Term Rewriting System R:
[X, Y, M, N, X1, X2, X3]
filter(cons(X, Y), 0, M) -> cons(0, nfilter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) -> cons(X, nfilter(activate(Y), N, M))
filter(X1, X2, X3) -> nfilter(X1, X2, X3)
sieve(cons(0, Y)) -> cons(0, nsieve(activate(Y)))
sieve(cons(s(N), Y)) -> cons(s(N), nsieve(filter(activate(Y), N, N)))
sieve(X) -> nsieve(X)
nats(N) -> cons(N, nnats(s(N)))
nats(X) -> nnats(X)
zprimes -> sieve(nats(s(s(0))))
activate(nfilter(X1, X2, X3)) -> filter(X1, X2, X3)
activate(nsieve(X)) -> sieve(X)
activate(nnats(X)) -> nats(X)
activate(X) -> X

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

FILTER(cons(X, Y), 0, M) -> ACTIVATE(Y)
FILTER(cons(X, Y), s(N), M) -> ACTIVATE(Y)
SIEVE(cons(0, Y)) -> ACTIVATE(Y)
SIEVE(cons(s(N), Y)) -> FILTER(activate(Y), N, N)
SIEVE(cons(s(N), Y)) -> ACTIVATE(Y)
ZPRIMES -> SIEVE(nats(s(s(0))))
ZPRIMES -> NATS(s(s(0)))
ACTIVATE(nfilter(X1, X2, X3)) -> FILTER(X1, X2, X3)
ACTIVATE(nsieve(X)) -> SIEVE(X)
ACTIVATE(nnats(X)) -> NATS(X)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

SIEVE(cons(s(N), Y)) -> ACTIVATE(Y)
SIEVE(cons(s(N), Y)) -> FILTER(activate(Y), N, N)
SIEVE(cons(0, Y)) -> ACTIVATE(Y)
ACTIVATE(nsieve(X)) -> SIEVE(X)
FILTER(cons(X, Y), s(N), M) -> ACTIVATE(Y)
ACTIVATE(nfilter(X1, X2, X3)) -> FILTER(X1, X2, X3)
FILTER(cons(X, Y), 0, M) -> ACTIVATE(Y)


Rules:


filter(cons(X, Y), 0, M) -> cons(0, nfilter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) -> cons(X, nfilter(activate(Y), N, M))
filter(X1, X2, X3) -> nfilter(X1, X2, X3)
sieve(cons(0, Y)) -> cons(0, nsieve(activate(Y)))
sieve(cons(s(N), Y)) -> cons(s(N), nsieve(filter(activate(Y), N, N)))
sieve(X) -> nsieve(X)
nats(N) -> cons(N, nnats(s(N)))
nats(X) -> nnats(X)
zprimes -> sieve(nats(s(s(0))))
activate(nfilter(X1, X2, X3)) -> filter(X1, X2, X3)
activate(nsieve(X)) -> sieve(X)
activate(nnats(X)) -> nats(X)
activate(X) -> X


Strategy:

innermost




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

SIEVE(cons(s(N), Y)) -> FILTER(activate(Y), N, N)
four new Dependency Pairs are created:

SIEVE(cons(s(N), nfilter(X1', X2', X3'))) -> FILTER(filter(X1', X2', X3'), N, N)
SIEVE(cons(s(N), nsieve(X'))) -> FILTER(sieve(X'), N, N)
SIEVE(cons(s(N), nnats(X'))) -> FILTER(nats(X'), N, N)
SIEVE(cons(s(N), Y')) -> FILTER(Y', N, N)

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

SIEVE(cons(s(N), Y')) -> FILTER(Y', N, N)
SIEVE(cons(s(N), nnats(X'))) -> FILTER(nats(X'), N, N)
SIEVE(cons(s(N), nsieve(X'))) -> FILTER(sieve(X'), N, N)
FILTER(cons(X, Y), s(N), M) -> ACTIVATE(Y)
SIEVE(cons(s(N), nfilter(X1', X2', X3'))) -> FILTER(filter(X1', X2', X3'), N, N)
SIEVE(cons(0, Y)) -> ACTIVATE(Y)
ACTIVATE(nsieve(X)) -> SIEVE(X)
FILTER(cons(X, Y), 0, M) -> ACTIVATE(Y)
ACTIVATE(nfilter(X1, X2, X3)) -> FILTER(X1, X2, X3)
SIEVE(cons(s(N), Y)) -> ACTIVATE(Y)


Rules:


filter(cons(X, Y), 0, M) -> cons(0, nfilter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) -> cons(X, nfilter(activate(Y), N, M))
filter(X1, X2, X3) -> nfilter(X1, X2, X3)
sieve(cons(0, Y)) -> cons(0, nsieve(activate(Y)))
sieve(cons(s(N), Y)) -> cons(s(N), nsieve(filter(activate(Y), N, N)))
sieve(X) -> nsieve(X)
nats(N) -> cons(N, nnats(s(N)))
nats(X) -> nnats(X)
zprimes -> sieve(nats(s(s(0))))
activate(nfilter(X1, X2, X3)) -> filter(X1, X2, X3)
activate(nsieve(X)) -> sieve(X)
activate(nnats(X)) -> nats(X)
activate(X) -> X


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:02 minutes