YES
We show the termination of the TRS R:

  a(a(divides(),|0|()),a(s(),y)) -> true()
  a(a(divides(),a(s(),x)),a(s(),y)) -> a(a(a(div2(),x),a(s(),y)),y)
  a(a(a(div2(),x),y),|0|()) -> a(a(divides(),x),y)
  a(a(a(div2(),|0|()),y),a(s(),z)) -> false()
  a(a(a(div2(),a(s(),x)),y),a(s(),z)) -> a(a(a(div2(),x),y),z)
  a(a(filter(),f),nil()) -> nil()
  a(a(filter(),f),a(a(cons(),x),xs)) -> a(a(a(if(),a(f,x)),x),a(a(filter(),f),xs))
  a(a(a(if(),true()),x),xs) -> a(a(cons(),x),xs)
  a(a(a(if(),false()),x),xs) -> xs
  a(a(not(),f),x) -> a(not2(),a(f,x))
  a(not2(),true()) -> false()
  a(not2(),false()) -> true()
  a(sieve(),nil()) -> nil()
  a(sieve(),a(a(cons(),x),xs)) -> a(a(cons(),x),a(sieve(),a(a(filter(),a(not(),a(divides(),x))),xs)))

Instead we prove termination of the uncurried system:

  a(cons_0(),y) -> cons_1(y)
  a(cons_1(x1),y) -> cons_2(x1,y)
  a(div2_0(),y) -> div2_1(y)
  a(div2_1(x1),y) -> div2_2(x1,y)
  a(div2_2(x1,x2),y) -> div2_3(x1,x2,y)
  a(divides_0(),y) -> divides_1(y)
  a(divides_1(x1),y) -> divides_2(x1,y)
  a(filter_0(),y) -> filter_1(y)
  a(filter_1(x1),y) -> filter_2(x1,y)
  a(if_0(),y) -> if_1(y)
  a(if_1(x1),y) -> if_2(x1,y)
  a(if_2(x1,x2),y) -> if_3(x1,x2,y)
  a(not_0(),y) -> not_1(y)
  a(not_1(x1),y) -> not_2(x1,y)
  a(not2_0(),y) -> not2_1(y)
  a(s_0(),y) -> s_1(y)
  a(sieve_0(),y) -> sieve_1(y)
  divides_2(|0|_0(),s_1(y)) -> true_0()
  divides_2(s_1(x),s_1(y)) -> div2_3(x,s_1(y),y)
  div2_3(x,y,|0|_0()) -> divides_2(x,y)
  div2_3(|0|_0(),y,s_1(z)) -> false_0()
  div2_3(s_1(x),y,s_1(z)) -> div2_3(x,y,z)
  filter_2(f,nil_0()) -> nil_0()
  filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs))
  if_3(true_0(),x,xs) -> cons_2(x,xs)
  if_3(false_0(),x,xs) -> xs
  not_2(f,x) -> not2_1(a(f,x))
  not2_1(true_0()) -> false_0()
  not2_1(false_0()) -> true_0()
  sieve_1(nil_0()) -> nil_0()
  sieve_1(cons_2(x,xs)) -> cons_2(x,sieve_1(filter_2(not_1(divides_1(x)),xs)))


-- SCC decomposition.

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

p1: a#(div2_2(x1,x2),y) -> div2_3#(x1,x2,y)
p2: a#(divides_1(x1),y) -> divides_2#(x1,y)
p3: a#(filter_1(x1),y) -> filter_2#(x1,y)
p4: a#(if_2(x1,x2),y) -> if_3#(x1,x2,y)
p5: a#(not_1(x1),y) -> not_2#(x1,y)
p6: a#(not2_0(),y) -> not2_1#(y)
p7: a#(sieve_0(),y) -> sieve_1#(y)
p8: divides_2#(s_1(x),s_1(y)) -> div2_3#(x,s_1(y),y)
p9: div2_3#(x,y,|0|_0()) -> divides_2#(x,y)
p10: div2_3#(s_1(x),y,s_1(z)) -> div2_3#(x,y,z)
p11: filter_2#(f,cons_2(x,xs)) -> if_3#(a(f,x),x,filter_2(f,xs))
p12: filter_2#(f,cons_2(x,xs)) -> a#(f,x)
p13: filter_2#(f,cons_2(x,xs)) -> filter_2#(f,xs)
p14: not_2#(f,x) -> not2_1#(a(f,x))
p15: not_2#(f,x) -> a#(f,x)
p16: sieve_1#(cons_2(x,xs)) -> sieve_1#(filter_2(not_1(divides_1(x)),xs))
p17: sieve_1#(cons_2(x,xs)) -> filter_2#(not_1(divides_1(x)),xs)

and R consists of:

r1: a(cons_0(),y) -> cons_1(y)
r2: a(cons_1(x1),y) -> cons_2(x1,y)
r3: a(div2_0(),y) -> div2_1(y)
r4: a(div2_1(x1),y) -> div2_2(x1,y)
r5: a(div2_2(x1,x2),y) -> div2_3(x1,x2,y)
r6: a(divides_0(),y) -> divides_1(y)
r7: a(divides_1(x1),y) -> divides_2(x1,y)
r8: a(filter_0(),y) -> filter_1(y)
r9: a(filter_1(x1),y) -> filter_2(x1,y)
r10: a(if_0(),y) -> if_1(y)
r11: a(if_1(x1),y) -> if_2(x1,y)
r12: a(if_2(x1,x2),y) -> if_3(x1,x2,y)
r13: a(not_0(),y) -> not_1(y)
r14: a(not_1(x1),y) -> not_2(x1,y)
r15: a(not2_0(),y) -> not2_1(y)
r16: a(s_0(),y) -> s_1(y)
r17: a(sieve_0(),y) -> sieve_1(y)
r18: divides_2(|0|_0(),s_1(y)) -> true_0()
r19: divides_2(s_1(x),s_1(y)) -> div2_3(x,s_1(y),y)
r20: div2_3(x,y,|0|_0()) -> divides_2(x,y)
r21: div2_3(|0|_0(),y,s_1(z)) -> false_0()
r22: div2_3(s_1(x),y,s_1(z)) -> div2_3(x,y,z)
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs))
r25: if_3(true_0(),x,xs) -> cons_2(x,xs)
r26: if_3(false_0(),x,xs) -> xs
r27: not_2(f,x) -> not2_1(a(f,x))
r28: not2_1(true_0()) -> false_0()
r29: not2_1(false_0()) -> true_0()
r30: sieve_1(nil_0()) -> nil_0()
r31: sieve_1(cons_2(x,xs)) -> cons_2(x,sieve_1(filter_2(not_1(divides_1(x)),xs)))

The estimated dependency graph contains the following SCCs:

  {p3, p5, p7, p12, p13, p15, p16, p17}
  {p8, p9, p10}


-- Reduction pair.

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

p1: not_2#(f,x) -> a#(f,x)
p2: a#(sieve_0(),y) -> sieve_1#(y)
p3: sieve_1#(cons_2(x,xs)) -> filter_2#(not_1(divides_1(x)),xs)
p4: filter_2#(f,cons_2(x,xs)) -> filter_2#(f,xs)
p5: filter_2#(f,cons_2(x,xs)) -> a#(f,x)
p6: a#(not_1(x1),y) -> not_2#(x1,y)
p7: a#(filter_1(x1),y) -> filter_2#(x1,y)
p8: sieve_1#(cons_2(x,xs)) -> sieve_1#(filter_2(not_1(divides_1(x)),xs))

and R consists of:

r1: a(cons_0(),y) -> cons_1(y)
r2: a(cons_1(x1),y) -> cons_2(x1,y)
r3: a(div2_0(),y) -> div2_1(y)
r4: a(div2_1(x1),y) -> div2_2(x1,y)
r5: a(div2_2(x1,x2),y) -> div2_3(x1,x2,y)
r6: a(divides_0(),y) -> divides_1(y)
r7: a(divides_1(x1),y) -> divides_2(x1,y)
r8: a(filter_0(),y) -> filter_1(y)
r9: a(filter_1(x1),y) -> filter_2(x1,y)
r10: a(if_0(),y) -> if_1(y)
r11: a(if_1(x1),y) -> if_2(x1,y)
r12: a(if_2(x1,x2),y) -> if_3(x1,x2,y)
r13: a(not_0(),y) -> not_1(y)
r14: a(not_1(x1),y) -> not_2(x1,y)
r15: a(not2_0(),y) -> not2_1(y)
r16: a(s_0(),y) -> s_1(y)
r17: a(sieve_0(),y) -> sieve_1(y)
r18: divides_2(|0|_0(),s_1(y)) -> true_0()
r19: divides_2(s_1(x),s_1(y)) -> div2_3(x,s_1(y),y)
r20: div2_3(x,y,|0|_0()) -> divides_2(x,y)
r21: div2_3(|0|_0(),y,s_1(z)) -> false_0()
r22: div2_3(s_1(x),y,s_1(z)) -> div2_3(x,y,z)
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs))
r25: if_3(true_0(),x,xs) -> cons_2(x,xs)
r26: if_3(false_0(),x,xs) -> xs
r27: not_2(f,x) -> not2_1(a(f,x))
r28: not2_1(true_0()) -> false_0()
r29: not2_1(false_0()) -> true_0()
r30: sieve_1(nil_0()) -> nil_0()
r31: sieve_1(cons_2(x,xs)) -> cons_2(x,sieve_1(filter_2(not_1(divides_1(x)),xs)))

The set of usable rules consists of

  r23, r24, r25, r26

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        not_2#_A(x1,x2) = 1
        a#_A(x1,x2) = 1
        sieve_0_A = 2
        sieve_1#_A(x1) = 1
        cons_2_A(x1,x2) = max{4, x1 - 8, x2}
        filter_2#_A(x1,x2) = 1
        not_1_A(x1) = max{5, x1 - 18}
        divides_1_A(x1) = max{17, x1 - 18}
        filter_1_A(x1) = max{2, x1 - 18}
        filter_2_A(x1,x2) = max{28, x1 - 6, x2 + 2}
        nil_0_A = 29
        if_3_A(x1,x2,x3) = max{5, x2 - 8, x3}
        a_A(x1,x2) = max{9, x2 - 8}
        true_0_A = 1
        false_0_A = 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          not_2#_A(x1,x2) = max{0, x1 - 30, x2 - 12}
          a#_A(x1,x2) = max{0, x1 - 30, x2 - 12}
          sieve_0_A = 29
          sieve_1#_A(x1) = max{0, x1 - 34}
          cons_2_A(x1,x2) = max{x1 + 22, x2}
          filter_2#_A(x1,x2) = max{0, x1 - 28, x2 - 34}
          not_1_A(x1) = max{27, x1 + 2}
          divides_1_A(x1) = x1 + 14
          filter_1_A(x1) = max{29, x1 + 2}
          filter_2_A(x1,x2) = max{23, x1 - 1, x2}
          nil_0_A = 24
          if_3_A(x1,x2,x3) = max{x2 + 22, x3}
          a_A(x1,x2) = max{0, x1 - 48, x2 - 48}
          true_0_A = 20
          false_0_A = 3
          not_2#!_A(x1,x2) = max{5, x1 - 16, x2 - 7}
          a#!_A(x1,x2) = max{2, x1 - 17, x2 - 7}
          sieve_0!_A = 1
          sieve_1#!_A(x1) = max{11, x1 - 22}
          cons_2!_A(x1,x2) = max{3, x1 - 6, x2 - 55}
          filter_2#!_A(x1,x2) = max{11, x1 - 16, x2 - 22}
          not_1!_A(x1) = max{4, x1 - 14}
          divides_1!_A(x1) = max{11, x1 - 1}
          filter_1!_A(x1) = max{13, x1 - 31}
          filter_2!_A(x1,x2) = max{14, x1 - 30, x2 - 25}
          nil_0!_A = 3
          if_3!_A(x1,x2,x3) = max{4, x2 - 4, x3 - 54}
          a!_A(x1,x2) = max{14, x1 - 30, x2 - 5}
          true_0!_A = 1
          false_0!_A = 18
  
    partial status:
  
      pi(not_2#) = []
      pi(a#) = []
      pi(sieve_0) = []
      pi(sieve_1#) = []
      pi(cons_2) = [2]
      pi(filter_2#) = []
      pi(not_1) = []
      pi(divides_1) = []
      pi(filter_1) = []
      pi(filter_2) = [2]
      pi(nil_0) = []
      pi(if_3) = [3]
      pi(a) = []
      pi(true_0) = []
      pi(false_0) = []

The following rules are removed:

  p2, p5, p7



-- SCC decomposition.

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

p1: not_2#(f,x) -> a#(f,x)
p2: sieve_1#(cons_2(x,xs)) -> filter_2#(not_1(divides_1(x)),xs)
p3: filter_2#(f,cons_2(x,xs)) -> filter_2#(f,xs)
p4: a#(not_1(x1),y) -> not_2#(x1,y)
p5: sieve_1#(cons_2(x,xs)) -> sieve_1#(filter_2(not_1(divides_1(x)),xs))

and R consists of:

r1: a(cons_0(),y) -> cons_1(y)
r2: a(cons_1(x1),y) -> cons_2(x1,y)
r3: a(div2_0(),y) -> div2_1(y)
r4: a(div2_1(x1),y) -> div2_2(x1,y)
r5: a(div2_2(x1,x2),y) -> div2_3(x1,x2,y)
r6: a(divides_0(),y) -> divides_1(y)
r7: a(divides_1(x1),y) -> divides_2(x1,y)
r8: a(filter_0(),y) -> filter_1(y)
r9: a(filter_1(x1),y) -> filter_2(x1,y)
r10: a(if_0(),y) -> if_1(y)
r11: a(if_1(x1),y) -> if_2(x1,y)
r12: a(if_2(x1,x2),y) -> if_3(x1,x2,y)
r13: a(not_0(),y) -> not_1(y)
r14: a(not_1(x1),y) -> not_2(x1,y)
r15: a(not2_0(),y) -> not2_1(y)
r16: a(s_0(),y) -> s_1(y)
r17: a(sieve_0(),y) -> sieve_1(y)
r18: divides_2(|0|_0(),s_1(y)) -> true_0()
r19: divides_2(s_1(x),s_1(y)) -> div2_3(x,s_1(y),y)
r20: div2_3(x,y,|0|_0()) -> divides_2(x,y)
r21: div2_3(|0|_0(),y,s_1(z)) -> false_0()
r22: div2_3(s_1(x),y,s_1(z)) -> div2_3(x,y,z)
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs))
r25: if_3(true_0(),x,xs) -> cons_2(x,xs)
r26: if_3(false_0(),x,xs) -> xs
r27: not_2(f,x) -> not2_1(a(f,x))
r28: not2_1(true_0()) -> false_0()
r29: not2_1(false_0()) -> true_0()
r30: sieve_1(nil_0()) -> nil_0()
r31: sieve_1(cons_2(x,xs)) -> cons_2(x,sieve_1(filter_2(not_1(divides_1(x)),xs)))

The estimated dependency graph contains the following SCCs:

  {p1, p4}
  {p5}
  {p3}


-- Reduction pair.

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

p1: not_2#(f,x) -> a#(f,x)
p2: a#(not_1(x1),y) -> not_2#(x1,y)

and R consists of:

r1: a(cons_0(),y) -> cons_1(y)
r2: a(cons_1(x1),y) -> cons_2(x1,y)
r3: a(div2_0(),y) -> div2_1(y)
r4: a(div2_1(x1),y) -> div2_2(x1,y)
r5: a(div2_2(x1,x2),y) -> div2_3(x1,x2,y)
r6: a(divides_0(),y) -> divides_1(y)
r7: a(divides_1(x1),y) -> divides_2(x1,y)
r8: a(filter_0(),y) -> filter_1(y)
r9: a(filter_1(x1),y) -> filter_2(x1,y)
r10: a(if_0(),y) -> if_1(y)
r11: a(if_1(x1),y) -> if_2(x1,y)
r12: a(if_2(x1,x2),y) -> if_3(x1,x2,y)
r13: a(not_0(),y) -> not_1(y)
r14: a(not_1(x1),y) -> not_2(x1,y)
r15: a(not2_0(),y) -> not2_1(y)
r16: a(s_0(),y) -> s_1(y)
r17: a(sieve_0(),y) -> sieve_1(y)
r18: divides_2(|0|_0(),s_1(y)) -> true_0()
r19: divides_2(s_1(x),s_1(y)) -> div2_3(x,s_1(y),y)
r20: div2_3(x,y,|0|_0()) -> divides_2(x,y)
r21: div2_3(|0|_0(),y,s_1(z)) -> false_0()
r22: div2_3(s_1(x),y,s_1(z)) -> div2_3(x,y,z)
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs))
r25: if_3(true_0(),x,xs) -> cons_2(x,xs)
r26: if_3(false_0(),x,xs) -> xs
r27: not_2(f,x) -> not2_1(a(f,x))
r28: not2_1(true_0()) -> false_0()
r29: not2_1(false_0()) -> true_0()
r30: sieve_1(nil_0()) -> nil_0()
r31: sieve_1(cons_2(x,xs)) -> cons_2(x,sieve_1(filter_2(not_1(divides_1(x)),xs)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        not_2#_A(x1,x2) = 1
        a#_A(x1,x2) = 1
        not_1_A(x1) = max{7, x1 - 68}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          not_2#_A(x1,x2) = max{0, x1 - 56, x2 - 1}
          a#_A(x1,x2) = max{0, x1 - 56, x2 - 1}
          not_1_A(x1) = max{55, x1 + 5}
          not_2#!_A(x1,x2) = max{3, x1 + 1}
          a#!_A(x1,x2) = max{2, x1 - 1}
          not_1!_A(x1) = max{2, x1 - 89}
  
    partial status:
  
      pi(not_2#) = []
      pi(a#) = []
      pi(not_1) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.

-- Reduction pair.

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

p1: sieve_1#(cons_2(x,xs)) -> sieve_1#(filter_2(not_1(divides_1(x)),xs))

and R consists of:

r1: a(cons_0(),y) -> cons_1(y)
r2: a(cons_1(x1),y) -> cons_2(x1,y)
r3: a(div2_0(),y) -> div2_1(y)
r4: a(div2_1(x1),y) -> div2_2(x1,y)
r5: a(div2_2(x1,x2),y) -> div2_3(x1,x2,y)
r6: a(divides_0(),y) -> divides_1(y)
r7: a(divides_1(x1),y) -> divides_2(x1,y)
r8: a(filter_0(),y) -> filter_1(y)
r9: a(filter_1(x1),y) -> filter_2(x1,y)
r10: a(if_0(),y) -> if_1(y)
r11: a(if_1(x1),y) -> if_2(x1,y)
r12: a(if_2(x1,x2),y) -> if_3(x1,x2,y)
r13: a(not_0(),y) -> not_1(y)
r14: a(not_1(x1),y) -> not_2(x1,y)
r15: a(not2_0(),y) -> not2_1(y)
r16: a(s_0(),y) -> s_1(y)
r17: a(sieve_0(),y) -> sieve_1(y)
r18: divides_2(|0|_0(),s_1(y)) -> true_0()
r19: divides_2(s_1(x),s_1(y)) -> div2_3(x,s_1(y),y)
r20: div2_3(x,y,|0|_0()) -> divides_2(x,y)
r21: div2_3(|0|_0(),y,s_1(z)) -> false_0()
r22: div2_3(s_1(x),y,s_1(z)) -> div2_3(x,y,z)
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs))
r25: if_3(true_0(),x,xs) -> cons_2(x,xs)
r26: if_3(false_0(),x,xs) -> xs
r27: not_2(f,x) -> not2_1(a(f,x))
r28: not2_1(true_0()) -> false_0()
r29: not2_1(false_0()) -> true_0()
r30: sieve_1(nil_0()) -> nil_0()
r31: sieve_1(cons_2(x,xs)) -> cons_2(x,sieve_1(filter_2(not_1(divides_1(x)),xs)))

The set of usable rules consists of

  r23, r24, r25, r26

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        sieve_1#_A(x1) = max{1, x1}
        cons_2_A(x1,x2) = max{18, x1 + 16, x2 + 10}
        filter_2_A(x1,x2) = max{9, x2 + 1}
        not_1_A(x1) = max{13, x1 - 2}
        divides_1_A(x1) = x1 + 15
        nil_0_A = 1
        if_3_A(x1,x2,x3) = max{18, x2 + 17, x3 + 10}
        a_A(x1,x2) = max{12, x1 - 16, x2 - 16}
        true_0_A = 1
        false_0_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          sieve_1#_A(x1) = max{0, x1 - 18}
          cons_2_A(x1,x2) = max{16, x1 - 1, x2}
          filter_2_A(x1,x2) = max{17, x1 - 1, x2}
          not_1_A(x1) = max{8, x1 - 6}
          divides_1_A(x1) = max{14, x1 + 6}
          nil_0_A = 18
          if_3_A(x1,x2,x3) = max{17, x2 - 1, x3}
          a_A(x1,x2) = max{3, x1 - 23, x2 - 23}
          true_0_A = 2
          false_0_A = 2
          sieve_1#!_A(x1) = 0
          cons_2!_A(x1,x2) = max{5, x1 - 34, x2 - 39}
          filter_2!_A(x1,x2) = max{9, x1 - 33, x2 - 19}
          not_1!_A(x1) = max{6, x1 - 37}
          divides_1!_A(x1) = x1 + 6
          nil_0!_A = 0
          if_3!_A(x1,x2,x3) = max{8, x2 - 21, x3 - 38}
          a!_A(x1,x2) = max{4, x1 - 15, x2 - 1}
          true_0!_A = 1
          false_0!_A = 2
  
    partial status:
  
      pi(sieve_1#) = [1]
      pi(cons_2) = [2]
      pi(filter_2) = []
      pi(not_1) = []
      pi(divides_1) = []
      pi(nil_0) = []
      pi(if_3) = [3]
      pi(a) = []
      pi(true_0) = []
      pi(false_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: filter_2#(f,cons_2(x,xs)) -> filter_2#(f,xs)

and R consists of:

r1: a(cons_0(),y) -> cons_1(y)
r2: a(cons_1(x1),y) -> cons_2(x1,y)
r3: a(div2_0(),y) -> div2_1(y)
r4: a(div2_1(x1),y) -> div2_2(x1,y)
r5: a(div2_2(x1,x2),y) -> div2_3(x1,x2,y)
r6: a(divides_0(),y) -> divides_1(y)
r7: a(divides_1(x1),y) -> divides_2(x1,y)
r8: a(filter_0(),y) -> filter_1(y)
r9: a(filter_1(x1),y) -> filter_2(x1,y)
r10: a(if_0(),y) -> if_1(y)
r11: a(if_1(x1),y) -> if_2(x1,y)
r12: a(if_2(x1,x2),y) -> if_3(x1,x2,y)
r13: a(not_0(),y) -> not_1(y)
r14: a(not_1(x1),y) -> not_2(x1,y)
r15: a(not2_0(),y) -> not2_1(y)
r16: a(s_0(),y) -> s_1(y)
r17: a(sieve_0(),y) -> sieve_1(y)
r18: divides_2(|0|_0(),s_1(y)) -> true_0()
r19: divides_2(s_1(x),s_1(y)) -> div2_3(x,s_1(y),y)
r20: div2_3(x,y,|0|_0()) -> divides_2(x,y)
r21: div2_3(|0|_0(),y,s_1(z)) -> false_0()
r22: div2_3(s_1(x),y,s_1(z)) -> div2_3(x,y,z)
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs))
r25: if_3(true_0(),x,xs) -> cons_2(x,xs)
r26: if_3(false_0(),x,xs) -> xs
r27: not_2(f,x) -> not2_1(a(f,x))
r28: not2_1(true_0()) -> false_0()
r29: not2_1(false_0()) -> true_0()
r30: sieve_1(nil_0()) -> nil_0()
r31: sieve_1(cons_2(x,xs)) -> cons_2(x,sieve_1(filter_2(not_1(divides_1(x)),xs)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        filter_2#_A(x1,x2) = x2 + 4
        cons_2_A(x1,x2) = max{x1 + 21, x2 + 5}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          filter_2#_A(x1,x2) = 0
          cons_2_A(x1,x2) = max{x1 + 9, x2 - 14}
          filter_2#!_A(x1,x2) = max{x1 + 12, x2 + 26}
          cons_2!_A(x1,x2) = max{10, x1 - 74, x2 - 109}
  
    partial status:
  
      pi(filter_2#) = [2]
      pi(cons_2) = [1, 2]

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: div2_3#(x,y,|0|_0()) -> divides_2#(x,y)
p2: divides_2#(s_1(x),s_1(y)) -> div2_3#(x,s_1(y),y)
p3: div2_3#(s_1(x),y,s_1(z)) -> div2_3#(x,y,z)

and R consists of:

r1: a(cons_0(),y) -> cons_1(y)
r2: a(cons_1(x1),y) -> cons_2(x1,y)
r3: a(div2_0(),y) -> div2_1(y)
r4: a(div2_1(x1),y) -> div2_2(x1,y)
r5: a(div2_2(x1,x2),y) -> div2_3(x1,x2,y)
r6: a(divides_0(),y) -> divides_1(y)
r7: a(divides_1(x1),y) -> divides_2(x1,y)
r8: a(filter_0(),y) -> filter_1(y)
r9: a(filter_1(x1),y) -> filter_2(x1,y)
r10: a(if_0(),y) -> if_1(y)
r11: a(if_1(x1),y) -> if_2(x1,y)
r12: a(if_2(x1,x2),y) -> if_3(x1,x2,y)
r13: a(not_0(),y) -> not_1(y)
r14: a(not_1(x1),y) -> not_2(x1,y)
r15: a(not2_0(),y) -> not2_1(y)
r16: a(s_0(),y) -> s_1(y)
r17: a(sieve_0(),y) -> sieve_1(y)
r18: divides_2(|0|_0(),s_1(y)) -> true_0()
r19: divides_2(s_1(x),s_1(y)) -> div2_3(x,s_1(y),y)
r20: div2_3(x,y,|0|_0()) -> divides_2(x,y)
r21: div2_3(|0|_0(),y,s_1(z)) -> false_0()
r22: div2_3(s_1(x),y,s_1(z)) -> div2_3(x,y,z)
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs))
r25: if_3(true_0(),x,xs) -> cons_2(x,xs)
r26: if_3(false_0(),x,xs) -> xs
r27: not_2(f,x) -> not2_1(a(f,x))
r28: not2_1(true_0()) -> false_0()
r29: not2_1(false_0()) -> true_0()
r30: sieve_1(nil_0()) -> nil_0()
r31: sieve_1(cons_2(x,xs)) -> cons_2(x,sieve_1(filter_2(not_1(divides_1(x)),xs)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        div2_3#_A(x1,x2,x3) = 0
        |0|_0_A = 3
        divides_2#_A(x1,x2) = 0
        s_1_A(x1) = max{0, x1 - 15}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          div2_3#_A(x1,x2,x3) = max{0, x1 - 5, x2 - 7, x3 - 7}
          |0|_0_A = 6
          divides_2#_A(x1,x2) = max{0, x1 - 11, x2 - 7}
          s_1_A(x1) = x1 + 6
          div2_3#!_A(x1,x2,x3) = max{133, x1 + 131}
          |0|_0!_A = 143
          divides_2#!_A(x1,x2) = max{131, x1 + 128}
          s_1!_A(x1) = max{135, x1 - 10}
  
    partial status:
  
      pi(div2_3#) = []
      pi(|0|_0) = []
      pi(divides_2#) = []
      pi(s_1) = []

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.