YES

We show the termination of the TRS R:

  active(filter(cons(X,Y),|0|(),M)) -> mark(cons(|0|(),filter(Y,M,M)))
  active(filter(cons(X,Y),s(N),M)) -> mark(cons(X,filter(Y,N,M)))
  active(sieve(cons(|0|(),Y))) -> mark(cons(|0|(),sieve(Y)))
  active(sieve(cons(s(N),Y))) -> mark(cons(s(N),sieve(filter(Y,N,N))))
  active(nats(N)) -> mark(cons(N,nats(s(N))))
  active(zprimes()) -> mark(sieve(nats(s(s(|0|())))))
  mark(filter(X1,X2,X3)) -> active(filter(mark(X1),mark(X2),mark(X3)))
  mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
  mark(|0|()) -> active(|0|())
  mark(s(X)) -> active(s(mark(X)))
  mark(sieve(X)) -> active(sieve(mark(X)))
  mark(nats(X)) -> active(nats(mark(X)))
  mark(zprimes()) -> active(zprimes())
  filter(mark(X1),X2,X3) -> filter(X1,X2,X3)
  filter(X1,mark(X2),X3) -> filter(X1,X2,X3)
  filter(X1,X2,mark(X3)) -> filter(X1,X2,X3)
  filter(active(X1),X2,X3) -> filter(X1,X2,X3)
  filter(X1,active(X2),X3) -> filter(X1,X2,X3)
  filter(X1,X2,active(X3)) -> filter(X1,X2,X3)
  cons(mark(X1),X2) -> cons(X1,X2)
  cons(X1,mark(X2)) -> cons(X1,X2)
  cons(active(X1),X2) -> cons(X1,X2)
  cons(X1,active(X2)) -> cons(X1,X2)
  s(mark(X)) -> s(X)
  s(active(X)) -> s(X)
  sieve(mark(X)) -> sieve(X)
  sieve(active(X)) -> sieve(X)
  nats(mark(X)) -> nats(X)
  nats(active(X)) -> nats(X)

-- SCC decomposition.

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

p1: active#(filter(cons(X,Y),|0|(),M)) -> mark#(cons(|0|(),filter(Y,M,M)))
p2: active#(filter(cons(X,Y),|0|(),M)) -> cons#(|0|(),filter(Y,M,M))
p3: active#(filter(cons(X,Y),|0|(),M)) -> filter#(Y,M,M)
p4: active#(filter(cons(X,Y),s(N),M)) -> mark#(cons(X,filter(Y,N,M)))
p5: active#(filter(cons(X,Y),s(N),M)) -> cons#(X,filter(Y,N,M))
p6: active#(filter(cons(X,Y),s(N),M)) -> filter#(Y,N,M)
p7: active#(sieve(cons(|0|(),Y))) -> mark#(cons(|0|(),sieve(Y)))
p8: active#(sieve(cons(|0|(),Y))) -> cons#(|0|(),sieve(Y))
p9: active#(sieve(cons(|0|(),Y))) -> sieve#(Y)
p10: active#(sieve(cons(s(N),Y))) -> mark#(cons(s(N),sieve(filter(Y,N,N))))
p11: active#(sieve(cons(s(N),Y))) -> cons#(s(N),sieve(filter(Y,N,N)))
p12: active#(sieve(cons(s(N),Y))) -> sieve#(filter(Y,N,N))
p13: active#(sieve(cons(s(N),Y))) -> filter#(Y,N,N)
p14: active#(nats(N)) -> mark#(cons(N,nats(s(N))))
p15: active#(nats(N)) -> cons#(N,nats(s(N)))
p16: active#(nats(N)) -> nats#(s(N))
p17: active#(nats(N)) -> s#(N)
p18: active#(zprimes()) -> mark#(sieve(nats(s(s(|0|())))))
p19: active#(zprimes()) -> sieve#(nats(s(s(|0|()))))
p20: active#(zprimes()) -> nats#(s(s(|0|())))
p21: active#(zprimes()) -> s#(s(|0|()))
p22: active#(zprimes()) -> s#(|0|())
p23: mark#(filter(X1,X2,X3)) -> active#(filter(mark(X1),mark(X2),mark(X3)))
p24: mark#(filter(X1,X2,X3)) -> filter#(mark(X1),mark(X2),mark(X3))
p25: mark#(filter(X1,X2,X3)) -> mark#(X1)
p26: mark#(filter(X1,X2,X3)) -> mark#(X2)
p27: mark#(filter(X1,X2,X3)) -> mark#(X3)
p28: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p29: mark#(cons(X1,X2)) -> cons#(mark(X1),X2)
p30: mark#(cons(X1,X2)) -> mark#(X1)
p31: mark#(|0|()) -> active#(|0|())
p32: mark#(s(X)) -> active#(s(mark(X)))
p33: mark#(s(X)) -> s#(mark(X))
p34: mark#(s(X)) -> mark#(X)
p35: mark#(sieve(X)) -> active#(sieve(mark(X)))
p36: mark#(sieve(X)) -> sieve#(mark(X))
p37: mark#(sieve(X)) -> mark#(X)
p38: mark#(nats(X)) -> active#(nats(mark(X)))
p39: mark#(nats(X)) -> nats#(mark(X))
p40: mark#(nats(X)) -> mark#(X)
p41: mark#(zprimes()) -> active#(zprimes())
p42: filter#(mark(X1),X2,X3) -> filter#(X1,X2,X3)
p43: filter#(X1,mark(X2),X3) -> filter#(X1,X2,X3)
p44: filter#(X1,X2,mark(X3)) -> filter#(X1,X2,X3)
p45: filter#(active(X1),X2,X3) -> filter#(X1,X2,X3)
p46: filter#(X1,active(X2),X3) -> filter#(X1,X2,X3)
p47: filter#(X1,X2,active(X3)) -> filter#(X1,X2,X3)
p48: cons#(mark(X1),X2) -> cons#(X1,X2)
p49: cons#(X1,mark(X2)) -> cons#(X1,X2)
p50: cons#(active(X1),X2) -> cons#(X1,X2)
p51: cons#(X1,active(X2)) -> cons#(X1,X2)
p52: s#(mark(X)) -> s#(X)
p53: s#(active(X)) -> s#(X)
p54: sieve#(mark(X)) -> sieve#(X)
p55: sieve#(active(X)) -> sieve#(X)
p56: nats#(mark(X)) -> nats#(X)
p57: nats#(active(X)) -> nats#(X)

and R consists of:

r1: active(filter(cons(X,Y),|0|(),M)) -> mark(cons(|0|(),filter(Y,M,M)))
r2: active(filter(cons(X,Y),s(N),M)) -> mark(cons(X,filter(Y,N,M)))
r3: active(sieve(cons(|0|(),Y))) -> mark(cons(|0|(),sieve(Y)))
r4: active(sieve(cons(s(N),Y))) -> mark(cons(s(N),sieve(filter(Y,N,N))))
r5: active(nats(N)) -> mark(cons(N,nats(s(N))))
r6: active(zprimes()) -> mark(sieve(nats(s(s(|0|())))))
r7: mark(filter(X1,X2,X3)) -> active(filter(mark(X1),mark(X2),mark(X3)))
r8: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r9: mark(|0|()) -> active(|0|())
r10: mark(s(X)) -> active(s(mark(X)))
r11: mark(sieve(X)) -> active(sieve(mark(X)))
r12: mark(nats(X)) -> active(nats(mark(X)))
r13: mark(zprimes()) -> active(zprimes())
r14: filter(mark(X1),X2,X3) -> filter(X1,X2,X3)
r15: filter(X1,mark(X2),X3) -> filter(X1,X2,X3)
r16: filter(X1,X2,mark(X3)) -> filter(X1,X2,X3)
r17: filter(active(X1),X2,X3) -> filter(X1,X2,X3)
r18: filter(X1,active(X2),X3) -> filter(X1,X2,X3)
r19: filter(X1,X2,active(X3)) -> filter(X1,X2,X3)
r20: cons(mark(X1),X2) -> cons(X1,X2)
r21: cons(X1,mark(X2)) -> cons(X1,X2)
r22: cons(active(X1),X2) -> cons(X1,X2)
r23: cons(X1,active(X2)) -> cons(X1,X2)
r24: s(mark(X)) -> s(X)
r25: s(active(X)) -> s(X)
r26: sieve(mark(X)) -> sieve(X)
r27: sieve(active(X)) -> sieve(X)
r28: nats(mark(X)) -> nats(X)
r29: nats(active(X)) -> nats(X)

The estimated dependency graph contains the following SCCs:

  {p1, p4, p7, p10, p14, p18, p23, p25, p26, p27, p28, p30, p32, p34, p35, p37, p38, p40, p41}
  {p48, p49, p50, p51}
  {p42, p43, p44, p45, p46, p47}
  {p54, p55}
  {p56, p57}
  {p52, p53}


-- Reduction pair.

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

p1: active#(filter(cons(X,Y),|0|(),M)) -> mark#(cons(|0|(),filter(Y,M,M)))
p2: mark#(zprimes()) -> active#(zprimes())
p3: active#(zprimes()) -> mark#(sieve(nats(s(s(|0|())))))
p4: mark#(sieve(X)) -> mark#(X)
p5: mark#(nats(X)) -> mark#(X)
p6: mark#(nats(X)) -> active#(nats(mark(X)))
p7: active#(nats(N)) -> mark#(cons(N,nats(s(N))))
p8: mark#(sieve(X)) -> active#(sieve(mark(X)))
p9: active#(sieve(cons(s(N),Y))) -> mark#(cons(s(N),sieve(filter(Y,N,N))))
p10: mark#(s(X)) -> mark#(X)
p11: mark#(s(X)) -> active#(s(mark(X)))
p12: active#(sieve(cons(|0|(),Y))) -> mark#(cons(|0|(),sieve(Y)))
p13: mark#(cons(X1,X2)) -> mark#(X1)
p14: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p15: active#(filter(cons(X,Y),s(N),M)) -> mark#(cons(X,filter(Y,N,M)))
p16: mark#(filter(X1,X2,X3)) -> mark#(X3)
p17: mark#(filter(X1,X2,X3)) -> mark#(X2)
p18: mark#(filter(X1,X2,X3)) -> mark#(X1)
p19: mark#(filter(X1,X2,X3)) -> active#(filter(mark(X1),mark(X2),mark(X3)))

and R consists of:

r1: active(filter(cons(X,Y),|0|(),M)) -> mark(cons(|0|(),filter(Y,M,M)))
r2: active(filter(cons(X,Y),s(N),M)) -> mark(cons(X,filter(Y,N,M)))
r3: active(sieve(cons(|0|(),Y))) -> mark(cons(|0|(),sieve(Y)))
r4: active(sieve(cons(s(N),Y))) -> mark(cons(s(N),sieve(filter(Y,N,N))))
r5: active(nats(N)) -> mark(cons(N,nats(s(N))))
r6: active(zprimes()) -> mark(sieve(nats(s(s(|0|())))))
r7: mark(filter(X1,X2,X3)) -> active(filter(mark(X1),mark(X2),mark(X3)))
r8: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r9: mark(|0|()) -> active(|0|())
r10: mark(s(X)) -> active(s(mark(X)))
r11: mark(sieve(X)) -> active(sieve(mark(X)))
r12: mark(nats(X)) -> active(nats(mark(X)))
r13: mark(zprimes()) -> active(zprimes())
r14: filter(mark(X1),X2,X3) -> filter(X1,X2,X3)
r15: filter(X1,mark(X2),X3) -> filter(X1,X2,X3)
r16: filter(X1,X2,mark(X3)) -> filter(X1,X2,X3)
r17: filter(active(X1),X2,X3) -> filter(X1,X2,X3)
r18: filter(X1,active(X2),X3) -> filter(X1,X2,X3)
r19: filter(X1,X2,active(X3)) -> filter(X1,X2,X3)
r20: cons(mark(X1),X2) -> cons(X1,X2)
r21: cons(X1,mark(X2)) -> cons(X1,X2)
r22: cons(active(X1),X2) -> cons(X1,X2)
r23: cons(X1,active(X2)) -> cons(X1,X2)
r24: s(mark(X)) -> s(X)
r25: s(active(X)) -> s(X)
r26: sieve(mark(X)) -> sieve(X)
r27: sieve(active(X)) -> sieve(X)
r28: nats(mark(X)) -> nats(X)
r29: nats(active(X)) -> nats(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = max{80, x1 - 87}
          filter_A(x1,x2,x3) = max{334, x1 + 151, x2 + 180, x3 + 122}
          cons_A(x1,x2) = max{88, x1 + 74}
          |0|_A = 8
          mark#_A(x1) = max{81, x1 + 29}
          zprimes_A = 996
          sieve_A(x1) = x1 + 452
          nats_A(x1) = max{297, x1 + 191}
          s_A(x1) = max{132, x1 + 105}
          mark_A(x1) = x1 + 102
          active_A(x1) = max{106, x1}
    
      precedence:
      
        mark > filter > active# = cons = |0| = zprimes = nats = s = active > mark# = sieve
      
      partial status:
    
        pi(active#) = []
        pi(filter) = [2]
        pi(cons) = []
        pi(|0|) = []
        pi(mark#) = [1]
        pi(zprimes) = []
        pi(sieve) = [1]
        pi(nats) = [1]
        pi(s) = [1]
        pi(mark) = []
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = 61
          filter_A(x1,x2,x3) = 19
          cons_A(x1,x2) = 9
          |0|_A = 155
          mark#_A(x1) = x1 + 38
          zprimes_A = 183
          sieve_A(x1) = x1 + 64
          nats_A(x1) = x1 + 123
          s_A(x1) = 60
          mark_A(x1) = 58
          active_A(x1) = 58
    
      precedence:
      
        |0| = zprimes > s > active# = sieve = mark = active > cons = mark# = nats > filter
      
      partial status:
    
        pi(active#) = []
        pi(filter) = []
        pi(cons) = []
        pi(|0|) = []
        pi(mark#) = [1]
        pi(zprimes) = []
        pi(sieve) = [1]
        pi(nats) = [1]
        pi(s) = []
        pi(mark) = []
        pi(active) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17, p18, p19

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

-- Reduction pair.

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

p1: cons#(X1,mark(X2)) -> cons#(X1,X2)
p2: cons#(X1,active(X2)) -> cons#(X1,X2)
p3: cons#(active(X1),X2) -> cons#(X1,X2)
p4: cons#(mark(X1),X2) -> cons#(X1,X2)

and R consists of:

r1: active(filter(cons(X,Y),|0|(),M)) -> mark(cons(|0|(),filter(Y,M,M)))
r2: active(filter(cons(X,Y),s(N),M)) -> mark(cons(X,filter(Y,N,M)))
r3: active(sieve(cons(|0|(),Y))) -> mark(cons(|0|(),sieve(Y)))
r4: active(sieve(cons(s(N),Y))) -> mark(cons(s(N),sieve(filter(Y,N,N))))
r5: active(nats(N)) -> mark(cons(N,nats(s(N))))
r6: active(zprimes()) -> mark(sieve(nats(s(s(|0|())))))
r7: mark(filter(X1,X2,X3)) -> active(filter(mark(X1),mark(X2),mark(X3)))
r8: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r9: mark(|0|()) -> active(|0|())
r10: mark(s(X)) -> active(s(mark(X)))
r11: mark(sieve(X)) -> active(sieve(mark(X)))
r12: mark(nats(X)) -> active(nats(mark(X)))
r13: mark(zprimes()) -> active(zprimes())
r14: filter(mark(X1),X2,X3) -> filter(X1,X2,X3)
r15: filter(X1,mark(X2),X3) -> filter(X1,X2,X3)
r16: filter(X1,X2,mark(X3)) -> filter(X1,X2,X3)
r17: filter(active(X1),X2,X3) -> filter(X1,X2,X3)
r18: filter(X1,active(X2),X3) -> filter(X1,X2,X3)
r19: filter(X1,X2,active(X3)) -> filter(X1,X2,X3)
r20: cons(mark(X1),X2) -> cons(X1,X2)
r21: cons(X1,mark(X2)) -> cons(X1,X2)
r22: cons(active(X1),X2) -> cons(X1,X2)
r23: cons(X1,active(X2)) -> cons(X1,X2)
r24: s(mark(X)) -> s(X)
r25: s(active(X)) -> s(X)
r26: sieve(mark(X)) -> sieve(X)
r27: sieve(active(X)) -> sieve(X)
r28: nats(mark(X)) -> nats(X)
r29: nats(active(X)) -> nats(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          cons#_A(x1,x2) = max{2, x1 + 1, x2}
          mark_A(x1) = max{1, x1}
          active_A(x1) = max{1, x1}
    
      precedence:
      
        cons# = mark = active
      
      partial status:
    
        pi(cons#) = [1, 2]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          cons#_A(x1,x2) = max{x1 + 1, x2 + 1}
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        cons# = mark = active
      
      partial status:
    
        pi(cons#) = [1, 2]
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3, p4

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

-- Reduction pair.

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

p1: filter#(mark(X1),X2,X3) -> filter#(X1,X2,X3)
p2: filter#(X1,X2,active(X3)) -> filter#(X1,X2,X3)
p3: filter#(X1,active(X2),X3) -> filter#(X1,X2,X3)
p4: filter#(active(X1),X2,X3) -> filter#(X1,X2,X3)
p5: filter#(X1,X2,mark(X3)) -> filter#(X1,X2,X3)
p6: filter#(X1,mark(X2),X3) -> filter#(X1,X2,X3)

and R consists of:

r1: active(filter(cons(X,Y),|0|(),M)) -> mark(cons(|0|(),filter(Y,M,M)))
r2: active(filter(cons(X,Y),s(N),M)) -> mark(cons(X,filter(Y,N,M)))
r3: active(sieve(cons(|0|(),Y))) -> mark(cons(|0|(),sieve(Y)))
r4: active(sieve(cons(s(N),Y))) -> mark(cons(s(N),sieve(filter(Y,N,N))))
r5: active(nats(N)) -> mark(cons(N,nats(s(N))))
r6: active(zprimes()) -> mark(sieve(nats(s(s(|0|())))))
r7: mark(filter(X1,X2,X3)) -> active(filter(mark(X1),mark(X2),mark(X3)))
r8: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r9: mark(|0|()) -> active(|0|())
r10: mark(s(X)) -> active(s(mark(X)))
r11: mark(sieve(X)) -> active(sieve(mark(X)))
r12: mark(nats(X)) -> active(nats(mark(X)))
r13: mark(zprimes()) -> active(zprimes())
r14: filter(mark(X1),X2,X3) -> filter(X1,X2,X3)
r15: filter(X1,mark(X2),X3) -> filter(X1,X2,X3)
r16: filter(X1,X2,mark(X3)) -> filter(X1,X2,X3)
r17: filter(active(X1),X2,X3) -> filter(X1,X2,X3)
r18: filter(X1,active(X2),X3) -> filter(X1,X2,X3)
r19: filter(X1,X2,active(X3)) -> filter(X1,X2,X3)
r20: cons(mark(X1),X2) -> cons(X1,X2)
r21: cons(X1,mark(X2)) -> cons(X1,X2)
r22: cons(active(X1),X2) -> cons(X1,X2)
r23: cons(X1,active(X2)) -> cons(X1,X2)
r24: s(mark(X)) -> s(X)
r25: s(active(X)) -> s(X)
r26: sieve(mark(X)) -> sieve(X)
r27: sieve(active(X)) -> sieve(X)
r28: nats(mark(X)) -> nats(X)
r29: nats(active(X)) -> nats(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          filter#_A(x1,x2,x3) = max{3, x1 + 1, x2 + 1, x3 + 1}
          mark_A(x1) = max{2, x1 + 1}
          active_A(x1) = max{1, x1}
    
      precedence:
      
        filter# = mark = active
      
      partial status:
    
        pi(filter#) = [1, 2]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          filter#_A(x1,x2,x3) = max{2, x1 + 1, x2 + 1}
          mark_A(x1) = max{1, x1}
          active_A(x1) = x1 + 1
    
      precedence:
      
        filter# = mark = active
      
      partial status:
    
        pi(filter#) = [1, 2]
        pi(mark) = [1]
        pi(active) = []
    

The next rules are strictly ordered:

  p1, p3, p4, p6

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: active(filter(cons(X,Y),|0|(),M)) -> mark(cons(|0|(),filter(Y,M,M)))
r2: active(filter(cons(X,Y),s(N),M)) -> mark(cons(X,filter(Y,N,M)))
r3: active(sieve(cons(|0|(),Y))) -> mark(cons(|0|(),sieve(Y)))
r4: active(sieve(cons(s(N),Y))) -> mark(cons(s(N),sieve(filter(Y,N,N))))
r5: active(nats(N)) -> mark(cons(N,nats(s(N))))
r6: active(zprimes()) -> mark(sieve(nats(s(s(|0|())))))
r7: mark(filter(X1,X2,X3)) -> active(filter(mark(X1),mark(X2),mark(X3)))
r8: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r9: mark(|0|()) -> active(|0|())
r10: mark(s(X)) -> active(s(mark(X)))
r11: mark(sieve(X)) -> active(sieve(mark(X)))
r12: mark(nats(X)) -> active(nats(mark(X)))
r13: mark(zprimes()) -> active(zprimes())
r14: filter(mark(X1),X2,X3) -> filter(X1,X2,X3)
r15: filter(X1,mark(X2),X3) -> filter(X1,X2,X3)
r16: filter(X1,X2,mark(X3)) -> filter(X1,X2,X3)
r17: filter(active(X1),X2,X3) -> filter(X1,X2,X3)
r18: filter(X1,active(X2),X3) -> filter(X1,X2,X3)
r19: filter(X1,X2,active(X3)) -> filter(X1,X2,X3)
r20: cons(mark(X1),X2) -> cons(X1,X2)
r21: cons(X1,mark(X2)) -> cons(X1,X2)
r22: cons(active(X1),X2) -> cons(X1,X2)
r23: cons(X1,active(X2)) -> cons(X1,X2)
r24: s(mark(X)) -> s(X)
r25: s(active(X)) -> s(X)
r26: sieve(mark(X)) -> sieve(X)
r27: sieve(active(X)) -> sieve(X)
r28: nats(mark(X)) -> nats(X)
r29: nats(active(X)) -> nats(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

r1: active(filter(cons(X,Y),|0|(),M)) -> mark(cons(|0|(),filter(Y,M,M)))
r2: active(filter(cons(X,Y),s(N),M)) -> mark(cons(X,filter(Y,N,M)))
r3: active(sieve(cons(|0|(),Y))) -> mark(cons(|0|(),sieve(Y)))
r4: active(sieve(cons(s(N),Y))) -> mark(cons(s(N),sieve(filter(Y,N,N))))
r5: active(nats(N)) -> mark(cons(N,nats(s(N))))
r6: active(zprimes()) -> mark(sieve(nats(s(s(|0|())))))
r7: mark(filter(X1,X2,X3)) -> active(filter(mark(X1),mark(X2),mark(X3)))
r8: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r9: mark(|0|()) -> active(|0|())
r10: mark(s(X)) -> active(s(mark(X)))
r11: mark(sieve(X)) -> active(sieve(mark(X)))
r12: mark(nats(X)) -> active(nats(mark(X)))
r13: mark(zprimes()) -> active(zprimes())
r14: filter(mark(X1),X2,X3) -> filter(X1,X2,X3)
r15: filter(X1,mark(X2),X3) -> filter(X1,X2,X3)
r16: filter(X1,X2,mark(X3)) -> filter(X1,X2,X3)
r17: filter(active(X1),X2,X3) -> filter(X1,X2,X3)
r18: filter(X1,active(X2),X3) -> filter(X1,X2,X3)
r19: filter(X1,X2,active(X3)) -> filter(X1,X2,X3)
r20: cons(mark(X1),X2) -> cons(X1,X2)
r21: cons(X1,mark(X2)) -> cons(X1,X2)
r22: cons(active(X1),X2) -> cons(X1,X2)
r23: cons(X1,active(X2)) -> cons(X1,X2)
r24: s(mark(X)) -> s(X)
r25: s(active(X)) -> s(X)
r26: sieve(mark(X)) -> sieve(X)
r27: sieve(active(X)) -> sieve(X)
r28: nats(mark(X)) -> nats(X)
r29: nats(active(X)) -> nats(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          filter#_A(x1,x2,x3) = max{2, x1, x2 + 1, x3 + 1}
          active_A(x1) = max{1, x1}
          mark_A(x1) = max{1, x1}
    
      precedence:
      
        filter# = active = mark
      
      partial status:
    
        pi(filter#) = [1, 3]
        pi(active) = [1]
        pi(mark) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          filter#_A(x1,x2,x3) = max{0, x3 - 2}
          active_A(x1) = max{3, x1 + 1}
          mark_A(x1) = x1
    
      precedence:
      
        filter# = active = mark
      
      partial status:
    
        pi(filter#) = []
        pi(active) = [1]
        pi(mark) = [1]
    

The next rules are strictly ordered:

  p1, p2

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

-- Reduction pair.

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

p1: sieve#(mark(X)) -> sieve#(X)
p2: sieve#(active(X)) -> sieve#(X)

and R consists of:

r1: active(filter(cons(X,Y),|0|(),M)) -> mark(cons(|0|(),filter(Y,M,M)))
r2: active(filter(cons(X,Y),s(N),M)) -> mark(cons(X,filter(Y,N,M)))
r3: active(sieve(cons(|0|(),Y))) -> mark(cons(|0|(),sieve(Y)))
r4: active(sieve(cons(s(N),Y))) -> mark(cons(s(N),sieve(filter(Y,N,N))))
r5: active(nats(N)) -> mark(cons(N,nats(s(N))))
r6: active(zprimes()) -> mark(sieve(nats(s(s(|0|())))))
r7: mark(filter(X1,X2,X3)) -> active(filter(mark(X1),mark(X2),mark(X3)))
r8: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r9: mark(|0|()) -> active(|0|())
r10: mark(s(X)) -> active(s(mark(X)))
r11: mark(sieve(X)) -> active(sieve(mark(X)))
r12: mark(nats(X)) -> active(nats(mark(X)))
r13: mark(zprimes()) -> active(zprimes())
r14: filter(mark(X1),X2,X3) -> filter(X1,X2,X3)
r15: filter(X1,mark(X2),X3) -> filter(X1,X2,X3)
r16: filter(X1,X2,mark(X3)) -> filter(X1,X2,X3)
r17: filter(active(X1),X2,X3) -> filter(X1,X2,X3)
r18: filter(X1,active(X2),X3) -> filter(X1,X2,X3)
r19: filter(X1,X2,active(X3)) -> filter(X1,X2,X3)
r20: cons(mark(X1),X2) -> cons(X1,X2)
r21: cons(X1,mark(X2)) -> cons(X1,X2)
r22: cons(active(X1),X2) -> cons(X1,X2)
r23: cons(X1,active(X2)) -> cons(X1,X2)
r24: s(mark(X)) -> s(X)
r25: s(active(X)) -> s(X)
r26: sieve(mark(X)) -> sieve(X)
r27: sieve(active(X)) -> sieve(X)
r28: nats(mark(X)) -> nats(X)
r29: nats(active(X)) -> nats(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sieve#_A(x1) = max{6, x1 + 4}
          mark_A(x1) = max{4, x1 + 3}
          active_A(x1) = max{2, x1 + 1}
    
      precedence:
      
        sieve# = mark = active
      
      partial status:
    
        pi(sieve#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sieve#_A(x1) = x1 + 1
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        sieve# = mark = active
      
      partial status:
    
        pi(sieve#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2

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

-- Reduction pair.

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

p1: nats#(mark(X)) -> nats#(X)
p2: nats#(active(X)) -> nats#(X)

and R consists of:

r1: active(filter(cons(X,Y),|0|(),M)) -> mark(cons(|0|(),filter(Y,M,M)))
r2: active(filter(cons(X,Y),s(N),M)) -> mark(cons(X,filter(Y,N,M)))
r3: active(sieve(cons(|0|(),Y))) -> mark(cons(|0|(),sieve(Y)))
r4: active(sieve(cons(s(N),Y))) -> mark(cons(s(N),sieve(filter(Y,N,N))))
r5: active(nats(N)) -> mark(cons(N,nats(s(N))))
r6: active(zprimes()) -> mark(sieve(nats(s(s(|0|())))))
r7: mark(filter(X1,X2,X3)) -> active(filter(mark(X1),mark(X2),mark(X3)))
r8: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r9: mark(|0|()) -> active(|0|())
r10: mark(s(X)) -> active(s(mark(X)))
r11: mark(sieve(X)) -> active(sieve(mark(X)))
r12: mark(nats(X)) -> active(nats(mark(X)))
r13: mark(zprimes()) -> active(zprimes())
r14: filter(mark(X1),X2,X3) -> filter(X1,X2,X3)
r15: filter(X1,mark(X2),X3) -> filter(X1,X2,X3)
r16: filter(X1,X2,mark(X3)) -> filter(X1,X2,X3)
r17: filter(active(X1),X2,X3) -> filter(X1,X2,X3)
r18: filter(X1,active(X2),X3) -> filter(X1,X2,X3)
r19: filter(X1,X2,active(X3)) -> filter(X1,X2,X3)
r20: cons(mark(X1),X2) -> cons(X1,X2)
r21: cons(X1,mark(X2)) -> cons(X1,X2)
r22: cons(active(X1),X2) -> cons(X1,X2)
r23: cons(X1,active(X2)) -> cons(X1,X2)
r24: s(mark(X)) -> s(X)
r25: s(active(X)) -> s(X)
r26: sieve(mark(X)) -> sieve(X)
r27: sieve(active(X)) -> sieve(X)
r28: nats(mark(X)) -> nats(X)
r29: nats(active(X)) -> nats(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          nats#_A(x1) = max{6, x1 + 4}
          mark_A(x1) = max{4, x1 + 3}
          active_A(x1) = max{2, x1 + 1}
    
      precedence:
      
        nats# = mark = active
      
      partial status:
    
        pi(nats#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          nats#_A(x1) = x1 + 1
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        nats# = mark = active
      
      partial status:
    
        pi(nats#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2

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

-- Reduction pair.

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

p1: s#(mark(X)) -> s#(X)
p2: s#(active(X)) -> s#(X)

and R consists of:

r1: active(filter(cons(X,Y),|0|(),M)) -> mark(cons(|0|(),filter(Y,M,M)))
r2: active(filter(cons(X,Y),s(N),M)) -> mark(cons(X,filter(Y,N,M)))
r3: active(sieve(cons(|0|(),Y))) -> mark(cons(|0|(),sieve(Y)))
r4: active(sieve(cons(s(N),Y))) -> mark(cons(s(N),sieve(filter(Y,N,N))))
r5: active(nats(N)) -> mark(cons(N,nats(s(N))))
r6: active(zprimes()) -> mark(sieve(nats(s(s(|0|())))))
r7: mark(filter(X1,X2,X3)) -> active(filter(mark(X1),mark(X2),mark(X3)))
r8: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r9: mark(|0|()) -> active(|0|())
r10: mark(s(X)) -> active(s(mark(X)))
r11: mark(sieve(X)) -> active(sieve(mark(X)))
r12: mark(nats(X)) -> active(nats(mark(X)))
r13: mark(zprimes()) -> active(zprimes())
r14: filter(mark(X1),X2,X3) -> filter(X1,X2,X3)
r15: filter(X1,mark(X2),X3) -> filter(X1,X2,X3)
r16: filter(X1,X2,mark(X3)) -> filter(X1,X2,X3)
r17: filter(active(X1),X2,X3) -> filter(X1,X2,X3)
r18: filter(X1,active(X2),X3) -> filter(X1,X2,X3)
r19: filter(X1,X2,active(X3)) -> filter(X1,X2,X3)
r20: cons(mark(X1),X2) -> cons(X1,X2)
r21: cons(X1,mark(X2)) -> cons(X1,X2)
r22: cons(active(X1),X2) -> cons(X1,X2)
r23: cons(X1,active(X2)) -> cons(X1,X2)
r24: s(mark(X)) -> s(X)
r25: s(active(X)) -> s(X)
r26: sieve(mark(X)) -> sieve(X)
r27: sieve(active(X)) -> sieve(X)
r28: nats(mark(X)) -> nats(X)
r29: nats(active(X)) -> nats(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          s#_A(x1) = max{6, x1 + 4}
          mark_A(x1) = max{4, x1 + 3}
          active_A(x1) = max{2, x1 + 1}
    
      precedence:
      
        s# = mark = active
      
      partial status:
    
        pi(s#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          s#_A(x1) = x1 + 1
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        s# = mark = active
      
      partial status:
    
        pi(s#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2

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