Termination w.r.t. Q of the given QTRS could be disproven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 QDPOrderProof (⇔)
↳6 QDP
↳7 DependencyGraphProof (⇔)
↳8 AND
  ↳9 QDP
    ↳10 Narrowing (⇔)
    ↳11 QDP
    ↳12 Narrowing (⇔)
    ↳13 QDP
    ↳14 DependencyGraphProof (⇔)
    ↳15 QDP
    ↳16 Narrowing (⇔)
    ↳17 QDP
    ↳18 DependencyGraphProof (⇔)
    ↳19 QDP
    ↳20 QDPOrderProof (⇔)
    ↳21 QDP
    ↳22 QDPOrderProof (⇔)
    ↳23 QDP
    ↳24 NonTerminationProof (⇔)
    ↳25 FALSE
  ↳26 QDP

(0) Obligation:

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

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) 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) → CONS(X, n__from(s(X)))
TAIL(cons(X, Y)) → ACTIVATE(Y)
IF(true, X, Y) → ACTIVATE(X)
IF(false, X, Y) → ACTIVATE(Y)
FILTER(s(s(X)), cons(Y, Z)) → IF(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
FILTER(s(s(X)), cons(Y, Z)) → ACTIVATE(Z)
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
SIEVE(cons(X, Y)) → CONS(X, n__filter(X, sieve(activate(Y))))
SIEVE(cons(X, Y)) → SIEVE(activate(Y))
SIEVE(cons(X, Y)) → ACTIVATE(Y)
ACTIVATE(n__from(X)) → FROM(X)
ACTIVATE(n__filter(X1, X2)) → FILTER(X1, X2)
ACTIVATE(n__cons(X1, X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 10 less nodes.

(4) Obligation:

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

ACTIVATE(n__filter(X1, X2)) → FILTER(X1, X2)
FILTER(s(s(X)), cons(Y, Z)) → ACTIVATE(Z)
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
SIEVE(cons(X, Y)) → SIEVE(activate(Y))
SIEVE(cons(X, Y)) → ACTIVATE(Y)

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SIEVE(cons(X, Y)) → ACTIVATE(Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(ACTIVATE(x1)) =
/0\
\0/
 +
/01\
\00/
·x1

POL(n__filter(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/00\
\10/
·x2

POL(FILTER(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/10\
\00/
·x2

POL(s(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(cons(x1, x2)) =
/0\
\1/
 +
/01\
\01/
·x1 +
/01\
\01/
·x2

POL(SIEVE(x1)) =
/0\
\0/
 +
/01\
\00/
·x1

POL(activate(x1)) =
/0\
\1/
 +
/10\
\01/
·x1

POL(from(x1)) =
/0\
\1/
 +
/01\
\01/
·x1

POL(n__from(x1)) =
/0\
\0/
 +
/01\
\01/
·x1

POL(filter(x1, x2)) =
/0\
\1/
 +
/00\
\00/
·x1 +
/00\
\10/
·x2

POL(if(x1, x2, x3)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/00\
\00/
·x2 +
/00\
\00/
·x3

POL(divides(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/00\
\00/
·x2

POL(n__cons(x1, x2)) =
/0\
\0/
 +
/01\
\01/
·x1 +
/01\
\01/
·x2

POL(sieve(x1)) =
/0\
\1/
 +
/01\
\01/
·x1

The following usable rules [FROCOS05] were oriented:

from(X) → cons(X, n__from(s(X)))
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)

(6) Obligation:

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

ACTIVATE(n__filter(X1, X2)) → FILTER(X1, X2)
FILTER(s(s(X)), cons(Y, Z)) → ACTIVATE(Z)
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
SIEVE(cons(X, Y)) → SIEVE(activate(Y))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(8) Complex Obligation (AND)

(9) Obligation:

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

SIEVE(cons(X, Y)) → SIEVE(activate(Y))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(10) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule SIEVE(cons(X, Y)) → SIEVE(activate(Y)) at position [0] we obtained the following new rules [LPAR04]:

SIEVE(cons(y0, n__from(x0))) → SIEVE(from(x0))
SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(filter(x0, x1))
SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
SIEVE(cons(y0, x0)) → SIEVE(x0)

(11) Obligation:

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

SIEVE(cons(y0, n__from(x0))) → SIEVE(from(x0))
SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(filter(x0, x1))
SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
SIEVE(cons(y0, x0)) → SIEVE(x0)

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(12) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule SIEVE(cons(y0, n__from(x0))) → SIEVE(from(x0)) at position [0] we obtained the following new rules [LPAR04]:

SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))
SIEVE(cons(y0, n__from(x0))) → SIEVE(n__from(x0))

(13) Obligation:

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

SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(filter(x0, x1))
SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
SIEVE(cons(y0, x0)) → SIEVE(x0)
SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))
SIEVE(cons(y0, n__from(x0))) → SIEVE(n__from(x0))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(14) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(15) Obligation:

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

SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(filter(x0, x1))
SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
SIEVE(cons(y0, x0)) → SIEVE(x0)
SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(16) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(filter(x0, x1)) at position [0] we obtained the following new rules [LPAR04]:

SIEVE(cons(y0, n__filter(s(s(x0)), cons(x1, x2)))) → SIEVE(if(divides(s(s(x0)), x1), n__filter(s(s(x0)), activate(x2)), n__cons(x1, n__filter(x0, sieve(x1)))))
SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(n__filter(x0, x1))

(17) Obligation:

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

SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
SIEVE(cons(y0, x0)) → SIEVE(x0)
SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))
SIEVE(cons(y0, n__filter(s(s(x0)), cons(x1, x2)))) → SIEVE(if(divides(s(s(x0)), x1), n__filter(s(s(x0)), activate(x2)), n__cons(x1, n__filter(x0, sieve(x1)))))
SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(n__filter(x0, x1))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(18) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(19) Obligation:

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

SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
SIEVE(cons(y0, x0)) → SIEVE(x0)
SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SIEVE(cons(y0, x0)) → SIEVE(x0)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(SIEVE(x1)) = x1   
POL(cons(x1, x2)) = 1 + x2   
POL(n__cons(x1, x2)) = x2   
POL(n__from(x1)) = x1   
POL(s(x1)) = 0   

The following usable rules [FROCOS05] were oriented:

cons(X1, X2) → n__cons(X1, X2)

(21) Obligation:

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

SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(SIEVE(x1)) = x1   
POL(cons(x1, x2)) = 1 + x2   
POL(n__cons(x1, x2)) = 1 + x2   
POL(n__from(x1)) = 0   
POL(s(x1)) = 0   

The following usable rules [FROCOS05] were oriented:

cons(X1, X2) → n__cons(X1, X2)

(23) Obligation:

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

SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(24) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = SIEVE(cons(y0, n__from(x0))) evaluates to t =SIEVE(cons(x0, n__from(s(x0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Semiunifier: [ ]
  • Matcher: [y0 / x0, x0 / s(x0)]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from SIEVE(cons(y0, n__from(x0))) to SIEVE(cons(x0, n__from(s(x0)))).



(25) FALSE

(26) Obligation:

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

FILTER(s(s(X)), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__filter(X1, X2)) → FILTER(X1, X2)

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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