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:

filter3(cons1(X), 0, M) -> cons1(0)
filter3(cons1(X), s1(N), M) -> cons1(X)
sieve1(cons1(0)) -> cons1(0)
sieve1(cons1(s1(N))) -> cons1(s1(N))
nats1(N) -> cons1(N)
zprimes -> sieve1(nats1(s1(s1(0))))

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

filter3(cons1(X), 0, M) -> cons1(0)
filter3(cons1(X), s1(N), M) -> cons1(X)
sieve1(cons1(0)) -> cons1(0)
sieve1(cons1(s1(N))) -> cons1(s1(N))
nats1(N) -> cons1(N)
zprimes -> sieve1(nats1(s1(s1(0))))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
QTRS
      ↳ DependencyPairsProof

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

filter3(cons1(X), 0, M) -> cons1(0)
filter3(cons1(X), s1(N), M) -> cons1(X)
sieve1(cons1(0)) -> cons1(0)
sieve1(cons1(s1(N))) -> cons1(s1(N))
nats1(N) -> cons1(N)
zprimes -> sieve1(nats1(s1(s1(0))))

The set Q consists of the following terms:

filter3(cons1(x0), 0, x1)
filter3(cons1(x0), s1(x1), x2)
sieve1(cons1(0))
sieve1(cons1(s1(x0)))
nats1(x0)
zprimes


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

ZPRIMES -> SIEVE1(nats1(s1(s1(0))))
ZPRIMES -> NATS1(s1(s1(0)))

The TRS R consists of the following rules:

filter3(cons1(X), 0, M) -> cons1(0)
filter3(cons1(X), s1(N), M) -> cons1(X)
sieve1(cons1(0)) -> cons1(0)
sieve1(cons1(s1(N))) -> cons1(s1(N))
nats1(N) -> cons1(N)
zprimes -> sieve1(nats1(s1(s1(0))))

The set Q consists of the following terms:

filter3(cons1(x0), 0, x1)
filter3(cons1(x0), s1(x1), x2)
sieve1(cons1(0))
sieve1(cons1(s1(x0)))
nats1(x0)
zprimes

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ DependencyGraphProof

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

ZPRIMES -> SIEVE1(nats1(s1(s1(0))))
ZPRIMES -> NATS1(s1(s1(0)))

The TRS R consists of the following rules:

filter3(cons1(X), 0, M) -> cons1(0)
filter3(cons1(X), s1(N), M) -> cons1(X)
sieve1(cons1(0)) -> cons1(0)
sieve1(cons1(s1(N))) -> cons1(s1(N))
nats1(N) -> cons1(N)
zprimes -> sieve1(nats1(s1(s1(0))))

The set Q consists of the following terms:

filter3(cons1(x0), 0, x1)
filter3(cons1(x0), s1(x1), x2)
sieve1(cons1(0))
sieve1(cons1(s1(x0)))
nats1(x0)
zprimes

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