Termination w.r.t. Q proof of /tmp/tmpeIj1XL/Ex9_BLR02

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

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 QDP

↳5 QDPOrderProof (⇔)

↳6 QDP

↳7 DependencyGraphProof (⇔)

↳8 QDP

↳9 QDPOrderProof (⇔)

↳10 QDP

↳11 PisEmptyProof (⇔)

↳12 TRUE

(0) Obligation:

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

filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(s(N)))
zprimes → sieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
activate(n__filter(X1, X2, X3)) → filter(X1, X2, X3)
activate(n__sieve(X)) → sieve(X)
activate(n__nats(X)) → nats(X)
activate(X) → 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:

FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)
FILTER(cons(X, Y), s(N), M) → ACTIVATE(Y)
SIEVE(cons(0, Y)) → ACTIVATE(Y)
SIEVE(cons(s(N), Y)) → FILTER(activate(Y), N, N)
SIEVE(cons(s(N), Y)) → ACTIVATE(Y)
ZPRIMES → SIEVE(nats(s(s(0))))
ZPRIMES → NATS(s(s(0)))
ACTIVATE(n__filter(X1, X2, X3)) → FILTER(X1, X2, X3)
ACTIVATE(n__sieve(X)) → SIEVE(X)
ACTIVATE(n__nats(X)) → NATS(X)

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(s(N)))
zprimes → sieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
activate(n__filter(X1, X2, X3)) → filter(X1, X2, X3)
activate(n__sieve(X)) → sieve(X)
activate(n__nats(X)) → nats(X)
activate(X) → 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 1 SCC with 3 less nodes.

(4) Obligation:

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

ACTIVATE(n__filter(X1, X2, X3)) → FILTER(X1, X2, X3)
FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)
ACTIVATE(n__sieve(X)) → SIEVE(X)
SIEVE(cons(0, Y)) → ACTIVATE(Y)
SIEVE(cons(s(N), Y)) → FILTER(activate(Y), N, N)
FILTER(cons(X, Y), s(N), M) → ACTIVATE(Y)
SIEVE(cons(s(N), Y)) → ACTIVATE(Y)

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(s(N)))
zprimes → sieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
activate(n__filter(X1, X2, X3)) → filter(X1, X2, X3)
activate(n__sieve(X)) → sieve(X)
activate(n__nats(X)) → nats(X)
activate(X) → X

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.

ACTIVATE(n__sieve(X)) → SIEVE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1) = x1
n__filter(x1, x2, x3) = x1
FILTER(x1, x2, x3) = x1
cons(x1, x2) = x2
0 = 0
n__sieve(x1) = n__sieve(x1)
SIEVE(x1) = x1
s(x1) = s
activate(x1) = x1
filter(x1, x2, x3) = x1
sieve(x1) = sieve(x1)
nats(x1) = x1
n__nats(x1) = x1
zprimes = zprimes

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

[0, zprimes] > [n_{_sieve₁}, sieve₁] > s

Status:

n_{_sieve₁}: [1]
s: []
0: []
sieve₁: [1]
zprimes: []

The following usable rules [FROCOS05] were oriented:

filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(s(N)))
zprimes → sieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
activate(n__filter(X1, X2, X3)) → filter(X1, X2, X3)
activate(n__sieve(X)) → sieve(X)
activate(n__nats(X)) → nats(X)
activate(X) → X

(6) Obligation:

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

ACTIVATE(n__filter(X1, X2, X3)) → FILTER(X1, X2, X3)
FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)
SIEVE(cons(0, Y)) → ACTIVATE(Y)
SIEVE(cons(s(N), Y)) → FILTER(activate(Y), N, N)
FILTER(cons(X, Y), s(N), M) → ACTIVATE(Y)
SIEVE(cons(s(N), Y)) → ACTIVATE(Y)

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(s(N)))
zprimes → sieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
activate(n__filter(X1, X2, X3)) → filter(X1, X2, X3)
activate(n__sieve(X)) → sieve(X)
activate(n__nats(X)) → nats(X)
activate(X) → X

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 3 less nodes.

(8) Obligation:

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

FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)
ACTIVATE(n__filter(X1, X2, X3)) → FILTER(X1, X2, X3)
FILTER(cons(X, Y), s(N), M) → ACTIVATE(Y)

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(s(N)))
zprimes → sieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
activate(n__filter(X1, X2, X3)) → filter(X1, X2, X3)
activate(n__sieve(X)) → sieve(X)
activate(n__nats(X)) → nats(X)
activate(X) → X

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.

FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)
ACTIVATE(n__filter(X1, X2, X3)) → FILTER(X1, X2, X3)
FILTER(cons(X, Y), s(N), M) → ACTIVATE(Y)

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)
cons(x1, x2) = x2
0 = 0
ACTIVATE(x1) = ACTIVATE(x1)
n__filter(x1, x2, x3) = n__filter(x1, x3)
s(x1) = s
filter(x1, x2, x3) = filter(x1, x3)
activate(x1) = activate(x1)
sieve(x1) = sieve
n__sieve(x1) = n__sieve
nats(x1) = nats(x1)
n__nats(x1) = n__nats(x1)
zprimes = zprimes

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

[nats₁, n_{_nats₁}] > s
zprimes > [filter₂, activate₁, sieve] > [0, n_{_sieve}] > s
zprimes > [filter₂, activate₁, sieve] > n_{_filter₂} > FILTER₁ > ACTIVATE₁ > s

Status:

activate₁: [1]
s: []
n_{_sieve}: []
0: []
sieve: []
FILTER₁: [1]
filter₂: [1,2]
nats₁: [1]
n_{_nats₁}: [1]
ACTIVATE₁: [1]
zprimes: []
n_{_filter₂}: [2,1]

The following usable rules [FROCOS05] were oriented:

filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(s(N)))
zprimes → sieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
activate(n__filter(X1, X2, X3)) → filter(X1, X2, X3)
activate(n__sieve(X)) → sieve(X)
activate(n__nats(X)) → nats(X)
activate(X) → X

(10) Obligation:

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

filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(s(N)))
zprimes → sieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
activate(n__filter(X1, X2, X3)) → filter(X1, X2, X3)
activate(n__sieve(X)) → sieve(X)
activate(n__nats(X)) → nats(X)
activate(X) → X

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

(11) 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) Obligation:

(5) QDPOrderProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) Obligation:

(9) QDPOrderProof (EQUIVALENT transformation)

(10) Obligation:

(11) PisEmptyProof (EQUIVALENT transformation)

(12) TRUE