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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 QReductionProof (⇔)
    ↳11 QDP
    ↳12 QDPOrderProof (⇔)
    ↳13 QDP
    ↳14 PisEmptyProof (⇔)
    ↳15 TRUE
  ↳16 QDP
    ↳17 UsableRulesProof (⇔)
    ↳18 QDP
    ↳19 QReductionProof (⇔)
    ↳20 QDP
      ↳21 Instantiation (⇔)
        ↳22 QDP
      ↳23 Instantiation (⇔)
        ↳24 QDP
        ↳25 Instantiation (⇔)
        ↳26 QDP
        ↳27 NonTerminationProof (⇔)
        ↳28 FALSE

(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].


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 interpretation [POLO]:

POL(FILTER(x1, x2)) = x2   
POL(SIEVE(x1)) = x1   
POL(cons(x1, x2)) = 1 + x1 + x2   
POL(divides(x1, x2)) = 0   
POL(filter(x1, x2)) = 0   
POL(if(x1, x2, x3)) = 0   
POL(s(x1)) = 0   
POL(sieve(x1)) = x1   

The following usable rules [FROCOS05] 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) TRUE

(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) Instantiation (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)))

(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) Instantiation (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)))

(24) 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.

(25) Instantiation (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))))

(26) 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.

(27) 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 = 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:
  • Semiunifier: [ ]
  • Matcher: [z0 / s(z0)]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from FROM(s(s(z0))) to FROM(s(s(s(z0)))).



(28) FALSE