Termination w.r.t. Q proof of /tmp/tmpC7tb-G/Ex9

Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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)))
zprimes → sieve(nats(s(s(0))))

The set Q consists of the following terms:

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

↳ 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)))
zprimes → sieve(nats(s(s(0))))

The set Q consists of the following terms:

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

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

SIEVE(cons(s(N), Y)) → FILTER(Y, N, N)
FILTER(cons(X, Y), 0, M) → FILTER(Y, M, M)
NATS(N) → NATS(s(N))
ZPRIMES → SIEVE(nats(s(s(0))))
ZPRIMES → NATS(s(s(0)))
SIEVE(cons(0, Y)) → SIEVE(Y)
FILTER(cons(X, Y), s(N), M) → FILTER(Y, N, M)
SIEVE(cons(s(N), Y)) → SIEVE(filter(Y, N, 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)))
zprimes → sieve(nats(s(s(0))))

The set Q consists of the following terms:

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

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:

SIEVE(cons(s(N), Y)) → FILTER(Y, N, N)
FILTER(cons(X, Y), 0, M) → FILTER(Y, M, M)
NATS(N) → NATS(s(N))
ZPRIMES → SIEVE(nats(s(s(0))))
ZPRIMES → NATS(s(s(0)))
SIEVE(cons(0, Y)) → SIEVE(Y)
FILTER(cons(X, Y), s(N), M) → FILTER(Y, N, M)
SIEVE(cons(s(N), Y)) → SIEVE(filter(Y, N, 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)))
zprimes → sieve(nats(s(s(0))))

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ 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)))
zprimes → sieve(nats(s(s(0))))

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

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

NATS(N) → NATS(s(N))

R is empty.
The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Instantiation

          ↳ QDP

          ↳ QDP

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

NATS(N) → NATS(s(N))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule NATS(N) → NATS(s(N)) we obtained the following new rules:

NATS(s(z0)) → NATS(s(s(z0)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Instantiation

                      ↳ QDP

                        ↳ Instantiation

          ↳ QDP

          ↳ QDP

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

NATS(s(z0)) → NATS(s(s(z0)))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule NATS(s(z0)) → NATS(s(s(z0))) we obtained the following new rules:

NATS(s(s(z0))) → NATS(s(s(s(z0))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Instantiation

                      ↳ QDP

                        ↳ Instantiation

                          ↳ QDP

                            ↳ NonTerminationProof

          ↳ QDP

          ↳ QDP

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

NATS(s(s(z0))) → NATS(s(s(s(z0))))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

NATS(s(s(z0))) → NATS(s(s(s(z0))))

The TRS R consists of the following rules:none

s = NATS(s(s(z0))) evaluates to t =NATS(s(s(s(z0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [z0 / s(z0)]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from NATS(s(s(z0))) to NATS(s(s(s(z0)))).

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ 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)))
zprimes → sieve(nats(s(s(0))))

The set Q consists of the following terms:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ 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)

R is empty.
The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ 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)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

FILTER(cons(X, Y), 0, M) → FILTER(Y, M, M)
The graph contains the following edges 1 > 1, 3 >= 2, 3 >= 3
FILTER(cons(X, Y), s(N), M) → FILTER(Y, N, M)
The graph contains the following edges 1 > 1, 2 > 2, 3 >= 3

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

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

SIEVE(cons(0, Y)) → SIEVE(Y)
SIEVE(cons(s(N), Y)) → SIEVE(filter(Y, N, 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)))
zprimes → sieve(nats(s(s(0))))

The set Q consists of the following terms:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

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

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

nats(x0)
sieve(cons(s(x0), x1))
sieve(cons(0, x0))
zprimes

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPOrderProof

                    ↳ QDPOrderProof

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

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

The set Q consists of the following terms:

filter(cons(x0, x1), s(x2), x3)
filter(cons(x0, x1), 0, x2)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation with max and min functions [25]:

POL(0) = 0
POL(SIEVE(x₁)) = x₁
POL(cons(x₁, x₂)) = 1 + x₂
POL(filter(x₁, x₂, x₃)) = x₁
POL(s(x₁)) = 0

The following usable rules [17] were oriented:

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))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                    ↳ QDPOrderProof

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))

The set Q consists of the following terms:

filter(cons(x0, x1), s(x2), x3)
filter(cons(x0, x1), 0, x2)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(SIEVE(x₁)) = x₁
POL(cons(x₁, x₂)) = 1 + x₁ + x₂
POL(filter(x₁, x₂, x₃)) = x₁
POL(s(x₁)) = 1

The following usable rules [17] were oriented:

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))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPOrderProof

                    ↳ QDPOrderProof

                      ↳ 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))

The set Q consists of the following terms:

filter(cons(x0, x1), s(x2), x3)
filter(cons(x0, x1), 0, x2)

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