Termination w.r.t. Q proof of /tmp/tmpTQB6bX/ExIntrod_GM99

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

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 QDP

↳5 QDPOrderProof (⇔)

↳6 QDP

↳7 DependencyGraphProof (⇔)

↳8 AND

    ↳9 QDP

        ↳10 Narrowing (⇔)

        ↳11 QDP

        ↳12 Narrowing (⇔)

        ↳13 QDP

        ↳14 DependencyGraphProof (⇔)

        ↳15 QDP

        ↳16 Narrowing (⇔)

        ↳17 QDP

        ↳18 DependencyGraphProof (⇔)

        ↳19 QDP

        ↳20 QDPOrderProof (⇔)

        ↳21 QDP

        ↳22 QDPOrderProof (⇔)

        ↳23 QDP

        ↳24 NonTerminationProof (⇔)

        ↳25 NO

    ↳26 QDP

        ↳27 UsableRulesProof (⇔)

        ↳28 QDP

        ↳29 QDPSizeChangeProof (⇔)

        ↳30 YES

(0) Obligation:

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

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
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:

PRIMES → SIEVE(from(s(s(0))))
PRIMES → FROM(s(s(0)))
FROM(X) → CONS(X, n__from(s(X)))
TAIL(cons(X, Y)) → ACTIVATE(Y)
IF(true, X, Y) → ACTIVATE(X)
IF(false, X, Y) → ACTIVATE(Y)
FILTER(s(s(X)), cons(Y, Z)) → IF(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
FILTER(s(s(X)), cons(Y, Z)) → ACTIVATE(Z)
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
SIEVE(cons(X, Y)) → CONS(X, n__filter(X, sieve(activate(Y))))
SIEVE(cons(X, Y)) → SIEVE(activate(Y))
SIEVE(cons(X, Y)) → ACTIVATE(Y)
ACTIVATE(n__from(X)) → FROM(X)
ACTIVATE(n__filter(X1, X2)) → FILTER(X1, X2)
ACTIVATE(n__cons(X1, X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
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 10 less nodes.

(4) Obligation:

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

ACTIVATE(n__filter(X1, X2)) → FILTER(X1, X2)
FILTER(s(s(X)), cons(Y, Z)) → ACTIVATE(Z)
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
SIEVE(cons(X, Y)) → SIEVE(activate(Y))
SIEVE(cons(X, Y)) → ACTIVATE(Y)

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
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.

SIEVE(cons(X, Y)) → ACTIVATE(Y)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( SIEVE(x₁) ) = x₁ + 2

POL( activate(x₁) ) = x₁ + 2

POL( n__from(x₁) ) = 2x₁

POL( from(x₁) ) = 2x₁ + 2

POL( n__filter(x₁, x₂) ) = max{0, x₂ - 2}

POL( filter(x₁, x₂) ) = x₂

POL( n__cons(x₁, x₂) ) = 2x₁ + 2x₂ + 2

POL( cons(x₁, x₂) ) = 2x₁ + 2x₂ + 2

POL( s(x₁) ) = max{0, -2}

POL( if(x₁, ..., x₃) ) = 2x₁ + 2x₂

POL( divides(x₁, x₂) ) = 2x₁ + x₂ + 1

POL( sieve(x₁) ) = max{0, 2x₁ - 2}

POL( ACTIVATE(x₁) ) = x₁ + 2

POL( FILTER(x₁, x₂) ) = x₂

The following usable rules [FROCOS05] were oriented:

activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X
filter(X1, X2) → n__filter(X1, X2)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
cons(X1, X2) → n__cons(X1, X2)
from(X) → cons(X, n__from(s(X)))
from(X) → n__from(X)

(6) Obligation:

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

ACTIVATE(n__filter(X1, X2)) → FILTER(X1, X2)
FILTER(s(s(X)), cons(Y, Z)) → ACTIVATE(Z)
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
SIEVE(cons(X, Y)) → SIEVE(activate(Y))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
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 2 SCCs with 1 less node.

(8) Complex Obligation (AND)

(9) Obligation:

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

SIEVE(cons(X, Y)) → SIEVE(activate(Y))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(10) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule SIEVE(cons(X, Y)) → SIEVE(activate(Y)) at position [0] we obtained the following new rules [LPAR04]:

SIEVE(cons(y0, n__from(x0))) → SIEVE(from(x0))
SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(filter(x0, x1))
SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
SIEVE(cons(y0, x0)) → SIEVE(x0)

(11) Obligation:

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

SIEVE(cons(y0, n__from(x0))) → SIEVE(from(x0))
SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(filter(x0, x1))
SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
SIEVE(cons(y0, x0)) → SIEVE(x0)

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(12) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule SIEVE(cons(y0, n__from(x0))) → SIEVE(from(x0)) at position [0] we obtained the following new rules [LPAR04]:

SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))
SIEVE(cons(y0, n__from(x0))) → SIEVE(n__from(x0))

(13) Obligation:

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

SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(filter(x0, x1))
SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
SIEVE(cons(y0, x0)) → SIEVE(x0)
SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))
SIEVE(cons(y0, n__from(x0))) → SIEVE(n__from(x0))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(14) DependencyGraphProof (EQUIVALENT transformation)

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

(15) Obligation:

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

SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(filter(x0, x1))
SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
SIEVE(cons(y0, x0)) → SIEVE(x0)
SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(16) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(filter(x0, x1)) at position [0] we obtained the following new rules [LPAR04]:

SIEVE(cons(y0, n__filter(s(s(x0)), cons(x1, x2)))) → SIEVE(if(divides(s(s(x0)), x1), n__filter(s(s(x0)), activate(x2)), n__cons(x1, n__filter(x0, sieve(x1)))))
SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(n__filter(x0, x1))

(17) Obligation:

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

SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
SIEVE(cons(y0, x0)) → SIEVE(x0)
SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))
SIEVE(cons(y0, n__filter(s(s(x0)), cons(x1, x2)))) → SIEVE(if(divides(s(s(x0)), x1), n__filter(s(s(x0)), activate(x2)), n__cons(x1, n__filter(x0, sieve(x1)))))
SIEVE(cons(y0, n__filter(x0, x1))) → SIEVE(n__filter(x0, x1))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(18) DependencyGraphProof (EQUIVALENT transformation)

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

(19) Obligation:

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

SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
SIEVE(cons(y0, x0)) → SIEVE(x0)
SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → 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.

SIEVE(cons(y0, x0)) → SIEVE(x0)

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

Knuth-Bendix order [KBO] with precedence:

trivial

and weight map:

cons_1=1
n__from=1

The following usable rules [FROCOS05] were oriented:

cons(X1, X2) → n__cons(X1, X2)

(21) Obligation:

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

SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))
SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → 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.

SIEVE(cons(y0, n__cons(x0, x1))) → SIEVE(cons(x0, x1))

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

Knuth-Bendix order [KBO] with precedence:

trivial

and weight map:

n__cons_1=1
n__from=1
cons_1=2

The following usable rules [FROCOS05] were oriented:

cons(X1, X2) → n__cons(X1, X2)

(23) Obligation:

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

SIEVE(cons(y0, n__from(x0))) → SIEVE(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(24) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = SIEVE(cons(y0, n__from(x0))) evaluates to t =SIEVE(cons(x0, n__from(s(x0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Matcher: [y0 / x0, x0 / s(x0)]
Semiunifier: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from SIEVE(cons(y0, n__from(x0))) to SIEVE(cons(x0, n__from(s(x0)))).

(25) NO

(26) Obligation:

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

FILTER(s(s(X)), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__filter(X1, X2)) → FILTER(X1, X2)

The TRS R consists of the following rules:

primes → sieve(from(s(s(0))))
from(X) → cons(X, n__from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → activate(Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), n__filter(s(s(X)), activate(Z)), n__cons(Y, n__filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, n__filter(X, sieve(activate(Y))))
from(X) → n__from(X)
filter(X1, X2) → n__filter(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(27) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(28) Obligation:

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

FILTER(s(s(X)), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__filter(X1, X2)) → FILTER(X1, X2)

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

(29) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

ACTIVATE(n__filter(X1, X2)) → FILTER(X1, X2)
The graph contains the following edges 1 > 1, 1 > 2
FILTER(s(s(X)), cons(Y, Z)) → ACTIVATE(Z)
The graph contains the following edges 2 > 1

(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) Complex Obligation (AND)

(9) Obligation:

(10) Narrowing (EQUIVALENT transformation)

(11) Obligation:

(12) Narrowing (EQUIVALENT transformation)

(13) Obligation:

(14) DependencyGraphProof (EQUIVALENT transformation)

(15) Obligation:

(16) Narrowing (EQUIVALENT transformation)

(17) Obligation:

(18) DependencyGraphProof (EQUIVALENT transformation)

(19) Obligation:

(20) QDPOrderProof (EQUIVALENT transformation)

(21) Obligation:

(22) QDPOrderProof (EQUIVALENT transformation)

(23) Obligation:

(24) NonTerminationProof (EQUIVALENT transformation)

(25) NO

(26) Obligation:

(27) UsableRulesProof (EQUIVALENT transformation)

(28) Obligation:

(29) QDPSizeChangeProof (EQUIVALENT transformation)

(30) YES