Termination Proof using AProVE

Term Rewriting System R:
[X, Y, M, N, X1, X2, X3]
a_{_filter}(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
a_{_filter}(cons(X, Y), s(N), M) -> cons(mark(X), filter(Y, N, M))
a_{_filter}(X1, X2, X3) -> filter(X1, X2, X3)
a_{_sieve}(cons(0, Y)) -> cons(0, sieve(Y))
a_{_sieve}(cons(s(N), Y)) -> cons(s(mark(N)), sieve(filter(Y, N, N)))
a_{_sieve}(X) -> sieve(X)
a_{_nats}(N) -> cons(mark(N), nats(s(N)))
a_{_nats}(X) -> nats(X)
a_{_zprimes} -> a_{_sieve}(a_{_nats}(s(s(0))))
a_{_zprimes} -> zprimes
mark(filter(X1, X2, X3)) -> a_{_filter}(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) -> a_{_sieve}(mark(X))
mark(nats(X)) -> a_{_nats}(mark(X))
mark(zprimes) -> a_{_zprimes}
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(0) -> 0
mark(s(X)) -> s(mark(X))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

A_{_FILTER}(cons(X, Y), s(N), M) -> MARK(X)
A_{_SIEVE}(cons(s(N), Y)) -> MARK(N)
A_{_NATS}(N) -> MARK(N)
A_{_ZPRIMES} -> A_{_SIEVE}(a_{_nats}(s(s(0))))
A_{_ZPRIMES} -> A_{_NATS}(s(s(0)))
MARK(filter(X1, X2, X3)) -> A_{_FILTER}(mark(X1), mark(X2), mark(X3))
MARK(filter(X1, X2, X3)) -> MARK(X1)
MARK(filter(X1, X2, X3)) -> MARK(X2)
MARK(filter(X1, X2, X3)) -> MARK(X3)
MARK(sieve(X)) -> A_{_SIEVE}(mark(X))
MARK(sieve(X)) -> MARK(X)
MARK(nats(X)) -> A_{_NATS}(mark(X))
MARK(nats(X)) -> MARK(X)
MARK(zprimes) -> A_{_ZPRIMES}
MARK(cons(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
A_{_ZPRIMES} -> A_{_NATS}(s(s(0)))
A_{_ZPRIMES} -> A_{_SIEVE}(a_{_nats}(s(s(0))))
MARK(zprimes) -> A_{_ZPRIMES}
MARK(nats(X)) -> MARK(X)
A_{_NATS}(N) -> MARK(N)
MARK(nats(X)) -> A_{_NATS}(mark(X))
MARK(sieve(X)) -> MARK(X)
A_{_SIEVE}(cons(s(N), Y)) -> MARK(N)
MARK(sieve(X)) -> A_{_SIEVE}(mark(X))
MARK(filter(X1, X2, X3)) -> MARK(X3)
MARK(filter(X1, X2, X3)) -> MARK(X2)
MARK(filter(X1, X2, X3)) -> MARK(X1)
MARK(filter(X1, X2, X3)) -> A_{_FILTER}(mark(X1), mark(X2), mark(X3))
A_{_FILTER}(cons(X, Y), s(N), M) -> MARK(X)

Rules:

a_{_filter}(cons(X, Y), 0, M) -> cons(0, filter(Y, M, M))
a_{_filter}(cons(X, Y), s(N), M) -> cons(mark(X), filter(Y, N, M))
a_{_filter}(X1, X2, X3) -> filter(X1, X2, X3)
a_{_sieve}(cons(0, Y)) -> cons(0, sieve(Y))
a_{_sieve}(cons(s(N), Y)) -> cons(s(mark(N)), sieve(filter(Y, N, N)))
a_{_sieve}(X) -> sieve(X)
a_{_nats}(N) -> cons(mark(N), nats(s(N)))
a_{_nats}(X) -> nats(X)
a_{_zprimes} -> a_{_sieve}(a_{_nats}(s(s(0))))
a_{_zprimes} -> zprimes
mark(filter(X1, X2, X3)) -> a_{_filter}(mark(X1), mark(X2), mark(X3))
mark(sieve(X)) -> a_{_sieve}(mark(X))
mark(nats(X)) -> a_{_nats}(mark(X))
mark(zprimes) -> a_{_zprimes}
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(0) -> 0
mark(s(X)) -> s(mark(X))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:05 minutes