Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

filter3(cons2(X, Y), 0, M) -> cons2(0, filter3(Y, M, M))
filter3(cons2(X, Y), s1(N), M) -> cons2(X, filter3(Y, N, M))
sieve1(cons2(0, Y)) -> cons2(0, sieve1(Y))
sieve1(cons2(s1(N), Y)) -> cons2(s1(N), sieve1(filter3(Y, N, N)))
nats1(N) -> cons2(N, nats1(s1(N)))
zprimes -> sieve1(nats1(s1(s1(0))))

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, filter3(Y, M, M))
filter3(cons2(X, Y), s1(N), M) -> cons2(X, filter3(Y, N, M))
sieve1(cons2(0, Y)) -> cons2(0, sieve1(Y))
sieve1(cons2(s1(N), Y)) -> cons2(s1(N), sieve1(filter3(Y, N, N)))
nats1(N) -> cons2(N, nats1(s1(N)))
zprimes -> sieve1(nats1(s1(s1(0))))

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:

SIEVE1(cons2(s1(N), Y)) -> SIEVE1(filter3(Y, N, N))
ZPRIMES -> SIEVE1(nats1(s1(s1(0))))
SIEVE1(cons2(s1(N), Y)) -> FILTER3(Y, N, N)
FILTER3(cons2(X, Y), s1(N), M) -> FILTER3(Y, N, M)
FILTER3(cons2(X, Y), 0, M) -> FILTER3(Y, M, M)
SIEVE1(cons2(0, Y)) -> SIEVE1(Y)
ZPRIMES -> NATS1(s1(s1(0)))
NATS1(N) -> NATS1(s1(N))

The TRS R consists of the following rules:

filter3(cons2(X, Y), 0, M) -> cons2(0, filter3(Y, M, M))
filter3(cons2(X, Y), s1(N), M) -> cons2(X, filter3(Y, N, M))
sieve1(cons2(0, Y)) -> cons2(0, sieve1(Y))
sieve1(cons2(s1(N), Y)) -> cons2(s1(N), sieve1(filter3(Y, N, N)))
nats1(N) -> cons2(N, nats1(s1(N)))
zprimes -> sieve1(nats1(s1(s1(0))))

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:

SIEVE1(cons2(s1(N), Y)) -> SIEVE1(filter3(Y, N, N))
ZPRIMES -> SIEVE1(nats1(s1(s1(0))))
SIEVE1(cons2(s1(N), Y)) -> FILTER3(Y, N, N)
FILTER3(cons2(X, Y), s1(N), M) -> FILTER3(Y, N, M)
FILTER3(cons2(X, Y), 0, M) -> FILTER3(Y, M, M)
SIEVE1(cons2(0, Y)) -> SIEVE1(Y)
ZPRIMES -> NATS1(s1(s1(0)))
NATS1(N) -> NATS1(s1(N))

The TRS R consists of the following rules:

filter3(cons2(X, Y), 0, M) -> cons2(0, filter3(Y, M, M))
filter3(cons2(X, Y), s1(N), M) -> cons2(X, filter3(Y, N, M))
sieve1(cons2(0, Y)) -> cons2(0, sieve1(Y))
sieve1(cons2(s1(N), Y)) -> cons2(s1(N), sieve1(filter3(Y, N, N)))
nats1(N) -> cons2(N, nats1(s1(N)))
zprimes -> sieve1(nats1(s1(s1(0))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
          ↳ QDP
          ↳ QDP

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

NATS1(N) -> NATS1(s1(N))

The TRS R consists of the following rules:

filter3(cons2(X, Y), 0, M) -> cons2(0, filter3(Y, M, M))
filter3(cons2(X, Y), s1(N), M) -> cons2(X, filter3(Y, N, M))
sieve1(cons2(0, Y)) -> cons2(0, sieve1(Y))
sieve1(cons2(s1(N), Y)) -> cons2(s1(N), sieve1(filter3(Y, N, N)))
nats1(N) -> cons2(N, nats1(s1(N)))
zprimes -> sieve1(nats1(s1(s1(0))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP
            ↳ QDPOrderProof
          ↳ QDP

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

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

The TRS R consists of the following rules:

filter3(cons2(X, Y), 0, M) -> cons2(0, filter3(Y, M, M))
filter3(cons2(X, Y), s1(N), M) -> cons2(X, filter3(Y, N, M))
sieve1(cons2(0, Y)) -> cons2(0, sieve1(Y))
sieve1(cons2(s1(N), Y)) -> cons2(s1(N), sieve1(filter3(Y, N, N)))
nats1(N) -> cons2(N, nats1(s1(N)))
zprimes -> sieve1(nats1(s1(s1(0))))

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), 0, M) -> FILTER3(Y, M, M)
FILTER3(cons2(X, Y), s1(N), M) -> FILTER3(Y, N, M)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

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

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

filter3(cons2(X, Y), 0, M) -> cons2(0, filter3(Y, M, M))
filter3(cons2(X, Y), s1(N), M) -> cons2(X, filter3(Y, N, M))
sieve1(cons2(0, Y)) -> cons2(0, sieve1(Y))
sieve1(cons2(s1(N), Y)) -> cons2(s1(N), sieve1(filter3(Y, N, N)))
nats1(N) -> cons2(N, nats1(s1(N)))
zprimes -> sieve1(nats1(s1(s1(0))))

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPOrderProof

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

SIEVE1(cons2(s1(N), Y)) -> SIEVE1(filter3(Y, N, N))
SIEVE1(cons2(0, Y)) -> SIEVE1(Y)

The TRS R consists of the following rules:

filter3(cons2(X, Y), 0, M) -> cons2(0, filter3(Y, M, M))
filter3(cons2(X, Y), s1(N), M) -> cons2(X, filter3(Y, N, M))
sieve1(cons2(0, Y)) -> cons2(0, sieve1(Y))
sieve1(cons2(s1(N), Y)) -> cons2(s1(N), sieve1(filter3(Y, N, N)))
nats1(N) -> cons2(N, nats1(s1(N)))
zprimes -> sieve1(nats1(s1(s1(0))))

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)) -> SIEVE1(filter3(Y, N, N))
The remaining pairs can at least be oriented weakly.

SIEVE1(cons2(0, Y)) -> SIEVE1(Y)
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(SIEVE1(x1)) = 2·x1   
POL(cons2(x1, x2)) = x1 + 2·x2   
POL(filter3(x1, x2, x3)) = x1   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented:

filter3(cons2(X, Y), 0, M) -> cons2(0, filter3(Y, M, M))
filter3(cons2(X, Y), s1(N), M) -> cons2(X, filter3(Y, N, M))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof

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

SIEVE1(cons2(0, Y)) -> SIEVE1(Y)

The TRS R consists of the following rules:

filter3(cons2(X, Y), 0, M) -> cons2(0, filter3(Y, M, M))
filter3(cons2(X, Y), s1(N), M) -> cons2(X, filter3(Y, N, M))
sieve1(cons2(0, Y)) -> cons2(0, sieve1(Y))
sieve1(cons2(s1(N), Y)) -> cons2(s1(N), sieve1(filter3(Y, N, N)))
nats1(N) -> cons2(N, nats1(s1(N)))
zprimes -> sieve1(nats1(s1(s1(0))))

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(0, Y)) -> SIEVE1(Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(SIEVE1(x1)) = 2·x1   
POL(cons2(x1, x2)) = 2 + x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ 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, filter3(Y, M, M))
filter3(cons2(X, Y), s1(N), M) -> cons2(X, filter3(Y, N, M))
sieve1(cons2(0, Y)) -> cons2(0, sieve1(Y))
sieve1(cons2(s1(N), Y)) -> cons2(s1(N), sieve1(filter3(Y, N, N)))
nats1(N) -> cons2(N, nats1(s1(N)))
zprimes -> sieve1(nats1(s1(s1(0))))

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.