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:

a__filter₃(cons₂(X, Y), 0, M) -> cons₂(0, filter₃(Y, M, M))
a__filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(mark₁(X), filter₃(Y, N, M))
a__sieve₁(cons₂(0, Y)) -> cons₂(0, sieve₁(Y))
a__sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(mark₁(N)), sieve₁(filter₃(Y, N, N)))
a__nats₁(N) -> cons₂(mark₁(N), nats₁(s₁(N)))
a__zprimes -> a__sieve₁(a__nats₁(s₁(s₁(0))))
mark₁(filter₃(X1, X2, X3)) -> a__filter₃(mark₁(X1), mark₁(X2), mark₁(X3))
mark₁(sieve₁(X)) -> a__sieve₁(mark₁(X))
mark₁(nats₁(X)) -> a__nats₁(mark₁(X))
mark₁(zprimes) -> a__zprimes
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
a__filter₃(X1, X2, X3) -> filter₃(X1, X2, X3)
a__sieve₁(X) -> sieve₁(X)
a__nats₁(X) -> nats₁(X)
a__zprimes -> zprimes

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__filter₃(cons₂(X, Y), 0, M) -> cons₂(0, filter₃(Y, M, M))
a__filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(mark₁(X), filter₃(Y, N, M))
a__sieve₁(cons₂(0, Y)) -> cons₂(0, sieve₁(Y))
a__sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(mark₁(N)), sieve₁(filter₃(Y, N, N)))
a__nats₁(N) -> cons₂(mark₁(N), nats₁(s₁(N)))
a__zprimes -> a__sieve₁(a__nats₁(s₁(s₁(0))))
mark₁(filter₃(X1, X2, X3)) -> a__filter₃(mark₁(X1), mark₁(X2), mark₁(X3))
mark₁(sieve₁(X)) -> a__sieve₁(mark₁(X))
mark₁(nats₁(X)) -> a__nats₁(mark₁(X))
mark₁(zprimes) -> a__zprimes
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
a__filter₃(X1, X2, X3) -> filter₃(X1, X2, X3)
a__sieve₁(X) -> sieve₁(X)
a__nats₁(X) -> nats₁(X)
a__zprimes -> zprimes

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:

A__ZPRIMES -> A__NATS₁(s₁(s₁(0)))
MARK₁(filter₃(X1, X2, X3)) -> MARK₁(X1)
MARK₁(filter₃(X1, X2, X3)) -> MARK₁(X3)
MARK₁(filter₃(X1, X2, X3)) -> MARK₁(X2)
A__NATS₁(N) -> MARK₁(N)
MARK₁(nats₁(X)) -> A__NATS₁(mark₁(X))
MARK₁(filter₃(X1, X2, X3)) -> A__FILTER₃(mark₁(X1), mark₁(X2), mark₁(X3))
MARK₁(s₁(X)) -> MARK₁(X)
A__ZPRIMES -> A__SIEVE₁(a__nats₁(s₁(s₁(0))))
MARK₁(sieve₁(X)) -> A__SIEVE₁(mark₁(X))
MARK₁(nats₁(X)) -> MARK₁(X)
MARK₁(zprimes) -> A__ZPRIMES
A__SIEVE₁(cons₂(s₁(N), Y)) -> MARK₁(N)
A__FILTER₃(cons₂(X, Y), s₁(N), M) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(sieve₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__filter₃(cons₂(X, Y), 0, M) -> cons₂(0, filter₃(Y, M, M))
a__filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(mark₁(X), filter₃(Y, N, M))
a__sieve₁(cons₂(0, Y)) -> cons₂(0, sieve₁(Y))
a__sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(mark₁(N)), sieve₁(filter₃(Y, N, N)))
a__nats₁(N) -> cons₂(mark₁(N), nats₁(s₁(N)))
a__zprimes -> a__sieve₁(a__nats₁(s₁(s₁(0))))
mark₁(filter₃(X1, X2, X3)) -> a__filter₃(mark₁(X1), mark₁(X2), mark₁(X3))
mark₁(sieve₁(X)) -> a__sieve₁(mark₁(X))
mark₁(nats₁(X)) -> a__nats₁(mark₁(X))
mark₁(zprimes) -> a__zprimes
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
a__filter₃(X1, X2, X3) -> filter₃(X1, X2, X3)
a__sieve₁(X) -> sieve₁(X)
a__nats₁(X) -> nats₁(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:

A__ZPRIMES -> A__NATS₁(s₁(s₁(0)))
MARK₁(filter₃(X1, X2, X3)) -> MARK₁(X1)
MARK₁(filter₃(X1, X2, X3)) -> MARK₁(X3)
MARK₁(filter₃(X1, X2, X3)) -> MARK₁(X2)
A__NATS₁(N) -> MARK₁(N)
MARK₁(nats₁(X)) -> A__NATS₁(mark₁(X))
MARK₁(filter₃(X1, X2, X3)) -> A__FILTER₃(mark₁(X1), mark₁(X2), mark₁(X3))
MARK₁(s₁(X)) -> MARK₁(X)
A__ZPRIMES -> A__SIEVE₁(a__nats₁(s₁(s₁(0))))
MARK₁(sieve₁(X)) -> A__SIEVE₁(mark₁(X))
MARK₁(nats₁(X)) -> MARK₁(X)
MARK₁(zprimes) -> A__ZPRIMES
A__SIEVE₁(cons₂(s₁(N), Y)) -> MARK₁(N)
A__FILTER₃(cons₂(X, Y), s₁(N), M) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(sieve₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__filter₃(cons₂(X, Y), 0, M) -> cons₂(0, filter₃(Y, M, M))
a__filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(mark₁(X), filter₃(Y, N, M))
a__sieve₁(cons₂(0, Y)) -> cons₂(0, sieve₁(Y))
a__sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(mark₁(N)), sieve₁(filter₃(Y, N, N)))
a__nats₁(N) -> cons₂(mark₁(N), nats₁(s₁(N)))
a__zprimes -> a__sieve₁(a__nats₁(s₁(s₁(0))))
mark₁(filter₃(X1, X2, X3)) -> a__filter₃(mark₁(X1), mark₁(X2), mark₁(X3))
mark₁(sieve₁(X)) -> a__sieve₁(mark₁(X))
mark₁(nats₁(X)) -> a__nats₁(mark₁(X))
mark₁(zprimes) -> a__zprimes
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
a__filter₃(X1, X2, X3) -> filter₃(X1, X2, X3)
a__sieve₁(X) -> sieve₁(X)
a__nats₁(X) -> nats₁(X)
a__zprimes -> zprimes

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.

A__ZPRIMES -> A__NATS₁(s₁(s₁(0)))
MARK₁(filter₃(X1, X2, X3)) -> MARK₁(X1)
MARK₁(filter₃(X1, X2, X3)) -> MARK₁(X3)
MARK₁(filter₃(X1, X2, X3)) -> MARK₁(X2)
MARK₁(filter₃(X1, X2, X3)) -> A__FILTER₃(mark₁(X1), mark₁(X2), mark₁(X3))
A__ZPRIMES -> A__SIEVE₁(a__nats₁(s₁(s₁(0))))
MARK₁(sieve₁(X)) -> A__SIEVE₁(mark₁(X))
MARK₁(zprimes) -> A__ZPRIMES
MARK₁(sieve₁(X)) -> MARK₁(X)

The remaining pairs can at least be oriented weakly.

A__NATS₁(N) -> MARK₁(N)
MARK₁(nats₁(X)) -> A__NATS₁(mark₁(X))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(nats₁(X)) -> MARK₁(X)
A__SIEVE₁(cons₂(s₁(N), Y)) -> MARK₁(N)
A__FILTER₃(cons₂(X, Y), s₁(N), M) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( A__NATS₁(x₁) ) = 2x₁

POL( filter₃(x₁, ..., x₃) ) = 2x₁ + 3x₂ + 3x₃ + 1

POL( mark₁(x₁) ) = x₁

POL( sieve₁(x₁) ) = x₁ + 2

POL( A__SIEVE₁(x₁) ) = 2x₁

POL( A__FILTER₃(x₁, ..., x₃) ) = x₁

POL( A__ZPRIMES ) = 1

POL( MARK₁(x₁) ) = 2x₁

POL( 0 ) = max{0, -3}

POL( a__sieve₁(x₁) ) = x₁ + 2

POL( cons₂(x₁, x₂) ) = 2x₁

POL( nats₁(x₁) ) = 3x₁

POL( a__zprimes ) = 3

POL( a__filter₃(x₁, ..., x₃) ) = 2x₁ + 3x₂ + 3x₃ + 1

POL( zprimes ) = 3

POL( s₁(x₁) ) = 2x₁

POL( a__nats₁(x₁) ) = 3x₁

The following usable rules [14] were oriented:

a__filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(mark₁(X), filter₃(Y, N, M))
mark₁(0) -> 0
a__zprimes -> a__sieve₁(a__nats₁(s₁(s₁(0))))
mark₁(sieve₁(X)) -> a__sieve₁(mark₁(X))
mark₁(s₁(X)) -> s₁(mark₁(X))
a__filter₃(cons₂(X, Y), 0, M) -> cons₂(0, filter₃(Y, M, M))
a__nats₁(N) -> cons₂(mark₁(N), nats₁(s₁(N)))
mark₁(nats₁(X)) -> a__nats₁(mark₁(X))
mark₁(filter₃(X1, X2, X3)) -> a__filter₃(mark₁(X1), mark₁(X2), mark₁(X3))
a__filter₃(X1, X2, X3) -> filter₃(X1, X2, X3)
a__zprimes -> zprimes
mark₁(zprimes) -> a__zprimes
a__sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(mark₁(N)), sieve₁(filter₃(Y, N, N)))
a__sieve₁(cons₂(0, Y)) -> cons₂(0, sieve₁(Y))
a__sieve₁(X) -> sieve₁(X)
a__nats₁(X) -> nats₁(X)
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(nats₁(X)) -> MARK₁(X)
A__SIEVE₁(cons₂(s₁(N), Y)) -> MARK₁(N)
A__NATS₁(N) -> MARK₁(N)
MARK₁(nats₁(X)) -> A__NATS₁(mark₁(X))
A__FILTER₃(cons₂(X, Y), s₁(N), M) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__filter₃(cons₂(X, Y), 0, M) -> cons₂(0, filter₃(Y, M, M))
a__filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(mark₁(X), filter₃(Y, N, M))
a__sieve₁(cons₂(0, Y)) -> cons₂(0, sieve₁(Y))
a__sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(mark₁(N)), sieve₁(filter₃(Y, N, N)))
a__nats₁(N) -> cons₂(mark₁(N), nats₁(s₁(N)))
a__zprimes -> a__sieve₁(a__nats₁(s₁(s₁(0))))
mark₁(filter₃(X1, X2, X3)) -> a__filter₃(mark₁(X1), mark₁(X2), mark₁(X3))
mark₁(sieve₁(X)) -> a__sieve₁(mark₁(X))
mark₁(nats₁(X)) -> a__nats₁(mark₁(X))
mark₁(zprimes) -> a__zprimes
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
a__filter₃(X1, X2, X3) -> filter₃(X1, X2, X3)
a__sieve₁(X) -> sieve₁(X)
a__nats₁(X) -> nats₁(X)
a__zprimes -> zprimes

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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

MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(nats₁(X)) -> MARK₁(X)
A__NATS₁(N) -> MARK₁(N)
MARK₁(nats₁(X)) -> A__NATS₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__filter₃(cons₂(X, Y), 0, M) -> cons₂(0, filter₃(Y, M, M))
a__filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(mark₁(X), filter₃(Y, N, M))
a__sieve₁(cons₂(0, Y)) -> cons₂(0, sieve₁(Y))
a__sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(mark₁(N)), sieve₁(filter₃(Y, N, N)))
a__nats₁(N) -> cons₂(mark₁(N), nats₁(s₁(N)))
a__zprimes -> a__sieve₁(a__nats₁(s₁(s₁(0))))
mark₁(filter₃(X1, X2, X3)) -> a__filter₃(mark₁(X1), mark₁(X2), mark₁(X3))
mark₁(sieve₁(X)) -> a__sieve₁(mark₁(X))
mark₁(nats₁(X)) -> a__nats₁(mark₁(X))
mark₁(zprimes) -> a__zprimes
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
a__filter₃(X1, X2, X3) -> filter₃(X1, X2, X3)
a__sieve₁(X) -> sieve₁(X)
a__nats₁(X) -> nats₁(X)
a__zprimes -> zprimes

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.

MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(nats₁(X)) -> MARK₁(X)
MARK₁(nats₁(X)) -> A__NATS₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The remaining pairs can at least be oriented weakly.

A__NATS₁(N) -> MARK₁(N)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( A__NATS₁(x₁) ) = 2x₁

POL( mark₁(x₁) ) = x₁

POL( filter₃(x₁, ..., x₃) ) = x₁ + 2x₂ + 2

POL( sieve₁(x₁) ) = x₁

POL( MARK₁(x₁) ) = max{0, 2x₁ - 1}

POL( 0 ) = 0

POL( cons₂(x₁, x₂) ) = x₁ + 1

POL( a__sieve₁(x₁) ) = x₁

POL( nats₁(x₁) ) = x₁ + 1

POL( a__zprimes ) = 3

POL( a__filter₃(x₁, ..., x₃) ) = x₁ + 2x₂ + 2

POL( zprimes ) = 3

POL( s₁(x₁) ) = x₁ + 1

POL( a__nats₁(x₁) ) = x₁ + 1

The following usable rules [14] were oriented:

mark₁(0) -> 0
a__filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(mark₁(X), filter₃(Y, N, M))
a__zprimes -> a__sieve₁(a__nats₁(s₁(s₁(0))))
mark₁(sieve₁(X)) -> a__sieve₁(mark₁(X))
mark₁(s₁(X)) -> s₁(mark₁(X))
a__filter₃(cons₂(X, Y), 0, M) -> cons₂(0, filter₃(Y, M, M))
a__nats₁(N) -> cons₂(mark₁(N), nats₁(s₁(N)))
mark₁(nats₁(X)) -> a__nats₁(mark₁(X))
mark₁(filter₃(X1, X2, X3)) -> a__filter₃(mark₁(X1), mark₁(X2), mark₁(X3))
a__filter₃(X1, X2, X3) -> filter₃(X1, X2, X3)
a__zprimes -> zprimes
mark₁(zprimes) -> a__zprimes
a__sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(mark₁(N)), sieve₁(filter₃(Y, N, N)))
a__sieve₁(cons₂(0, Y)) -> cons₂(0, sieve₁(Y))
a__nats₁(X) -> nats₁(X)
a__sieve₁(X) -> sieve₁(X)
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

A__NATS₁(N) -> MARK₁(N)

The TRS R consists of the following rules:

a__filter₃(cons₂(X, Y), 0, M) -> cons₂(0, filter₃(Y, M, M))
a__filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(mark₁(X), filter₃(Y, N, M))
a__sieve₁(cons₂(0, Y)) -> cons₂(0, sieve₁(Y))
a__sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(mark₁(N)), sieve₁(filter₃(Y, N, N)))
a__nats₁(N) -> cons₂(mark₁(N), nats₁(s₁(N)))
a__zprimes -> a__sieve₁(a__nats₁(s₁(s₁(0))))
mark₁(filter₃(X1, X2, X3)) -> a__filter₃(mark₁(X1), mark₁(X2), mark₁(X3))
mark₁(sieve₁(X)) -> a__sieve₁(mark₁(X))
mark₁(nats₁(X)) -> a__nats₁(mark₁(X))
mark₁(zprimes) -> a__zprimes
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
a__filter₃(X1, X2, X3) -> filter₃(X1, X2, X3)
a__sieve₁(X) -> sieve₁(X)
a__nats₁(X) -> nats₁(X)
a__zprimes -> zprimes

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 1 less node.