Termination w.r.t. Q proof of /tmp/aprove_benchmark_example

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:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MARK(zprimes) → A__ZPRIMES
MARK(s(X)) → MARK(X)
MARK(filter(X1, X2, X3)) → MARK(X3)
A__SIEVE(cons(s(N), Y)) → MARK(N)
MARK(cons(X1, X2)) → MARK(X1)
MARK(filter(X1, X2, X3)) → A__FILTER(mark(X1), mark(X2), mark(X3))
MARK(nats(X)) → A__NATS(mark(X))
MARK(filter(X1, X2, X3)) → MARK(X2)
A__FILTER(cons(X, Y), s(N), M) → MARK(X)
MARK(nats(X)) → MARK(X)
MARK(sieve(X)) → A__SIEVE(mark(X))
A__NATS(N) → MARK(N)
A__ZPRIMES → A__NATS(s(s(0)))
A__ZPRIMES → A__SIEVE(a__nats(s(s(0))))
MARK(sieve(X)) → MARK(X)
MARK(filter(X1, X2, X3)) → MARK(X1)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

MARK(zprimes) → A__ZPRIMES
MARK(s(X)) → MARK(X)
MARK(filter(X1, X2, X3)) → MARK(X3)
A__SIEVE(cons(s(N), Y)) → MARK(N)
MARK(cons(X1, X2)) → MARK(X1)
MARK(filter(X1, X2, X3)) → A__FILTER(mark(X1), mark(X2), mark(X3))
MARK(nats(X)) → A__NATS(mark(X))
MARK(filter(X1, X2, X3)) → MARK(X2)
A__FILTER(cons(X, Y), s(N), M) → MARK(X)
MARK(nats(X)) → MARK(X)
MARK(sieve(X)) → A__SIEVE(mark(X))
A__NATS(N) → MARK(N)
A__ZPRIMES → A__NATS(s(s(0)))
A__ZPRIMES → A__SIEVE(a__nats(s(s(0))))
MARK(sieve(X)) → MARK(X)
MARK(filter(X1, X2, X3)) → MARK(X1)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

A__ZPRIMES → A__NATS(s(s(0)))
A__ZPRIMES → A__SIEVE(a__nats(s(s(0))))

The remaining pairs can at least be oriented weakly.

MARK(zprimes) → A__ZPRIMES
MARK(s(X)) → MARK(X)
MARK(filter(X1, X2, X3)) → MARK(X3)
A__SIEVE(cons(s(N), Y)) → MARK(N)
MARK(cons(X1, X2)) → MARK(X1)
MARK(filter(X1, X2, X3)) → A__FILTER(mark(X1), mark(X2), mark(X3))
MARK(nats(X)) → A__NATS(mark(X))
MARK(filter(X1, X2, X3)) → MARK(X2)
A__FILTER(cons(X, Y), s(N), M) → MARK(X)
MARK(nats(X)) → MARK(X)
MARK(sieve(X)) → A__SIEVE(mark(X))
A__NATS(N) → MARK(N)
MARK(sieve(X)) → MARK(X)
MARK(filter(X1, X2, X3)) → MARK(X1)

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(A__FILTER(x₁, x₂, x₃)) = x₁
POL(A__NATS(x₁)) = x₁
POL(A__SIEVE(x₁)) = x₁
POL(A__ZPRIMES) = 1
POL(MARK(x₁)) = x₁
POL(a__filter(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(a__nats(x₁)) = x₁
POL(a__sieve(x₁)) = x₁
POL(a__zprimes) = 1
POL(cons(x₁, x₂)) = x₁
POL(filter(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(mark(x₁)) = x₁
POL(nats(x₁)) = x₁
POL(s(x₁)) = x₁
POL(sieve(x₁)) = x₁
POL(zprimes) = 1

The following usable rules [17] were oriented:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
mark(zprimes) → a__zprimes
mark(nats(X)) → a__nats(mark(X))
mark(sieve(X)) → a__sieve(mark(X))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__zprimes → zprimes
a__nats(X) → nats(X)
a__sieve(X) → sieve(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

MARK(zprimes) → A__ZPRIMES
MARK(s(X)) → MARK(X)
MARK(filter(X1, X2, X3)) → MARK(X3)
A__SIEVE(cons(s(N), Y)) → MARK(N)
MARK(cons(X1, X2)) → MARK(X1)
MARK(filter(X1, X2, X3)) → A__FILTER(mark(X1), mark(X2), mark(X3))
MARK(nats(X)) → A__NATS(mark(X))
MARK(filter(X1, X2, X3)) → MARK(X2)
A__FILTER(cons(X, Y), s(N), M) → MARK(X)
MARK(nats(X)) → MARK(X)
MARK(sieve(X)) → A__SIEVE(mark(X))
A__NATS(N) → MARK(N)
MARK(sieve(X)) → MARK(X)
MARK(filter(X1, X2, X3)) → MARK(X1)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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

MARK(filter(X1, X2, X3)) → MARK(X3)
MARK(s(X)) → MARK(X)
A__SIEVE(cons(s(N), Y)) → MARK(N)
MARK(cons(X1, X2)) → MARK(X1)
MARK(filter(X1, X2, X3)) → A__FILTER(mark(X1), mark(X2), mark(X3))
MARK(nats(X)) → A__NATS(mark(X))
MARK(filter(X1, X2, X3)) → MARK(X2)
A__FILTER(cons(X, Y), s(N), M) → MARK(X)
MARK(nats(X)) → MARK(X)
MARK(sieve(X)) → A__SIEVE(mark(X))
A__NATS(N) → MARK(N)
MARK(sieve(X)) → MARK(X)
MARK(filter(X1, X2, X3)) → MARK(X1)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

MARK(nats(X)) → MARK(X)
A__NATS(N) → MARK(N)

The remaining pairs can at least be oriented weakly.

MARK(filter(X1, X2, X3)) → MARK(X3)
MARK(s(X)) → MARK(X)
A__SIEVE(cons(s(N), Y)) → MARK(N)
MARK(cons(X1, X2)) → MARK(X1)
MARK(filter(X1, X2, X3)) → A__FILTER(mark(X1), mark(X2), mark(X3))
MARK(nats(X)) → A__NATS(mark(X))
MARK(filter(X1, X2, X3)) → MARK(X2)
A__FILTER(cons(X, Y), s(N), M) → MARK(X)
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(sieve(X)) → MARK(X)
MARK(filter(X1, X2, X3)) → MARK(X1)

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(A__FILTER(x₁, x₂, x₃)) = x₁
POL(A__NATS(x₁)) = 1 + x₁
POL(A__SIEVE(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(a__filter(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(a__nats(x₁)) = 1 + x₁
POL(a__sieve(x₁)) = x₁
POL(a__zprimes) = 1
POL(cons(x₁, x₂)) = x₁
POL(filter(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(mark(x₁)) = x₁
POL(nats(x₁)) = 1 + x₁
POL(s(x₁)) = x₁
POL(sieve(x₁)) = x₁
POL(zprimes) = 1

The following usable rules [17] were oriented:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
mark(zprimes) → a__zprimes
mark(nats(X)) → a__nats(mark(X))
mark(sieve(X)) → a__sieve(mark(X))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__zprimes → zprimes
a__nats(X) → nats(X)
a__sieve(X) → sieve(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

MARK(filter(X1, X2, X3)) → A__FILTER(mark(X1), mark(X2), mark(X3))
MARK(nats(X)) → A__NATS(mark(X))
MARK(filter(X1, X2, X3)) → MARK(X2)
A__FILTER(cons(X, Y), s(N), M) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(filter(X1, X2, X3)) → MARK(X3)
MARK(sieve(X)) → A__SIEVE(mark(X))
A__SIEVE(cons(s(N), Y)) → MARK(N)
MARK(cons(X1, X2)) → MARK(X1)
MARK(sieve(X)) → MARK(X)
MARK(filter(X1, X2, X3)) → MARK(X1)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPOrderProof

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

MARK(filter(X1, X2, X3)) → A__FILTER(mark(X1), mark(X2), mark(X3))
MARK(filter(X1, X2, X3)) → MARK(X2)
MARK(s(X)) → MARK(X)
MARK(filter(X1, X2, X3)) → MARK(X3)
A__FILTER(cons(X, Y), s(N), M) → MARK(X)
MARK(sieve(X)) → A__SIEVE(mark(X))
A__SIEVE(cons(s(N), Y)) → MARK(N)
MARK(cons(X1, X2)) → MARK(X1)
MARK(sieve(X)) → MARK(X)
MARK(filter(X1, X2, X3)) → MARK(X1)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

MARK(filter(X1, X2, X3)) → A__FILTER(mark(X1), mark(X2), mark(X3))
MARK(filter(X1, X2, X3)) → MARK(X2)
MARK(filter(X1, X2, X3)) → MARK(X3)
A__SIEVE(cons(s(N), Y)) → MARK(N)
MARK(sieve(X)) → MARK(X)
MARK(filter(X1, X2, X3)) → MARK(X1)

The remaining pairs can at least be oriented weakly.

MARK(s(X)) → MARK(X)
A__FILTER(cons(X, Y), s(N), M) → MARK(X)
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(cons(X1, X2)) → MARK(X1)

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(A__FILTER(x₁, x₂, x₃)) = x₁
POL(A__SIEVE(x₁)) = 1 + x₁
POL(MARK(x₁)) = x₁
POL(a__filter(x₁, x₂, x₃)) = 1 + x₁ + x₂ + x₃
POL(a__nats(x₁)) = x₁
POL(a__sieve(x₁)) = 1 + x₁
POL(a__zprimes) = 1
POL(cons(x₁, x₂)) = x₁
POL(filter(x₁, x₂, x₃)) = 1 + x₁ + x₂ + x₃
POL(mark(x₁)) = x₁
POL(nats(x₁)) = x₁
POL(s(x₁)) = x₁
POL(sieve(x₁)) = 1 + x₁
POL(zprimes) = 1

The following usable rules [17] were oriented:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
mark(zprimes) → a__zprimes
mark(nats(X)) → a__nats(mark(X))
mark(sieve(X)) → a__sieve(mark(X))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__zprimes → zprimes
a__nats(X) → nats(X)
a__sieve(X) → sieve(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ DependencyGraphProof

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

A__FILTER(cons(X, Y), s(N), M) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

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

MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QDPSizeChangeProof

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

MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

MARK(s(X)) → MARK(X)
The graph contains the following edges 1 > 1
MARK(cons(X1, X2)) → MARK(X1)
The graph contains the following edges 1 > 1