Termination w.r.t. Q proof of /tmp/tmpguk5NY/ExIntrod_GM99

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

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 QDPOrderProof (⇔)

    ↳11 QDP

    ↳12 PisEmptyProof (⇔)

    ↳13 TRUE

  ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 QDPOrderProof (⇔)

    ↳18 QDP

    ↳19 QDPOrderProof (⇔)

    ↳20 QDP

    ↳21 PisEmptyProof (⇔)

    ↳22 TRUE

  ↳23 QDP

    ↳24 QDPOrderProof (⇔)

    ↳25 QDP

    ↳26 QDPOrderProof (⇔)

    ↳27 QDP

    ↳28 PisEmptyProof (⇔)

    ↳29 TRUE

  ↳30 QDP

    ↳31 QDPOrderProof (⇔)

    ↳32 QDP

    ↳33 QDPOrderProof (⇔)

    ↳34 QDP

    ↳35 PisEmptyProof (⇔)

    ↳36 TRUE

  ↳37 QDP

    ↳38 QDPOrderProof (⇔)

    ↳39 QDP

    ↳40 QDPOrderProof (⇔)

    ↳41 QDP

    ↳42 PisEmptyProof (⇔)

    ↳43 TRUE

  ↳44 QDP

    ↳45 QDPOrderProof (⇔)

    ↳46 QDP

    ↳47 QDPOrderProof (⇔)

    ↳48 QDP

    ↳49 PisEmptyProof (⇔)

    ↳50 TRUE

  ↳51 QDP

    ↳52 QDPOrderProof (⇔)

    ↳53 QDP

    ↳54 QDPOrderProof (⇔)

    ↳55 QDP

    ↳56 PisEmptyProof (⇔)

    ↳57 TRUE

  ↳58 QDP

    ↳59 QDPOrderProof (⇔)

    ↳60 QDP

    ↳61 QDPOrderProof (⇔)

    ↳62 QDP

    ↳63 PisEmptyProof (⇔)

    ↳64 TRUE

  ↳65 QDP

    ↳66 QDPOrderProof (⇔)

    ↳67 QDP

    ↳68 QDPOrderProof (⇔)

    ↳69 QDP

    ↳70 PisEmptyProof (⇔)

    ↳71 TRUE

  ↳72 QDP

    ↳73 QDPOrderProof (⇔)

    ↳74 QDP

    ↳75 QDPOrderProof (⇔)

    ↳76 QDP

    ↳77 QDPOrderProof (⇔)

    ↳78 QDP

    ↳79 QDPOrderProof (⇔)

    ↳80 QDP

    ↳81 QDPOrderProof (⇔)

    ↳82 QDP

    ↳83 QDPOrderProof (⇔)

    ↳84 QDP

    ↳85 PisEmptyProof (⇔)

    ↳86 TRUE

  ↳87 QDP

    ↳88 QDPOrderProof (⇔)

    ↳89 QDP

    ↳90 QDPOrderProof (⇔)

    ↳91 QDP

    ↳92 QDPOrderProof (⇔)

    ↳93 QDP

    ↳94 QDPOrderProof (⇔)

    ↳95 QDP

    ↳96 QDPOrderProof (⇔)

    ↳97 QDP

    ↳98 QDPOrderProof (⇔)

    ↳99 QDP

    ↳100 QDPOrderProof (⇔)

    ↳101 QDP

    ↳102 QDPOrderProof (⇔)

    ↳103 QDP

    ↳104 PisEmptyProof (⇔)

    ↳105 TRUE

  ↳106 QDP

(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(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) = DIVIDES(x2)
mark(x1) = mark(x1)
ok(x1) = 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(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.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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) = x2
mark(x1) = x1
ok(x1) = ok(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

DIVIDES(mark(X1), 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.

(10) 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)

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)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(11) 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.

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) TRUE

(14) 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.

(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) = FILTER(x2)
mark(x1) = mark(x1)
ok(x1) = x1

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

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.

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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) = x2
mark(x1) = x1
ok(x1) = ok(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

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

FILTER(mark(X1), 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.

(19) 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)

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)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(20) 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.

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE

(23) 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.

(24) 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)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

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

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.

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(mark(X1), X2, X3) → IF(X1, X2, X3)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(27) 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.

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) TRUE

(30) 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.

(31) 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

(32) 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.

(33) 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

(34) 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.

(35) PisEmptyProof (EQUIVALENT transformation)

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

(36) TRUE

(37) 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.

(38) 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

(39) 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.

(40) 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

(41) 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.

(42) PisEmptyProof (EQUIVALENT transformation)

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

(43) TRUE

(44) 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.

(45) 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)
mark(x1) = mark(x1)

Lexicographic Path Order [LPO].
Precedence:

[CONS₁, mark₁]

The following usable rules [FROCOS05] were oriented: none

(46) Obligation:

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

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.

(47) 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
mark(x1) = mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(48) 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.

(49) PisEmptyProof (EQUIVALENT transformation)

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

(50) TRUE

(51) 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.

(52) 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

(53) 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.

(54) 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

(55) 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.

(56) PisEmptyProof (EQUIVALENT transformation)

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

(57) TRUE

(58) 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.

(59) 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

(60) 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.

(61) 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

(62) 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.

(63) PisEmptyProof (EQUIVALENT transformation)

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

(64) TRUE

(65) 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.

(66) 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

(67) 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.

(68) 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

(69) 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.

(70) PisEmptyProof (EQUIVALENT transformation)

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

(71) TRUE

(72) 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.

(73) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
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) = PROPER(x1)
from(x1) = x1
sieve(x1) = x1
s(x1) = x1
cons(x1, x2) = cons(x1, x2)
head(x1) = 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

(74) 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(head(X)) → PROPER(X)
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.

(75) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(sieve(X)) → PROPER(X)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(76) Obligation:

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

PROPER(from(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(head(X)) → PROPER(X)
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.

(77) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(s(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
from(x1) = x1
s(x1) = s(x1)
head(x1) = x1
tail(x1) = x1

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(78) Obligation:

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

PROPER(from(X)) → PROPER(X)
PROPER(head(X)) → PROPER(X)
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.

(79) 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)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(80) Obligation:

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

PROPER(head(X)) → PROPER(X)
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.

(81) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(head(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
head(x1) = head(x1)
tail(x1) = x1

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(82) 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.

(83) 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

(84) 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.

(85) PisEmptyProof (EQUIVALENT transformation)

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

(86) TRUE

(87) 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.

(88) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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) = ACTIVE(x1)
from(x1) = x1
sieve(x1) = x1
s(x1) = x1
cons(x1, x2) = x1
head(x1) = x1
tail(x1) = 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

(89) 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)

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.

(90) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(head(X)) → ACTIVE(X)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(91) 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(tail(X)) → ACTIVE(X)
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.

(92) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(tail(X)) → ACTIVE(X)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(93) 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(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.

(94) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(sieve(X)) → ACTIVE(X)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(95) Obligation:

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

ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
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.

(96) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(s(X)) → ACTIVE(X)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(97) Obligation:

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

ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
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.

(98) 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)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(99) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
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.

(100) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(cons(X1, X2)) → 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)
cons(x1, x2) = cons(x1, x2)
if(x1, x2, x3) = x1

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(101) 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.

(102) 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) = x1
if(x1, x2, x3) = if(x1, x2, x3)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(103) 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.

(104) PisEmptyProof (EQUIVALENT transformation)

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

(105) TRUE

(106) 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.