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:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimessieve(nats(s(s(0))))

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

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

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimessieve(nats(s(s(0))))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
QTRS
      ↳ DependencyPairsProof

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

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimessieve(nats(s(s(0))))

The set Q consists of the following terms:

filter(cons(x0, x1), 0, x2)
filter(cons(x0, x1), s(x2), x3)
sieve(cons(0, x0))
sieve(cons(s(x0), x1))
nats(x0)
zprimes


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

SIEVE(cons(s(N), Y)) → SIEVE(filter(Y, N, N))
ZPRIMESSIEVE(nats(s(s(0))))
SIEVE(cons(s(N), Y)) → FILTER(Y, N, N)
FILTER(cons(X, Y), s(N), M) → FILTER(Y, N, M)
FILTER(cons(X, Y), 0, M) → FILTER(Y, M, M)
SIEVE(cons(0, Y)) → SIEVE(Y)
ZPRIMESNATS(s(s(0)))
NATS(N) → NATS(s(N))

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimessieve(nats(s(s(0))))

The set Q consists of the following terms:

filter(cons(x0, x1), 0, x2)
filter(cons(x0, x1), s(x2), x3)
sieve(cons(0, x0))
sieve(cons(s(x0), x1))
nats(x0)
zprimes

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ EdgeDeletionProof

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

SIEVE(cons(s(N), Y)) → SIEVE(filter(Y, N, N))
ZPRIMESSIEVE(nats(s(s(0))))
SIEVE(cons(s(N), Y)) → FILTER(Y, N, N)
FILTER(cons(X, Y), s(N), M) → FILTER(Y, N, M)
FILTER(cons(X, Y), 0, M) → FILTER(Y, M, M)
SIEVE(cons(0, Y)) → SIEVE(Y)
ZPRIMESNATS(s(s(0)))
NATS(N) → NATS(s(N))

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimessieve(nats(s(s(0))))

The set Q consists of the following terms:

filter(cons(x0, x1), 0, x2)
filter(cons(x0, x1), s(x2), x3)
sieve(cons(0, x0))
sieve(cons(s(x0), x1))
nats(x0)
zprimes

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
QDP
              ↳ DependencyGraphProof

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

SIEVE(cons(s(N), Y)) → SIEVE(filter(Y, N, N))
FILTER(cons(X, Y), 0, M) → FILTER(Y, M, M)
FILTER(cons(X, Y), s(N), M) → FILTER(Y, N, M)
SIEVE(cons(s(N), Y)) → FILTER(Y, N, N)
ZPRIMESSIEVE(nats(s(s(0))))
SIEVE(cons(0, Y)) → SIEVE(Y)
NATS(N) → NATS(s(N))
ZPRIMESNATS(s(s(0)))

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimessieve(nats(s(s(0))))

The set Q consists of the following terms:

filter(cons(x0, x1), 0, x2)
filter(cons(x0, x1), s(x2), x3)
sieve(cons(0, x0))
sieve(cons(s(x0), x1))
nats(x0)
zprimes

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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
QDP
                  ↳ QDP
                  ↳ QDP

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

NATS(N) → NATS(s(N))

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimessieve(nats(s(s(0))))

The set Q consists of the following terms:

filter(cons(x0, x1), 0, x2)
filter(cons(x0, x1), s(x2), x3)
sieve(cons(0, x0))
sieve(cons(s(x0), x1))
nats(x0)
zprimes

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

FILTER(cons(X, Y), 0, M) → FILTER(Y, M, M)
FILTER(cons(X, Y), s(N), M) → FILTER(Y, N, M)

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimessieve(nats(s(s(0))))

The set Q consists of the following terms:

filter(cons(x0, x1), 0, x2)
filter(cons(x0, x1), s(x2), x3)
sieve(cons(0, x0))
sieve(cons(s(x0), x1))
nats(x0)
zprimes

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.


FILTER(cons(X, Y), 0, M) → FILTER(Y, M, M)
FILTER(cons(X, Y), s(N), M) → FILTER(Y, N, M)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
FILTER(x1, x2, x3)  =  x1
cons(x1, x2)  =  cons(x2)

Lexicographic path order with status [19].
Precedence:
trivial

Status:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
                  ↳ QDP

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

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimessieve(nats(s(s(0))))

The set Q consists of the following terms:

filter(cons(x0, x1), 0, x2)
filter(cons(x0, x1), s(x2), x3)
sieve(cons(0, x0))
sieve(cons(s(x0), x1))
nats(x0)
zprimes

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
QDP
                    ↳ QDPOrderProof

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

SIEVE(cons(s(N), Y)) → SIEVE(filter(Y, N, N))
SIEVE(cons(0, Y)) → SIEVE(Y)

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimessieve(nats(s(s(0))))

The set Q consists of the following terms:

filter(cons(x0, x1), 0, x2)
filter(cons(x0, x1), s(x2), x3)
sieve(cons(0, x0))
sieve(cons(s(x0), x1))
nats(x0)
zprimes

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.


SIEVE(cons(s(N), Y)) → SIEVE(filter(Y, N, N))
SIEVE(cons(0, Y)) → SIEVE(Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
SIEVE(x1)  =  x1
cons(x1, x2)  =  cons(x2)
filter(x1, x2, x3)  =  x1

Lexicographic path order with status [19].
Precedence:
trivial

Status:
trivial

The following usable rules [14] were oriented:

filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof

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

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimessieve(nats(s(s(0))))

The set Q consists of the following terms:

filter(cons(x0, x1), 0, x2)
filter(cons(x0, x1), s(x2), x3)
sieve(cons(0, x0))
sieve(cons(s(x0), x1))
nats(x0)
zprimes

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.