Termination w.r.t. Q proof of TRS/TRCSRExIntrod_GM99

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, n__from₁(s₁(X)))
head₁(cons₂(X, Y)) -> X
tail₁(cons₂(X, Y)) -> activate₁(Y)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
filter₂(s₁(s₁(X)), cons₂(Y, Z)) -> if₃(divides₂(s₁(s₁(X)), Y), n__filter₂(s₁(s₁(X)), activate₁(Z)), n__cons₂(Y, n__filter₂(X, sieve₁(Y))))
sieve₁(cons₂(X, Y)) -> cons₂(X, n__filter₂(X, sieve₁(activate₁(Y))))
from₁(X) -> n__from₁(X)
filter₂(X1, X2) -> n__filter₂(X1, X2)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__filter₂(X1, X2)) -> filter₂(X1, X2)
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(X) -> X

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, n__from₁(s₁(X)))
head₁(cons₂(X, Y)) -> X
tail₁(cons₂(X, Y)) -> activate₁(Y)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
filter₂(s₁(s₁(X)), cons₂(Y, Z)) -> if₃(divides₂(s₁(s₁(X)), Y), n__filter₂(s₁(s₁(X)), activate₁(Z)), n__cons₂(Y, n__filter₂(X, sieve₁(Y))))
sieve₁(cons₂(X, Y)) -> cons₂(X, n__filter₂(X, sieve₁(activate₁(Y))))
from₁(X) -> n__from₁(X)
filter₂(X1, X2) -> n__filter₂(X1, X2)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__filter₂(X1, X2)) -> filter₂(X1, X2)
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(X) -> X

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)) -> IF₃(divides₂(s₁(s₁(X)), Y), n__filter₂(s₁(s₁(X)), activate₁(Z)), n__cons₂(Y, n__filter₂(X, sieve₁(Y))))
SIEVE₁(cons₂(X, Y)) -> CONS₂(X, n__filter₂(X, sieve₁(activate₁(Y))))
IF₃(true, X, Y) -> ACTIVATE₁(X)
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> SIEVE₁(Y)
SIEVE₁(cons₂(X, Y)) -> SIEVE₁(activate₁(Y))
ACTIVATE₁(n__cons₂(X1, X2)) -> CONS₂(X1, X2)
ACTIVATE₁(n__filter₂(X1, X2)) -> FILTER₂(X1, X2)
PRIMES -> SIEVE₁(from₁(s₁(s₁(0))))
SIEVE₁(cons₂(X, Y)) -> ACTIVATE₁(Y)
PRIMES -> FROM₁(s₁(s₁(0)))
FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> ACTIVATE₁(Z)
IF₃(false, X, Y) -> ACTIVATE₁(Y)
FROM₁(X) -> CONS₂(X, n__from₁(s₁(X)))
TAIL₁(cons₂(X, Y)) -> ACTIVATE₁(Y)

The TRS R consists of the following rules:

primes -> sieve₁(from₁(s₁(s₁(0))))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
head₁(cons₂(X, Y)) -> X
tail₁(cons₂(X, Y)) -> activate₁(Y)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
filter₂(s₁(s₁(X)), cons₂(Y, Z)) -> if₃(divides₂(s₁(s₁(X)), Y), n__filter₂(s₁(s₁(X)), activate₁(Z)), n__cons₂(Y, n__filter₂(X, sieve₁(Y))))
sieve₁(cons₂(X, Y)) -> cons₂(X, n__filter₂(X, sieve₁(activate₁(Y))))
from₁(X) -> n__from₁(X)
filter₂(X1, X2) -> n__filter₂(X1, X2)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__filter₂(X1, X2)) -> filter₂(X1, X2)
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(X) -> X

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)) -> IF₃(divides₂(s₁(s₁(X)), Y), n__filter₂(s₁(s₁(X)), activate₁(Z)), n__cons₂(Y, n__filter₂(X, sieve₁(Y))))
SIEVE₁(cons₂(X, Y)) -> CONS₂(X, n__filter₂(X, sieve₁(activate₁(Y))))
IF₃(true, X, Y) -> ACTIVATE₁(X)
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> SIEVE₁(Y)
SIEVE₁(cons₂(X, Y)) -> SIEVE₁(activate₁(Y))
ACTIVATE₁(n__cons₂(X1, X2)) -> CONS₂(X1, X2)
ACTIVATE₁(n__filter₂(X1, X2)) -> FILTER₂(X1, X2)
PRIMES -> SIEVE₁(from₁(s₁(s₁(0))))
SIEVE₁(cons₂(X, Y)) -> ACTIVATE₁(Y)
PRIMES -> FROM₁(s₁(s₁(0)))
FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> ACTIVATE₁(Z)
IF₃(false, X, Y) -> ACTIVATE₁(Y)
FROM₁(X) -> CONS₂(X, n__from₁(s₁(X)))
TAIL₁(cons₂(X, Y)) -> ACTIVATE₁(Y)

The TRS R consists of the following rules:

primes -> sieve₁(from₁(s₁(s₁(0))))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
head₁(cons₂(X, Y)) -> X
tail₁(cons₂(X, Y)) -> activate₁(Y)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
filter₂(s₁(s₁(X)), cons₂(Y, Z)) -> if₃(divides₂(s₁(s₁(X)), Y), n__filter₂(s₁(s₁(X)), activate₁(Z)), n__cons₂(Y, n__filter₂(X, sieve₁(Y))))
sieve₁(cons₂(X, Y)) -> cons₂(X, n__filter₂(X, sieve₁(activate₁(Y))))
from₁(X) -> n__from₁(X)
filter₂(X1, X2) -> n__filter₂(X1, X2)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__filter₂(X1, X2)) -> filter₂(X1, X2)
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 10 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> SIEVE₁(Y)
SIEVE₁(cons₂(X, Y)) -> SIEVE₁(activate₁(Y))
ACTIVATE₁(n__filter₂(X1, X2)) -> FILTER₂(X1, X2)
SIEVE₁(cons₂(X, Y)) -> ACTIVATE₁(Y)
FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> ACTIVATE₁(Z)

The TRS R consists of the following rules:

primes -> sieve₁(from₁(s₁(s₁(0))))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
head₁(cons₂(X, Y)) -> X
tail₁(cons₂(X, Y)) -> activate₁(Y)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
filter₂(s₁(s₁(X)), cons₂(Y, Z)) -> if₃(divides₂(s₁(s₁(X)), Y), n__filter₂(s₁(s₁(X)), activate₁(Z)), n__cons₂(Y, n__filter₂(X, sieve₁(Y))))
sieve₁(cons₂(X, Y)) -> cons₂(X, n__filter₂(X, sieve₁(activate₁(Y))))
from₁(X) -> n__from₁(X)
filter₂(X1, X2) -> n__filter₂(X1, X2)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__filter₂(X1, X2)) -> filter₂(X1, X2)
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(X) -> X

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.

ACTIVATE₁(n__filter₂(X1, X2)) -> FILTER₂(X1, X2)

The remaining pairs can at least be oriented weakly.

FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> SIEVE₁(Y)
SIEVE₁(cons₂(X, Y)) -> SIEVE₁(activate₁(Y))
SIEVE₁(cons₂(X, Y)) -> ACTIVATE₁(Y)
FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> ACTIVATE₁(Z)

Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE₁(x₁)) = x₁
POL(FILTER₂(x₁, x₂)) = x₂
POL(SIEVE₁(x₁)) = 2·x₁
POL(activate₁(x₁)) = 2·x₁
POL(cons₂(x₁, x₂)) = 3·x₁ + 2·x₂
POL(divides₂(x₁, x₂)) = 3·x₂
POL(filter₂(x₁, x₂)) = 1 + 2·x₂
POL(from₁(x₁)) = 3·x₁
POL(if₃(x₁, x₂, x₃)) = x₁ + x₂
POL(n__cons₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(n__filter₂(x₁, x₂)) = 1 + x₂
POL(n__from₁(x₁)) = 2·x₁
POL(s₁(x₁)) = 0
POL(sieve₁(x₁)) = 0

The following usable rules [14] were oriented:

from₁(X) -> n__from₁(X)
activate₁(n__filter₂(X1, X2)) -> filter₂(X1, X2)
cons₂(X1, X2) -> n__cons₂(X1, X2)
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
filter₂(s₁(s₁(X)), cons₂(Y, Z)) -> if₃(divides₂(s₁(s₁(X)), Y), n__filter₂(s₁(s₁(X)), activate₁(Z)), n__cons₂(Y, n__filter₂(X, sieve₁(Y))))
activate₁(X) -> X
filter₂(X1, X2) -> n__filter₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> SIEVE₁(Y)
SIEVE₁(cons₂(X, Y)) -> SIEVE₁(activate₁(Y))
SIEVE₁(cons₂(X, Y)) -> ACTIVATE₁(Y)
FILTER₂(s₁(s₁(X)), cons₂(Y, Z)) -> ACTIVATE₁(Z)

The TRS R consists of the following rules:

primes -> sieve₁(from₁(s₁(s₁(0))))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
head₁(cons₂(X, Y)) -> X
tail₁(cons₂(X, Y)) -> activate₁(Y)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
filter₂(s₁(s₁(X)), cons₂(Y, Z)) -> if₃(divides₂(s₁(s₁(X)), Y), n__filter₂(s₁(s₁(X)), activate₁(Z)), n__cons₂(Y, n__filter₂(X, sieve₁(Y))))
sieve₁(cons₂(X, Y)) -> cons₂(X, n__filter₂(X, sieve₁(activate₁(Y))))
from₁(X) -> n__from₁(X)
filter₂(X1, X2) -> n__filter₂(X1, X2)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__filter₂(X1, X2)) -> filter₂(X1, X2)
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

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

SIEVE₁(cons₂(X, Y)) -> SIEVE₁(activate₁(Y))

The TRS R consists of the following rules:

primes -> sieve₁(from₁(s₁(s₁(0))))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
head₁(cons₂(X, Y)) -> X
tail₁(cons₂(X, Y)) -> activate₁(Y)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
filter₂(s₁(s₁(X)), cons₂(Y, Z)) -> if₃(divides₂(s₁(s₁(X)), Y), n__filter₂(s₁(s₁(X)), activate₁(Z)), n__cons₂(Y, n__filter₂(X, sieve₁(Y))))
sieve₁(cons₂(X, Y)) -> cons₂(X, n__filter₂(X, sieve₁(activate₁(Y))))
from₁(X) -> n__from₁(X)
filter₂(X1, X2) -> n__filter₂(X1, X2)
cons₂(X1, X2) -> n__cons₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__filter₂(X1, X2)) -> filter₂(X1, X2)
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(X) -> X

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