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 Order [17,21] with Interpretation:

POL( s1(x1) ) = max{0, -3}


POL( 0 ) = max{0, -2}


POL( FILTER3(x1, ..., x3) ) = 3x1 + x3 + 2


POL( cons2(x1, x2) ) = x1 + x2 + 3



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))
SIEVE1(cons2(0, Y)) -> SIEVE1(Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( 0 ) = 0


POL( s1(x1) ) = 3x1 + 1


POL( filter3(x1, ..., x3) ) = x1 + 3


POL( cons2(x1, x2) ) = x1 + x2 + 3


POL( SIEVE1(x1) ) = max{0, 2x1 - 2}



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
                ↳ 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.