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:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

ACTIVE(primes) → SIEVE(from(s(s(0))))
ACTIVE(primes) → FROM(s(s(0)))
ACTIVE(primes) → S(s(0))
ACTIVE(primes) → S(0)
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(filter(s(s(X)), cons(Y, Z))) → IF(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
ACTIVE(filter(s(s(X)), cons(Y, Z))) → DIVIDES(s(s(X)), Y)
ACTIVE(filter(s(s(X)), cons(Y, Z))) → FILTER(s(s(X)), Z)
ACTIVE(filter(s(s(X)), cons(Y, Z))) → CONS(Y, filter(X, sieve(Y)))
ACTIVE(filter(s(s(X)), cons(Y, Z))) → FILTER(X, sieve(Y))
ACTIVE(filter(s(s(X)), cons(Y, Z))) → SIEVE(Y)
ACTIVE(sieve(cons(X, Y))) → CONS(X, filter(X, sieve(Y)))
ACTIVE(sieve(cons(X, Y))) → FILTER(X, sieve(Y))
ACTIVE(sieve(cons(X, Y))) → SIEVE(Y)
ACTIVE(sieve(X)) → SIEVE(active(X))
ACTIVE(sieve(X)) → ACTIVE(X)
ACTIVE(from(X)) → FROM(active(X))
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(head(X)) → HEAD(active(X))
ACTIVE(head(X)) → ACTIVE(X)
ACTIVE(tail(X)) → TAIL(active(X))
ACTIVE(tail(X)) → ACTIVE(X)
ACTIVE(if(X1, X2, X3)) → IF(active(X1), X2, X3)
ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(filter(X1, X2)) → FILTER(active(X1), X2)
ACTIVE(filter(X1, X2)) → ACTIVE(X1)
ACTIVE(filter(X1, X2)) → FILTER(X1, active(X2))
ACTIVE(filter(X1, X2)) → ACTIVE(X2)
ACTIVE(divides(X1, X2)) → DIVIDES(active(X1), X2)
ACTIVE(divides(X1, X2)) → ACTIVE(X1)
ACTIVE(divides(X1, X2)) → DIVIDES(X1, active(X2))
ACTIVE(divides(X1, X2)) → ACTIVE(X2)
SIEVE(mark(X)) → SIEVE(X)
FROM(mark(X)) → FROM(X)
S(mark(X)) → S(X)
CONS(mark(X1), X2) → CONS(X1, X2)
HEAD(mark(X)) → HEAD(X)
TAIL(mark(X)) → TAIL(X)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
FILTER(mark(X1), X2) → FILTER(X1, X2)
FILTER(X1, mark(X2)) → FILTER(X1, X2)
DIVIDES(mark(X1), X2) → DIVIDES(X1, X2)
DIVIDES(X1, mark(X2)) → DIVIDES(X1, X2)
PROPER(sieve(X)) → SIEVE(proper(X))
PROPER(sieve(X)) → PROPER(X)
PROPER(from(X)) → FROM(proper(X))
PROPER(from(X)) → PROPER(X)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(head(X)) → HEAD(proper(X))
PROPER(head(X)) → PROPER(X)
PROPER(tail(X)) → TAIL(proper(X))
PROPER(tail(X)) → PROPER(X)
PROPER(if(X1, X2, X3)) → IF(proper(X1), proper(X2), proper(X3))
PROPER(if(X1, X2, X3)) → PROPER(X1)
PROPER(if(X1, X2, X3)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → PROPER(X3)
PROPER(filter(X1, X2)) → FILTER(proper(X1), proper(X2))
PROPER(filter(X1, X2)) → PROPER(X1)
PROPER(filter(X1, X2)) → PROPER(X2)
PROPER(divides(X1, X2)) → DIVIDES(proper(X1), proper(X2))
PROPER(divides(X1, X2)) → PROPER(X1)
PROPER(divides(X1, X2)) → PROPER(X2)
SIEVE(ok(X)) → SIEVE(X)
FROM(ok(X)) → FROM(X)
S(ok(X)) → S(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
HEAD(ok(X)) → HEAD(X)
TAIL(ok(X)) → TAIL(X)
IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)
FILTER(ok(X1), ok(X2)) → FILTER(X1, X2)
DIVIDES(ok(X1), ok(X2)) → DIVIDES(X1, X2)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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 12 SCCs with 38 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

DIVIDES(X1, mark(X2)) → DIVIDES(X1, X2)
DIVIDES(mark(X1), X2) → DIVIDES(X1, X2)
DIVIDES(ok(X1), ok(X2)) → DIVIDES(X1, X2)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DIVIDES(mark(X1), X2) → DIVIDES(X1, X2)
DIVIDES(ok(X1), ok(X2)) → DIVIDES(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DIVIDES(x1, x2)  =  x1
mark(x1)  =  mark(x1)
ok(x1)  =  ok(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

DIVIDES(X1, mark(X2)) → DIVIDES(X1, X2)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
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.

DIVIDES(X1, mark(X2)) → DIVIDES(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DIVIDES(x1, x2)  =  x2
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
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:

FILTER(X1, mark(X2)) → FILTER(X1, X2)
FILTER(mark(X1), X2) → FILTER(X1, X2)
FILTER(ok(X1), ok(X2)) → FILTER(X1, X2)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FILTER(mark(X1), X2) → FILTER(X1, X2)
FILTER(ok(X1), ok(X2)) → FILTER(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FILTER(x1, x2)  =  x1
mark(x1)  =  mark(x1)
ok(x1)  =  ok(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

FILTER(X1, mark(X2)) → FILTER(X1, X2)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FILTER(X1, mark(X2)) → FILTER(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FILTER(x1, x2)  =  x2
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) Obligation:

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

IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

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

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(22) PisEmptyProof (EQUIVALENT transformation)

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

(23) TRUE

(24) Obligation:

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

TAIL(ok(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(X)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(25) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TAIL(ok(X)) → TAIL(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAIL(x1)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(26) Obligation:

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

TAIL(mark(X)) → TAIL(X)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(27) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TAIL(mark(X)) → TAIL(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAIL(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(28) Obligation:

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

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(29) PisEmptyProof (EQUIVALENT transformation)

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

(30) TRUE

(31) Obligation:

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

HEAD(ok(X)) → HEAD(X)
HEAD(mark(X)) → HEAD(X)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(32) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

HEAD(ok(X)) → HEAD(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
HEAD(x1)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(33) Obligation:

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

HEAD(mark(X)) → HEAD(X)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(34) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

HEAD(mark(X)) → HEAD(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
HEAD(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(35) Obligation:

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

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(36) PisEmptyProof (EQUIVALENT transformation)

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

(37) TRUE

(38) Obligation:

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(39) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(mark(X1), X2) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2)  =  x1
ok(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(40) Obligation:

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(41) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2)  =  CONS(x2)
ok(x1)  =  ok(x1)

Lexicographic Path Order [LPO].
Precedence:

ok₁ > CONS₁

The following usable rules [FROCOS05] were oriented: none

(42) Obligation:

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

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(43) PisEmptyProof (EQUIVALENT transformation)

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

(44) TRUE

(45) Obligation:

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

S(ok(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(46) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(ok(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(47) Obligation:

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

S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(48) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(mark(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(49) Obligation:

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

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(50) PisEmptyProof (EQUIVALENT transformation)

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

(51) TRUE

(52) Obligation:

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

FROM(ok(X)) → FROM(X)
FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(53) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FROM(ok(X)) → FROM(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FROM(x1)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(54) Obligation:

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

FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(55) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FROM(mark(X)) → FROM(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FROM(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(56) Obligation:

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

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(57) PisEmptyProof (EQUIVALENT transformation)

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

(58) TRUE

(59) Obligation:

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

SIEVE(ok(X)) → SIEVE(X)
SIEVE(mark(X)) → SIEVE(X)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(60) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SIEVE(ok(X)) → SIEVE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SIEVE(x1)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(61) Obligation:

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

SIEVE(mark(X)) → SIEVE(X)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(62) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SIEVE(mark(X)) → SIEVE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SIEVE(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(63) Obligation:

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

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(64) PisEmptyProof (EQUIVALENT transformation)

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

(65) TRUE

(66) Obligation:

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

PROPER(from(X)) → PROPER(X)
PROPER(sieve(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(head(X)) → PROPER(X)
PROPER(tail(X)) → PROPER(X)
PROPER(if(X1, X2, X3)) → PROPER(X1)
PROPER(if(X1, X2, X3)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → PROPER(X3)
PROPER(filter(X1, X2)) → PROPER(X1)
PROPER(filter(X1, X2)) → PROPER(X2)
PROPER(divides(X1, X2)) → PROPER(X1)
PROPER(divides(X1, X2)) → PROPER(X2)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(67) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(from(X)) → PROPER(X)
PROPER(sieve(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(head(X)) → PROPER(X)
PROPER(if(X1, X2, X3)) → PROPER(X1)
PROPER(if(X1, X2, X3)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → PROPER(X3)
PROPER(filter(X1, X2)) → PROPER(X1)
PROPER(filter(X1, X2)) → PROPER(X2)
PROPER(divides(X1, X2)) → PROPER(X1)
PROPER(divides(X1, X2)) → PROPER(X2)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(68) Obligation:

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

PROPER(tail(X)) → PROPER(X)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(69) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(tail(X)) → PROPER(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  x1
tail(x1)  =  tail(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(70) Obligation:

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

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(71) PisEmptyProof (EQUIVALENT transformation)

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

(72) TRUE

(73) Obligation:

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

ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(sieve(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(head(X)) → ACTIVE(X)
ACTIVE(tail(X)) → ACTIVE(X)
ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(filter(X1, X2)) → ACTIVE(X1)
ACTIVE(filter(X1, X2)) → ACTIVE(X2)
ACTIVE(divides(X1, X2)) → ACTIVE(X1)
ACTIVE(divides(X1, X2)) → ACTIVE(X2)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(74) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(sieve(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(head(X)) → ACTIVE(X)
ACTIVE(tail(X)) → ACTIVE(X)
ACTIVE(filter(X1, X2)) → ACTIVE(X1)
ACTIVE(filter(X1, X2)) → ACTIVE(X2)
ACTIVE(divides(X1, X2)) → ACTIVE(X1)
ACTIVE(divides(X1, X2)) → ACTIVE(X2)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(75) Obligation:

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

ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(76) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  ACTIVE(x1)
if(x1, x2, x3)  =  if(x1, x2)

Lexicographic Path Order [LPO].
Precedence:

[ACTIVE₁, if₂]

The following usable rules [FROCOS05] were oriented: none

(77) Obligation:

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

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(78) PisEmptyProof (EQUIVALENT transformation)

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

(79) TRUE

(80) Obligation:

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

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

active(primes) → mark(sieve(from(s(s(0)))))
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, Y))) → mark(X)
active(tail(cons(X, Y))) → mark(Y)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(filter(s(s(X)), cons(Y, Z))) → mark(if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y)))))
active(sieve(cons(X, Y))) → mark(cons(X, filter(X, sieve(Y))))
active(sieve(X)) → sieve(active(X))
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(filter(X1, X2)) → filter(active(X1), X2)
active(filter(X1, X2)) → filter(X1, active(X2))
active(divides(X1, X2)) → divides(active(X1), X2)
active(divides(X1, X2)) → divides(X1, active(X2))
sieve(mark(X)) → mark(sieve(X))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
filter(mark(X1), X2) → mark(filter(X1, X2))
filter(X1, mark(X2)) → mark(filter(X1, X2))
divides(mark(X1), X2) → mark(divides(X1, X2))
divides(X1, mark(X2)) → mark(divides(X1, X2))
proper(primes) → ok(primes)
proper(sieve(X)) → sieve(proper(X))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(true) → ok(true)
proper(false) → ok(false)
proper(filter(X1, X2)) → filter(proper(X1), proper(X2))
proper(divides(X1, X2)) → divides(proper(X1), proper(X2))
sieve(ok(X)) → ok(sieve(X))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
filter(ok(X1), ok(X2)) → ok(filter(X1, X2))
divides(ok(X1), ok(X2)) → ok(divides(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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