YES Termination w.r.t. Q proof of Transformed_CSR_04_ExIntrod_GM99_GM.ari

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

0 QTRS
↳1 DependencyPairsProof (⇔, 1 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 QDP
↳5 QDPOrderProof (⇔, 190 ms)
↳6 QDP
↳7 QDPOrderProof (⇔, 40 ms)
↳8 QDP
↳9 QDPOrderProof (⇔, 115 ms)
↳10 QDP
↳11 DependencyGraphProof (⇔, 0 ms)
↳12 QDP
↳13 QDPOrderProof (⇔, 135 ms)
↳14 QDP
↳15 QDPOrderProof (⇔, 0 ms)
↳16 QDP
↳17 DependencyGraphProof (⇔, 0 ms)
↳18 QDP
↳19 QDPOrderProof (⇔, 0 ms)
↳20 QDP
↳21 DependencyGraphProof (⇔, 0 ms)
↳22 QDP
↳23 QDPOrderProof (⇔, 0 ms)
↳24 QDP
↳25 QDPOrderProof (⇔, 0 ms)
↳26 QDP
↳27 DependencyGraphProof (⇔, 0 ms)
↳28 QDP
↳29 QDPSizeChangeProof (⇔, 0 ms)
↳30 YES

(0) Obligation:

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

a__primes → a__sieve(a__from(s(s(0))))
a__from(X) → cons(mark(X), from(s(X)))
a__head(cons(X, Y)) → mark(X)
a__tail(cons(X, Y)) → mark(Y)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__primes → primes
a__sieve(X) → sieve(X)
a__from(X) → from(X)
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)

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__PRIMES → A__SIEVE(a__from(s(s(0))))
A__PRIMES → A__FROM(s(s(0)))
A__FROM(X) → MARK(X)
A__HEAD(cons(X, Y)) → MARK(X)
A__TAIL(cons(X, Y)) → MARK(Y)
A__IF(true, X, Y) → MARK(X)
A__IF(false, X, Y) → MARK(Y)
A__FILTER(s(s(X)), cons(Y, Z)) → A__IF(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
A__FILTER(s(s(X)), cons(Y, Z)) → MARK(X)
A__FILTER(s(s(X)), cons(Y, Z)) → MARK(Y)
A__SIEVE(cons(X, Y)) → MARK(X)
MARK(primes) → A__PRIMES
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(sieve(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(head(X)) → A__HEAD(mark(X))
MARK(head(X)) → MARK(X)
MARK(tail(X)) → A__TAIL(mark(X))
MARK(tail(X)) → MARK(X)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2)) → A__FILTER(mark(X1), mark(X2))
MARK(filter(X1, X2)) → MARK(X1)
MARK(filter(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

a__primes → a__sieve(a__from(s(s(0))))
a__from(X) → cons(mark(X), from(s(X)))
a__head(cons(X, Y)) → mark(X)
a__tail(cons(X, Y)) → mark(Y)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__primes → primes
a__sieve(X) → sieve(X)
a__from(X) → from(X)
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)

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 1 less node.

(4) Obligation:

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

A__SIEVE(cons(X, Y)) → MARK(X)
MARK(primes) → A__PRIMES
A__PRIMES → A__SIEVE(a__from(s(s(0))))
A__PRIMES → A__FROM(s(s(0)))
A__FROM(X) → MARK(X)
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(sieve(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(head(X)) → A__HEAD(mark(X))
A__HEAD(cons(X, Y)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(tail(X)) → A__TAIL(mark(X))
A__TAIL(cons(X, Y)) → MARK(Y)
MARK(tail(X)) → MARK(X)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__IF(true, X, Y) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2)) → A__FILTER(mark(X1), mark(X2))
A__FILTER(s(s(X)), cons(Y, Z)) → MARK(X)
MARK(filter(X1, X2)) → MARK(X1)
MARK(filter(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X2)
A__FILTER(s(s(X)), cons(Y, Z)) → MARK(Y)
A__IF(false, X, Y) → MARK(Y)

The TRS R consists of the following rules:

a__primes → a__sieve(a__from(s(s(0))))
a__from(X) → cons(mark(X), from(s(X)))
a__head(cons(X, Y)) → mark(X)
a__tail(cons(X, Y)) → mark(Y)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__primes → primes
a__sieve(X) → sieve(X)
a__from(X) → from(X)
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


A__PRIMES → A__FROM(s(s(0)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(A__SIEVE(x1)) = 4A + 0A·x1

POL(cons(x1, x2)) = -I + 0A·x1 + 0A·x2

POL(MARK(x1)) = 4A + 0A·x1

POL(primes) = 5A

POL(A__PRIMES) = 5A

POL(a__from(x1)) = -I + 1A·x1

POL(s(x1)) = -I + 0A·x1

POL(0) = 4A

POL(A__FROM(x1)) = 4A + 0A·x1

POL(sieve(x1)) = 2A + 0A·x1

POL(mark(x1)) = -I + 0A·x1

POL(from(x1)) = -I + 1A·x1

POL(head(x1)) = -I + 2A·x1

POL(A__HEAD(x1)) = 4A + 1A·x1

POL(tail(x1)) = -I + 0A·x1

POL(A__TAIL(x1)) = 4A + 0A·x1

POL(if(x1, x2, x3)) = 1A + 0A·x1 + 0A·x2 + 0A·x3

POL(A__IF(x1, x2, x3)) = 4A + -I·x1 + 0A·x2 + 0A·x3

POL(true) = 0A

POL(filter(x1, x2)) = 2A + 0A·x1 + 0A·x2

POL(A__FILTER(x1, x2)) = 4A + 0A·x1 + 0A·x2

POL(divides(x1, x2)) = 1A + 0A·x1 + 0A·x2

POL(false) = 2A

POL(a__primes) = 5A

POL(a__sieve(x1)) = 2A + 0A·x1

POL(a__head(x1)) = -I + 2A·x1

POL(a__tail(x1)) = -I + 0A·x1

POL(a__if(x1, x2, x3)) = 1A + 0A·x1 + 0A·x2 + 0A·x3

POL(a__filter(x1, x2)) = 2A + 0A·x1 + 0A·x2

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

a__from(X) → cons(mark(X), from(s(X)))
a__from(X) → from(X)
mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
a__head(cons(X, Y)) → mark(X)
mark(tail(X)) → a__tail(mark(X))
a__tail(cons(X, Y)) → mark(Y)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__primes → primes
a__primes → a__sieve(a__from(s(s(0))))
a__sieve(X) → sieve(X)
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))

(6) Obligation:

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

A__SIEVE(cons(X, Y)) → MARK(X)
MARK(primes) → A__PRIMES
A__PRIMES → A__SIEVE(a__from(s(s(0))))
A__FROM(X) → MARK(X)
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(sieve(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(head(X)) → A__HEAD(mark(X))
A__HEAD(cons(X, Y)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(tail(X)) → A__TAIL(mark(X))
A__TAIL(cons(X, Y)) → MARK(Y)
MARK(tail(X)) → MARK(X)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__IF(true, X, Y) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2)) → A__FILTER(mark(X1), mark(X2))
A__FILTER(s(s(X)), cons(Y, Z)) → MARK(X)
MARK(filter(X1, X2)) → MARK(X1)
MARK(filter(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X2)
A__FILTER(s(s(X)), cons(Y, Z)) → MARK(Y)
A__IF(false, X, Y) → MARK(Y)

The TRS R consists of the following rules:

a__primes → a__sieve(a__from(s(s(0))))
a__from(X) → cons(mark(X), from(s(X)))
a__head(cons(X, Y)) → mark(X)
a__tail(cons(X, Y)) → mark(Y)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__primes → primes
a__sieve(X) → sieve(X)
a__from(X) → from(X)
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


MARK(from(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(A__SIEVE(x1)) = 4A + 1A·x1

POL(cons(x1, x2)) = 1A + 0A·x1 + 0A·x2

POL(MARK(x1)) = 4A + 1A·x1

POL(primes) = 4A

POL(A__PRIMES) = 5A

POL(a__from(x1)) = 4A + 1A·x1

POL(s(x1)) = 3A + 0A·x1

POL(0) = 2A

POL(A__FROM(x1)) = 4A + 2A·x1

POL(sieve(x1)) = 2A + 0A·x1

POL(mark(x1)) = 1A + 0A·x1

POL(from(x1)) = 4A + 1A·x1

POL(head(x1)) = -I + 0A·x1

POL(A__HEAD(x1)) = 4A + 1A·x1

POL(tail(x1)) = 3A + 0A·x1

POL(A__TAIL(x1)) = 4A + 1A·x1

POL(if(x1, x2, x3)) = -I + 0A·x1 + 0A·x2 + 0A·x3

POL(A__IF(x1, x2, x3)) = 4A + -I·x1 + 1A·x2 + 1A·x3

POL(true) = 1A

POL(filter(x1, x2)) = -I + 0A·x1 + 0A·x2

POL(A__FILTER(x1, x2)) = 2A + 1A·x1 + 1A·x2

POL(divides(x1, x2)) = -I + 0A·x1 + 0A·x2

POL(false) = 1A

POL(a__primes) = 4A

POL(a__sieve(x1)) = 2A + 0A·x1

POL(a__head(x1)) = -I + 0A·x1

POL(a__tail(x1)) = 3A + 0A·x1

POL(a__if(x1, x2, x3)) = 1A + 0A·x1 + 0A·x2 + 0A·x3

POL(a__filter(x1, x2)) = 1A + 0A·x1 + 0A·x2

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

a__from(X) → cons(mark(X), from(s(X)))
a__from(X) → from(X)
mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
a__head(cons(X, Y)) → mark(X)
mark(tail(X)) → a__tail(mark(X))
a__tail(cons(X, Y)) → mark(Y)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__primes → primes
a__primes → a__sieve(a__from(s(s(0))))
a__sieve(X) → sieve(X)
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))

(8) Obligation:

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

A__SIEVE(cons(X, Y)) → MARK(X)
MARK(primes) → A__PRIMES
A__PRIMES → A__SIEVE(a__from(s(s(0))))
A__FROM(X) → MARK(X)
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(sieve(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(head(X)) → A__HEAD(mark(X))
A__HEAD(cons(X, Y)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(tail(X)) → A__TAIL(mark(X))
A__TAIL(cons(X, Y)) → MARK(Y)
MARK(tail(X)) → MARK(X)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__IF(true, X, Y) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2)) → A__FILTER(mark(X1), mark(X2))
A__FILTER(s(s(X)), cons(Y, Z)) → MARK(X)
MARK(filter(X1, X2)) → MARK(X1)
MARK(filter(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X2)
A__FILTER(s(s(X)), cons(Y, Z)) → MARK(Y)
A__IF(false, X, Y) → MARK(Y)

The TRS R consists of the following rules:

a__primes → a__sieve(a__from(s(s(0))))
a__from(X) → cons(mark(X), from(s(X)))
a__head(cons(X, Y)) → mark(X)
a__tail(cons(X, Y)) → mark(Y)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__primes → primes
a__sieve(X) → sieve(X)
a__from(X) → from(X)
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


MARK(primes) → A__PRIMES
A__TAIL(cons(X, Y)) → MARK(Y)
MARK(tail(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(A__SIEVE(x1)) = 3A + 0A·x1

POL(cons(x1, x2)) = -I + 1A·x1 + 0A·x2

POL(MARK(x1)) = 3A + 1A·x1

POL(primes) = 5A

POL(A__PRIMES) = 5A

POL(a__from(x1)) = 0A + 3A·x1

POL(s(x1)) = -I + 0A·x1

POL(0) = 2A

POL(A__FROM(x1)) = 3A + 1A·x1

POL(sieve(x1)) = -I + 0A·x1

POL(mark(x1)) = -I + 0A·x1

POL(from(x1)) = 0A + 3A·x1

POL(head(x1)) = -I + 0A·x1

POL(A__HEAD(x1)) = 3A + 0A·x1

POL(tail(x1)) = 3A + 5A·x1

POL(A__TAIL(x1)) = 4A + 5A·x1

POL(if(x1, x2, x3)) = -I + 1A·x1 + 0A·x2 + 0A·x3

POL(A__IF(x1, x2, x3)) = 3A + -I·x1 + 1A·x2 + 1A·x3

POL(true) = 1A

POL(filter(x1, x2)) = -I + 1A·x1 + 0A·x2

POL(A__FILTER(x1, x2)) = 3A + 2A·x1 + 0A·x2

POL(divides(x1, x2)) = -I + 0A·x1 + 0A·x2

POL(false) = 0A

POL(a__primes) = 5A

POL(a__sieve(x1)) = -I + 0A·x1

POL(a__head(x1)) = -I + 0A·x1

POL(a__tail(x1)) = 3A + 5A·x1

POL(a__if(x1, x2, x3)) = -I + 1A·x1 + 0A·x2 + 0A·x3

POL(a__filter(x1, x2)) = -I + 1A·x1 + 0A·x2

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

a__from(X) → cons(mark(X), from(s(X)))
a__from(X) → from(X)
mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
a__head(cons(X, Y)) → mark(X)
mark(tail(X)) → a__tail(mark(X))
a__tail(cons(X, Y)) → mark(Y)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__primes → primes
a__primes → a__sieve(a__from(s(s(0))))
a__sieve(X) → sieve(X)
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))

(10) Obligation:

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

A__SIEVE(cons(X, Y)) → MARK(X)
A__PRIMES → A__SIEVE(a__from(s(s(0))))
A__FROM(X) → MARK(X)
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(sieve(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(head(X)) → A__HEAD(mark(X))
A__HEAD(cons(X, Y)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(tail(X)) → A__TAIL(mark(X))
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__IF(true, X, Y) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2)) → A__FILTER(mark(X1), mark(X2))
A__FILTER(s(s(X)), cons(Y, Z)) → MARK(X)
MARK(filter(X1, X2)) → MARK(X1)
MARK(filter(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X2)
A__FILTER(s(s(X)), cons(Y, Z)) → MARK(Y)
A__IF(false, X, Y) → MARK(Y)

The TRS R consists of the following rules:

a__primes → a__sieve(a__from(s(s(0))))
a__from(X) → cons(mark(X), from(s(X)))
a__head(cons(X, Y)) → mark(X)
a__tail(cons(X, Y)) → mark(Y)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__primes → primes
a__sieve(X) → sieve(X)
a__from(X) → from(X)
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)

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 2 less nodes.

(12) Obligation:

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

MARK(sieve(X)) → A__SIEVE(mark(X))
A__SIEVE(cons(X, Y)) → MARK(X)
MARK(sieve(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
MARK(head(X)) → A__HEAD(mark(X))
A__HEAD(cons(X, Y)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__IF(true, X, Y) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2)) → A__FILTER(mark(X1), mark(X2))
A__FILTER(s(s(X)), cons(Y, Z)) → MARK(X)
MARK(filter(X1, X2)) → MARK(X1)
MARK(filter(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X2)
A__FILTER(s(s(X)), cons(Y, Z)) → MARK(Y)
A__IF(false, X, Y) → MARK(Y)

The TRS R consists of the following rules:

a__primes → a__sieve(a__from(s(s(0))))
a__from(X) → cons(mark(X), from(s(X)))
a__head(cons(X, Y)) → mark(X)
a__tail(cons(X, Y)) → mark(Y)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__primes → primes
a__sieve(X) → sieve(X)
a__from(X) → from(X)
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


A__FILTER(s(s(X)), cons(Y, Z)) → MARK(Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(MARK(x1)) = 0A + 1A·x1

POL(sieve(x1)) = -I + 0A·x1

POL(A__SIEVE(x1)) = 0A + 0A·x1

POL(mark(x1)) = -I + 0A·x1

POL(cons(x1, x2)) = -I + 1A·x1 + 0A·x2

POL(from(x1)) = 4A + 4A·x1

POL(A__FROM(x1)) = 0A + 5A·x1

POL(head(x1)) = -I + 0A·x1

POL(A__HEAD(x1)) = 0A + 0A·x1

POL(if(x1, x2, x3)) = 0A + 0A·x1 + 0A·x2 + 0A·x3

POL(A__IF(x1, x2, x3)) = 1A + -I·x1 + 1A·x2 + 1A·x3

POL(true) = 0A

POL(filter(x1, x2)) = -I + 0A·x1 + 0A·x2

POL(A__FILTER(x1, x2)) = 0A + 1A·x1 + 1A·x2

POL(s(x1)) = 0A + 0A·x1

POL(divides(x1, x2)) = -I + 0A·x1 + 0A·x2

POL(false) = 0A

POL(primes) = 5A

POL(a__primes) = 5A

POL(a__sieve(x1)) = -I + 0A·x1

POL(a__from(x1)) = 4A + 4A·x1

POL(a__head(x1)) = -I + 0A·x1

POL(tail(x1)) = 4A + 0A·x1

POL(a__tail(x1)) = 4A + 0A·x1

POL(a__if(x1, x2, x3)) = 0A + 0A·x1 + 0A·x2 + 0A·x3

POL(a__filter(x1, x2)) = -I + 0A·x1 + 0A·x2

POL(0) = 1A

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
a__head(cons(X, Y)) → mark(X)
mark(tail(X)) → a__tail(mark(X))
a__tail(cons(X, Y)) → mark(Y)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__from(X) → cons(mark(X), from(s(X)))
a__primes → primes
a__primes → a__sieve(a__from(s(s(0))))
a__from(X) → from(X)
a__sieve(X) → sieve(X)
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))

(14) Obligation:

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

MARK(sieve(X)) → A__SIEVE(mark(X))
A__SIEVE(cons(X, Y)) → MARK(X)
MARK(sieve(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
MARK(head(X)) → A__HEAD(mark(X))
A__HEAD(cons(X, Y)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__IF(true, X, Y) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2)) → A__FILTER(mark(X1), mark(X2))
A__FILTER(s(s(X)), cons(Y, Z)) → MARK(X)
MARK(filter(X1, X2)) → MARK(X1)
MARK(filter(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X2)
A__IF(false, X, Y) → MARK(Y)

The TRS R consists of the following rules:

a__primes → a__sieve(a__from(s(s(0))))
a__from(X) → cons(mark(X), from(s(X)))
a__head(cons(X, Y)) → mark(X)
a__tail(cons(X, Y)) → mark(Y)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__primes → primes
a__sieve(X) → sieve(X)
a__from(X) → from(X)
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


MARK(from(X)) → A__FROM(mark(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(MARK(x1)) = 1A + 0A·x1

POL(sieve(x1)) = 4A + 0A·x1

POL(A__SIEVE(x1)) = 4A + 0A·x1

POL(mark(x1)) = 0A + 0A·x1

POL(cons(x1, x2)) = 0A + 0A·x1 + 0A·x2

POL(from(x1)) = 3A + 1A·x1

POL(A__FROM(x1)) = 1A + 0A·x1

POL(head(x1)) = 1A + 1A·x1

POL(A__HEAD(x1)) = 1A + 0A·x1

POL(if(x1, x2, x3)) = -I + 0A·x1 + 0A·x2 + 0A·x3

POL(A__IF(x1, x2, x3)) = 1A + -I·x1 + 0A·x2 + 0A·x3

POL(true) = 0A

POL(filter(x1, x2)) = 4A + 0A·x1 + 0A·x2

POL(A__FILTER(x1, x2)) = 4A + 0A·x1 + 0A·x2

POL(s(x1)) = 1A + 0A·x1

POL(divides(x1, x2)) = 1A + 0A·x1 + 0A·x2

POL(false) = 0A

POL(primes) = 4A

POL(a__primes) = 4A

POL(a__sieve(x1)) = 4A + 0A·x1

POL(a__from(x1)) = 3A + 1A·x1

POL(a__head(x1)) = 1A + 1A·x1

POL(tail(x1)) = 5A + 4A·x1

POL(a__tail(x1)) = 5A + 4A·x1

POL(a__if(x1, x2, x3)) = 0A + 0A·x1 + 0A·x2 + 0A·x3

POL(a__filter(x1, x2)) = 4A + 0A·x1 + 0A·x2

POL(0) = 0A

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
a__head(cons(X, Y)) → mark(X)
mark(tail(X)) → a__tail(mark(X))
a__tail(cons(X, Y)) → mark(Y)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__from(X) → cons(mark(X), from(s(X)))
a__primes → primes
a__primes → a__sieve(a__from(s(s(0))))
a__from(X) → from(X)
a__sieve(X) → sieve(X)
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))

(16) Obligation:

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

MARK(sieve(X)) → A__SIEVE(mark(X))
A__SIEVE(cons(X, Y)) → MARK(X)
MARK(sieve(X)) → MARK(X)
A__FROM(X) → MARK(X)
MARK(head(X)) → A__HEAD(mark(X))
A__HEAD(cons(X, Y)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__IF(true, X, Y) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2)) → A__FILTER(mark(X1), mark(X2))
A__FILTER(s(s(X)), cons(Y, Z)) → MARK(X)
MARK(filter(X1, X2)) → MARK(X1)
MARK(filter(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X2)
A__IF(false, X, Y) → MARK(Y)

The TRS R consists of the following rules:

a__primes → a__sieve(a__from(s(s(0))))
a__from(X) → cons(mark(X), from(s(X)))
a__head(cons(X, Y)) → mark(X)
a__tail(cons(X, Y)) → mark(Y)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__primes → primes
a__sieve(X) → sieve(X)
a__from(X) → from(X)
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)

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

(17) DependencyGraphProof (EQUIVALENT transformation)

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

(18) Obligation:

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

A__SIEVE(cons(X, Y)) → MARK(X)
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(sieve(X)) → MARK(X)
MARK(head(X)) → A__HEAD(mark(X))
A__HEAD(cons(X, Y)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__IF(true, X, Y) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2)) → A__FILTER(mark(X1), mark(X2))
A__FILTER(s(s(X)), cons(Y, Z)) → MARK(X)
MARK(filter(X1, X2)) → MARK(X1)
MARK(filter(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X2)
A__IF(false, X, Y) → MARK(Y)

The TRS R consists of the following rules:

a__primes → a__sieve(a__from(s(s(0))))
a__from(X) → cons(mark(X), from(s(X)))
a__head(cons(X, Y)) → mark(X)
a__tail(cons(X, Y)) → mark(Y)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__primes → primes
a__sieve(X) → sieve(X)
a__from(X) → from(X)
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


A__HEAD(cons(X, Y)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(A__SIEVE(x1)) = 5A + 1A·x1

POL(cons(x1, x2)) = 1A + 0A·x1 + 0A·x2

POL(MARK(x1)) = 4A + 1A·x1

POL(sieve(x1)) = 4A + 0A·x1

POL(mark(x1)) = -I + 0A·x1

POL(head(x1)) = -I + 3A·x1

POL(A__HEAD(x1)) = 3A + 4A·x1

POL(if(x1, x2, x3)) = -I + 0A·x1 + 0A·x2 + 0A·x3

POL(A__IF(x1, x2, x3)) = 4A + -I·x1 + 1A·x2 + 1A·x3

POL(true) = 2A

POL(filter(x1, x2)) = 4A + 0A·x1 + 0A·x2

POL(A__FILTER(x1, x2)) = 4A + 1A·x1 + -I·x2

POL(s(x1)) = -I + 0A·x1

POL(divides(x1, x2)) = -I + 0A·x1 + 0A·x2

POL(false) = 0A

POL(primes) = 5A

POL(a__primes) = 5A

POL(a__sieve(x1)) = 4A + 0A·x1

POL(from(x1)) = 1A + 5A·x1

POL(a__from(x1)) = 1A + 5A·x1

POL(a__head(x1)) = -I + 3A·x1

POL(tail(x1)) = 3A + 2A·x1

POL(a__tail(x1)) = 3A + 2A·x1

POL(a__if(x1, x2, x3)) = -I + 0A·x1 + 0A·x2 + 0A·x3

POL(a__filter(x1, x2)) = 4A + 0A·x1 + 0A·x2

POL(0) = 0A

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
a__head(cons(X, Y)) → mark(X)
mark(tail(X)) → a__tail(mark(X))
a__tail(cons(X, Y)) → mark(Y)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__from(X) → cons(mark(X), from(s(X)))
a__primes → primes
a__primes → a__sieve(a__from(s(s(0))))
a__from(X) → from(X)
a__sieve(X) → sieve(X)
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))

(20) Obligation:

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

A__SIEVE(cons(X, Y)) → MARK(X)
MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(sieve(X)) → MARK(X)
MARK(head(X)) → A__HEAD(mark(X))
MARK(head(X)) → MARK(X)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__IF(true, X, Y) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2)) → A__FILTER(mark(X1), mark(X2))
A__FILTER(s(s(X)), cons(Y, Z)) → MARK(X)
MARK(filter(X1, X2)) → MARK(X1)
MARK(filter(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X2)
A__IF(false, X, Y) → MARK(Y)

The TRS R consists of the following rules:

a__primes → a__sieve(a__from(s(s(0))))
a__from(X) → cons(mark(X), from(s(X)))
a__head(cons(X, Y)) → mark(X)
a__tail(cons(X, Y)) → mark(Y)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__primes → primes
a__sieve(X) → sieve(X)
a__from(X) → from(X)
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)

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

(21) DependencyGraphProof (EQUIVALENT transformation)

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

(22) Obligation:

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

MARK(sieve(X)) → A__SIEVE(mark(X))
A__SIEVE(cons(X, Y)) → MARK(X)
MARK(sieve(X)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__IF(true, X, Y) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2)) → A__FILTER(mark(X1), mark(X2))
A__FILTER(s(s(X)), cons(Y, Z)) → MARK(X)
MARK(filter(X1, X2)) → MARK(X1)
MARK(filter(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X2)
A__IF(false, X, Y) → MARK(Y)

The TRS R consists of the following rules:

a__primes → a__sieve(a__from(s(s(0))))
a__from(X) → cons(mark(X), from(s(X)))
a__head(cons(X, Y)) → mark(X)
a__tail(cons(X, Y)) → mark(Y)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__primes → primes
a__sieve(X) → sieve(X)
a__from(X) → from(X)
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


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: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(MARK(x1)) = 5A + 5A·x1

POL(sieve(x1)) = 0A + 0A·x1

POL(A__SIEVE(x1)) = 5A + 2A·x1

POL(mark(x1)) = 0A + 0A·x1

POL(cons(x1, x2)) = 3A + 3A·x1 + 0A·x2

POL(head(x1)) = 5A + 0A·x1

POL(if(x1, x2, x3)) = -I + 0A·x1 + 0A·x2 + 0A·x3

POL(A__IF(x1, x2, x3)) = 5A + -I·x1 + 5A·x2 + 5A·x3

POL(true) = 2A

POL(filter(x1, x2)) = -I + 0A·x1 + 0A·x2

POL(A__FILTER(x1, x2)) = 5A + 5A·x1 + 2A·x2

POL(s(x1)) = -I + 0A·x1

POL(divides(x1, x2)) = -I + 0A·x1 + 0A·x2

POL(false) = 0A

POL(primes) = 5A

POL(a__primes) = 5A

POL(a__sieve(x1)) = 0A + 0A·x1

POL(from(x1)) = 5A + 5A·x1

POL(a__from(x1)) = 5A + 5A·x1

POL(a__head(x1)) = 5A + 0A·x1

POL(tail(x1)) = 4A + 2A·x1

POL(a__tail(x1)) = 4A + 2A·x1

POL(a__if(x1, x2, x3)) = 0A + 0A·x1 + 0A·x2 + 0A·x3

POL(a__filter(x1, x2)) = 0A + 0A·x1 + 0A·x2

POL(0) = 0A

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
a__head(cons(X, Y)) → mark(X)
mark(tail(X)) → a__tail(mark(X))
a__tail(cons(X, Y)) → mark(Y)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__from(X) → cons(mark(X), from(s(X)))
a__primes → primes
a__primes → a__sieve(a__from(s(s(0))))
a__from(X) → from(X)
a__sieve(X) → sieve(X)
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))

(24) Obligation:

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

MARK(sieve(X)) → A__SIEVE(mark(X))
A__SIEVE(cons(X, Y)) → MARK(X)
MARK(sieve(X)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__IF(true, X, Y) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2)) → A__FILTER(mark(X1), mark(X2))
A__FILTER(s(s(X)), cons(Y, Z)) → MARK(X)
MARK(filter(X1, X2)) → MARK(X1)
MARK(filter(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
MARK(divides(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X2)
A__IF(false, X, Y) → MARK(Y)

The TRS R consists of the following rules:

a__primes → a__sieve(a__from(s(s(0))))
a__from(X) → cons(mark(X), from(s(X)))
a__head(cons(X, Y)) → mark(X)
a__tail(cons(X, Y)) → mark(Y)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__primes → primes
a__sieve(X) → sieve(X)
a__from(X) → from(X)
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)

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

(25) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


MARK(sieve(X)) → A__SIEVE(mark(X))
MARK(sieve(X)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(filter(X1, X2)) → A__FILTER(mark(X1), mark(X2))
MARK(filter(X1, X2)) → MARK(X1)
MARK(divides(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(MARK(x1)) = 2A + 2A·x1

POL(sieve(x1)) = 3A + 2A·x1

POL(A__SIEVE(x1)) = -I + 0A·x1

POL(mark(x1)) = 0A + 0A·x1

POL(cons(x1, x2)) = 3A + 2A·x1 + 0A·x2

POL(head(x1)) = 4A + 4A·x1

POL(if(x1, x2, x3)) = 1A + 0A·x1 + 0A·x2 + 0A·x3

POL(A__IF(x1, x2, x3)) = 3A + -I·x1 + 2A·x2 + 2A·x3

POL(true) = 0A

POL(filter(x1, x2)) = 3A + 3A·x1 + 0A·x2

POL(A__FILTER(x1, x2)) = 0A + 2A·x1 + -I·x2

POL(s(x1)) = 0A + 0A·x1

POL(divides(x1, x2)) = 2A + 2A·x1 + 0A·x2

POL(false) = 1A

POL(primes) = 5A

POL(a__primes) = 5A

POL(a__sieve(x1)) = 3A + 2A·x1

POL(from(x1)) = 3A + 2A·x1

POL(a__from(x1)) = 3A + 2A·x1

POL(a__head(x1)) = 4A + 4A·x1

POL(tail(x1)) = 3A + 1A·x1

POL(a__tail(x1)) = 3A + 1A·x1

POL(a__if(x1, x2, x3)) = 1A + 0A·x1 + 0A·x2 + 0A·x3

POL(a__filter(x1, x2)) = 3A + 3A·x1 + 0A·x2

POL(0) = 1A

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
a__head(cons(X, Y)) → mark(X)
mark(tail(X)) → a__tail(mark(X))
a__tail(cons(X, Y)) → mark(Y)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__from(X) → cons(mark(X), from(s(X)))
a__primes → primes
a__primes → a__sieve(a__from(s(s(0))))
a__from(X) → from(X)
a__sieve(X) → sieve(X)
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))

(26) Obligation:

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

A__SIEVE(cons(X, Y)) → MARK(X)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__IF(true, X, Y) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X1)
A__FILTER(s(s(X)), cons(Y, Z)) → MARK(X)
MARK(filter(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
MARK(divides(X1, X2)) → MARK(X2)
A__IF(false, X, Y) → MARK(Y)

The TRS R consists of the following rules:

a__primes → a__sieve(a__from(s(s(0))))
a__from(X) → cons(mark(X), from(s(X)))
a__head(cons(X, Y)) → mark(X)
a__tail(cons(X, Y)) → mark(Y)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__primes → primes
a__sieve(X) → sieve(X)
a__from(X) → from(X)
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)

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

(27) DependencyGraphProof (EQUIVALENT transformation)

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

(28) Obligation:

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

A__IF(true, X, Y) → MARK(X)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__IF(false, X, Y) → MARK(Y)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(filter(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
MARK(divides(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

a__primes → a__sieve(a__from(s(s(0))))
a__from(X) → cons(mark(X), from(s(X)))
a__head(cons(X, Y)) → mark(X)
a__tail(cons(X, Y)) → mark(Y)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__filter(s(s(X)), cons(Y, Z)) → a__if(divides(s(s(mark(X))), mark(Y)), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
a__sieve(cons(X, Y)) → cons(mark(X), filter(X, sieve(Y)))
mark(primes) → a__primes
mark(sieve(X)) → a__sieve(mark(X))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(filter(X1, X2)) → a__filter(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(true) → true
mark(false) → false
mark(divides(X1, X2)) → divides(mark(X1), mark(X2))
a__primes → primes
a__sieve(X) → sieve(X)
a__from(X) → from(X)
a__head(X) → head(X)
a__tail(X) → tail(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__filter(X1, X2) → filter(X1, X2)

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:

  • MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
    The graph contains the following edges 1 > 2, 1 > 3

  • MARK(if(X1, X2, X3)) → MARK(X1)
    The graph contains the following edges 1 > 1

  • MARK(filter(X1, X2)) → MARK(X2)
    The graph contains the following edges 1 > 1

  • MARK(s(X)) → MARK(X)
    The graph contains the following edges 1 > 1

  • MARK(divides(X1, X2)) → MARK(X2)
    The graph contains the following edges 1 > 1

  • A__IF(true, X, Y) → MARK(X)
    The graph contains the following edges 2 >= 1

  • A__IF(false, X, Y) → MARK(Y)
    The graph contains the following edges 3 >= 1

(30) YES