(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: Matrix interpretation [MATRO]:
POL(ACTIVATE(x1)) = | | + | | · | x1 |
POL(n__filter(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(FILTER(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(cons(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(activate(x1)) = | | + | | · | x1 |
POL(n__cons(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(filter(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(if(x1, x2, x3)) = | | + | | · | x1 | + | | · | x2 | + | | · | x3 |
POL(divides(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
from(X) → cons(X, n__from(s(X)))
activate(X) → X
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__filter(X1, X2)) → filter(X1, X2)
activate(n__from(X)) → from(X)
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(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))))
(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: Polynomial interpretation [POLO]:
POL(SIEVE(x1)) = x1
POL(cons(x1, x2)) = 1 + x2
POL(n__cons(x1, x2)) = x2
POL(n__from(x1)) = x1
POL(s(x1)) = 0
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: Matrix interpretation [MATRO]:
POL(cons(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(n__cons(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
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:
- Semiunifier: [ ]
- Matcher: [y0 / x0, x0 / s(x0)]
Rewriting sequenceThe 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) FALSE
(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.