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:

  max/plus interpretations on natural numbers:
  
    active#_A(x1) = max{85, x1 + 2}
    filter_A(x1,x2,x3) = 84
    cons_A(x1,x2) = 83
    |0|_A = 91
    mark#_A(x1) = 86
    zprimes_A = 84
    sieve_A(x1) = 84
    nats_A(x1) = 84
    s_A(x1) = 84
    mark_A(x1) = max{57, x1 - 35}
    active_A(x1) = 57

The next rules are strictly ordered:

  p14

We remove them from the problem.

-- 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: 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: active#(filter(cons(X,Y),s(N),M)) -> mark#(cons(X,filter(Y,N,M)))
p15: mark#(filter(X1,X2,X3)) -> mark#(X3)
p16: mark#(filter(X1,X2,X3)) -> mark#(X2)
p17: mark#(filter(X1,X2,X3)) -> mark#(X1)
p18: 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 estimated dependency graph contains the following SCCs:

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


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

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:

  max/plus interpretations on natural numbers:
  
    active#_A(x1) = x1 + 289
    filter_A(x1,x2,x3) = 89
    cons_A(x1,x2) = 78
    |0|_A = 31
    mark#_A(x1) = 378
    mark_A(x1) = max{87, x1 - 3}
    s_A(x1) = 20
    sieve_A(x1) = 89
    nats_A(x1) = 89
    zprimes_A = 89
    active_A(x1) = 87

The next rules are strictly ordered:

  p8

We remove them from the problem.

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

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, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17}


-- 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#(cons(X1,X2)) -> mark#(X1)
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)) -> active#(filter(mark(X1),mark(X2),mark(X3)))
p16: active#(sieve(cons(|0|(),Y))) -> mark#(cons(|0|(),sieve(Y)))
p17: active#(filter(cons(X,Y),s(N),M)) -> mark#(cons(X,filter(Y,N,M)))

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:

  max/plus interpretations on natural numbers:
  
    active#_A(x1) = x1 + 4
    filter_A(x1,x2,x3) = max{x1 + 41, x2 + 41, x3 + 41}
    cons_A(x1,x2) = max{26, x1 + 1}
    |0|_A = 14
    mark#_A(x1) = x1 + 29
    zprimes_A = 169
    sieve_A(x1) = max{61, x1 + 37}
    nats_A(x1) = x1 + 51
    s_A(x1) = max{34, x1 + 21}
    mark_A(x1) = x1 + 25
    active_A(x1) = max{26, x1}

The next rules are strictly ordered:

  p1, p2, p4, p5, p9, p10, p11, p12, p13, p14, p16, p17

We remove them from the problem.

-- SCC decomposition.

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

p1: active#(zprimes()) -> mark#(sieve(nats(s(s(|0|())))))
p2: mark#(nats(X)) -> active#(nats(mark(X)))
p3: active#(nats(N)) -> mark#(cons(N,nats(s(N))))
p4: mark#(sieve(X)) -> active#(sieve(mark(X)))
p5: 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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

p1: active#(zprimes()) -> mark#(sieve(nats(s(s(|0|())))))
p2: mark#(sieve(X)) -> active#(sieve(mark(X)))
p3: active#(nats(N)) -> mark#(cons(N,nats(s(N))))
p4: mark#(filter(X1,X2,X3)) -> active#(filter(mark(X1),mark(X2),mark(X3)))
p5: mark#(nats(X)) -> active#(nats(mark(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

  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:

  max/plus interpretations on natural numbers:
  
    active#_A(x1) = max{51, x1 - 15}
    zprimes_A = 39
    mark#_A(x1) = x1 + 11
    sieve_A(x1) = 40
    nats_A(x1) = 66
    s_A(x1) = 41
    |0|_A = 39
    mark_A(x1) = max{29, x1 - 11}
    cons_A(x1,x2) = 40
    filter_A(x1,x2,x3) = 45
    active_A(x1) = 29

The next rules are strictly ordered:

  p4, p5

We remove them from the problem.

-- SCC decomposition.

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

p1: active#(zprimes()) -> mark#(sieve(nats(s(s(|0|())))))
p2: mark#(sieve(X)) -> active#(sieve(mark(X)))
p3: active#(nats(N)) -> mark#(cons(N,nats(s(N))))

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, p3}


-- Reduction pair.

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

p1: active#(zprimes()) -> mark#(sieve(nats(s(s(|0|())))))
p2: mark#(sieve(X)) -> active#(sieve(mark(X)))
p3: active#(nats(N)) -> mark#(cons(N,nats(s(N))))

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:

  max/plus interpretations on natural numbers:
  
    active#_A(x1) = x1 + 6
    zprimes_A = 11
    mark#_A(x1) = x1 + 17
    sieve_A(x1) = 0
    nats_A(x1) = 15
    s_A(x1) = 80
    |0|_A = 22
    mark_A(x1) = max{10, x1 - 2}
    cons_A(x1,x2) = 4
    active_A(x1) = 10
    filter_A(x1,x2,x3) = 192

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: active#(zprimes()) -> mark#(sieve(nats(s(s(|0|())))))
p2: active#(nats(N)) -> mark#(cons(N,nats(s(N))))

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:

  (no SCCs)

-- 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 reduction pair:

  max/plus interpretations on natural numbers:
  
    cons#_A(x1,x2) = x2 + 2
    mark_A(x1) = max{2, x1}
    active_A(x1) = x1 + 1

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: cons#(X1,mark(X2)) -> cons#(X1,X2)
p2: cons#(active(X1),X2) -> cons#(X1,X2)
p3: 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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

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

  max/plus interpretations on natural numbers:
  
    cons#_A(x1,x2) = x1
    mark_A(x1) = x1
    active_A(x1) = x1 + 1

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

p1: cons#(X1,mark(X2)) -> cons#(X1,X2)
p2: 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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: cons#(X1,mark(X2)) -> cons#(X1,X2)
p2: 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 reduction pair:

  max/plus interpretations on natural numbers:
  
    cons#_A(x1,x2) = x1 + 1
    mark_A(x1) = x1 + 1

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: cons#(X1,mark(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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: cons#(X1,mark(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 reduction pair:

  max/plus interpretations on natural numbers:
  
    cons#_A(x1,x2) = x2
    mark_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1

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:

  max/plus interpretations on natural numbers:
  
    filter#_A(x1,x2,x3) = x3 + 1
    mark_A(x1) = x1 + 1
    active_A(x1) = x1

The next rules are strictly ordered:

  p5

We remove them from the problem.

-- SCC decomposition.

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,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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


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

  max/plus interpretations on natural numbers:
  
    filter#_A(x1,x2,x3) = x3
    mark_A(x1) = x1 + 2
    active_A(x1) = x1 + 1

The next rules are strictly ordered:

  p5

We remove them from the problem.

-- SCC decomposition.

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

p1: filter#(mark(X1),X2,X3) -> filter#(X1,X2,X3)
p2: filter#(X1,mark(X2),X3) -> filter#(X1,X2,X3)
p3: filter#(active(X1),X2,X3) -> filter#(X1,X2,X3)
p4: filter#(X1,active(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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- 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,active(X2),X3) -> filter#(X1,X2,X3)
p3: filter#(active(X1),X2,X3) -> filter#(X1,X2,X3)
p4: 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:

  max/plus interpretations on natural numbers:
  
    filter#_A(x1,x2,x3) = max{0, x2 - 1}
    mark_A(x1) = x1 + 2
    active_A(x1) = x1 + 1

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

p1: filter#(mark(X1),X2,X3) -> filter#(X1,X2,X3)
p2: filter#(X1,active(X2),X3) -> filter#(X1,X2,X3)
p3: filter#(active(X1),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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- 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#(active(X1),X2,X3) -> filter#(X1,X2,X3)
p3: filter#(X1,active(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:

  max/plus interpretations on natural numbers:
  
    filter#_A(x1,x2,x3) = x1 + 1
    mark_A(x1) = x1
    active_A(x1) = x1 + 1

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: filter#(mark(X1),X2,X3) -> filter#(X1,X2,X3)
p2: filter#(X1,active(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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- 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,active(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:

  max/plus interpretations on natural numbers:
  
    filter#_A(x1,x2,x3) = x1
    mark_A(x1) = x1 + 1
    active_A(x1) = max{0, x1 - 1}

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: filter#(X1,active(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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: filter#(X1,active(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:

  max/plus interpretations on natural numbers:
  
    filter#_A(x1,x2,x3) = x2
    active_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1

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 reduction pair:

  max/plus interpretations on natural numbers:
  
    sieve#_A(x1) = max{0, x1 - 1}
    mark_A(x1) = x1 + 1
    active_A(x1) = x1 + 2

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: sieve#(mark(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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: sieve#(mark(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:

  max/plus interpretations on natural numbers:
  
    sieve#_A(x1) = x1
    mark_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1

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 reduction pair:

  max/plus interpretations on natural numbers:
  
    nats#_A(x1) = max{0, x1 - 1}
    mark_A(x1) = x1 + 1
    active_A(x1) = x1 + 2

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: nats#(mark(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}


-- Reduction pair.

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

p1: nats#(mark(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:

  max/plus interpretations on natural numbers:
  
    nats#_A(x1) = x1
    mark_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1

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 reduction pair:

  max/plus interpretations on natural numbers:
  
    s#_A(x1) = max{0, x1 - 1}
    mark_A(x1) = x1 + 1
    active_A(x1) = x1 + 2

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: s#(mark(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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: s#(mark(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:

  max/plus interpretations on natural numbers:
  
    s#_A(x1) = x1
    mark_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1

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