Termination w.r.t. Q proof of TRS/TRCSREx9_BLR02

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

  ↳ 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))))

Q is empty.

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 -> 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))))

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:

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))))

Q is empty.
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.