YES

We show the termination of the TRS R:

  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__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
  a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
  a__nats(N) -> cons(mark(N),nats(s(N)))
  a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
  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))
  a__filter(X1,X2,X3) -> filter(X1,X2,X3)
  a__sieve(X) -> sieve(X)
  a__nats(X) -> nats(X)
  a__zprimes() -> zprimes()

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__filter#(cons(X,Y),s(N),M) -> mark#(X)
p2: a__sieve#(cons(s(N),Y)) -> mark#(N)
p3: a__nats#(N) -> mark#(N)
p4: a__zprimes#() -> a__sieve#(a__nats(s(s(|0|()))))
p5: a__zprimes#() -> a__nats#(s(s(|0|())))
p6: mark#(filter(X1,X2,X3)) -> a__filter#(mark(X1),mark(X2),mark(X3))
p7: mark#(filter(X1,X2,X3)) -> mark#(X1)
p8: mark#(filter(X1,X2,X3)) -> mark#(X2)
p9: mark#(filter(X1,X2,X3)) -> mark#(X3)
p10: mark#(sieve(X)) -> a__sieve#(mark(X))
p11: mark#(sieve(X)) -> mark#(X)
p12: mark#(nats(X)) -> a__nats#(mark(X))
p13: mark#(nats(X)) -> mark#(X)
p14: mark#(zprimes()) -> a__zprimes#()
p15: mark#(cons(X1,X2)) -> mark#(X1)
p16: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: a__filter#(cons(X,Y),s(N),M) -> mark#(X)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(zprimes()) -> a__zprimes#()
p5: a__zprimes#() -> a__nats#(s(s(|0|())))
p6: a__nats#(N) -> mark#(N)
p7: mark#(nats(X)) -> mark#(X)
p8: mark#(nats(X)) -> a__nats#(mark(X))
p9: mark#(sieve(X)) -> mark#(X)
p10: mark#(sieve(X)) -> a__sieve#(mark(X))
p11: a__sieve#(cons(s(N),Y)) -> mark#(N)
p12: mark#(filter(X1,X2,X3)) -> mark#(X3)
p13: mark#(filter(X1,X2,X3)) -> mark#(X2)
p14: mark#(filter(X1,X2,X3)) -> mark#(X1)
p15: mark#(filter(X1,X2,X3)) -> a__filter#(mark(X1),mark(X2),mark(X3))
p16: a__zprimes#() -> a__sieve#(a__nats(s(s(|0|()))))

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        a__filter#_A(x1,x2,x3) = ((1,0,0),(0,1,0),(0,1,1)) x1 + ((1,0,0),(0,1,0),(0,1,1)) x2 + ((1,0,0),(0,1,0),(1,1,1)) x3
        cons_A(x1,x2) = x1 + (4,1,1)
        s_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (1,1,11)
        mark#_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (4,3,31)
        zprimes_A() = (11,1,1)
        a__zprimes#_A() = (9,16,45)
        a__nats#_A(x1) = ((1,0,0),(1,0,0),(0,0,0)) x1 + (5,14,55)
        |0|_A() = (1,1,15)
        nats_A(x1) = ((1,0,0),(1,1,0),(0,0,0)) x1 + (5,6,13)
        mark_A(x1) = ((1,0,0),(1,1,0),(1,1,0)) x1 + (0,8,12)
        sieve_A(x1) = ((1,0,0),(1,1,0),(0,0,1)) x1 + (2,3,13)
        a__sieve#_A(x1) = x1 + (0,0,30)
        filter_A(x1,x2,x3) = ((1,0,0),(1,1,0),(1,1,1)) x1 + ((1,0,0),(0,1,0),(0,1,1)) x2 + ((1,0,0),(1,1,0),(1,0,1)) x3 + (26,1,1)
        a__nats_A(x1) = ((1,0,0),(1,1,0),(1,0,1)) x1 + (5,10,12)
        a__filter_A(x1,x2,x3) = ((1,0,0),(1,1,0),(0,0,1)) x1 + x2 + ((1,0,0),(1,1,0),(1,0,1)) x3 + (26,2,2)
        a__sieve_A(x1) = ((1,0,0),(1,1,0),(1,0,0)) x1 + (2,4,12)
        a__zprimes_A() = (11,20,2)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        filter > mark# > a__nats > mark > a__nats# > a__zprimes > a__filter# > nats > a__sieve# > a__sieve > a__zprimes# > a__filter > sieve > |0| > cons > zprimes > s
      
      argument filter:
    
        pi(a__filter#) = [1, 2, 3]
        pi(cons) = [1]
        pi(s) = []
        pi(mark#) = 1
        pi(zprimes) = []
        pi(a__zprimes#) = []
        pi(a__nats#) = []
        pi(|0|) = []
        pi(nats) = []
        pi(mark) = []
        pi(sieve) = []
        pi(a__sieve#) = []
        pi(filter) = [2, 3]
        pi(a__nats) = 1
        pi(a__filter) = [1, 2, 3]
        pi(a__sieve) = []
        pi(a__zprimes) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16

We remove them from the problem.  Then no dependency pair remains.