Termination w.r.t. Q proof of /tmp/tmp1QydsV/Ex9_BLR02

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 PisEmptyProof (⇔)

    ↳41 TRUE

  ↳42 QDP

    ↳43 QDPOrderProof (⇔)

    ↳44 QDP

    ↳45 QDPOrderProof (⇔)

    ↳46 QDP

    ↳47 QDPOrderProof (⇔)

    ↳48 QDP

    ↳49 PisEmptyProof (⇔)

    ↳50 TRUE

  ↳51 QDP

    ↳52 QDPOrderProof (⇔)

    ↳53 QDP

    ↳54 QDPOrderProof (⇔)

    ↳55 QDP

    ↳56 QDPOrderProof (⇔)

    ↳57 QDP

    ↳58 PisEmptyProof (⇔)

    ↳59 TRUE

  ↳60 QDP

    ↳61 QDPOrderProof (⇔)

    ↳62 QDP

    ↳63 QDPOrderProof (⇔)

    ↳64 QDP

    ↳65 PisEmptyProof (⇔)

    ↳66 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(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) = NATS(x1)
ok(x1) = ok(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Quasi-Precedence:

[NATS₁, ok₁]

Status:

ok₁: [1]
NATS₁: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

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.

(8) 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) = x1
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

mark₁: [1]

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[SIEVE₁, ok₁]

Status:

SIEVE₁: [1]
ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(14) 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(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(mark(X)) → SIEVE(X)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

mark₁: [1]

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[S₁, ok₁]

Status:

ok₁: [1]
S₁: [1]

The following usable rules [FROCOS05] were oriented: none

(21) 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(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(mark(X)) → S(X)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

mark₁: [1]

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

ok₁ > [CONS₁, mark]

Status:

CONS₁: [1]
mark: []
ok₁: [1]

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: Lexicographic path order with status [LPO].
Quasi-Precedence:

mark₁ > CONS₂

Status:

CONS₂: [2,1]
mark₁: [1]

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, mark(X2), X3) → FILTER(X1, X2, X3)
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(x2, x3)
mark(x1) = mark(x1)
ok(x1) = x1

Lexicographic path order with status [LPO].
Quasi-Precedence:

[FILTER₂, mark₁]

Status:

mark₁: [1]
FILTER₂: [2,1]

The following usable rules [FROCOS05] were oriented: none

(35) Obligation:

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

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(mark(X1), 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) = FILTER(x1, x2, x3)
mark(x1) = mark(x1)
ok(x1) = x1

Lexicographic path order with status [LPO].
Quasi-Precedence:

[FILTER₃, mark₁]

Status:

FILTER₃: [3,2,1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(37) Obligation:

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

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.

(38) 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: Lexicographic path order with status [LPO].
Quasi-Precedence:

ok₁ > FILTER₃

Status:

FILTER₃: [2,1,3]
ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(39) 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.

(40) PisEmptyProof (EQUIVALENT transformation)

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

(41) TRUE

(42) 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.

(43) 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)
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)
filter(x1, x2, x3) = filter(x1, x2, x3)
cons(x1, x2) = cons(x1, x2)
s(x1) = x1
sieve(x1) = sieve(x1)
nats(x1) = x1

Lexicographic path order with status [LPO].
Quasi-Precedence:

[PROPER₁, filter₃]

Status:

cons₂: [2,1]
PROPER₁: [1]
filter₃: [3,2,1]
sieve₁: [1]

The following usable rules [FROCOS05] were oriented: none

(44) Obligation:

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

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[PROPER₁, s₁]

Status:

PROPER₁: [1]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(46) Obligation:

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

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) = x1
nats(x1) = nats(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

nats₁: [1]

The following usable rules [FROCOS05] were oriented: none

(48) Obligation:

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

active(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) PisEmptyProof (EQUIVALENT transformation)

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

(50) TRUE

(51) Obligation:

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

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.

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[ACTIVE₁, nats₁]

Status:

cons₂: [2,1]
filter₃: [1,3,2]
nats₁: [1]
ACTIVE₁: [1]

The following usable rules [FROCOS05] were oriented: none

(53) Obligation:

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

ACTIVE(s(X)) → ACTIVE(X)
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.

(54) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(s(X)) → ACTIVE(X)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[ACTIVE₁, s₁]

Status:

s₁: [1]
ACTIVE₁: [1]

The following usable rules [FROCOS05] were oriented: none

(55) 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.

(56) 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 with status [LPO].
Quasi-Precedence:

trivial

Status:

sieve₁: [1]

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(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) 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:

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.

(61) 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
filter(x1, x2, x3) = filter(x1, x2, x3)
cons(x1, x2) = x1
s(x1) = s(x1)
sieve(x1) = sieve(x1)
nats(x1) = nats(x1)
zprimes = zprimes
0 = 0

Lexicographic path order with status [LPO].
Quasi-Precedence:

TOP₁ > [mark₁, sieve₁]
zprimes > nats₁ > s₁ > [filter₃, 0] > [mark₁, sieve₁]

Status:

filter₃: [1,3,2]
nats₁: [1]
mark₁: [1]
s₁: [1]
TOP₁: [1]
0: []
zprimes: []
sieve₁: [1]

The following usable rules [FROCOS05] were oriented:

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

(62) 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.

(63) 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) = x3
cons(x1, x2) = x2
s(x1) = x1
sieve(x1) = x1
mark(x1) = mark
nats(x1) = nats(x1)
zprimes = zprimes
0 = 0

Lexicographic path order with status [LPO].
Quasi-Precedence:

zprimes > [TOP₁, ok₁, nats₁] > [mark, 0]

Status:

nats₁: [1]
mark: []
ok₁: [1]
TOP₁: [1]
0: []
zprimes: []

The following usable rules [FROCOS05] were oriented:

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

(64) 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.

(65) PisEmptyProof (EQUIVALENT transformation)

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

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Complex Obligation (AND)

(5) Obligation:

(6) QDPOrderProof (EQUIVALENT transformation)

(7) Obligation:

(8) QDPOrderProof (EQUIVALENT transformation)

(9) Obligation:

(10) PisEmptyProof (EQUIVALENT transformation)

(11) TRUE

(12) Obligation:

(13) QDPOrderProof (EQUIVALENT transformation)

(14) Obligation:

(15) QDPOrderProof (EQUIVALENT transformation)

(16) Obligation:

(17) PisEmptyProof (EQUIVALENT transformation)

(18) TRUE

(19) Obligation:

(20) QDPOrderProof (EQUIVALENT transformation)

(21) Obligation:

(22) QDPOrderProof (EQUIVALENT transformation)

(23) Obligation:

(24) PisEmptyProof (EQUIVALENT transformation)

(25) TRUE

(26) Obligation:

(27) QDPOrderProof (EQUIVALENT transformation)

(28) Obligation:

(29) QDPOrderProof (EQUIVALENT transformation)

(30) Obligation:

(31) PisEmptyProof (EQUIVALENT transformation)

(32) TRUE

(33) Obligation:

(34) QDPOrderProof (EQUIVALENT transformation)

(35) Obligation:

(36) QDPOrderProof (EQUIVALENT transformation)

(37) Obligation:

(38) QDPOrderProof (EQUIVALENT transformation)

(39) Obligation:

(40) PisEmptyProof (EQUIVALENT transformation)

(41) TRUE

(42) Obligation:

(43) QDPOrderProof (EQUIVALENT transformation)

(44) Obligation:

(45) QDPOrderProof (EQUIVALENT transformation)

(46) Obligation:

(47) QDPOrderProof (EQUIVALENT transformation)

(48) Obligation:

(49) PisEmptyProof (EQUIVALENT transformation)

(50) TRUE

(51) Obligation:

(52) QDPOrderProof (EQUIVALENT transformation)

(53) Obligation:

(54) QDPOrderProof (EQUIVALENT transformation)

(55) Obligation:

(56) QDPOrderProof (EQUIVALENT transformation)

(57) Obligation:

(58) PisEmptyProof (EQUIVALENT transformation)

(59) TRUE

(60) Obligation:

(61) QDPOrderProof (EQUIVALENT transformation)

(62) Obligation:

(63) QDPOrderProof (EQUIVALENT transformation)

(64) Obligation:

(65) PisEmptyProof (EQUIVALENT transformation)

(66) TRUE