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)))
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 [19] 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 [1,15] 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))
ZPRIMESSIEVE(nats(s(s(0))))
ZPRIMESNATS(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)))
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
          ↳ 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))
ZPRIMESSIEVE(nats(s(s(0))))
ZPRIMESNATS(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)))
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 [15,17,22] contains 3 SCCs with 3 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ 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)))
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.
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
  ↳ Overlay + Local Confluence
    ↳ 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), 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 the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ 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
  ↳ Overlay + Local Confluence
    ↳ 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
  ↳ Overlay + Local Confluence
    ↳ 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:




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
  ↳ Overlay + Local Confluence
    ↳ 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)))
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.
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
  ↳ Overlay + Local Confluence
    ↳ 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), 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 the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ 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:



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ 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)))
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.
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
  ↳ Overlay + Local Confluence
    ↳ 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), 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 the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
QDP
                        ↳ 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), 0, x2)
filter(cons(x0, x1), s(x2), x3)

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 [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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ 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))

The set Q consists of the following terms:

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

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