(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
primes → sieve(from(s(s(0))))
from(X) → cons(X, from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → Y
if(true, X, Y) → X
if(false, X, Y) → Y
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))
Q is empty.
(1) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
primes → sieve(from(s(s(0))))
from(X) → cons(X, from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → Y
if(true, X, Y) → X
if(false, X, Y) → Y
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))
The set Q consists of the following terms:
primes
from(x0)
head(cons(x0, x1))
tail(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) 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) → FROM(s(X))
FILTER(s(s(X)), cons(Y, Z)) → IF(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
FILTER(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)
FILTER(s(s(X)), cons(Y, Z)) → FILTER(X, sieve(Y))
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
SIEVE(cons(X, Y)) → FILTER(X, sieve(Y))
SIEVE(cons(X, Y)) → SIEVE(Y)
The TRS R consists of the following rules:
primes → sieve(from(s(s(0))))
from(X) → cons(X, from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → Y
if(true, X, Y) → X
if(false, X, Y) → Y
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))
The set Q consists of the following terms:
primes
from(x0)
head(cons(x0, x1))
tail(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FILTER(s(s(X)), cons(Y, Z)) → FILTER(X, sieve(Y))
FILTER(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
SIEVE(cons(X, Y)) → FILTER(X, sieve(Y))
SIEVE(cons(X, Y)) → SIEVE(Y)
The TRS R consists of the following rules:
primes → sieve(from(s(s(0))))
from(X) → cons(X, from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → Y
if(true, X, Y) → X
if(false, X, Y) → Y
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))
The set Q consists of the following terms:
primes
from(x0)
head(cons(x0, x1))
tail(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(8) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FILTER(s(s(X)), cons(Y, Z)) → FILTER(X, sieve(Y))
FILTER(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
SIEVE(cons(X, Y)) → FILTER(X, sieve(Y))
SIEVE(cons(X, Y)) → SIEVE(Y)
The TRS R consists of the following rules:
sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
The set Q consists of the following terms:
primes
from(x0)
head(cons(x0, x1))
tail(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(10) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
primes
from(x0)
head(cons(x0, x1))
tail(cons(x0, x1))
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FILTER(s(s(X)), cons(Y, Z)) → FILTER(X, sieve(Y))
FILTER(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
SIEVE(cons(X, Y)) → FILTER(X, sieve(Y))
SIEVE(cons(X, Y)) → SIEVE(Y)
The TRS R consists of the following rules:
sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
The set Q consists of the following terms:
if(true, x0, x1)
if(false, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(12) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
FILTER(s(s(X)), cons(Y, Z)) → FILTER(X, sieve(Y))
FILTER(s(s(X)), cons(Y, Z)) → FILTER(s(s(X)), Z)
FILTER(s(s(X)), cons(Y, Z)) → SIEVE(Y)
SIEVE(cons(X, Y)) → FILTER(X, sieve(Y))
SIEVE(cons(X, Y)) → SIEVE(Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( FILTER(x1, x2) ) = x2 + 1 |
POL( cons(x1, x2) ) = 2x1 + 2x2 + 1 |
POL( filter(x1, x2) ) = max{0, -2} |
POL( if(x1, ..., x3) ) = 0 |
POL( divides(x1, x2) ) = 2x1 |
POL( SIEVE(x1) ) = x1 + 1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
(13) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
The set Q consists of the following terms:
if(true, x0, x1)
if(false, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(14) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(15) YES
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FROM(X) → FROM(s(X))
The TRS R consists of the following rules:
primes → sieve(from(s(s(0))))
from(X) → cons(X, from(s(X)))
head(cons(X, Y)) → X
tail(cons(X, Y)) → Y
if(true, X, Y) → X
if(false, X, Y) → Y
filter(s(s(X)), cons(Y, Z)) → if(divides(s(s(X)), Y), filter(s(s(X)), Z), cons(Y, filter(X, sieve(Y))))
sieve(cons(X, Y)) → cons(X, filter(X, sieve(Y)))
The set Q consists of the following terms:
primes
from(x0)
head(cons(x0, x1))
tail(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(17) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FROM(X) → FROM(s(X))
R is empty.
The set Q consists of the following terms:
primes
from(x0)
head(cons(x0, x1))
tail(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(19) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
primes
from(x0)
head(cons(x0, x1))
tail(cons(x0, x1))
if(true, x0, x1)
if(false, x0, x1)
filter(s(s(x0)), cons(x1, x2))
sieve(cons(x0, x1))
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FROM(X) → FROM(s(X))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) TransformationProof (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
FROM(
X) →
FROM(
s(
X)) we obtained the following new rules [LPAR04]:
FROM(s(z0)) → FROM(s(s(z0))) → FROM(s(z0)) → FROM(s(s(z0)))
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FROM(s(z0)) → FROM(s(s(z0)))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(23) TransformationProof (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
FROM(
s(
z0)) →
FROM(
s(
s(
z0))) we obtained the following new rules [LPAR04]:
FROM(s(s(z0))) → FROM(s(s(s(z0)))) → FROM(s(s(z0))) → FROM(s(s(s(z0))))
(24) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FROM(s(s(z0))) → FROM(s(s(s(z0))))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(25) NonTerminationLoopProof (COMPLETE 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 =
FROM(
s(
s(
z0))) evaluates to t =
FROM(
s(
s(
s(
z0))))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [z0 / s(z0)]
- Semiunifier: [ ]
Rewriting sequenceThe DP semiunifies directly so there is only one rewrite step from FROM(s(s(z0))) to FROM(s(s(s(z0)))).
(26) NO