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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__filter#_A(x1,x2,x3) = ((1,0),(1,0)) x1 + ((0,0),(1,0)) x2 + ((0,0),(1,0)) x3 + (1,4)
            cons_A(x1,x2) = ((1,0),(0,0)) x1 + (16,0)
            s_A(x1) = x1 + (17,2)
            mark#_A(x1) = ((1,0),(1,0)) x1 + (1,7)
            zprimes_A() = (72,0)
            a__zprimes#_A() = (69,3)
            a__nats#_A(x1) = x1 + (2,7)
            |0|_A() = (17,4)
            nats_A(x1) = ((1,0),(0,0)) x1 + (16,14)
            mark_A(x1) = ((1,0),(1,0)) x1 + (0,6)
            sieve_A(x1) = ((1,0),(0,0)) x1 + (4,1)
            a__sieve#_A(x1) = ((1,0),(1,0)) x1 + (1,2)
            filter_A(x1,x2,x3) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + ((1,0),(0,0)) x3 + (0,3)
            a__nats_A(x1) = x1 + (16,15)
            a__filter_A(x1,x2,x3) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + ((1,0),(0,0)) x3 + (0,5)
            a__sieve_A(x1) = ((1,0),(0,0)) x1 + (4,9)
            a__zprimes_A() = (72,24)
    
      precedence:
      
        a__sieve# > mark > a__filter > zprimes = a__zprimes > a__nats > cons = sieve = a__sieve > filter > mark# = a__nats# > a__filter# = s = |0| > a__zprimes# = nats
      
      partial status:
    
        pi(a__filter#) = []
        pi(cons) = []
        pi(s) = [1]
        pi(mark#) = []
        pi(zprimes) = []
        pi(a__zprimes#) = []
        pi(a__nats#) = []
        pi(|0|) = []
        pi(nats) = []
        pi(mark) = []
        pi(sieve) = []
        pi(a__sieve#) = []
        pi(filter) = []
        pi(a__nats) = []
        pi(a__filter) = []
        pi(a__sieve) = []
        pi(a__zprimes) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__filter#_A(x1,x2,x3) = (4,22)
            cons_A(x1,x2) = (0,0)
            s_A(x1) = ((1,0),(1,1)) x1 + (6,0)
            mark#_A(x1) = (4,22)
            zprimes_A() = (20,19)
            a__zprimes#_A() = (19,24)
            a__nats#_A(x1) = (4,22)
            |0|_A() = (6,0)
            nats_A(x1) = (0,1)
            mark_A(x1) = (3,21)
            sieve_A(x1) = (1,1)
            a__sieve#_A(x1) = (5,23)
            filter_A(x1,x2,x3) = (5,23)
            a__nats_A(x1) = (1,2)
            a__filter_A(x1,x2,x3) = (1,1)
            a__sieve_A(x1) = (2,20)
            a__zprimes_A() = (20,19)
    
      precedence:
      
        a__zprimes > a__sieve# > sieve > s = nats = mark > a__nats > cons = a__filter = a__sieve > zprimes = a__zprimes# = |0| > a__filter# = mark# = a__nats# = filter
      
      partial status:
    
        pi(a__filter#) = []
        pi(cons) = []
        pi(s) = []
        pi(mark#) = []
        pi(zprimes) = []
        pi(a__zprimes#) = []
        pi(a__nats#) = []
        pi(|0|) = []
        pi(nats) = []
        pi(mark) = []
        pi(sieve) = []
        pi(a__sieve#) = []
        pi(filter) = []
        pi(a__nats) = []
        pi(a__filter) = []
        pi(a__sieve) = []
        pi(a__zprimes) = []
    

The next rules are strictly ordered:

  p2, p3, p4, p5, p7, p10, p16

We remove them from the problem.

-- 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__nats#(N) -> mark#(N)
p3: mark#(nats(X)) -> a__nats#(mark(X))
p4: mark#(sieve(X)) -> mark#(X)
p5: a__sieve#(cons(s(N),Y)) -> mark#(N)
p6: mark#(filter(X1,X2,X3)) -> mark#(X3)
p7: mark#(filter(X1,X2,X3)) -> mark#(X2)
p8: mark#(filter(X1,X2,X3)) -> mark#(X1)
p9: mark#(filter(X1,X2,X3)) -> a__filter#(mark(X1),mark(X2),mark(X3))

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, p6, p7, p8, p9}


-- 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#(filter(X1,X2,X3)) -> a__filter#(mark(X1),mark(X2),mark(X3))
p3: mark#(filter(X1,X2,X3)) -> mark#(X1)
p4: mark#(filter(X1,X2,X3)) -> mark#(X2)
p5: mark#(filter(X1,X2,X3)) -> mark#(X3)
p6: mark#(sieve(X)) -> mark#(X)
p7: mark#(nats(X)) -> a__nats#(mark(X))
p8: a__nats#(N) -> mark#(N)

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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__filter#_A(x1,x2,x3) = x1 + x2 + x3 + (1,3)
            cons_A(x1,x2) = x1 + (0,3)
            s_A(x1) = (6,0)
            mark#_A(x1) = x1 + (0,2)
            filter_A(x1,x2,x3) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + x3 + (2,0)
            mark_A(x1) = ((1,0),(1,1)) x1
            sieve_A(x1) = x1 + (3,1)
            nats_A(x1) = ((1,0),(1,1)) x1 + (2,4)
            a__nats#_A(x1) = x1 + (1,1)
            a__filter_A(x1,x2,x3) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + x3 + (2,2)
            |0|_A() = (3,4)
            a__sieve_A(x1) = x1 + (3,4)
            a__nats_A(x1) = ((1,0),(1,1)) x1 + (2,4)
            a__zprimes_A() = (12,15)
            zprimes_A() = (12,3)
    
      precedence:
      
        a__filter# = mark# = mark = a__nats# = a__nats > cons = a__filter = |0| > nats > filter > a__zprimes > s = a__sieve > sieve > zprimes
      
      partial status:
    
        pi(a__filter#) = [1, 2, 3]
        pi(cons) = [1]
        pi(s) = []
        pi(mark#) = [1]
        pi(filter) = [1, 2, 3]
        pi(mark) = [1]
        pi(sieve) = [1]
        pi(nats) = [1]
        pi(a__nats#) = [1]
        pi(a__filter) = [3]
        pi(|0|) = []
        pi(a__sieve) = [1]
        pi(a__nats) = [1]
        pi(a__zprimes) = []
        pi(zprimes) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__filter#_A(x1,x2,x3) = ((0,0),(1,0)) x1 + x2 + ((0,0),(1,0)) x3
            cons_A(x1,x2) = x1 + (8,13)
            s_A(x1) = (0,3)
            mark#_A(x1) = x1 + (1,12)
            filter_A(x1,x2,x3) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + ((1,0),(1,1)) x3 + (17,4)
            mark_A(x1) = ((1,0),(1,1)) x1 + (7,3)
            sieve_A(x1) = x1 + (8,1)
            nats_A(x1) = ((1,0),(1,1)) x1 + (22,2)
            a__nats#_A(x1) = ((1,0),(0,0)) x1 + (2,13)
            a__filter_A(x1,x2,x3) = ((0,0),(1,0)) x3 + (25,18)
            |0|_A() = (8,19)
            a__sieve_A(x1) = x1 + (8,2)
            a__nats_A(x1) = ((1,0),(1,1)) x1 + (15,16)
            a__zprimes_A() = (23,22)
            zprimes_A() = (17,1)
    
      precedence:
      
        zprimes > filter > s = a__zprimes > nats > a__filter# = cons = mark# = mark = a__nats# = a__filter = |0| = a__sieve = a__nats > sieve
      
      partial status:
    
        pi(a__filter#) = []
        pi(cons) = [1]
        pi(s) = []
        pi(mark#) = [1]
        pi(filter) = [1, 3]
        pi(mark) = [1]
        pi(sieve) = []
        pi(nats) = [1]
        pi(a__nats#) = []
        pi(a__filter) = []
        pi(|0|) = []
        pi(a__sieve) = []
        pi(a__nats) = []
        pi(a__zprimes) = []
        pi(zprimes) = []
    

The next rules are strictly ordered:

  p1, p2, p4, p6, p7, p8

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(filter(X1,X2,X3)) -> mark#(X1)
p2: mark#(filter(X1,X2,X3)) -> mark#(X3)

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}


-- Reduction pair.

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

p1: mark#(filter(X1,X2,X3)) -> mark#(X1)
p2: mark#(filter(X1,X2,X3)) -> mark#(X3)

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

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = x1 + (1,1)
            filter_A(x1,x2,x3) = x1 + x2 + ((1,0),(1,1)) x3 + (2,2)
    
      precedence:
      
        mark# > filter
      
      partial status:
    
        pi(mark#) = [1]
        pi(filter) = [1, 3]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = ((1,0),(1,1)) x1
            filter_A(x1,x2,x3) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x3 + (1,1)
    
      precedence:
      
        filter > mark#
      
      partial status:
    
        pi(mark#) = [1]
        pi(filter) = []
    

The next rules are strictly ordered:

  p1, p2

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