YES

Problem:
 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()))))

Proof:
 DP Processor:
  DPs:
   zprimes#() -> nats#(s(s(0())))
   zprimes#() -> sieve#(nats(s(s(0()))))
  TRS:
   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()))))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [zprimes#] = 3,
    
    [nats#](x0) = x0 + 0,
    
    [sieve#](x0) = -7x0 + 0,
    
    [zprimes] = 2,
    
    [nats](x0) = 1,
    
    [sieve](x0) = x0 + 1,
    
    [s](x0) = x0 + 0,
    
    [filter](x0, x1, x2) = x0 + 5x1 + x2 + -16,
    
    [0] = 0,
    
    [cons](x0) = 0
   orientation:
    zprimes#() = 3 >= 0 = nats#(s(s(0())))
    
    zprimes#() = 3 >= 0 = sieve#(nats(s(s(0()))))
    
    filter(cons(X),0(),M) = M + 5 >= 0 = cons(0())
    
    filter(cons(X),s(N),M) = M + 5N + 5 >= 0 = cons(X)
    
    sieve(cons(0())) = 1 >= 0 = cons(0())
    
    sieve(cons(s(N))) = 1 >= 0 = cons(s(N))
    
    nats(N) = 1 >= 0 = cons(N)
    
    zprimes() = 2 >= 1 = sieve(nats(s(s(0()))))
   problem:
    DPs:
     
    TRS:
     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()))))
   Qed