Termination Proof using AProVE

Term Rewriting System R:
[X, Y, M, N]
filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

FILTER(cons(X, Y), 0, M) -> FILTER(Y, M, M)
FILTER(cons(X, Y), s(N), M) -> FILTER(Y, N, M)
SIEVE(cons(0, Y)) -> SIEVE(Y)
SIEVE(cons(s(N), Y)) -> SIEVE(filter(Y, N, N))
SIEVE(cons(s(N), Y)) -> FILTER(Y, N, N)
NATS(N) -> NATS(s(N))
ZPRIMES -> SIEVE(nats(s(s(0))))
ZPRIMES -> NATS(s(s(0)))

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

Dependency Pairs:

FILTER(cons(X, Y), s(N), M) -> FILTER(Y, N, M)
FILTER(cons(X, Y), 0, M) -> FILTER(Y, M, M)

Rules:

filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))

The following dependency pairs can be strictly oriented:

FILTER(cons(X, Y), s(N), M) -> FILTER(Y, N, M)
FILTER(cons(X, Y), 0, M) -> FILTER(Y, M, M)

There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

FILTER(x₁, x₂, x₃) -> FILTER(x₁, x₂, x₃)
cons(x₁, x₂) -> cons(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

Dependency Pair:

Rules:

filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
NATS(N) -> NATS(s(N))

Rules:

filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))
Dependency Pairs:
SIEVE(cons(s(N), Y)) -> SIEVE(filter(Y, N, N))
SIEVE(cons(0, Y)) -> SIEVE(Y)

Rules:

filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
NATS(N) -> NATS(s(N))

Rules:

filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))
Dependency Pairs:
SIEVE(cons(s(N), Y)) -> SIEVE(filter(Y, N, N))
SIEVE(cons(0, Y)) -> SIEVE(Y)

Rules:

filter(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) -> cons(X, filter(Y, N, M))
sieve(cons(0, Y)) -> cons(0, sieve(Y))
sieve(cons(s(N), Y)) -> cons(s(N), sieve(filter(Y, N, N)))
nats(N) -> cons(N, nats(s(N)))
zprimes -> sieve(nats(s(s(0))))

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