Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

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

(2) Obligation:

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)))

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))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

PRIMES → SIEVE(from(s(s(0))))
PRIMES → FROM(s(s(0)))
FROM(X) → FROM(s(X))
FILTER(s(s(X)), cons(Y, Z)) → IF(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, 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))
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
SIEVE(cons(X, Y)) → FILTER(X, sieve(Y))
SIEVE(cons(X, Y)) → 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)))

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.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

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

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.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FILTER(x1, x2)  =  FILTER(x2)
s(x1)  =  s(x1)
cons(x1, x2)  =  cons(x1, x2)
sieve(x1)  =  x1
SIEVE(x1)  =  SIEVE(x1)
filter(x1, x2)  =  x2
if(x1, x2, x3)  =  if(x1, x2)
divides(x1, x2)  =  divides

Recursive Path Order [RPO].
Precedence:

[s₁, cons₂] > [FILTER₁, SIEVE₁, if₂, divides]

The following usable rules [FROCOS05] 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))))

(9) Obligation:

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)))

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.

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

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)))

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.