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:

primessieve(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
  ↳ Overlay + Local Confluence

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

primessieve(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.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
QTRS
      ↳ DependencyPairsProof

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

primessieve(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)))

The set Q consists of the following terms:

primes
from(x0)
head(cons(x0, x1))
tail(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) → FILTER(X, sieve(Y))
FILTER(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)
PRIMESSIEVE(from(s(s(0))))
SIEVE(cons(X, Y)) → SIEVE(Y)
PRIMESFROM(s(s(0)))
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)) → FILTER(X, sieve(Y))
FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

primessieve(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)))

The set Q consists of the following terms:

primes
from(x0)
head(cons(x0, x1))
tail(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ EdgeDeletionProof

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

FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) → FILTER(X, sieve(Y))
FILTER(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)
PRIMESSIEVE(from(s(s(0))))
SIEVE(cons(X, Y)) → SIEVE(Y)
PRIMESFROM(s(s(0)))
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)) → FILTER(X, sieve(Y))
FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

primessieve(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)))

The set Q consists of the following terms:

primes
from(x0)
head(cons(x0, x1))
tail(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
QDP
              ↳ DependencyGraphProof

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

FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) → FILTER(X, sieve(Y))
FILTER(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)
PRIMESSIEVE(from(s(s(0))))
SIEVE(cons(X, Y)) → SIEVE(Y)
PRIMESFROM(s(s(0)))
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)) → FILTER(X, sieve(Y))
FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

primessieve(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)))

The set Q consists of the following terms:

primes
from(x0)
head(cons(x0, x1))
tail(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))

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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) → FILTER(X, sieve(Y))
FILTER(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)
SIEVE(cons(X, Y)) → SIEVE(Y)
SIEVE(cons(X, Y)) → FILTER(X, sieve(Y))

The TRS R consists of the following rules:

primessieve(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)))

The set Q consists of the following terms:

primes
from(x0)
head(cons(x0, x1))
tail(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))

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.


FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) → FILTER(X, sieve(Y))
FILTER(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)
SIEVE(cons(X, Y)) → SIEVE(Y)
SIEVE(cons(X, Y)) → FILTER(X, sieve(Y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
FILTER(x1, x2)  =  x2
cons(x1, x2)  =  cons(x1, x2)
SIEVE(x1)  =  x1
sieve(x1)  =  x1
filter(x1, x2)  =  filter
if(x1, x2, x3)  =  x2

Recursive Path Order [2].
Precedence:
cons2 > filter

The following usable rules [14] were oriented:

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)))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
                  ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

primessieve(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)))

The set Q consists of the following terms:

primes
from(x0)
head(cons(x0, x1))
tail(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ 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:

primessieve(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)))

The set Q consists of the following terms:

primes
from(x0)
head(cons(x0, x1))
tail(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))

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