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)
zprimes → sieve(nats(s(s(0))))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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)
zprimes → sieve(nats(s(s(0))))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ 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)
zprimes → sieve(nats(s(s(0))))

The set Q consists of the following terms:

filter(cons(x0), 0, x1)
filter(cons(x0), s(x1), x2)
sieve(cons(0))
sieve(cons(s(x0)))
nats(x0)
zprimes


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ZPRIMES → SIEVE(nats(s(s(0))))
ZPRIMES → NATS(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)
zprimes → sieve(nats(s(s(0))))

The set Q consists of the following terms:

filter(cons(x0), 0, x1)
filter(cons(x0), s(x1), x2)
sieve(cons(0))
sieve(cons(s(x0)))
nats(x0)
zprimes

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

ZPRIMES → SIEVE(nats(s(s(0))))
ZPRIMES → NATS(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)
zprimes → sieve(nats(s(s(0))))

The set Q consists of the following terms:

filter(cons(x0), 0, x1)
filter(cons(x0), s(x1), x2)
sieve(cons(0))
sieve(cons(s(x0)))
nats(x0)
zprimes

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

ZPRIMES → SIEVE(nats(s(s(0))))
ZPRIMES → NATS(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)
zprimes → sieve(nats(s(s(0))))

The set Q consists of the following terms:

filter(cons(x0), 0, x1)
filter(cons(x0), s(x1), x2)
sieve(cons(0))
sieve(cons(s(x0)))
nats(x0)
zprimes

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.