YES

We show the termination of the TRS R:

  filter(cons(X),|0|(),M) -> cons(|0|())
  filter(cons(X),s(N),M) -> cons(X)
  sieve(cons(|0|())) -> cons(|0|())
  sieve(cons(s(N))) -> cons(s(N))
  nats(N) -> cons(N)
  zprimes() -> sieve(nats(s(s(|0|()))))

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: zprimes#() -> sieve#(nats(s(s(|0|()))))
p2: zprimes#() -> nats#(s(s(|0|())))

and R consists of:

r1: filter(cons(X),|0|(),M) -> cons(|0|())
r2: filter(cons(X),s(N),M) -> cons(X)
r3: sieve(cons(|0|())) -> cons(|0|())
r4: sieve(cons(s(N))) -> cons(s(N))
r5: nats(N) -> cons(N)
r6: zprimes() -> sieve(nats(s(s(|0|()))))

The estimated dependency graph contains the following SCCs:

  (no SCCs)