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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 QDPOrderProof (⇔)
    ↳16 QDP
    ↳17 PisEmptyProof (⇔)
    ↳18 TRUE
  ↳19 QDP
    ↳20 QDPOrderProof (⇔)
    ↳21 QDP
    ↳22 QDPOrderProof (⇔)
    ↳23 QDP
    ↳24 PisEmptyProof (⇔)
    ↳25 TRUE
  ↳26 QDP
    ↳27 QDPOrderProof (⇔)
    ↳28 QDP
    ↳29 QDPOrderProof (⇔)
    ↳30 QDP
    ↳31 PisEmptyProof (⇔)
    ↳32 TRUE
  ↳33 QDP
    ↳34 QDPOrderProof (⇔)
    ↳35 QDP
    ↳36 QDPOrderProof (⇔)
    ↳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 QDPOrderProof (⇔)
    ↳52 QDP
    ↳53 PisEmptyProof (⇔)
    ↳54 TRUE
  ↳55 QDP
    ↳56 QDPOrderProof (⇔)
    ↳57 QDP
    ↳58 QDPOrderProof (⇔)
    ↳59 QDP
    ↳60 QDPOrderProof (⇔)
    ↳61 QDP
    ↳62 QDPOrderProof (⇔)
    ↳63 QDP
    ↳64 PisEmptyProof (⇔)
    ↳65 TRUE
  ↳66 QDP
    ↳67 QDPOrderProof (⇔)
    ↳68 QDP
    ↳69 QDPOrderProof (⇔)
    ↳70 QDP
    ↳71 PisEmptyProof (⇔)
    ↳72 TRUE

(0) Obligation:

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

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
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(filter(cons(X, Y), 0, M)) → CONS(0, filter(Y, M, M))
ACTIVE(filter(cons(X, Y), 0, M)) → FILTER(Y, M, M)
ACTIVE(filter(cons(X, Y), s(N), M)) → CONS(X, filter(Y, N, M))
ACTIVE(filter(cons(X, Y), s(N), M)) → FILTER(Y, N, M)
ACTIVE(sieve(cons(0, Y))) → CONS(0, sieve(Y))
ACTIVE(sieve(cons(0, Y))) → SIEVE(Y)
ACTIVE(sieve(cons(s(N), Y))) → CONS(s(N), sieve(filter(Y, N, N)))
ACTIVE(sieve(cons(s(N), Y))) → SIEVE(filter(Y, N, N))
ACTIVE(sieve(cons(s(N), Y))) → FILTER(Y, N, N)
ACTIVE(nats(N)) → CONS(N, nats(s(N)))
ACTIVE(nats(N)) → NATS(s(N))
ACTIVE(nats(N)) → S(N)
ACTIVE(zprimes) → SIEVE(nats(s(s(0))))
ACTIVE(zprimes) → NATS(s(s(0)))
ACTIVE(zprimes) → S(s(0))
ACTIVE(zprimes) → S(0)
ACTIVE(filter(X1, X2, X3)) → FILTER(active(X1), X2, X3)
ACTIVE(filter(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(filter(X1, X2, X3)) → FILTER(X1, active(X2), X3)
ACTIVE(filter(X1, X2, X3)) → ACTIVE(X2)
ACTIVE(filter(X1, X2, X3)) → FILTER(X1, X2, active(X3))
ACTIVE(filter(X1, X2, X3)) → ACTIVE(X3)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(sieve(X)) → SIEVE(active(X))
ACTIVE(sieve(X)) → ACTIVE(X)
ACTIVE(nats(X)) → NATS(active(X))
ACTIVE(nats(X)) → ACTIVE(X)
FILTER(mark(X1), X2, X3) → FILTER(X1, X2, X3)
FILTER(X1, mark(X2), X3) → FILTER(X1, X2, X3)
FILTER(X1, X2, mark(X3)) → FILTER(X1, X2, X3)
CONS(mark(X1), X2) → CONS(X1, X2)
S(mark(X)) → S(X)
SIEVE(mark(X)) → SIEVE(X)
NATS(mark(X)) → NATS(X)
PROPER(filter(X1, X2, X3)) → FILTER(proper(X1), proper(X2), proper(X3))
PROPER(filter(X1, X2, X3)) → PROPER(X1)
PROPER(filter(X1, X2, X3)) → PROPER(X2)
PROPER(filter(X1, X2, X3)) → PROPER(X3)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(sieve(X)) → SIEVE(proper(X))
PROPER(sieve(X)) → PROPER(X)
PROPER(nats(X)) → NATS(proper(X))
PROPER(nats(X)) → PROPER(X)
FILTER(ok(X1), ok(X2), ok(X3)) → FILTER(X1, X2, X3)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
S(ok(X)) → S(X)
SIEVE(ok(X)) → SIEVE(X)
NATS(ok(X)) → NATS(X)
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(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
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 8 SCCs with 30 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

NATS(ok(X)) → NATS(X)
NATS(mark(X)) → NATS(X)

The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
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.


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

Lexicographic Path Order [LPO].
Precedence:
mark1 > NATS1

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

NATS(ok(X)) → NATS(X)

The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
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.


NATS(ok(X)) → NATS(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
NATS(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:
P is empty.
The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

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

The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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)  =  SIEVE(x1)
ok(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
mark1 > SIEVE1

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

SIEVE(ok(X)) → SIEVE(X)

The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
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.


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)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) Obligation:

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

S(ok(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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)  =  S(x1)
ok(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
mark1 > S1

The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

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

S(ok(X)) → S(X)

The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(22) 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)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

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

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
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) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE

(26) 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(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(27) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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:
trivial

The following usable rules [FROCOS05] were oriented: none

(28) 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(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(29) 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)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(30) Obligation:

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

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
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) PisEmptyProof (EQUIVALENT transformation)

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

(32) TRUE

(33) Obligation:

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

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

The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(34) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FILTER(X1, X2, mark(X3)) → FILTER(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FILTER(x1, x2, x3)  =  FILTER(x3)
mark(x1)  =  mark(x1)
ok(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
mark1 > FILTER1

The following usable rules [FROCOS05] were oriented: none

(35) Obligation:

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

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

The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(36) 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), ok(X3)) → FILTER(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FILTER(x1, x2, x3)  =  x3
mark(x1)  =  mark
ok(x1)  =  ok(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(37) Obligation:

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

FILTER(X1, mark(X2), X3) → FILTER(X1, X2, X3)
FILTER(mark(X1), X2, X3) → FILTER(X1, X2, X3)

The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
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.


FILTER(X1, mark(X2), X3) → FILTER(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FILTER(x1, x2, x3)  =  x2
mark(x1)  =  mark(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:

FILTER(mark(X1), X2, X3) → FILTER(X1, X2, X3)

The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
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.


FILTER(mark(X1), X2, X3) → FILTER(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(41) Obligation:

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

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
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:

PROPER(filter(X1, X2, X3)) → PROPER(X2)
PROPER(filter(X1, X2, X3)) → PROPER(X1)
PROPER(filter(X1, X2, X3)) → PROPER(X3)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(s(X)) → PROPER(X)
PROPER(sieve(X)) → PROPER(X)
PROPER(nats(X)) → PROPER(X)

The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
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.


PROPER(filter(X1, X2, X3)) → PROPER(X2)
PROPER(filter(X1, X2, X3)) → PROPER(X1)
PROPER(filter(X1, X2, X3)) → PROPER(X3)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  x1
filter(x1, x2, x3)  =  filter(x1, x2, x3)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
sieve(x1)  =  x1
nats(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(46) Obligation:

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

PROPER(s(X)) → PROPER(X)
PROPER(sieve(X)) → PROPER(X)
PROPER(nats(X)) → PROPER(X)

The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
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.


PROPER(nats(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)
s(x1)  =  x1
sieve(x1)  =  x1
nats(x1)  =  nats(x1)

Lexicographic Path Order [LPO].
Precedence:
nats1 > PROPER1

The following usable rules [FROCOS05] were oriented: none

(48) Obligation:

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

PROPER(s(X)) → PROPER(X)
PROPER(sieve(X)) → PROPER(X)

The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
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) 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)
s(x1)  =  x1
sieve(x1)  =  sieve(x1)

Lexicographic Path Order [LPO].
Precedence:
sieve1 > PROPER1

The following usable rules [FROCOS05] were oriented: none

(50) Obligation:

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

PROPER(s(X)) → PROPER(X)

The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
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.

(51) 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
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(52) Obligation:

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

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(53) PisEmptyProof (EQUIVALENT transformation)

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

(54) TRUE

(55) Obligation:

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

ACTIVE(filter(X1, X2, X3)) → ACTIVE(X2)
ACTIVE(filter(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(filter(X1, X2, X3)) → ACTIVE(X3)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(sieve(X)) → ACTIVE(X)
ACTIVE(nats(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(filter(X1, X2, X3)) → ACTIVE(X2)
ACTIVE(filter(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(filter(X1, X2, X3)) → ACTIVE(X3)
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)  =  x1
filter(x1, x2, x3)  =  filter(x1, x2, x3)
cons(x1, x2)  =  x1
s(x1)  =  s(x1)
sieve(x1)  =  x1
nats(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(57) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(sieve(X)) → ACTIVE(X)
ACTIVE(nats(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
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.

(58) 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)
sieve(x1)  =  x1
nats(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
cons2 > ACTIVE1

The following usable rules [FROCOS05] were oriented: none

(59) Obligation:

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

ACTIVE(sieve(X)) → ACTIVE(X)
ACTIVE(nats(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(60) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(nats(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)
sieve(x1)  =  x1
nats(x1)  =  nats(x1)

Lexicographic Path Order [LPO].
Precedence:
nats1 > ACTIVE1

The following usable rules [FROCOS05] were oriented: none

(61) Obligation:

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

ACTIVE(sieve(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(62) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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)  =  x1
sieve(x1)  =  sieve(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(63) Obligation:

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

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(64) PisEmptyProof (EQUIVALENT transformation)

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

(65) TRUE

(66) Obligation:

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

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

The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(67) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TOP(mark(X)) → TOP(proper(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TOP(x1)  =  TOP(x1)
ok(x1)  =  x1
active(x1)  =  x1
mark(x1)  =  mark(x1)
proper(x1)  =  x1
s(x1)  =  x1
sieve(x1)  =  sieve(x1)
nats(x1)  =  nats(x1)
zprimes  =  zprimes
filter(x1, x2, x3)  =  filter(x1, x2, x3)
cons(x1, x2)  =  cons(x1)
0  =  0

Lexicographic Path Order [LPO].
Precedence:
zprimes > sieve1 > cons1 > mark1
zprimes > sieve1 > 0 > mark1
zprimes > nats1 > cons1 > mark1
filter3 > cons1 > mark1
filter3 > 0 > mark1

The following usable rules [FROCOS05] were oriented:

proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(nats(X)) → nats(active(X))
active(sieve(X)) → sieve(active(X))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
sieve(mark(X)) → mark(sieve(X))
s(mark(X)) → mark(s(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
nats(mark(X)) → mark(nats(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))

(68) Obligation:

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

TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(69) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TOP(ok(X)) → TOP(active(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TOP(x1)  =  TOP(x1)
ok(x1)  =  ok(x1)
active(x1)  =  x1
filter(x1, x2, x3)  =  x2
cons(x1, x2)  =  x1
s(x1)  =  x1
sieve(x1)  =  x1
nats(x1)  =  x1
mark(x1)  =  mark
zprimes  =  zprimes
0  =  0

Lexicographic Path Order [LPO].
Precedence:
TOP1 > mark
ok1 > mark
zprimes > mark
0 > mark

The following usable rules [FROCOS05] were oriented:

filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(nats(X)) → nats(active(X))
active(sieve(X)) → sieve(active(X))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
sieve(mark(X)) → mark(sieve(X))
s(mark(X)) → mark(s(X))
nats(mark(X)) → mark(nats(X))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))

(70) Obligation:

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

active(filter(cons(X, Y), 0, M)) → mark(cons(0, filter(Y, M, M)))
active(filter(cons(X, Y), s(N), M)) → mark(cons(X, filter(Y, N, M)))
active(sieve(cons(0, Y))) → mark(cons(0, sieve(Y)))
active(sieve(cons(s(N), Y))) → mark(cons(s(N), sieve(filter(Y, N, N))))
active(nats(N)) → mark(cons(N, nats(s(N))))
active(zprimes) → mark(sieve(nats(s(s(0)))))
active(filter(X1, X2, X3)) → filter(active(X1), X2, X3)
active(filter(X1, X2, X3)) → filter(X1, active(X2), X3)
active(filter(X1, X2, X3)) → filter(X1, X2, active(X3))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sieve(X)) → sieve(active(X))
active(nats(X)) → nats(active(X))
filter(mark(X1), X2, X3) → mark(filter(X1, X2, X3))
filter(X1, mark(X2), X3) → mark(filter(X1, X2, X3))
filter(X1, X2, mark(X3)) → mark(filter(X1, X2, X3))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sieve(mark(X)) → mark(sieve(X))
nats(mark(X)) → mark(nats(X))
proper(filter(X1, X2, X3)) → filter(proper(X1), proper(X2), proper(X3))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sieve(X)) → sieve(proper(X))
proper(nats(X)) → nats(proper(X))
proper(zprimes) → ok(zprimes)
filter(ok(X1), ok(X2), ok(X3)) → ok(filter(X1, X2, X3))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sieve(ok(X)) → ok(sieve(X))
nats(ok(X)) → ok(nats(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(71) PisEmptyProof (EQUIVALENT transformation)

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

(72) TRUE