Termination w.r.t. Q proof of /tmp/tmpORZKoo/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 QDPOrderProof (⇔)

    ↳50 QDP

    ↳51 PisEmptyProof (⇔)

    ↳52 TRUE

  ↳53 QDP

    ↳54 QDPOrderProof (⇔)

    ↳55 QDP

    ↳56 QDPOrderProof (⇔)

    ↳57 QDP

    ↳58 PisEmptyProof (⇔)

    ↳59 TRUE

  ↳60 QDP

    ↳61 QDPOrderProof (⇔)

    ↳62 QDP

    ↳63 QDPOrderProof (⇔)

    ↳64 QDP

    ↳65 PisEmptyProof (⇔)

    ↳66 TRUE

  ↳67 QDP

    ↳68 QDPOrderProof (⇔)

    ↳69 QDP

    ↳70 QDPOrderProof (⇔)

    ↳71 QDP

    ↳72 PisEmptyProof (⇔)

    ↳73 TRUE

  ↳74 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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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) → MARK(sieve(from(s(s(0)))))
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)) → MARK(cons(X, from(s(X))))
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → 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(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))) → MARK(cons(X, filter(X, 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)
MARK(primes) → ACTIVE(primes)
MARK(sieve(X)) → ACTIVE(sieve(mark(X)))
MARK(sieve(X)) → SIEVE(mark(X))
MARK(sieve(X)) → MARK(X)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(from(X)) → FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(0) → ACTIVE(0)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(head(X)) → HEAD(mark(X))
MARK(head(X)) → MARK(X)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(tail(X)) → TAIL(mark(X))
MARK(tail(X)) → MARK(X)
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))
MARK(if(X1, X2, X3)) → IF(mark(X1), X2, X3)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(true) → ACTIVE(true)
MARK(false) → ACTIVE(false)
MARK(filter(X1, X2)) → ACTIVE(filter(mark(X1), mark(X2)))
MARK(filter(X1, X2)) → FILTER(mark(X1), mark(X2))
MARK(filter(X1, X2)) → MARK(X1)
MARK(filter(X1, X2)) → MARK(X2)
MARK(divides(X1, X2)) → ACTIVE(divides(mark(X1), mark(X2)))
MARK(divides(X1, X2)) → DIVIDES(mark(X1), mark(X2))
MARK(divides(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X2)
SIEVE(mark(X)) → SIEVE(X)
SIEVE(active(X)) → SIEVE(X)
FROM(mark(X)) → FROM(X)
FROM(active(X)) → FROM(X)
S(mark(X)) → S(X)
S(active(X)) → S(X)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
HEAD(mark(X)) → HEAD(X)
HEAD(active(X)) → HEAD(X)
TAIL(mark(X)) → TAIL(X)
TAIL(active(X)) → TAIL(X)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
IF(X1, mark(X2), X3) → IF(X1, X2, X3)
IF(X1, X2, mark(X3)) → IF(X1, X2, X3)
IF(active(X1), X2, X3) → IF(X1, X2, X3)
IF(X1, active(X2), X3) → IF(X1, X2, X3)
IF(X1, X2, active(X3)) → IF(X1, X2, X3)
FILTER(mark(X1), X2) → FILTER(X1, X2)
FILTER(X1, mark(X2)) → FILTER(X1, X2)
FILTER(active(X1), X2) → FILTER(X1, X2)
FILTER(X1, active(X2)) → FILTER(X1, X2)
DIVIDES(mark(X1), X2) → DIVIDES(X1, X2)
DIVIDES(X1, mark(X2)) → DIVIDES(X1, X2)
DIVIDES(active(X1), X2) → DIVIDES(X1, X2)
DIVIDES(X1, active(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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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 10 SCCs with 28 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(active(X1), X2) → DIVIDES(X1, X2)
DIVIDES(X1, active(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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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)
DIVIDES(X1, active(X2)) → DIVIDES(X1, X2)

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

Recursive path order with status [RPO].
Precedence:

trivial

Status:

mark₁: multiset
active₁: multiset

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

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(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) = DIVIDES(x1)
mark(x1) = mark(x1)
active(x1) = x1

Recursive path order with status [RPO].
Precedence:

mark₁ > DIVIDES₁

Status:

DIVIDES₁: multiset
mark₁: multiset

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

DIVIDES(active(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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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(active(X1), X2) → DIVIDES(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Precedence:

active₁ > DIVIDES₂

Status:

DIVIDES₂: [1,2]
active₁: multiset

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

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(active(X1), X2) → FILTER(X1, X2)
FILTER(X1, active(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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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)
FILTER(X1, active(X2)) → FILTER(X1, X2)

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

Recursive path order with status [RPO].
Precedence:

trivial

Status:

mark₁: multiset
active₁: multiset

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

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(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) = FILTER(x1)
mark(x1) = mark(x1)
active(x1) = x1

Recursive path order with status [RPO].
Precedence:

mark₁ > FILTER₁

Status:

FILTER₁: multiset
mark₁: multiset

The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

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

FILTER(active(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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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(active(X1), X2) → FILTER(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Precedence:

active₁ > FILTER₂

Status:

FILTER₂: [1,2]
active₁: multiset

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

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(X1, mark(X2), X3) → IF(X1, X2, X3)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
IF(X1, X2, mark(X3)) → IF(X1, X2, X3)
IF(active(X1), X2, X3) → IF(X1, X2, X3)
IF(X1, active(X2), X3) → IF(X1, X2, X3)
IF(X1, X2, active(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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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(mark(X1), X2, X3) → IF(X1, X2, X3)
IF(active(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)
active(x1) = active(x1)

Recursive path order with status [RPO].
Precedence:

trivial

Status:

mark₁: multiset
active₁: multiset

The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

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

IF(X1, mark(X2), X3) → IF(X1, X2, X3)
IF(X1, X2, mark(X3)) → IF(X1, X2, X3)
IF(X1, active(X2), X3) → IF(X1, X2, X3)
IF(X1, X2, active(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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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(X1, mark(X2), X3) → IF(X1, X2, X3)
IF(X1, X2, mark(X3)) → IF(X1, X2, X3)
IF(X1, active(X2), X3) → IF(X1, X2, X3)
IF(X1, X2, active(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(x2, x3)
mark(x1) = mark(x1)
active(x1) = active(x1)

Recursive path order with status [RPO].
Precedence:

mark₁ > IF₂
active₁ > IF₂

Status:

IF₂: multiset
mark₁: multiset
active₁: multiset

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

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

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(active(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
active(x1) = active(x1)
mark(x1) = x1

Recursive path order with status [RPO].
Precedence:

trivial

Status:

active₁: multiset

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

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: Recursive path order with status [RPO].
Precedence:

mark₁ > TAIL₁

Status:

TAIL₁: [1]
mark₁: multiset

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

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

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(active(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
active(x1) = active(x1)
mark(x1) = x1

Recursive path order with status [RPO].
Precedence:

trivial

Status:

active₁: multiset

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

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: Recursive path order with status [RPO].
Precedence:

mark₁ > HEAD₁

Status:

HEAD₁: [1]
mark₁: multiset

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

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(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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(X1, mark(X2)) → CONS(X1, X2)
CONS(X1, active(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) = x2
mark(x1) = mark(x1)
active(x1) = active(x1)

Recursive path order with status [RPO].
Precedence:

trivial

Status:

mark₁: multiset
active₁: multiset

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)
CONS(active(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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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) = CONS(x1)
mark(x1) = mark(x1)
active(x1) = x1

Recursive path order with status [RPO].
Precedence:

mark₁ > CONS₁

Status:

CONS₁: multiset
mark₁: multiset

The following usable rules [FROCOS05] were oriented: none

(48) Obligation:

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

CONS(active(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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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

(49) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(active(X1), X2) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Precedence:

active₁ > CONS₂

Status:

CONS₂: [1,2]
active₁: multiset

The following usable rules [FROCOS05] were oriented: none

(50) 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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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

(51) PisEmptyProof (EQUIVALENT transformation)

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

(52) TRUE

(53) Obligation:

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

S(active(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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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(active(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
active(x1) = active(x1)
mark(x1) = x1

Recursive path order with status [RPO].
Precedence:

trivial

Status:

active₁: multiset

The following usable rules [FROCOS05] were oriented: none

(55) 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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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

(56) 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: Recursive path order with status [RPO].
Precedence:

mark₁ > S₁

Status:

S₁: [1]
mark₁: multiset

The following usable rules [FROCOS05] were oriented: none

(57) 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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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

(58) PisEmptyProof (EQUIVALENT transformation)

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

(59) TRUE

(60) Obligation:

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

FROM(active(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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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(active(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
active(x1) = active(x1)
mark(x1) = x1

Recursive path order with status [RPO].
Precedence:

trivial

Status:

active₁: multiset

The following usable rules [FROCOS05] were oriented: none

(62) 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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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

(63) 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: Recursive path order with status [RPO].
Precedence:

mark₁ > FROM₁

Status:

FROM₁: [1]
mark₁: multiset

The following usable rules [FROCOS05] were oriented: none

(64) 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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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

(65) PisEmptyProof (EQUIVALENT transformation)

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

(66) TRUE

(67) Obligation:

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

SIEVE(active(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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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(active(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
active(x1) = active(x1)
mark(x1) = x1

Recursive path order with status [RPO].
Precedence:

trivial

Status:

active₁: multiset

The following usable rules [FROCOS05] were oriented: none

(69) 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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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

(70) 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: Recursive path order with status [RPO].
Precedence:

mark₁ > SIEVE₁

Status:

SIEVE₁: [1]
mark₁: multiset

The following usable rules [FROCOS05] were oriented: none

(71) 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))))
mark(primes) → active(primes)
mark(sieve(X)) → active(sieve(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(true) → active(true)
mark(false) → active(false)
mark(filter(X1, X2)) → active(filter(mark(X1), mark(X2)))
mark(divides(X1, X2)) → active(divides(mark(X1), mark(X2)))
sieve(mark(X)) → sieve(X)
sieve(active(X)) → sieve(X)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
filter(mark(X1), X2) → filter(X1, X2)
filter(X1, mark(X2)) → filter(X1, X2)
filter(active(X1), X2) → filter(X1, X2)
filter(X1, active(X2)) → filter(X1, X2)
divides(mark(X1), X2) → divides(X1, X2)
divides(X1, mark(X2)) → divides(X1, X2)
divides(active(X1), X2) → divides(X1, X2)
divides(X1, active(X2)) → divides(X1, X2)

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

(72) PisEmptyProof (EQUIVALENT transformation)

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

(73) TRUE

(74) Obligation:

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

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

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