Termination w.r.t. Q proof of TRS/nontermin/CSREx9

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

  ↳ 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))))

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:

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))))

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:

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))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ 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))))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(FILTER₃(x₁, x₂, x₃)) = 3·x₁
POL(cons₂(x₁, x₂)) = 3 + x₂
POL(s₁(x₁)) = 0

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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))))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(SIEVE₁(x₁)) = 2·x₁
POL(cons₂(x₁, x₂)) = 1 + 2·x₂
POL(filter₃(x₁, x₂, x₃)) = 2·x₁
POL(s₁(x₁)) = 0

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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))))

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