,M))
  sieve(cons(0(),Y)) -> cons(0(),n__sieve(activate(Y)))
  sieve(cons(s(N),Y)) -> cons(s(N),n__sieve(filter(activate(Y),N,N)))
  nats(N) -> cons(N,n__nats(s(N)))
  zprimes() -> sieve(nats(s(s(0()))))
  filter(X1,X2,X3) -> n__filter(X1,X2,X3)
  sieve(X) -> n__sieve(X)
  nats(X) -> n__nats(X)
  activate(n__filter(X1,X2,X3)) -> filter(X1,X2,X3)
  activate(n__sieve(X)) -> sieve(X)
  activate(n__nats(X)) -> nats(X)
  activate(X) -> X

linear polynomial interpretations on N:

  filter_A(x1,x2,x3) = x1
  filter#_A(x1,x2,x3) = x1 + 2
  cons_A(x1,x2) = x2
  cons#_A(x1,x2) = 1
  0_A = 1
  0#_A = 0
  n__filter_A(x1,x2,x3) = x1
  n__filter#_A(x1,x2,x3) = 0
  activate_A(x1) = x1
  activate#_A(x1) = x1 + 2
  s_A(x1) = 1
  s#_A(x1) = 0
  sieve_A(x1) = x1 + 2
  sieve#_A(x1) = x1 + 3
  n__sieve_A(x1) = x1 + 2
  n__sieve#_A(x1) = x1
  nats_A(x1) = 1
  nats#_A(x1) = 2
  n__nats_A(x1) = 1
  n__nats#_A(x1) = 0
  zprimes_A = 3
  zprimes#_A = 5

precedence:

  n__sieve = zprimes > sieve > filter = activate = n__nats > nats > s > cons = n__filter > 0