Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/TRCSRSLASHEx9_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

  ↳ 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.