Termination w.r.t. Q proof of /tmp/tmp2sXPC0/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 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 PisEmptyProof (⇔)

    ↳63 TRUE

  ↳64 QDP

    ↳65 QDPOrderProof (⇔)

    ↳66 QDP

    ↳67 QDPOrderProof (⇔)

    ↳68 QDP

    ↳69 PisEmptyProof (⇔)

    ↳70 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)
active(x1) = active(x1)
filter(x1, x2, x3) = filter(x1, x2, x3)
cons(x1, x2) = x1
0 = 0
s(x1) = s(x1)
sieve(x1) = x1
nats(x1) = x1
zprimes = zprimes
proper(x1) = x1
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

active₁ > filter₃ > 0 > mark₁
active₁ > s₁ > mark₁

Status:

active₁: [1]
filter₃: [2,3,1]
mark₁: [1]
s₁: [1]
NATS₁: [1]
top: []
0: []
zprimes: []

The following usable rules [FROCOS05] were oriented:

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))

(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) = NATS(x1)
ok(x1) = ok(x1)
active(x1) = active(x1)
filter(x1, x2, x3) = x1
cons(x1, x2) = x1
0 = 0
mark(x1) = x1
s(x1) = x1
sieve(x1) = x1
nats(x1) = nats(x1)
zprimes = zprimes
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

top > active₁ > 0
top > active₁ > nats₁ > ok₁
top > proper₁ > 0
top > proper₁ > nats₁ > ok₁
top > proper₁ > zprimes

Status:

active₁: [1]
nats₁: [1]
ok₁: [1]
NATS₁: [1]
proper₁: [1]
top: []
0: []
zprimes: []

The following usable rules [FROCOS05] were oriented:

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))

(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)
active(x1) = active(x1)
filter(x1, x2, x3) = filter(x1, x2, x3)
cons(x1, x2) = x1
0 = 0
s(x1) = s(x1)
sieve(x1) = x1
nats(x1) = x1
zprimes = zprimes
proper(x1) = x1
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

active₁ > filter₃ > 0 > mark₁
active₁ > s₁ > mark₁

Status:

SIEVE₁: [1]
active₁: [1]
filter₃: [2,3,1]
mark₁: [1]
s₁: [1]
top: []
0: []
zprimes: []

The following usable rules [FROCOS05] were oriented:

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))

(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) = SIEVE(x1)
ok(x1) = ok(x1)
active(x1) = active(x1)
filter(x1, x2, x3) = x1
cons(x1, x2) = x1
0 = 0
mark(x1) = x1
s(x1) = x1
sieve(x1) = x1
nats(x1) = nats(x1)
zprimes = zprimes
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

top > active₁ > 0
top > active₁ > nats₁ > ok₁
top > proper₁ > 0
top > proper₁ > nats₁ > ok₁
top > proper₁ > zprimes

Status:

SIEVE₁: [1]
active₁: [1]
nats₁: [1]
ok₁: [1]
proper₁: [1]
top: []
0: []
zprimes: []

The following usable rules [FROCOS05] were oriented:

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))

(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)
active(x1) = active(x1)
filter(x1, x2, x3) = filter(x1, x2, x3)
cons(x1, x2) = x1
0 = 0
s(x1) = s(x1)
sieve(x1) = x1
nats(x1) = x1
zprimes = zprimes
proper(x1) = x1
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

active₁ > filter₃ > 0 > mark₁
active₁ > s₁ > mark₁

Status:

active₁: [1]
filter₃: [2,3,1]
mark₁: [1]
s₁: [1]
top: []
0: []
zprimes: []
S₁: [1]

The following usable rules [FROCOS05] were oriented:

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))

(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) = S(x1)
ok(x1) = ok(x1)
active(x1) = active(x1)
filter(x1, x2, x3) = x1
cons(x1, x2) = x1
0 = 0
mark(x1) = x1
s(x1) = x1
sieve(x1) = x1
nats(x1) = nats(x1)
zprimes = zprimes
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

top > active₁ > 0
top > active₁ > nats₁ > ok₁
top > proper₁ > 0
top > proper₁ > nats₁ > ok₁
top > proper₁ > zprimes

Status:

active₁: [1]
nats₁: [1]
ok₁: [1]
proper₁: [1]
top: []
0: []
zprimes: []
S₁: [1]

The following usable rules [FROCOS05] were oriented:

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))

(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) = x1
active(x1) = active(x1)
filter(x1, x2, x3) = x1
cons(x1, x2) = x1
0 = 0
s(x1) = x1
sieve(x1) = x1
nats(x1) = x1
zprimes = zprimes
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

top > proper₁ > ok₁ > active₁ > 0
top > proper₁ > zprimes

Status:

active₁: [1]
CONS₁: [1]
ok₁: [1]
proper₁: [1]
top: []
0: []
zprimes: []

The following usable rules [FROCOS05] were oriented:

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))

(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)
active(x1) = active(x1)
filter(x1, x2, x3) = filter(x1, x2, x3)
cons(x1, x2) = cons(x1)
0 = 0
s(x1) = s(x1)
sieve(x1) = x1
nats(x1) = x1
zprimes = zprimes
proper(x1) = x1
ok(x1) = x1
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

CONS₁ > mark₁
zprimes > mark₁
top > active₁ > 0 > filter₃ > mark₁
top > active₁ > 0 > cons₁ > mark₁
top > active₁ > s₁ > filter₃ > mark₁
top > active₁ > s₁ > cons₁ > mark₁

Status:

active₁: [1]
cons₁: [1]
filter₃: [3,2,1]
CONS₁: [1]
mark₁: [1]
s₁: [1]
top: []
0: []
zprimes: []

The following usable rules [FROCOS05] were oriented:

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))

(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(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) = x1
ok(x1) = ok(x1)
active(x1) = active(x1)
filter(x1, x2, x3) = filter(x1, x2, x3)
cons(x1, x2) = cons(x1)
0 = 0
s(x1) = s(x1)
sieve(x1) = x1
nats(x1) = nats(x1)
zprimes = zprimes
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

active₁ > s₁ > cons₁ > 0 > filter₃ > ok₁
active₁ > nats₁ > ok₁
proper₁ > s₁ > cons₁ > 0 > filter₃ > ok₁
proper₁ > zprimes > 0 > filter₃ > ok₁
proper₁ > zprimes > nats₁ > ok₁
top > ok₁

Status:

active₁: [1]
cons₁: [1]
filter₃: [2,1,3]
nats₁: [1]
ok₁: [1]
s₁: [1]
proper₁: [1]
top: []
0: []
zprimes: []

The following usable rules [FROCOS05] were oriented:

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))

(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(X1, X2, mark(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(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)
active(x1) = active(x1)
filter(x1, x2, x3) = filter(x1, x2, x3)
cons(x1, x2) = x1
0 = 0
s(x1) = s(x1)
sieve(x1) = x1
nats(x1) = nats(x1)
zprimes = zprimes
proper(x1) = x1
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

active₁ > 0 > filter₃ > mark₁
active₁ > 0 > ok
active₁ > nats₁ > s₁ > mark₁
active₁ > nats₁ > ok
zprimes > 0 > filter₃ > mark₁
zprimes > 0 > ok
zprimes > nats₁ > s₁ > mark₁
zprimes > nats₁ > ok

Status:

active₁: [1]
filter₃: [1,2,3]
nats₁: [1]
mark₁: [1]
s₁: [1]
ok: []
top: []
0: []
FILTER₁: [1]
zprimes: []

The following usable rules [FROCOS05] were oriented:

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))

(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) = FILTER(x2, x3)
mark(x1) = mark(x1)
active(x1) = x1
filter(x1, x2, x3) = filter(x1, x2, x3)
cons(x1, x2) = x1
0 = 0
s(x1) = x1
sieve(x1) = sieve(x1)
nats(x1) = nats(x1)
zprimes = zprimes
proper(x1) = x1
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

filter₃ > 0 > mark₁
filter₃ > 0 > ok
zprimes > 0 > mark₁
zprimes > 0 > ok
zprimes > sieve₁ > mark₁
zprimes > sieve₁ > ok
zprimes > nats₁ > mark₁

Status:

filter₃: [1,2,3]
nats₁: [1]
mark₁: [1]
ok: []
top: []
0: []
sieve₁: [1]
zprimes: []
FILTER₂: [1,2]

The following usable rules [FROCOS05] were oriented:

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))

(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: Combined order from the following AFS and order.
FILTER(x1, x2, x3) = FILTER(x1, x2, x3)
mark(x1) = mark(x1)
active(x1) = active(x1)
filter(x1, x2, x3) = filter(x1, x2, x3)
cons(x1, x2) = cons(x1, x2)
0 = 0
s(x1) = x1
sieve(x1) = x1
nats(x1) = x1
zprimes = zprimes
proper(x1) = x1
ok(x1) = x1
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

FILTER₃ > mark₁
zprimes > mark₁
top > active₁ > 0 > filter₃ > mark₁
top > active₁ > 0 > cons₂ > mark₁

Status:

cons₂: [1,2]
active₁: [1]
FILTER₃: [1,3,2]
filter₃: [1,2,3]
mark₁: [1]
top: []
0: []
zprimes: []

The following usable rules [FROCOS05] were oriented:

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))

(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) = PROPER(x1)
filter(x1, x2, x3) = filter(x1, x2, x3)
cons(x1, x2) = cons(x1, x2)
s(x1) = x1
sieve(x1) = x1
nats(x1) = x1
active(x1) = x1
0 = 0
mark(x1) = mark
zprimes = zprimes
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

PROPER₁ > mark
proper₁ > zprimes > mark
proper₁ > ok > filter₃ > cons₂ > mark
proper₁ > ok > filter₃ > 0 > mark
proper₁ > ok > top > mark

Status:

PROPER₁: [1]
cons₂: [1,2]
filter₃: [2,3,1]
mark: []
ok: []
proper₁: [1]
top: []
0: []
zprimes: []

The following usable rules [FROCOS05] were oriented:

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))

(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(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)
sieve(x1) = x1
nats(x1) = x1
active(x1) = x1
filter(x1, x2, x3) = x3
cons(x1, x2) = cons
0 = 0
mark(x1) = x1
zprimes = zprimes
proper(x1) = x1
ok(x1) = x1
top(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

PROPER₁ > cons
zprimes > s₁ > cons
zprimes > 0 > cons

Status:

PROPER₁: [1]
cons: []
s₁: [1]
0: []
zprimes: []

The following usable rules [FROCOS05] were oriented:

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))

(48) Obligation:

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

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.

(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)
sieve(x1) = sieve(x1)
nats(x1) = x1
active(x1) = active(x1)
filter(x1, x2, x3) = filter(x1, x2, x3)
cons(x1, x2) = cons(x1, x2)
0 = 0
mark(x1) = mark
s(x1) = s
zprimes = zprimes
proper(x1) = x1
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

0 > cons₂ > filter₃
0 > cons₂ > ok > sieve₁
0 > mark > sieve₁
0 > mark > filter₃
s > active₁ > cons₂ > filter₃
s > active₁ > cons₂ > ok > sieve₁
s > active₁ > mark > sieve₁
s > active₁ > mark > filter₃
zprimes > ok > sieve₁
top > active₁ > cons₂ > filter₃
top > active₁ > cons₂ > ok > sieve₁
top > active₁ > mark > sieve₁
top > active₁ > mark > filter₃

Status:

PROPER₁: [1]
active₁: [1]
cons₂: [1,2]
filter₃: [3,1,2]
mark: []
ok: []
s: []
top: []
0: []
sieve₁: [1]
zprimes: []

The following usable rules [FROCOS05] were oriented:

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))

(50) 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.

(51) 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)
nats(x1) = nats(x1)
active(x1) = active(x1)
filter(x1, x2, x3) = filter
cons(x1, x2) = cons
0 = 0
mark(x1) = mark
s(x1) = s
sieve(x1) = x1
zprimes = zprimes
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

PROPER₁ > ok
filter > active₁ > nats₁ > mark > ok
filter > active₁ > cons > 0 > ok
filter > active₁ > cons > mark > ok
filter > proper₁ > nats₁ > mark > ok
filter > proper₁ > cons > 0 > ok
filter > proper₁ > cons > mark > ok
filter > proper₁ > zprimes > ok
s > active₁ > nats₁ > mark > ok
s > active₁ > cons > 0 > ok
s > active₁ > cons > mark > ok
top > active₁ > nats₁ > mark > ok
top > active₁ > cons > 0 > ok
top > active₁ > cons > mark > ok
top > proper₁ > nats₁ > mark > ok
top > proper₁ > cons > 0 > ok
top > proper₁ > cons > mark > ok
top > proper₁ > zprimes > ok

Status:

PROPER₁: [1]
s: []
0: []
filter: []
active₁: [1]
cons: []
nats₁: [1]
mark: []
ok: []
proper₁: [1]
top: []
zprimes: []

The following usable rules [FROCOS05] were oriented:

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))

(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)
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
filter(x1, x2, x3) = filter(x1, x2, x3)
cons(x1, x2) = x1
s(x1) = s(x1)
sieve(x1) = sieve(x1)
nats(x1) = x1
active(x1) = active(x1)
0 = 0
mark(x1) = mark(x1)
zprimes = zprimes
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

proper₁ > zprimes > s₁ > mark₁
proper₁ > zprimes > 0 > sieve₁ > mark₁
proper₁ > ok > active₁ > filter₃ > mark₁
proper₁ > ok > active₁ > s₁ > mark₁
proper₁ > ok > active₁ > 0 > sieve₁ > mark₁
top > active₁ > filter₃ > mark₁
top > active₁ > s₁ > mark₁
top > active₁ > 0 > sieve₁ > mark₁

Status:

active₁: [1]
filter₃: [1,2,3]
mark₁: [1]
s₁: [1]
ok: []
proper₁: [1]
top: []
0: []
sieve₁: [1]
zprimes: []

The following usable rules [FROCOS05] were oriented:

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))

(57) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
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(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)
cons(x1, x2) = x1
nats(x1) = nats(x1)
active(x1) = x1
filter(x1, x2, x3) = filter(x1, x2, x3)
0 = 0
mark(x1) = mark(x1)
s(x1) = x1
sieve(x1) = sieve(x1)
zprimes = zprimes
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

zprimes > nats₁ > mark₁
zprimes > nats₁ > ok
zprimes > sieve₁ > 0
zprimes > sieve₁ > mark₁
zprimes > sieve₁ > ok
proper₁ > nats₁ > mark₁
proper₁ > nats₁ > ok
proper₁ > filter₃ > 0
proper₁ > filter₃ > mark₁
proper₁ > filter₃ > ok
proper₁ > sieve₁ > 0
proper₁ > sieve₁ > mark₁
proper₁ > sieve₁ > ok

Status:

filter₃: [1,2,3]
nats₁: [1]
mark₁: [1]
ok: []
proper₁: [1]
top: []
0: []
ACTIVE₁: [1]
sieve₁: [1]
zprimes: []

The following usable rules [FROCOS05] were oriented:

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))

(59) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)

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(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)
active(x1) = active(x1)
filter(x1, x2, x3) = filter(x1, x2, x3)
0 = 0
mark(x1) = x1
s(x1) = s(x1)
sieve(x1) = x1
nats(x1) = x1
zprimes = zprimes
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

active₁ > 0 > ok > s₁ > cons₁
active₁ > 0 > ok > s₁ > filter₃
zprimes > 0 > ok > s₁ > cons₁
zprimes > 0 > ok > s₁ > filter₃
top > proper₁ > s₁ > cons₁
top > proper₁ > s₁ > filter₃

Status:

active₁: [1]
cons₁: [1]
filter₃: [2,3,1]
s₁: [1]
ok: []
proper₁: [1]
top: []
0: []
ACTIVE₁: [1]
zprimes: []

The following usable rules [FROCOS05] were oriented:

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))

(61) 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.

(62) PisEmptyProof (EQUIVALENT transformation)

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

(63) TRUE

(64) 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.

(65) 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
0 = 0
s(x1) = s(x1)
sieve(x1) = sieve(x1)
nats(x1) = nats(x1)
zprimes = zprimes
top(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

filter₃ > mark₁
filter₃ > 0
zprimes > s₁ > mark₁
zprimes > sieve₁ > mark₁
zprimes > sieve₁ > 0
zprimes > nats₁ > mark₁

Status:

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

The following usable rules [FROCOS05] were oriented:

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))

(66) 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.

(67) 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) = filter(x1, x2, x3)
cons(x1, x2) = cons(x1)
0 = 0
mark(x1) = mark(x1)
s(x1) = s(x1)
sieve(x1) = sieve(x1)
nats(x1) = nats(x1)
zprimes = zprimes
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

TOP₁ > 0
zprimes > s₁ > filter₃ > ok₁ > 0
zprimes > s₁ > filter₃ > mark₁ > 0
zprimes > s₁ > cons₁ > ok₁ > 0
zprimes > s₁ > cons₁ > mark₁ > 0
zprimes > s₁ > sieve₁ > ok₁ > 0
zprimes > s₁ > sieve₁ > mark₁ > 0
zprimes > nats₁ > cons₁ > ok₁ > 0
zprimes > nats₁ > cons₁ > mark₁ > 0
proper₁ > s₁ > filter₃ > ok₁ > 0
proper₁ > s₁ > filter₃ > mark₁ > 0
proper₁ > s₁ > cons₁ > ok₁ > 0
proper₁ > s₁ > cons₁ > mark₁ > 0
proper₁ > s₁ > sieve₁ > ok₁ > 0
proper₁ > s₁ > sieve₁ > mark₁ > 0
proper₁ > nats₁ > cons₁ > ok₁ > 0
proper₁ > nats₁ > cons₁ > mark₁ > 0
top > 0

Status:

cons₁: [1]
filter₃: [3,2,1]
ok₁: [1]
mark₁: [1]
0: []
nats₁: [1]
TOP₁: [1]
s₁: [1]
proper₁: [1]
top: []
sieve₁: [1]
zprimes: []

The following usable rules [FROCOS05] were oriented:

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))

(68) 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.

(69) 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) QDPOrderProof (EQUIVALENT transformation)

(41) Obligation:

(42) PisEmptyProof (EQUIVALENT transformation)

(43) TRUE

(44) Obligation:

(45) QDPOrderProof (EQUIVALENT transformation)

(46) Obligation:

(47) QDPOrderProof (EQUIVALENT transformation)

(48) Obligation:

(49) QDPOrderProof (EQUIVALENT transformation)

(50) Obligation:

(51) QDPOrderProof (EQUIVALENT transformation)

(52) Obligation:

(53) PisEmptyProof (EQUIVALENT transformation)

(54) TRUE

(55) Obligation:

(56) QDPOrderProof (EQUIVALENT transformation)

(57) Obligation:

(58) QDPOrderProof (EQUIVALENT transformation)

(59) Obligation:

(60) QDPOrderProof (EQUIVALENT transformation)

(61) Obligation:

(62) PisEmptyProof (EQUIVALENT transformation)

(63) TRUE

(64) Obligation:

(65) QDPOrderProof (EQUIVALENT transformation)

(66) Obligation:

(67) QDPOrderProof (EQUIVALENT transformation)

(68) Obligation:

(69) PisEmptyProof (EQUIVALENT transformation)

(70) TRUE