Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/nonterminSLASHCSRSLASHEx9

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

filter₃(cons₂(X, Y), 0, M) -> cons₂(0, filter₃(Y, M, M))
filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(X, filter₃(Y, N, M))
sieve₁(cons₂(0, Y)) -> cons₂(0, sieve₁(Y))
sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(N), sieve₁(filter₃(Y, N, N)))
nats₁(N) -> cons₂(N, nats₁(s₁(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(cons₂(X, Y), 0, M) -> cons₂(0, filter₃(Y, M, M))
filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(X, filter₃(Y, N, M))
sieve₁(cons₂(0, Y)) -> cons₂(0, sieve₁(Y))
sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(N), sieve₁(filter₃(Y, N, N)))
nats₁(N) -> cons₂(N, nats₁(s₁(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(cons₂(X, Y), 0, M) -> cons₂(0, filter₃(Y, M, M))
filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(X, filter₃(Y, N, M))
sieve₁(cons₂(0, Y)) -> cons₂(0, sieve₁(Y))
sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(N), sieve₁(filter₃(Y, N, N)))
nats₁(N) -> cons₂(N, nats₁(s₁(N)))
zprimes -> sieve₁(nats₁(s₁(s₁(0))))

The set Q consists of the following terms:

filter₃(cons₂(x0, x1), 0, x2)
filter₃(cons₂(x0, x1), s₁(x2), x3)
sieve₁(cons₂(0, x0))
sieve₁(cons₂(s₁(x0), x1))
nats₁(x0)
zprimes

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

SIEVE₁(cons₂(s₁(N), Y)) -> SIEVE₁(filter₃(Y, N, N))
ZPRIMES -> SIEVE₁(nats₁(s₁(s₁(0))))
SIEVE₁(cons₂(s₁(N), Y)) -> FILTER₃(Y, N, N)
FILTER₃(cons₂(X, Y), s₁(N), M) -> FILTER₃(Y, N, M)
FILTER₃(cons₂(X, Y), 0, M) -> FILTER₃(Y, M, M)
SIEVE₁(cons₂(0, Y)) -> SIEVE₁(Y)
ZPRIMES -> NATS₁(s₁(s₁(0)))
NATS₁(N) -> NATS₁(s₁(N))

The TRS R consists of the following rules:

filter₃(cons₂(X, Y), 0, M) -> cons₂(0, filter₃(Y, M, M))
filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(X, filter₃(Y, N, M))
sieve₁(cons₂(0, Y)) -> cons₂(0, sieve₁(Y))
sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(N), sieve₁(filter₃(Y, N, N)))
nats₁(N) -> cons₂(N, nats₁(s₁(N)))
zprimes -> sieve₁(nats₁(s₁(s₁(0))))

The set Q consists of the following terms:

filter₃(cons₂(x0, x1), 0, x2)
filter₃(cons₂(x0, x1), s₁(x2), x3)
sieve₁(cons₂(0, x0))
sieve₁(cons₂(s₁(x0), x1))
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:

SIEVE₁(cons₂(s₁(N), Y)) -> SIEVE₁(filter₃(Y, N, N))
ZPRIMES -> SIEVE₁(nats₁(s₁(s₁(0))))
SIEVE₁(cons₂(s₁(N), Y)) -> FILTER₃(Y, N, N)
FILTER₃(cons₂(X, Y), s₁(N), M) -> FILTER₃(Y, N, M)
FILTER₃(cons₂(X, Y), 0, M) -> FILTER₃(Y, M, M)
SIEVE₁(cons₂(0, Y)) -> SIEVE₁(Y)
ZPRIMES -> NATS₁(s₁(s₁(0)))
NATS₁(N) -> NATS₁(s₁(N))

The TRS R consists of the following rules:

filter₃(cons₂(X, Y), 0, M) -> cons₂(0, filter₃(Y, M, M))
filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(X, filter₃(Y, N, M))
sieve₁(cons₂(0, Y)) -> cons₂(0, sieve₁(Y))
sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(N), sieve₁(filter₃(Y, N, N)))
nats₁(N) -> cons₂(N, nats₁(s₁(N)))
zprimes -> sieve₁(nats₁(s₁(s₁(0))))

The set Q consists of the following terms:

filter₃(cons₂(x0, x1), 0, x2)
filter₃(cons₂(x0, x1), s₁(x2), x3)
sieve₁(cons₂(0, x0))
sieve₁(cons₂(s₁(x0), x1))
nats₁(x0)
zprimes

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

NATS₁(N) -> NATS₁(s₁(N))

The TRS R consists of the following rules:

filter₃(cons₂(X, Y), 0, M) -> cons₂(0, filter₃(Y, M, M))
filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(X, filter₃(Y, N, M))
sieve₁(cons₂(0, Y)) -> cons₂(0, sieve₁(Y))
sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(N), sieve₁(filter₃(Y, N, N)))
nats₁(N) -> cons₂(N, nats₁(s₁(N)))
zprimes -> sieve₁(nats₁(s₁(s₁(0))))

The set Q consists of the following terms:

filter₃(cons₂(x0, x1), 0, x2)
filter₃(cons₂(x0, x1), s₁(x2), x3)
sieve₁(cons₂(0, x0))
sieve₁(cons₂(s₁(x0), x1))
nats₁(x0)
zprimes

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

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

FILTER₃(cons₂(X, Y), 0, M) -> FILTER₃(Y, M, M)
FILTER₃(cons₂(X, Y), s₁(N), M) -> FILTER₃(Y, N, M)

The TRS R consists of the following rules:

filter₃(cons₂(X, Y), 0, M) -> cons₂(0, filter₃(Y, M, M))
filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(X, filter₃(Y, N, M))
sieve₁(cons₂(0, Y)) -> cons₂(0, sieve₁(Y))
sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(N), sieve₁(filter₃(Y, N, N)))
nats₁(N) -> cons₂(N, nats₁(s₁(N)))
zprimes -> sieve₁(nats₁(s₁(s₁(0))))

The set Q consists of the following terms:

filter₃(cons₂(x0, x1), 0, x2)
filter₃(cons₂(x0, x1), s₁(x2), x3)
sieve₁(cons₂(0, x0))
sieve₁(cons₂(s₁(x0), x1))
nats₁(x0)
zprimes

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

FILTER₃(cons₂(X, Y), 0, M) -> FILTER₃(Y, M, M)
FILTER₃(cons₂(X, Y), s₁(N), M) -> FILTER₃(Y, N, M)

Used argument filtering: FILTER₃(x1, x2, x3) = x1
cons₂(x1, x2) = cons₁(x2)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

filter₃(cons₂(X, Y), 0, M) -> cons₂(0, filter₃(Y, M, M))
filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(X, filter₃(Y, N, M))
sieve₁(cons₂(0, Y)) -> cons₂(0, sieve₁(Y))
sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(N), sieve₁(filter₃(Y, N, N)))
nats₁(N) -> cons₂(N, nats₁(s₁(N)))
zprimes -> sieve₁(nats₁(s₁(s₁(0))))

The set Q consists of the following terms:

filter₃(cons₂(x0, x1), 0, x2)
filter₃(cons₂(x0, x1), s₁(x2), x3)
sieve₁(cons₂(0, x0))
sieve₁(cons₂(s₁(x0), x1))
nats₁(x0)
zprimes

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

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

SIEVE₁(cons₂(s₁(N), Y)) -> SIEVE₁(filter₃(Y, N, N))
SIEVE₁(cons₂(0, Y)) -> SIEVE₁(Y)

The TRS R consists of the following rules:

filter₃(cons₂(X, Y), 0, M) -> cons₂(0, filter₃(Y, M, M))
filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(X, filter₃(Y, N, M))
sieve₁(cons₂(0, Y)) -> cons₂(0, sieve₁(Y))
sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(N), sieve₁(filter₃(Y, N, N)))
nats₁(N) -> cons₂(N, nats₁(s₁(N)))
zprimes -> sieve₁(nats₁(s₁(s₁(0))))

The set Q consists of the following terms:

filter₃(cons₂(x0, x1), 0, x2)
filter₃(cons₂(x0, x1), s₁(x2), x3)
sieve₁(cons₂(0, x0))
sieve₁(cons₂(s₁(x0), x1))
nats₁(x0)
zprimes

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SIEVE₁(cons₂(s₁(N), Y)) -> SIEVE₁(filter₃(Y, N, N))
SIEVE₁(cons₂(0, Y)) -> SIEVE₁(Y)

Used argument filtering: SIEVE₁(x1) = x1
cons₂(x1, x2) = cons₁(x2)
filter₃(x1, x2, x3) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

filter₃(cons₂(X, Y), 0, M) -> cons₂(0, filter₃(Y, M, M))
filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(X, filter₃(Y, N, M))
sieve₁(cons₂(0, Y)) -> cons₂(0, sieve₁(Y))
sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(N), sieve₁(filter₃(Y, N, N)))
nats₁(N) -> cons₂(N, nats₁(s₁(N)))
zprimes -> sieve₁(nats₁(s₁(s₁(0))))

The set Q consists of the following terms:

filter₃(cons₂(x0, x1), 0, x2)
filter₃(cons₂(x0, x1), s₁(x2), x3)
sieve₁(cons₂(0, x0))
sieve₁(cons₂(s₁(x0), x1))
nats₁(x0)
zprimes

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.