Termination w.r.t. Q proof of TRS/nontermin/CSRExIntrod

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:

primes -> sieve₁(from₁(s₁(s₁(0))))
from₁(X) -> cons₂(X, from₁(s₁(X)))
head₁(cons₂(X, Y)) -> X
tail₁(cons₂(X, Y)) -> Y
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y
filter₂(s₁(s₁(X)), cons₂(Y, Z)) -> if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y))))
sieve₁(cons₂(X, Y)) -> cons₂(X, filter₂(X, sieve₁(Y)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

primes -> sieve₁(from₁(s₁(s₁(0))))
from₁(X) -> cons₂(X, from₁(s₁(X)))
head₁(cons₂(X, Y)) -> X
tail₁(cons₂(X, Y)) -> Y
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y
filter₂(s₁(s₁(X)), cons₂(Y, Z)) -> if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y))))
sieve₁(cons₂(X, Y)) -> cons₂(X, filter₂(X, sieve₁(Y)))

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:

FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> SIEVE₁(Y)
FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> FILTER₂(X, sieve₁(Y))
FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> FILTER₂(s₁(s₁(X)), Z)
PRIMES -> SIEVE₁(from₁(s₁(s₁(0))))
SIEVE₁(cons₂(X, Y)) -> SIEVE₁(Y)
PRIMES -> FROM₁(s₁(s₁(0)))
FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> IF₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y))))
SIEVE₁(cons₂(X, Y)) -> FILTER₂(X, sieve₁(Y))
FROM₁(X) -> FROM₁(s₁(X))

The TRS R consists of the following rules:

primes -> sieve₁(from₁(s₁(s₁(0))))
from₁(X) -> cons₂(X, from₁(s₁(X)))
head₁(cons₂(X, Y)) -> X
tail₁(cons₂(X, Y)) -> Y
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y
filter₂(s₁(s₁(X)), cons₂(Y, Z)) -> if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y))))
sieve₁(cons₂(X, Y)) -> cons₂(X, filter₂(X, sieve₁(Y)))

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:

FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> SIEVE₁(Y)
FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> FILTER₂(X, sieve₁(Y))
FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> FILTER₂(s₁(s₁(X)), Z)
PRIMES -> SIEVE₁(from₁(s₁(s₁(0))))
SIEVE₁(cons₂(X, Y)) -> SIEVE₁(Y)
PRIMES -> FROM₁(s₁(s₁(0)))
FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> IF₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y))))
SIEVE₁(cons₂(X, Y)) -> FILTER₂(X, sieve₁(Y))
FROM₁(X) -> FROM₁(s₁(X))

The TRS R consists of the following rules:

primes -> sieve₁(from₁(s₁(s₁(0))))
from₁(X) -> cons₂(X, from₁(s₁(X)))
head₁(cons₂(X, Y)) -> X
tail₁(cons₂(X, Y)) -> Y
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y
filter₂(s₁(s₁(X)), cons₂(Y, Z)) -> if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y))))
sieve₁(cons₂(X, Y)) -> cons₂(X, filter₂(X, sieve₁(Y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> SIEVE₁(Y)
FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> FILTER₂(X, sieve₁(Y))
FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> FILTER₂(s₁(s₁(X)), Z)
SIEVE₁(cons₂(X, Y)) -> SIEVE₁(Y)
SIEVE₁(cons₂(X, Y)) -> FILTER₂(X, sieve₁(Y))

The TRS R consists of the following rules:

primes -> sieve₁(from₁(s₁(s₁(0))))
from₁(X) -> cons₂(X, from₁(s₁(X)))
head₁(cons₂(X, Y)) -> X
tail₁(cons₂(X, Y)) -> Y
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y
filter₂(s₁(s₁(X)), cons₂(Y, Z)) -> if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y))))
sieve₁(cons₂(X, Y)) -> cons₂(X, filter₂(X, sieve₁(Y)))

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.

FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> FILTER₂(s₁(s₁(X)), Z)

The remaining pairs can at least be oriented weakly.

FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> SIEVE₁(Y)
FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> FILTER₂(X, sieve₁(Y))
SIEVE₁(cons₂(X, Y)) -> SIEVE₁(Y)
SIEVE₁(cons₂(X, Y)) -> FILTER₂(X, sieve₁(Y))

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( FILTER₂(x₁, x₂) ) = x₂

POL( if₃(x₁, ..., x₃) ) = max{0, -1}

POL( sieve₁(x₁) ) = 1

POL( filter₂(x₁, x₂) ) = max{0, -1}

POL( SIEVE₁(x₁) ) = 1

POL( s₁(x₁) ) = 1

POL( divides₂(x₁, x₂) ) = max{0, -1}

POL( cons₂(x₁, x₂) ) = x₂ + 1

The following usable rules [14] were oriented:

filter₂(s₁(s₁(X)), cons₂(Y, Z)) -> if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y))))
sieve₁(cons₂(X, Y)) -> cons₂(X, filter₂(X, sieve₁(Y)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> SIEVE₁(Y)
FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> FILTER₂(X, sieve₁(Y))
SIEVE₁(cons₂(X, Y)) -> SIEVE₁(Y)
SIEVE₁(cons₂(X, Y)) -> FILTER₂(X, sieve₁(Y))

The TRS R consists of the following rules:

primes -> sieve₁(from₁(s₁(s₁(0))))
from₁(X) -> cons₂(X, from₁(s₁(X)))
head₁(cons₂(X, Y)) -> X
tail₁(cons₂(X, Y)) -> Y
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y
filter₂(s₁(s₁(X)), cons₂(Y, Z)) -> if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y))))
sieve₁(cons₂(X, Y)) -> cons₂(X, filter₂(X, sieve₁(Y)))

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.

FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> FILTER₂(X, sieve₁(Y))
SIEVE₁(cons₂(X, Y)) -> SIEVE₁(Y)
SIEVE₁(cons₂(X, Y)) -> FILTER₂(X, sieve₁(Y))

The remaining pairs can at least be oriented weakly.

FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> SIEVE₁(Y)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( FILTER₂(x₁, x₂) ) = x₂

POL( if₃(x₁, ..., x₃) ) = max{0, -1}

POL( sieve₁(x₁) ) = x₁

POL( filter₂(x₁, x₂) ) = max{0, -1}

POL( SIEVE₁(x₁) ) = x₁ + 1

POL( s₁(x₁) ) = x₁ + 1

POL( divides₂(x₁, x₂) ) = max{0, -1}

POL( cons₂(x₁, x₂) ) = x₁ + x₂ + 1

The following usable rules [14] were oriented:

filter₂(s₁(s₁(X)), cons₂(Y, Z)) -> if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y))))
sieve₁(cons₂(X, Y)) -> cons₂(X, filter₂(X, sieve₁(Y)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

          ↳ QDP

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

FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> SIEVE₁(Y)

The TRS R consists of the following rules:

primes -> sieve₁(from₁(s₁(s₁(0))))
from₁(X) -> cons₂(X, from₁(s₁(X)))
head₁(cons₂(X, Y)) -> X
tail₁(cons₂(X, Y)) -> Y
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y
filter₂(s₁(s₁(X)), cons₂(Y, Z)) -> if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y))))
sieve₁(cons₂(X, Y)) -> cons₂(X, filter₂(X, sieve₁(Y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

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

FROM₁(X) -> FROM₁(s₁(X))

The TRS R consists of the following rules:

primes -> sieve₁(from₁(s₁(s₁(0))))
from₁(X) -> cons₂(X, from₁(s₁(X)))
head₁(cons₂(X, Y)) -> X
tail₁(cons₂(X, Y)) -> Y
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y
filter₂(s₁(s₁(X)), cons₂(Y, Z)) -> if₃(divides₂(s₁(s₁(X)), Y), filter₂(s₁(s₁(X)), Z), cons₂(Y, filter₂(X, sieve₁(Y))))
sieve₁(cons₂(X, Y)) -> cons₂(X, filter₂(X, sieve₁(Y)))

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