Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

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)
zprimessieve(nats(s(s(0))))

Q is empty.


QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

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)
zprimessieve(nats(s(s(0))))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ZPRIMESSIEVE(nats(s(s(0))))
ZPRIMESNATS(s(s(0)))

The TRS R consists of the following rules:

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)
zprimessieve(nats(s(s(0))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

ZPRIMESSIEVE(nats(s(s(0))))
ZPRIMESNATS(s(s(0)))

The TRS R consists of the following rules:

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)
zprimessieve(nats(s(s(0))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.