Termination Proof using AProVE

Term Rewriting System R:
[X, Y, Z, X1, X2]
primes -> sieve(from(s(s(0))))
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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))))
filter(X1, X2) -> n_{_filter}(X1, X2)
sieve(cons(X, Y)) -> cons(X, n_{_filter}(X, sieve(activate(Y))))
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

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

FILTER(s(s(X)), cons(Y, Z)) -> SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_filter}(X1, X2)) -> FILTER(X1, X2)
SIEVE(cons(X, Y)) -> ACTIVATE(Y)
SIEVE(cons(X, Y)) -> SIEVE(activate(Y))

Rules:

primes -> sieve(from(s(s(0))))
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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))))
filter(X1, X2) -> n_{_filter}(X1, X2)
sieve(cons(X, Y)) -> cons(X, n_{_filter}(X, sieve(activate(Y))))
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

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

SIEVE(cons(X, Y)) -> SIEVE(activate(Y))

four new Dependency Pairs are created:

SIEVE(cons(X, n_{_from}(X''))) -> SIEVE(from(X''))
SIEVE(cons(X, n_{_filter}(X1', X2'))) -> SIEVE(filter(X1', X2'))
SIEVE(cons(X, n_{_cons}(X1', X2'))) -> SIEVE(cons(X1', X2'))
SIEVE(cons(X, Y')) -> SIEVE(Y')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

SIEVE(cons(X, Y')) -> SIEVE(Y')
SIEVE(cons(X, n_{_cons}(X1', X2'))) -> SIEVE(cons(X1', X2'))
SIEVE(cons(X, n_{_filter}(X1', X2'))) -> SIEVE(filter(X1', X2'))
SIEVE(cons(X, n_{_from}(X''))) -> SIEVE(from(X''))
FILTER(s(s(X)), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_filter}(X1, X2)) -> FILTER(X1, X2)
SIEVE(cons(X, Y)) -> ACTIVATE(Y)
FILTER(s(s(X)), cons(Y, Z)) -> SIEVE(Y)

Rules:

primes -> sieve(from(s(s(0))))
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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))))
filter(X1, X2) -> n_{_filter}(X1, X2)
sieve(cons(X, Y)) -> cons(X, n_{_filter}(X, sieve(activate(Y))))
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

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

SIEVE(cons(X, n_{_from}(X''))) -> SIEVE(from(X''))

two new Dependency Pairs are created:

SIEVE(cons(X, n_{_from}(X'''))) -> SIEVE(cons(X''', n_{_from}(s(X'''))))
SIEVE(cons(X, n_{_from}(X'''))) -> SIEVE(n_{_from}(X'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Narrowing Transformation

Dependency Pairs:

SIEVE(cons(X, n_{_from}(X'''))) -> SIEVE(cons(X''', n_{_from}(s(X'''))))
SIEVE(cons(X, n_{_cons}(X1', X2'))) -> SIEVE(cons(X1', X2'))
SIEVE(cons(X, n_{_filter}(X1', X2'))) -> SIEVE(filter(X1', X2'))
FILTER(s(s(X)), cons(Y, Z)) -> SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_filter}(X1, X2)) -> FILTER(X1, X2)
SIEVE(cons(X, Y)) -> ACTIVATE(Y)
SIEVE(cons(X, Y')) -> SIEVE(Y')

Rules:

primes -> sieve(from(s(s(0))))
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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))))
filter(X1, X2) -> n_{_filter}(X1, X2)
sieve(cons(X, Y)) -> cons(X, n_{_filter}(X, sieve(activate(Y))))
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

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

SIEVE(cons(X, n_{_filter}(X1', X2'))) -> SIEVE(filter(X1', X2'))

two new Dependency Pairs are created:

SIEVE(cons(X, n_{_filter}(s(s(X'')), cons(Y', Z')))) -> SIEVE(if(divides(s(s(X'')), Y'), n_{_filter}(s(s(X'')), activate(Z')), n_{_cons}(Y', n_{_filter}(X'', sieve(Y')))))
SIEVE(cons(X, n_{_filter}(X1'', X2''))) -> SIEVE(n_{_filter}(X1'', X2''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

SIEVE(cons(X, n_{_filter}(s(s(X'')), cons(Y', Z')))) -> SIEVE(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')) -> SIEVE(Y')
SIEVE(cons(X, n_{_cons}(X1', X2'))) -> SIEVE(cons(X1', X2'))
FILTER(s(s(X)), cons(Y, Z)) -> SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_filter}(X1, X2)) -> FILTER(X1, X2)
SIEVE(cons(X, Y)) -> ACTIVATE(Y)
SIEVE(cons(X, n_{_from}(X'''))) -> SIEVE(cons(X''', n_{_from}(s(X'''))))

Rules:

primes -> sieve(from(s(s(0))))
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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))))
filter(X1, X2) -> n_{_filter}(X1, X2)
sieve(cons(X, Y)) -> cons(X, n_{_filter}(X, sieve(activate(Y))))
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

Termination of R could not be shown.
Duration:
0:00 minutes