Termination w.r.t. Q proof of /tmp/tmpO5TqJ1/Ex9_BLR02

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

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 QDPOrderProof (⇔)

↳4 QDP

↳5 QDPOrderProof (⇔)

↳6 QDP

↳7 DependencyGraphProof (⇔)

↳8 QDP

↳9 QDPOrderProof (⇔)

↳10 QDP

↳11 DependencyGraphProof (⇔)

↳12 QDP

↳13 QDPOrderProof (⇔)

↳14 QDP

↳15 QDPOrderProof (⇔)

↳16 QDP

↳17 PisEmptyProof (⇔)

↳18 TRUE

(0) Obligation:

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

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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:

A__FILTER(cons(X, Y), s(N), M) → MARK(X)
A__SIEVE(cons(s(N), Y)) → MARK(N)
A__NATS(N) → MARK(N)
A__ZPRIMES → A__SIEVE(a__nats(s(s(0))))
A__ZPRIMES → A__NATS(s(s(0)))
MARK(filter(X1, X2, X3)) → A__FILTER(mark(X1), mark(X2), mark(X3))
MARK(filter(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2, X3)) → MARK(X2)
MARK(filter(X1, X2, X3)) → MARK(X3)
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(sieve(X)) → MARK(X)
MARK(nats(X)) → A__NATS(mark(X))
MARK(nats(X)) → MARK(X)
MARK(zprimes) → A__ZPRIMES
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

A__FILTER(cons(X, Y), s(N), M) → MARK(X)
MARK(filter(X1, X2, X3)) → A__FILTER(mark(X1), mark(X2), mark(X3))
MARK(filter(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2, X3)) → MARK(X2)
MARK(filter(X1, X2, X3)) → MARK(X3)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A__FILTER(x1, x2, x3) = A__FILTER(x1, x3)
cons(x1, x2) = x1
s(x1) = x1
MARK(x1) = x1
A__SIEVE(x1) = x1
A__NATS(x1) = x1
A__ZPRIMES = A__ZPRIMES
a__nats(x1) = x1
0 = 0
filter(x1, x2, x3) = filter(x1, x2, x3)
mark(x1) = x1
sieve(x1) = x1
nats(x1) = x1
zprimes = zprimes
a__filter(x1, x2, x3) = a__filter(x1, x2, x3)
a__sieve(x1) = x1
a__zprimes = a__zprimes

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

[A_{_ZPRIMES}, 0, filter₃, zprimes, a_{_filter₃}, a_{_zprimes}] > A_{_FILTER₂}

Status:

A_{_FILTER₂}: [2,1]
A_{_ZPRIMES}: []
0: []
filter₃: [1,2,3]
zprimes: []
a_{_filter₃}: [1,2,3]
a_{_zprimes}: []

The following usable rules [FROCOS05] were oriented:

a__nats(N) → cons(mark(N), nats(s(N)))
a__nats(X) → nats(X)
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(X) → sieve(X)
a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__zprimes → zprimes
a__zprimes → a__sieve(a__nats(s(s(0))))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))

(4) Obligation:

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

A__SIEVE(cons(s(N), Y)) → MARK(N)
A__NATS(N) → MARK(N)
A__ZPRIMES → A__SIEVE(a__nats(s(s(0))))
A__ZPRIMES → A__NATS(s(s(0)))
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(sieve(X)) → MARK(X)
MARK(nats(X)) → A__NATS(mark(X))
MARK(nats(X)) → MARK(X)
MARK(zprimes) → A__ZPRIMES
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(zprimes) → A__ZPRIMES

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A__SIEVE(x1) = x1
cons(x1, x2) = x1
s(x1) = x1
MARK(x1) = x1
A__NATS(x1) = x1
A__ZPRIMES = A__ZPRIMES
a__nats(x1) = x1
0 = 0
sieve(x1) = x1
mark(x1) = x1
nats(x1) = x1
zprimes = zprimes
filter(x1, x2, x3) = x1
a__filter(x1, x2, x3) = x1
a__sieve(x1) = x1
a__zprimes = a__zprimes

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

[zprimes, a_{_zprimes}] > [A_{_ZPRIMES}, 0]

Status:

A_{_ZPRIMES}: []
0: []
zprimes: []
a_{_zprimes}: []

The following usable rules [FROCOS05] were oriented:

a__nats(N) → cons(mark(N), nats(s(N)))
a__nats(X) → nats(X)
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(X) → sieve(X)
a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__zprimes → zprimes
a__zprimes → a__sieve(a__nats(s(s(0))))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))

(6) Obligation:

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

A__SIEVE(cons(s(N), Y)) → MARK(N)
A__NATS(N) → MARK(N)
A__ZPRIMES → A__SIEVE(a__nats(s(s(0))))
A__ZPRIMES → A__NATS(s(s(0)))
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(sieve(X)) → MARK(X)
MARK(nats(X)) → A__NATS(mark(X))
MARK(nats(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(8) Obligation:

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

MARK(sieve(X)) → A__SIEVE(mark(X))
A__SIEVE(cons(s(N), Y)) → MARK(N)
MARK(sieve(X)) → MARK(X)
MARK(nats(X)) → A__NATS(mark(X))
A__NATS(N) → MARK(N)
MARK(nats(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

A__SIEVE(cons(s(N), Y)) → MARK(N)
MARK(sieve(X)) → MARK(X)
MARK(nats(X)) → MARK(X)
MARK(s(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1) = MARK(x1)
sieve(x1) = sieve(x1)
A__SIEVE(x1) = A__SIEVE(x1)
mark(x1) = mark(x1)
cons(x1, x2) = x1
s(x1) = s(x1)
nats(x1) = nats(x1)
A__NATS(x1) = A__NATS(x1)
filter(x1, x2, x3) = filter(x1, x2, x3)
a__filter(x1, x2, x3) = a__filter(x1, x2, x3)
a__sieve(x1) = a__sieve(x1)
a__nats(x1) = a__nats(x1)
zprimes = zprimes
a__zprimes = a__zprimes
0 = 0

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

[zprimes, a_{_zprimes}] > [sieve₁, mark₁, nats₁, filter₃, a_{_filter₃}, a_{_sieve₁}, a_{_nats₁}] > [MARK₁, A_{_SIEVE₁}, A_{_NATS₁}]
[zprimes, a_{_zprimes}] > [sieve₁, mark₁, nats₁, filter₃, a_{_filter₃}, a_{_sieve₁}, a_{_nats₁}] > s₁
[zprimes, a_{_zprimes}] > [sieve₁, mark₁, nats₁, filter₃, a_{_filter₃}, a_{_sieve₁}, a_{_nats₁}] > 0

Status:

MARK₁: [1]
sieve₁: [1]
A_{_SIEVE₁}: [1]
mark₁: [1]
s₁: [1]
nats₁: [1]
A_{_NATS₁}: [1]
filter₃: [1,3,2]
a_{_filter₃}: [1,3,2]
a_{_sieve₁}: [1]
a_{_nats₁}: [1]
zprimes: []
a_{_zprimes}: []
0: []

The following usable rules [FROCOS05] were oriented:

mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(X) → sieve(X)
a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__nats(X) → nats(X)
a__zprimes → zprimes
a__zprimes → a__sieve(a__nats(s(s(0))))
a__nats(N) → cons(mark(N), nats(s(N)))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))

(10) Obligation:

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

MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(nats(X)) → A__NATS(mark(X))
A__NATS(N) → MARK(N)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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

(11) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(12) Obligation:

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

MARK(nats(X)) → A__NATS(mark(X))
A__NATS(N) → MARK(N)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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.

MARK(nats(X)) → A__NATS(mark(X))
A__NATS(N) → MARK(N)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1) = x1
nats(x1) = nats(x1)
A__NATS(x1) = A__NATS(x1)
mark(x1) = x1
cons(x1, x2) = x1
filter(x1, x2, x3) = filter(x1, x2, x3)
a__filter(x1, x2, x3) = a__filter(x1, x2, x3)
sieve(x1) = sieve
a__sieve(x1) = a__sieve
a__nats(x1) = a__nats(x1)
zprimes = zprimes
a__zprimes = a__zprimes
0 = 0
s(x1) = s

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

[nats₁, a_{_nats₁}] > A_{_NATS₁}
[nats₁, a_{_nats₁}] > s
[filter₃, a_{_filter₃}] > 0
[zprimes, a_{_zprimes}] > [sieve, a_{_sieve}] > 0
[zprimes, a_{_zprimes}] > [sieve, a_{_sieve}] > s

Status:

nats₁: [1]
A_{_NATS₁}: [1]
filter₃: [2,1,3]
a_{_filter₃}: [2,1,3]
sieve: []
a_{_sieve}: []
a_{_nats₁}: [1]
zprimes: []
a_{_zprimes}: []
0: []
s: []

The following usable rules [FROCOS05] were oriented:

mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(X) → sieve(X)
a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__nats(X) → nats(X)
a__zprimes → zprimes
a__zprimes → a__sieve(a__nats(s(s(0))))
a__nats(N) → cons(mark(N), nats(s(N)))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))

(14) Obligation:

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

MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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.

MARK(cons(X1, X2)) → MARK(X1)

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

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

trivial

Status:

cons₂: [2,1]

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

a__filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
a__filter(cons(X, Y), s(N), M) → cons(mark(X), filter(Y, N, M))
a__sieve(cons(0, Y)) → cons(0, sieve(Y))
a__sieve(cons(s(N), Y)) → cons(s(mark(N)), sieve(filter(Y, N, N)))
a__nats(N) → cons(mark(N), nats(s(N)))
a__zprimes → a__sieve(a__nats(s(s(0))))
mark(filter(X1, X2, X3)) → a__filter(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) → a__sieve(mark(X))
mark(nats(X)) → a__nats(mark(X))
mark(zprimes) → a__zprimes
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
a__filter(X1, X2, X3) → filter(X1, X2, X3)
a__sieve(X) → sieve(X)
a__nats(X) → nats(X)
a__zprimes → zprimes

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.

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) QDPOrderProof (EQUIVALENT transformation)

(4) Obligation:

(5) QDPOrderProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) Obligation:

(9) QDPOrderProof (EQUIVALENT transformation)

(10) Obligation:

(11) DependencyGraphProof (EQUIVALENT transformation)

(12) Obligation:

(13) QDPOrderProof (EQUIVALENT transformation)

(14) Obligation:

(15) QDPOrderProof (EQUIVALENT transformation)

(16) Obligation:

(17) PisEmptyProof (EQUIVALENT transformation)

(18) TRUE