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

         ↳Narrowing Transformation

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

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

A_{_ZPRIMES} -> A_{_SIEVE}(a_{_nats}(s(s(0))))

two new Dependency Pairs are created:

A_{_ZPRIMES} -> A_{_SIEVE}(cons(mark(s(s(0))), nats(s(s(s(0))))))
A_{_ZPRIMES} -> A_{_SIEVE}(nats(s(s(0))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

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

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

MARK(filter(X1, X2, X3)) -> A_{_FILTER}(mark(X1), mark(X2), mark(X3))

21 new Dependency Pairs are created:

MARK(filter(filter(X1'', X2'', X3''), X2, X3)) -> A_{_FILTER}(a_{_filter}(mark(X1''), mark(X2''), mark(X3'')), mark(X2), mark(X3))
MARK(filter(sieve(X'), X2, X3)) -> A_{_FILTER}(a_{_sieve}(mark(X')), mark(X2), mark(X3))
MARK(filter(nats(X'), X2, X3)) -> A_{_FILTER}(a_{_nats}(mark(X')), mark(X2), mark(X3))
MARK(filter(zprimes, X2, X3)) -> A_{_FILTER}(a_{_zprimes}, mark(X2), mark(X3))
MARK(filter(cons(X1'', X2''), X2, X3)) -> A_{_FILTER}(cons(mark(X1''), X2''), mark(X2), mark(X3))
MARK(filter(0, X2, X3)) -> A_{_FILTER}(0, mark(X2), mark(X3))
MARK(filter(s(X'), X2, X3)) -> A_{_FILTER}(s(mark(X')), mark(X2), mark(X3))
MARK(filter(X1, filter(X1'', X2'', X3''), X3)) -> A_{_FILTER}(mark(X1), a_{_filter}(mark(X1''), mark(X2''), mark(X3'')), mark(X3))
MARK(filter(X1, sieve(X'), X3)) -> A_{_FILTER}(mark(X1), a_{_sieve}(mark(X')), mark(X3))
MARK(filter(X1, nats(X'), X3)) -> A_{_FILTER}(mark(X1), a_{_nats}(mark(X')), mark(X3))
MARK(filter(X1, zprimes, X3)) -> A_{_FILTER}(mark(X1), a_{_zprimes}, mark(X3))
MARK(filter(X1, cons(X1'', X2''), X3)) -> A_{_FILTER}(mark(X1), cons(mark(X1''), X2''), mark(X3))
MARK(filter(X1, 0, X3)) -> A_{_FILTER}(mark(X1), 0, mark(X3))
MARK(filter(X1, s(X'), X3)) -> A_{_FILTER}(mark(X1), s(mark(X')), mark(X3))
MARK(filter(X1, X2, filter(X1'', X2'', X3''))) -> A_{_FILTER}(mark(X1), mark(X2), a_{_filter}(mark(X1''), mark(X2''), mark(X3'')))
MARK(filter(X1, X2, sieve(X'))) -> A_{_FILTER}(mark(X1), mark(X2), a_{_sieve}(mark(X')))
MARK(filter(X1, X2, nats(X'))) -> A_{_FILTER}(mark(X1), mark(X2), a_{_nats}(mark(X')))
MARK(filter(X1, X2, zprimes)) -> A_{_FILTER}(mark(X1), mark(X2), a_{_zprimes})
MARK(filter(X1, X2, cons(X1'', X2''))) -> A_{_FILTER}(mark(X1), mark(X2), cons(mark(X1''), X2''))
MARK(filter(X1, X2, 0)) -> A_{_FILTER}(mark(X1), mark(X2), 0)
MARK(filter(X1, X2, s(X'))) -> A_{_FILTER}(mark(X1), mark(X2), s(mark(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:

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

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

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

MARK(sieve(X)) -> A_{_SIEVE}(mark(X))

seven new Dependency Pairs are created:

MARK(sieve(filter(X1', X2', X3'))) -> A_{_SIEVE}(a_{_filter}(mark(X1'), mark(X2'), mark(X3')))
MARK(sieve(sieve(X''))) -> A_{_SIEVE}(a_{_sieve}(mark(X'')))
MARK(sieve(nats(X''))) -> A_{_SIEVE}(a_{_nats}(mark(X'')))
MARK(sieve(zprimes)) -> A_{_SIEVE}(a_{_zprimes})
MARK(sieve(cons(X1', X2'))) -> A_{_SIEVE}(cons(mark(X1'), X2'))
MARK(sieve(0)) -> A_{_SIEVE}(0)
MARK(sieve(s(X''))) -> A_{_SIEVE}(s(mark(X'')))

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:

MARK(sieve(cons(X1', X2'))) -> A_{_SIEVE}(cons(mark(X1'), X2'))
MARK(sieve(zprimes)) -> A_{_SIEVE}(a_{_zprimes})
MARK(sieve(nats(X''))) -> A_{_SIEVE}(a_{_nats}(mark(X'')))
MARK(sieve(sieve(X''))) -> A_{_SIEVE}(a_{_sieve}(mark(X'')))
MARK(sieve(filter(X1', X2', X3'))) -> A_{_SIEVE}(a_{_filter}(mark(X1'), mark(X2'), mark(X3')))
MARK(filter(X1, X2, 0)) -> A_{_FILTER}(mark(X1), mark(X2), 0)
MARK(filter(X1, X2, cons(X1'', X2''))) -> A_{_FILTER}(mark(X1), mark(X2), cons(mark(X1''), X2''))
MARK(filter(X1, X2, zprimes)) -> A_{_FILTER}(mark(X1), mark(X2), a_{_zprimes})
MARK(filter(X1, X2, nats(X'))) -> A_{_FILTER}(mark(X1), mark(X2), a_{_nats}(mark(X')))
MARK(filter(X1, X2, sieve(X'))) -> A_{_FILTER}(mark(X1), mark(X2), a_{_sieve}(mark(X')))
MARK(filter(X1, X2, filter(X1'', X2'', X3''))) -> A_{_FILTER}(mark(X1), mark(X2), a_{_filter}(mark(X1''), mark(X2''), mark(X3'')))
MARK(filter(X1, s(X'), X3)) -> A_{_FILTER}(mark(X1), s(mark(X')), mark(X3))
MARK(filter(X1, zprimes, X3)) -> A_{_FILTER}(mark(X1), a_{_zprimes}, mark(X3))
MARK(filter(X1, nats(X'), X3)) -> A_{_FILTER}(mark(X1), a_{_nats}(mark(X')), mark(X3))
MARK(filter(X1, sieve(X'), X3)) -> A_{_FILTER}(mark(X1), a_{_sieve}(mark(X')), mark(X3))
MARK(filter(X1, filter(X1'', X2'', X3''), X3)) -> A_{_FILTER}(mark(X1), a_{_filter}(mark(X1''), mark(X2''), mark(X3'')), mark(X3))
MARK(filter(cons(X1'', X2''), X2, X3)) -> A_{_FILTER}(cons(mark(X1''), X2''), mark(X2), mark(X3))
MARK(filter(zprimes, X2, X3)) -> A_{_FILTER}(a_{_zprimes}, mark(X2), mark(X3))
MARK(filter(nats(X'), X2, X3)) -> A_{_FILTER}(a_{_nats}(mark(X')), mark(X2), mark(X3))
MARK(filter(sieve(X'), X2, X3)) -> A_{_FILTER}(a_{_sieve}(mark(X')), mark(X2), mark(X3))
MARK(filter(filter(X1'', X2'', X3''), X2, X3)) -> A_{_FILTER}(a_{_filter}(mark(X1''), mark(X2''), mark(X3'')), mark(X2), mark(X3))
MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
A_{_SIEVE}(cons(s(N), Y)) -> MARK(N)
A_{_ZPRIMES} -> A_{_SIEVE}(cons(mark(s(s(0))), nats(s(s(s(0))))))
A_{_ZPRIMES} -> 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)
MARK(filter(X1, X2, X3)) -> MARK(X3)
MARK(filter(X1, X2, X3)) -> MARK(X2)
MARK(filter(X1, X2, X3)) -> MARK(X1)
A_{_FILTER}(cons(X, Y), s(N), M) -> MARK(X)
MARK(filter(X1, X2, s(X'))) -> A_{_FILTER}(mark(X1), mark(X2), s(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:
0:56 minutes