Termination w.r.t. Q proof of /tmp/tmpIZR0DX/ExIntrod

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 → 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)))

Q is empty.

↳ 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)))

Q is empty.

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)))

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:

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)))

Q is empty.
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

            ↳ QDPOrderProof

          ↳ 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)))

Q is empty.
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)) → FILTER(s(s(X)), Z)

The remaining pairs can at least be oriented weakly.

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(X, sieve(Y))

Used ordering: Polynomial interpretation [25,35]:

POL(SIEVE(x₁)) = 4
POL(cons(x₁, x₂)) = 2 + x_2
POL(if(x₁, x₂, x₃)) = 0
POL(filter(x₁, x₂)) = 0
POL(s(x₁)) = 0
POL(divides(x₁, x₂)) = 0
POL(sieve(x₁)) = 2
POL(FILTER(x₁, x₂)) = (2)x_2

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented:

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))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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)) → 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:

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)))

Q is empty.
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(X, Y)) → FILTER(X, sieve(Y))
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 remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(SIEVE(x₁)) = 1/4 + (4)x_1
POL(cons(x₁, x₂)) = 2 + (4)x_1 + x_2
POL(if(x₁, x₂, x₃)) = 0
POL(filter(x₁, x₂)) = 0
POL(s(x₁)) = 0
POL(divides(x₁, x₂)) = 0
POL(sieve(x₁)) = 1/2 + x_1
POL(FILTER(x₁, x₂)) = 4 + (4)x_2

The value of delta used in the strict ordering is 9/4.
The following usable rules [17] were oriented:

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))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
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)))

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:

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)))

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