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:
 Bounds Processor:
  bound: 2
  enrichment: match
  automaton:
   final states: {7,6,5,4}
   transitions:
    sieve1(12) -> 7*
    nats1(11) -> 12*
    s1(10) -> 11*
    s1(2) -> 9*
    s1(9) -> 10*
    s1(1) -> 9*
    s1(3) -> 9*
    01() -> 9*
    cons1(2) -> 6,4
    cons1(9) -> 5,4
    cons1(1) -> 6,4
    cons1(3) -> 6,4
    cons2(16) -> 7*
    cons2(11) -> 12*
    s2(10) -> 16*
    filter0(1,1,1) -> 4*
    filter0(1,3,1) -> 4*
    filter0(1,2,3) -> 4*
    filter0(2,2,1) -> 4*
    filter0(2,1,3) -> 4*
    filter0(2,3,3) -> 4*
    filter0(3,1,1) -> 4*
    filter0(1,1,2) -> 4*
    filter0(3,3,1) -> 4*
    filter0(1,3,2) -> 4*
    filter0(3,2,3) -> 4*
    filter0(2,2,2) -> 4*
    filter0(3,1,2) -> 4*
    filter0(3,3,2) -> 4*
    filter0(1,2,1) -> 4*
    filter0(1,1,3) -> 4*
    filter0(1,3,3) -> 4*
    filter0(2,1,1) -> 4*
    filter0(2,3,1) -> 4*
    filter0(2,2,3) -> 4*
    filter0(3,2,1) -> 4*
    filter0(1,2,2) -> 4*
    filter0(3,1,3) -> 4*
    filter0(3,3,3) -> 4*
    filter0(2,1,2) -> 4*
    filter0(2,3,2) -> 4*
    filter0(3,2,2) -> 4*
    cons0(2) -> 1*
    cons0(1) -> 1*
    cons0(3) -> 1*
    00() -> 2*
    s0(2) -> 3*
    s0(1) -> 3*
    s0(3) -> 3*
    sieve0(2) -> 5*
    sieve0(1) -> 5*
    sieve0(3) -> 5*
    nats0(2) -> 6*
    nats0(1) -> 6*
    nats0(3) -> 6*
    zprimes0() -> 7*
  problem:
   
  Qed