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. 



   R
     ↳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. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial 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: Polynomial ordering with Polynomial interpretation:
  POL(FILTER(x1, x2, x3))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  1 + x2  
  POL(s(x1))=  0  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →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.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)




The following remains to be proven:

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