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:

filter₃(cons₂(X, Y), 0, M) -> cons₂(0, n__filter₃(activate₁(Y), M, M))
filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(X, n__filter₃(activate₁(Y), N, M))
sieve₁(cons₂(0, Y)) -> cons₂(0, n__sieve₁(activate₁(Y)))
sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(N), n__sieve₁(filter₃(activate₁(Y), N, N)))
nats₁(N) -> cons₂(N, n__nats₁(s₁(N)))
zprimes -> sieve₁(nats₁(s₁(s₁(0))))
filter₃(X1, X2, X3) -> n__filter₃(X1, X2, X3)
sieve₁(X) -> n__sieve₁(X)
nats₁(X) -> n__nats₁(X)
activate₁(n__filter₃(X1, X2, X3)) -> filter₃(X1, X2, X3)
activate₁(n__sieve₁(X)) -> sieve₁(X)
activate₁(n__nats₁(X)) -> nats₁(X)
activate₁(X) -> X

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, n__filter₃(activate₁(Y), M, M))
filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(X, n__filter₃(activate₁(Y), N, M))
sieve₁(cons₂(0, Y)) -> cons₂(0, n__sieve₁(activate₁(Y)))
sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(N), n__sieve₁(filter₃(activate₁(Y), N, N)))
nats₁(N) -> cons₂(N, n__nats₁(s₁(N)))
zprimes -> sieve₁(nats₁(s₁(s₁(0))))
filter₃(X1, X2, X3) -> n__filter₃(X1, X2, X3)
sieve₁(X) -> n__sieve₁(X)
nats₁(X) -> n__nats₁(X)
activate₁(n__filter₃(X1, X2, X3)) -> filter₃(X1, X2, X3)
activate₁(n__sieve₁(X)) -> sieve₁(X)
activate₁(n__nats₁(X)) -> nats₁(X)
activate₁(X) -> X

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:

ACTIVATE₁(n__nats₁(X)) -> NATS₁(X)
SIEVE₁(cons₂(s₁(N), Y)) -> ACTIVATE₁(Y)
FILTER₃(cons₂(X, Y), s₁(N), M) -> ACTIVATE₁(Y)
ZPRIMES -> SIEVE₁(nats₁(s₁(s₁(0))))
SIEVE₁(cons₂(0, Y)) -> ACTIVATE₁(Y)
SIEVE₁(cons₂(s₁(N), Y)) -> FILTER₃(activate₁(Y), N, N)
ZPRIMES -> NATS₁(s₁(s₁(0)))
FILTER₃(cons₂(X, Y), 0, M) -> ACTIVATE₁(Y)
ACTIVATE₁(n__filter₃(X1, X2, X3)) -> FILTER₃(X1, X2, X3)
ACTIVATE₁(n__sieve₁(X)) -> SIEVE₁(X)

The TRS R consists of the following rules:

filter₃(cons₂(X, Y), 0, M) -> cons₂(0, n__filter₃(activate₁(Y), M, M))
filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(X, n__filter₃(activate₁(Y), N, M))
sieve₁(cons₂(0, Y)) -> cons₂(0, n__sieve₁(activate₁(Y)))
sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(N), n__sieve₁(filter₃(activate₁(Y), N, N)))
nats₁(N) -> cons₂(N, n__nats₁(s₁(N)))
zprimes -> sieve₁(nats₁(s₁(s₁(0))))
filter₃(X1, X2, X3) -> n__filter₃(X1, X2, X3)
sieve₁(X) -> n__sieve₁(X)
nats₁(X) -> n__nats₁(X)
activate₁(n__filter₃(X1, X2, X3)) -> filter₃(X1, X2, X3)
activate₁(n__sieve₁(X)) -> sieve₁(X)
activate₁(n__nats₁(X)) -> nats₁(X)
activate₁(X) -> X

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:

ACTIVATE₁(n__nats₁(X)) -> NATS₁(X)
SIEVE₁(cons₂(s₁(N), Y)) -> ACTIVATE₁(Y)
FILTER₃(cons₂(X, Y), s₁(N), M) -> ACTIVATE₁(Y)
ZPRIMES -> SIEVE₁(nats₁(s₁(s₁(0))))
SIEVE₁(cons₂(0, Y)) -> ACTIVATE₁(Y)
SIEVE₁(cons₂(s₁(N), Y)) -> FILTER₃(activate₁(Y), N, N)
ZPRIMES -> NATS₁(s₁(s₁(0)))
FILTER₃(cons₂(X, Y), 0, M) -> ACTIVATE₁(Y)
ACTIVATE₁(n__filter₃(X1, X2, X3)) -> FILTER₃(X1, X2, X3)
ACTIVATE₁(n__sieve₁(X)) -> SIEVE₁(X)

The TRS R consists of the following rules:

filter₃(cons₂(X, Y), 0, M) -> cons₂(0, n__filter₃(activate₁(Y), M, M))
filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(X, n__filter₃(activate₁(Y), N, M))
sieve₁(cons₂(0, Y)) -> cons₂(0, n__sieve₁(activate₁(Y)))
sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(N), n__sieve₁(filter₃(activate₁(Y), N, N)))
nats₁(N) -> cons₂(N, n__nats₁(s₁(N)))
zprimes -> sieve₁(nats₁(s₁(s₁(0))))
filter₃(X1, X2, X3) -> n__filter₃(X1, X2, X3)
sieve₁(X) -> n__sieve₁(X)
nats₁(X) -> n__nats₁(X)
activate₁(n__filter₃(X1, X2, X3)) -> filter₃(X1, X2, X3)
activate₁(n__sieve₁(X)) -> sieve₁(X)
activate₁(n__nats₁(X)) -> nats₁(X)
activate₁(X) -> X

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 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

SIEVE₁(cons₂(s₁(N), Y)) -> ACTIVATE₁(Y)
FILTER₃(cons₂(X, Y), s₁(N), M) -> ACTIVATE₁(Y)
SIEVE₁(cons₂(0, Y)) -> ACTIVATE₁(Y)
SIEVE₁(cons₂(s₁(N), Y)) -> FILTER₃(activate₁(Y), N, N)
FILTER₃(cons₂(X, Y), 0, M) -> ACTIVATE₁(Y)
ACTIVATE₁(n__filter₃(X1, X2, X3)) -> FILTER₃(X1, X2, X3)
ACTIVATE₁(n__sieve₁(X)) -> SIEVE₁(X)

The TRS R consists of the following rules:

filter₃(cons₂(X, Y), 0, M) -> cons₂(0, n__filter₃(activate₁(Y), M, M))
filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(X, n__filter₃(activate₁(Y), N, M))
sieve₁(cons₂(0, Y)) -> cons₂(0, n__sieve₁(activate₁(Y)))
sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(N), n__sieve₁(filter₃(activate₁(Y), N, N)))
nats₁(N) -> cons₂(N, n__nats₁(s₁(N)))
zprimes -> sieve₁(nats₁(s₁(s₁(0))))
filter₃(X1, X2, X3) -> n__filter₃(X1, X2, X3)
sieve₁(X) -> n__sieve₁(X)
nats₁(X) -> n__nats₁(X)
activate₁(n__filter₃(X1, X2, X3)) -> filter₃(X1, X2, X3)
activate₁(n__sieve₁(X)) -> sieve₁(X)
activate₁(n__nats₁(X)) -> nats₁(X)
activate₁(X) -> X

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)) -> ACTIVATE₁(Y)
SIEVE₁(cons₂(0, Y)) -> ACTIVATE₁(Y)
SIEVE₁(cons₂(s₁(N), Y)) -> FILTER₃(activate₁(Y), N, N)
ACTIVATE₁(n__sieve₁(X)) -> SIEVE₁(X)

The remaining pairs can at least be oriented weakly.

FILTER₃(cons₂(X, Y), s₁(N), M) -> ACTIVATE₁(Y)
FILTER₃(cons₂(X, Y), 0, M) -> ACTIVATE₁(Y)
ACTIVATE₁(n__filter₃(X1, X2, X3)) -> FILTER₃(X1, X2, X3)

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(ACTIVATE₁(x₁)) = x₁
POL(FILTER₃(x₁, x₂, x₃)) = x₁
POL(SIEVE₁(x₁)) = 1 + 2·x₁
POL(activate₁(x₁)) = x₁
POL(cons₂(x₁, x₂)) = x₂
POL(filter₃(x₁, x₂, x₃)) = x₁
POL(n__filter₃(x₁, x₂, x₃)) = x₁
POL(n__nats₁(x₁)) = 1
POL(n__sieve₁(x₁)) = 2 + 2·x₁
POL(nats₁(x₁)) = 1
POL(s₁(x₁)) = 0
POL(sieve₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented:

activate₁(X) -> X
filter₃(X1, X2, X3) -> n__filter₃(X1, X2, X3)
filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(X, n__filter₃(activate₁(Y), N, M))
sieve₁(X) -> n__sieve₁(X)
sieve₁(cons₂(0, Y)) -> cons₂(0, n__sieve₁(activate₁(Y)))
activate₁(n__nats₁(X)) -> nats₁(X)
filter₃(cons₂(X, Y), 0, M) -> cons₂(0, n__filter₃(activate₁(Y), M, M))
nats₁(X) -> n__nats₁(X)
sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(N), n__sieve₁(filter₃(activate₁(Y), N, N)))
activate₁(n__sieve₁(X)) -> sieve₁(X)
nats₁(N) -> cons₂(N, n__nats₁(s₁(N)))
activate₁(n__filter₃(X1, X2, X3)) -> filter₃(X1, X2, X3)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

FILTER₃(cons₂(X, Y), s₁(N), M) -> ACTIVATE₁(Y)
FILTER₃(cons₂(X, Y), 0, M) -> ACTIVATE₁(Y)
ACTIVATE₁(n__filter₃(X1, X2, X3)) -> FILTER₃(X1, X2, X3)

The TRS R consists of the following rules:

filter₃(cons₂(X, Y), 0, M) -> cons₂(0, n__filter₃(activate₁(Y), M, M))
filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(X, n__filter₃(activate₁(Y), N, M))
sieve₁(cons₂(0, Y)) -> cons₂(0, n__sieve₁(activate₁(Y)))
sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(N), n__sieve₁(filter₃(activate₁(Y), N, N)))
nats₁(N) -> cons₂(N, n__nats₁(s₁(N)))
zprimes -> sieve₁(nats₁(s₁(s₁(0))))
filter₃(X1, X2, X3) -> n__filter₃(X1, X2, X3)
sieve₁(X) -> n__sieve₁(X)
nats₁(X) -> n__nats₁(X)
activate₁(n__filter₃(X1, X2, X3)) -> filter₃(X1, X2, X3)
activate₁(n__sieve₁(X)) -> sieve₁(X)
activate₁(n__nats₁(X)) -> nats₁(X)
activate₁(X) -> X

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), s₁(N), M) -> ACTIVATE₁(Y)
FILTER₃(cons₂(X, Y), 0, M) -> ACTIVATE₁(Y)
ACTIVATE₁(n__filter₃(X1, X2, X3)) -> FILTER₃(X1, X2, X3)

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

POL(0) = 2
POL(ACTIVATE₁(x₁)) = 1 + 2·x₁
POL(FILTER₃(x₁, x₂, x₃)) = 2 + 2·x₁ + x₂ + x₃
POL(cons₂(x₁, x₂)) = 1 + 2·x₂
POL(n__filter₃(x₁, x₂, x₃)) = 2 + 2·x₁ + x₂ + x₃
POL(s₁(x₁)) = 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ 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, n__filter₃(activate₁(Y), M, M))
filter₃(cons₂(X, Y), s₁(N), M) -> cons₂(X, n__filter₃(activate₁(Y), N, M))
sieve₁(cons₂(0, Y)) -> cons₂(0, n__sieve₁(activate₁(Y)))
sieve₁(cons₂(s₁(N), Y)) -> cons₂(s₁(N), n__sieve₁(filter₃(activate₁(Y), N, N)))
nats₁(N) -> cons₂(N, n__nats₁(s₁(N)))
zprimes -> sieve₁(nats₁(s₁(s₁(0))))
filter₃(X1, X2, X3) -> n__filter₃(X1, X2, X3)
sieve₁(X) -> n__sieve₁(X)
nats₁(X) -> n__nats₁(X)
activate₁(n__filter₃(X1, X2, X3)) -> filter₃(X1, X2, X3)
activate₁(n__sieve₁(X)) -> sieve₁(X)
activate₁(n__nats₁(X)) -> nats₁(X)
activate₁(X) -> X

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.