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)) -> 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 remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(FILTER₂(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(SIEVE₁(x₁)) = 2 + 3·x₁
POL(cons₂(x₁, x₂)) = 1 + 3·x₁ + x₂
POL(divides₂(x₁, x₂)) = 0
POL(filter₂(x₁, x₂)) = 0
POL(if₃(x₁, x₂, x₃)) = 0
POL(s₁(x₁)) = 2 + x₁
POL(sieve₁(x₁)) = 3 + 3·x₁

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

                ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
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 TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ 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.