Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/TRCSRSLASHEx9_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 strictly oriented and are deleted.

ACTIVATE₁(n__sieve₁(X)) -> SIEVE₁(X)

The remaining pairs can at least by weakly be oriented.

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)

Used ordering: Combined order from the following AFS and order.
SIEVE₁(x1) = SIEVE₁(x1)
cons₂(x1, x2) = x2
s₁(x1) = s
ACTIVATE₁(x1) = ACTIVATE₁(x1)
FILTER₃(x1, x2, x3) = FILTER₁(x1)
0 = 0
activate₁(x1) = x1
n__filter₃(x1, x2, x3) = x1
n__sieve₁(x1) = n__sieve₁(x1)
filter₃(x1, x2, x3) = x1
sieve₁(x1) = sieve₁(x1)
n__nats₁(x1) = n__nats₁(x1)
nats₁(x1) = nats₁(x1)

Lexicographic Path Order [19].
Precedence:

[SIEVE₁, ACTIVATE₁, FILTER₁, 0, n_{_sieve₁}, sieve_1] > s
[n_{_nats₁}, nats_1] > s

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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)

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

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ 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 strictly oriented and are deleted.

FILTER₃(cons₂(X, Y), s₁(N), M) -> ACTIVATE₁(Y)
FILTER₃(cons₂(X, Y), 0, M) -> ACTIVATE₁(Y)

The remaining pairs can at least by weakly be oriented.

ACTIVATE₁(n__filter₃(X1, X2, X3)) -> FILTER₃(X1, X2, X3)

Used ordering: Combined order from the following AFS and order.
FILTER₃(x1, x2, x3) = FILTER₂(x1, x3)
cons₂(x1, x2) = cons₁(x2)
s₁(x1) = s₁(x1)
ACTIVATE₁(x1) = x1
0 = 0
n__filter₃(x1, x2, x3) = n__filter₂(x1, x3)

Lexicographic Path Order [19].
Precedence:

[FILTER₂, n_{_{filter_2]}}

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

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

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.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.