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.

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 contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVATE1(n__sieve1(X)) -> SIEVE1(X)
Used argument filtering: SIEVE1(x1)  =  x1
cons2(x1, x2)  =  x2
ACTIVATE1(x1)  =  x1
FILTER3(x1, x2, x3)  =  x1
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__nats
nats1(x1)  =  nats
Used ordering: Quasi Precedence: [n__sieve_1, sieve_1] [n__nats, nats] 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ 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 contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVATE1(n__filter3(X1, X2, X3)) -> FILTER3(X1, X2, X3)
Used argument filtering: FILTER3(x1, x2, x3)  =  x1
cons2(x1, x2)  =  x2
ACTIVATE1(x1)  =  x1
n__filter3(x1, x2, x3)  =  n__filter1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPAfsSolverProof
                    ↳ QDP
                      ↳ DependencyGraphProof

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)

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 contains 0 SCCs with 2 less nodes.