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

         ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

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

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

FILTER(s(s(X)), cons(Y, Z)) -> FILTER(X, sieve(Y))

one new Dependency Pair is created:

FILTER(s(s(X)), cons(cons(X'', Y''), Z)) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

FILTER(s(s(X)), cons(cons(X'', Y''), Z)) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))
FILTER(s(s(X)), cons(Y, Z)) -> SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) -> FILTER(s(s(X)), Z)
SIEVE(cons(X, Y)) -> FILTER(X, sieve(Y))
SIEVE(cons(X, Y)) -> 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

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

SIEVE(cons(X, Y)) -> FILTER(X, sieve(Y))

one new Dependency Pair is created:

SIEVE(cons(X, cons(X'', Y''))) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Nar

...

               →DP Problem 4

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

SIEVE(cons(X, cons(X'', Y''))) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))
SIEVE(cons(X, Y)) -> SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) -> SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) -> FILTER(s(s(X)), Z)
FILTER(s(s(X)), cons(cons(X'', Y''), Z)) -> FILTER(X, cons(X'', 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

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

FILTER(s(s(X)), cons(Y, Z)) -> FILTER(s(s(X)), Z)

two new Dependency Pairs are created:

FILTER(s(s(X'')), cons(Y, cons(Y'', Z''))) -> FILTER(s(s(X'')), cons(Y'', Z''))
FILTER(s(s(X'')), cons(Y, cons(cons(X'''', Y''''), Z''))) -> FILTER(s(s(X'')), cons(cons(X'''', Y''''), Z''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Nar

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

FILTER(s(s(X'')), cons(Y, cons(cons(X'''', Y''''), Z''))) -> FILTER(s(s(X'')), cons(cons(X'''', Y''''), Z''))
FILTER(s(s(X'')), cons(Y, cons(Y'', Z''))) -> FILTER(s(s(X'')), cons(Y'', Z''))
FILTER(s(s(X)), cons(cons(X'', Y''), Z)) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))
SIEVE(cons(X, Y)) -> SIEVE(Y)
FILTER(s(s(X)), cons(Y, Z)) -> SIEVE(Y)
SIEVE(cons(X, cons(X'', Y''))) -> FILTER(X, cons(X'', 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

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

FILTER(s(s(X)), cons(Y, Z)) -> SIEVE(Y)

two new Dependency Pairs are created:

FILTER(s(s(X)), cons(cons(X'', Y''), Z)) -> SIEVE(cons(X'', Y''))
FILTER(s(s(X)), cons(cons(X'', cons(X'''', Y'''')), Z)) -> SIEVE(cons(X'', cons(X'''', Y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 3

             ↳Nar

...

               →DP Problem 6

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Remaining

Dependency Pairs:

FILTER(s(s(X)), cons(cons(X'', cons(X'''', Y'''')), Z)) -> SIEVE(cons(X'', cons(X'''', Y'''')))
SIEVE(cons(X, cons(X'', Y''))) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))
SIEVE(cons(X, Y)) -> SIEVE(Y)
FILTER(s(s(X)), cons(cons(X'', Y''), Z)) -> SIEVE(cons(X'', Y''))
FILTER(s(s(X'')), cons(Y, cons(Y'', Z''))) -> FILTER(s(s(X'')), cons(Y'', Z''))
FILTER(s(s(X)), cons(cons(X'', Y''), Z)) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))
FILTER(s(s(X'')), cons(Y, cons(cons(X'''', Y''''), Z''))) -> FILTER(s(s(X'')), cons(cons(X'''', Y''''), Z''))

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

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

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

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

         ↳Nar

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
SIEVE(cons(X, cons(X'', cons(X'''', Y'''')))) -> SIEVE(cons(X'', cons(X'''', Y'''')))
SIEVE(cons(X, cons(X'', Y''))) -> SIEVE(cons(X'', Y''))
FILTER(s(s(X)), cons(cons(X'', Y''), Z)) -> SIEVE(cons(X'', Y''))
FILTER(s(s(X'')), cons(Y, cons(cons(X'''', Y''''), Z''))) -> FILTER(s(s(X'')), cons(cons(X'''', Y''''), Z''))
FILTER(s(s(X'')), cons(Y, cons(Y'', Z''))) -> FILTER(s(s(X'')), cons(Y'', Z''))
FILTER(s(s(X)), cons(cons(X'', Y''), Z)) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))
SIEVE(cons(X, cons(X'', Y''))) -> FILTER(X, cons(X'', filter(X'', sieve(Y''))))
FILTER(s(s(X)), cons(cons(X'', cons(X'''', Y'''')), Z)) -> SIEVE(cons(X'', cons(X'''', 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:
0:01 minutes