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

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
         ↳Negative Polynomial Order


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





The following Dependency Pairs can be strictly oriented using the given order.

SIEVE(cons(s(N), Y)) -> ACTIVATE(Y)
SIEVE(cons(s(N), Y)) -> FILTER(activate(Y), N, N)
SIEVE(cons(0, Y)) -> ACTIVATE(Y)


Moreover, the following usable rules (regarding the implicit AFS) are oriented.

activate(nfilter(X1, X2, X3)) -> filter(X1, X2, X3)
activate(nsieve(X)) -> sieve(X)
activate(nnats(X)) -> nats(X)
activate(X) -> X
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)


Used ordering:
Polynomial Order with Interpretation:

POL( SIEVE(x1) ) = x1 + 1

POL( cons(x1, x2) ) = x2

POL( ACTIVATE(x1) ) = x1

POL( nsieve(x1) ) = x1 + 1

POL( FILTER(x1, ..., x3) ) = x1

POL( nfilter(x1, ..., x3) ) = x1

POL( activate(x1) ) = x1

POL( filter(x1, ..., x3) ) = x1

POL( sieve(x1) ) = x1 + 1

POL( nnats(x1) ) = 0

POL( nats(x1) ) = 0


This results in one new DP problem. 


   R
     ↳DPs
       →DP Problem 1
         ↳Neg POLO
           →DP Problem 2
             ↳Dependency Graph


Dependency Pairs:

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





Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Neg POLO
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 3
                 ↳Size-Change Principle


Dependency Pairs:

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





We number the DPs as follows:
  1. FILTER(cons(X, Y), 0, M) -> ACTIVATE(Y)
  2. ACTIVATE(nfilter(X1, X2, X3)) -> FILTER(X1, X2, X3)
  3. FILTER(cons(X, Y), s(N), M) -> ACTIVATE(Y)
and get the following Size-Change Graph(s):
{1, 3} , {1, 3}
1>1
{2} , {2}
1>1
1>2
1>3

which lead(s) to this/these maximal multigraph(s):
{2} , {1, 3}
1>1
{1, 3} , {2}
1>1
1>2
1>3

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
nfilter(x1, x2, x3) -> nfilter(x1, x2, x3)
cons(x1, x2) -> cons(x1, x2)
s(x1) -> s(x1)

We obtain no new DP problems.

Termination of R successfully shown.
Duration:
0:01 minutes