Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/nonterminSLASHCSRSLASHExIntrod

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

  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)))

The set Q consists of the following terms:

primes
from₁(x0)
head₁(cons₂(x0, x1))
tail₁(cons₂(x0, x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
filter₂(s₁(s₁(x0)), cons₂(x1, x2))
sieve₁(cons₂(x0, x1))

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)))

The set Q consists of the following terms:

primes
from₁(x0)
head₁(cons₂(x0, x1))
tail₁(cons₂(x0, x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
filter₂(s₁(s₁(x0)), cons₂(x1, x2))
sieve₁(cons₂(x0, x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)))

The set Q consists of the following terms:

primes
from₁(x0)
head₁(cons₂(x0, x1))
tail₁(cons₂(x0, x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
filter₂(s₁(s₁(x0)), cons₂(x1, x2))
sieve₁(cons₂(x0, x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ 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)))

The set Q consists of the following terms:

primes
from₁(x0)
head₁(cons₂(x0, x1))
tail₁(cons₂(x0, x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
filter₂(s₁(s₁(x0)), cons₂(x1, x2))
sieve₁(cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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))

Used argument filtering: FILTER₂(x1, x2) = x2
cons₂(x1, x2) = cons₂(x1, x2)
SIEVE₁(x1) = x1
sieve₁(x1) = x1
filter₂(x1, x2) = filter
if₃(x1, x2, x3) = if
Used ordering: Quasi Precedence: cons_2 > [filter, if]

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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)))

The set Q consists of the following terms:

primes
from₁(x0)
head₁(cons₂(x0, x1))
tail₁(cons₂(x0, x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
filter₂(s₁(s₁(x0)), cons₂(x1, x2))
sieve₁(cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)))

The set Q consists of the following terms:

primes
from₁(x0)
head₁(cons₂(x0, x1))
tail₁(cons₂(x0, x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
filter₂(s₁(s₁(x0)), cons₂(x1, x2))
sieve₁(cons₂(x0, x1))

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