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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__filter#_A(x1,x2,x3) = max{42, x1 + 26, x2 - 16, x3 + 15}
          cons_A(x1,x2) = x1 + 16
          s_A(x1) = max{34, x1 + 33}
          mark#_A(x1) = x1 + 16
          zprimes_A = 123
          a__zprimes#_A = 122
          a__nats#_A(x1) = max{94, x1 + 18}
          |0|_A = 9
          nats_A(x1) = max{93, x1 + 32}
          mark_A(x1) = x1 + 15
          sieve_A(x1) = x1 + 28
          a__sieve#_A(x1) = max{27, x1 + 13}
          filter_A(x1,x2,x3) = max{x1 + 26, x2, x3 + 26}
          a__nats_A(x1) = max{108, x1 + 32}
          a__filter_A(x1,x2,x3) = max{41, x1 + 26, x2, x3 + 26}
          a__sieve_A(x1) = max{35, x1 + 28}
          a__zprimes_A = 136
    
      precedence:
      
        a__zprimes# = a__nats# = |0| = a__sieve# = a__zprimes > s = zprimes = mark = a__nats = a__filter > cons > a__filter# = mark# = nats = filter = a__sieve > sieve
      
      partial status:
    
        pi(a__filter#) = [1, 3]
        pi(cons) = [1]
        pi(s) = [1]
        pi(mark#) = [1]
        pi(zprimes) = []
        pi(a__zprimes#) = []
        pi(a__nats#) = [1]
        pi(|0|) = []
        pi(nats) = [1]
        pi(mark) = [1]
        pi(sieve) = [1]
        pi(a__sieve#) = [1]
        pi(filter) = [1, 2, 3]
        pi(a__nats) = [1]
        pi(a__filter) = [1, 2, 3]
        pi(a__sieve) = [1]
        pi(a__zprimes) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__filter#_A(x1,x2,x3) = max{x1 + 89, x3 + 53}
          cons_A(x1,x2) = x1 + 13
          s_A(x1) = x1 + 12
          mark#_A(x1) = max{52, x1 + 51}
          zprimes_A = 59
          a__zprimes#_A = 109
          a__nats#_A(x1) = max{56, x1 + 53}
          |0|_A = 12
          nats_A(x1) = x1 + 9
          mark_A(x1) = max{7, x1}
          sieve_A(x1) = max{28, x1 + 21}
          a__sieve#_A(x1) = max{71, x1 + 64}
          filter_A(x1,x2,x3) = max{x1 + 46, x2 + 50, x3 + 48}
          a__nats_A(x1) = max{10, x1 + 8}
          a__filter_A(x1,x2,x3) = max{x1 + 47, x2 + 51, x3 + 49}
          a__sieve_A(x1) = max{29, x1 + 22}
          a__zprimes_A = 8
    
      precedence:
      
        a__filter# = cons = s = mark# = zprimes = a__zprimes# = a__nats# = |0| = nats = mark = sieve = a__sieve# = filter = a__nats = a__filter = a__sieve = a__zprimes
      
      partial status:
    
        pi(a__filter#) = []
        pi(cons) = [1]
        pi(s) = [1]
        pi(mark#) = [1]
        pi(zprimes) = []
        pi(a__zprimes#) = []
        pi(a__nats#) = []
        pi(|0|) = []
        pi(nats) = [1]
        pi(mark) = [1]
        pi(sieve) = [1]
        pi(a__sieve#) = [1]
        pi(filter) = [3]
        pi(a__nats) = [1]
        pi(a__filter) = []
        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.