Termination w.r.t. Q proof of /tmp/tmpfbjKMC/ExIntrod

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:

primes → sieve(from(s(s(0))))
from(X) → cons(X, from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → Y
if(true, X, Y) → X
if(false, X, Y) → Y
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))

The set Q consists of the following terms:

if(true, x0, x1)
from(x0)
filter(s(s(x0)), cons(x1, x2))
primes
tail(cons(x0, x1))
sieve(cons(x0, x1))
head(cons(x0, x1))
if(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

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

primes → sieve(from(s(s(0))))
from(X) → cons(X, from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → Y
if(true, X, Y) → X
if(false, X, Y) → Y
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))

The set Q consists of the following terms:

if(true, x0, x1)
from(x0)
filter(s(s(x0)), cons(x1, x2))
primes
tail(cons(x0, x1))
sieve(cons(x0, x1))
head(cons(x0, x1))
if(false, x0, x1)

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(X, Y)) → FILTER(X, sieve(Y))
FILTER(s(s(X)), cons(Y, Z)) → IF(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
SIEVE(cons(X, Y)) → SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)
FILTER(s(s(X)), cons(Y, Z)) → FILTER(X, sieve(Y))
PRIMES → FROM(s(s(0)))
PRIMES → SIEVE(from(s(s(0))))
FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → Y
if(true, X, Y) → X
if(false, X, Y) → Y
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))

The set Q consists of the following terms:

if(true, x0, x1)
from(x0)
filter(s(s(x0)), cons(x1, x2))
primes
tail(cons(x0, x1))
sieve(cons(x0, x1))
head(cons(x0, x1))
if(false, x0, x1)

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(X, Y)) → FILTER(X, sieve(Y))
FILTER(s(s(X)), cons(Y, Z)) → IF(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
SIEVE(cons(X, Y)) → SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)
FILTER(s(s(X)), cons(Y, Z)) → FILTER(X, sieve(Y))
PRIMES → FROM(s(s(0)))
PRIMES → SIEVE(from(s(s(0))))
FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → Y
if(true, X, Y) → X
if(false, X, Y) → Y
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))

The set Q consists of the following terms:

if(true, x0, x1)
from(x0)
filter(s(s(x0)), cons(x1, x2))
primes
tail(cons(x0, x1))
sieve(cons(x0, x1))
head(cons(x0, x1))
if(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

SIEVE(cons(X, Y)) → FILTER(X, sieve(Y))
SIEVE(cons(X, Y)) → SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)
FILTER(s(s(X)), cons(Y, Z)) → FILTER(X, sieve(Y))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → Y
if(true, X, Y) → X
if(false, X, Y) → Y
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))

The set Q consists of the following terms:

if(true, x0, x1)
from(x0)
filter(s(s(x0)), cons(x1, x2))
primes
tail(cons(x0, x1))
sieve(cons(x0, x1))
head(cons(x0, x1))
if(false, x0, x1)

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

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

SIEVE(cons(X, Y)) → FILTER(X, sieve(Y))
SIEVE(cons(X, Y)) → SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)
FILTER(s(s(X)), cons(Y, Z)) → FILTER(X, sieve(Y))

The TRS R consists of the following rules:

sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))

The set Q consists of the following terms:

if(true, x0, x1)
from(x0)
filter(s(s(x0)), cons(x1, x2))
primes
tail(cons(x0, x1))
sieve(cons(x0, x1))
head(cons(x0, x1))
if(false, x0, x1)

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.

from(x0)
primes
tail(cons(x0, x1))
head(cons(x0, x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPOrderProof

          ↳ QDP

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

SIEVE(cons(X, Y)) → FILTER(X, sieve(Y))
FILTER(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
SIEVE(cons(X, Y)) → SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) → FILTER(X, sieve(Y))

The TRS R consists of the following rules:

sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))

The set Q consists of the following terms:

if(true, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))
if(false, x0, x1)

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.

FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) → FILTER(X, sieve(Y))

The remaining pairs can at least be oriented weakly.

SIEVE(cons(X, Y)) → FILTER(X, sieve(Y))
FILTER(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)
SIEVE(cons(X, Y)) → SIEVE(Y)

Used ordering: Polynomial interpretation [25]:

POL(FILTER(x₁, x₂)) = x₁ + x₂
POL(SIEVE(x₁)) = x₁
POL(cons(x₁, x₂)) = x₁ + x₂
POL(divides(x₁, x₂)) = 0
POL(filter(x₁, x₂)) = x₂
POL(if(x₁, x₂, x₃)) = 0
POL(s(x₁)) = 1 + x₁
POL(sieve(x₁)) = x₁

The following usable rules [17] were oriented:

filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

          ↳ QDP

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

SIEVE(cons(X, Y)) → FILTER(X, sieve(Y))
SIEVE(cons(X, Y)) → SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)

The TRS R consists of the following rules:

sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))

The set Q consists of the following terms:

if(true, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))
if(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ AND

                            ↳ QDP

                              ↳ UsableRulesProof

                            ↳ QDP

          ↳ QDP

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

FILTER(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)

The TRS R consists of the following rules:

sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))

The set Q consists of the following terms:

if(true, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))
if(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ AND

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QReductionProof

                            ↳ QDP

          ↳ QDP

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

FILTER(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)

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

if(true, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))
if(false, x0, x1)

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.

if(true, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))
if(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ AND

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QReductionProof

                                    ↳ QDP

                                      ↳ QDPSizeChangeProof

                            ↳ QDP

          ↳ QDP

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

FILTER(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)

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(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)
The graph contains the following edges 1 >= 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ AND

                            ↳ QDP

                            ↳ QDP

                              ↳ UsableRulesProof

          ↳ QDP

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

SIEVE(cons(X, Y)) → SIEVE(Y)

The TRS R consists of the following rules:

sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))

The set Q consists of the following terms:

if(true, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))
if(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ AND

                            ↳ QDP

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QReductionProof

          ↳ QDP

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

SIEVE(cons(X, Y)) → SIEVE(Y)

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

if(true, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))
if(false, x0, x1)

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.

if(true, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))
if(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ AND

                            ↳ QDP

                            ↳ QDP

                              ↳ UsableRulesProof

                                ↳ QDP

                                  ↳ QReductionProof

                                    ↳ QDP

                                      ↳ QDPSizeChangeProof

          ↳ QDP

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

SIEVE(cons(X, Y)) → SIEVE(Y)

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

SIEVE(cons(X, Y)) → SIEVE(Y)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

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

FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → Y
if(true, X, Y) → X
if(false, X, Y) → Y
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))

The set Q consists of the following terms:

if(true, x0, x1)
from(x0)
filter(s(s(x0)), cons(x1, x2))
primes
tail(cons(x0, x1))
sieve(cons(x0, x1))
head(cons(x0, x1))
if(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

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

FROM(X) → FROM(s(X))

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

if(true, x0, x1)
from(x0)
filter(s(s(x0)), cons(x1, x2))
primes
tail(cons(x0, x1))
sieve(cons(x0, x1))
head(cons(x0, x1))
if(false, x0, x1)

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.

if(true, x0, x1)
from(x0)
filter(s(s(x0)), cons(x1, x2))
primes
tail(cons(x0, x1))
sieve(cons(x0, x1))
head(cons(x0, x1))
if(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Instantiation

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

FROM(X) → FROM(s(X))

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

FROM(s(z0)) → FROM(s(s(z0)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Instantiation

                      ↳ QDP

                        ↳ Instantiation

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

FROM(s(z0)) → FROM(s(s(z0)))

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

FROM(s(s(z0))) → FROM(s(s(s(z0))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Instantiation

                      ↳ QDP

                        ↳ Instantiation

                          ↳ QDP

                            ↳ NonTerminationProof

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

FROM(s(s(z0))) → FROM(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:

FROM(s(s(z0))) → FROM(s(s(s(z0))))

The TRS R consists of the following rules:none

s = FROM(s(s(z0))) evaluates to t =FROM(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 FROM(s(s(z0))) to FROM(s(s(s(z0)))).