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, nsieve1, sieve1] > s
[nnats1, nats1] > 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, nfilter2]


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.