Termination Proof using AProVE

Term Rewriting System R:
[X, Y, Z]
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)))

Innermost 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) -> 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)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
SIEVE(cons(X, Y)) -> SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) -> SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) -> FILTER(X, sieve(Y))
FILTER(s(s(X)), cons(Y, Z)) -> FILTER(s(s(X)), Z)
SIEVE(cons(X, Y)) -> FILTER(X, sieve(Y))

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)))

Strategy:
innermost
Dependency Pair:
FROM(X) -> FROM(s(X))

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)))

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
SIEVE(cons(X, Y)) -> SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) -> SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) -> FILTER(X, sieve(Y))
FILTER(s(s(X)), cons(Y, Z)) -> FILTER(s(s(X)), Z)
SIEVE(cons(X, Y)) -> FILTER(X, sieve(Y))

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)))

Strategy:
innermost
Dependency Pair:
FROM(X) -> FROM(s(X))

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)))

Strategy:
innermost

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