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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              not_2#_A(x1,x2) = x1
              a#_A(x1,x2) = x1
              sieve_0_A() = 13
              sieve_1#_A(x1) = 10
              cons_2_A(x1,x2) = x1 + 11
              filter_2#_A(x1,x2) = x1
              not_1_A(x1) = x1 + 2
              divides_1_A(x1) = 8
              filter_1_A(x1) = x1 + 2
              filter_2_A(x1,x2) = x1 + 2
              not_2#!_A(x1,x2) = x1 + 1
              a#!_A(x1,x2) = x1
              sieve_0!_A() = 13
              sieve_1#!_A(x1) = 12
              cons_2!_A(x1,x2) = 27
              filter_2#!_A(x1,x2) = x1 + 1
              not_1!_A(x1) = 13
              divides_1!_A(x1) = 29
              filter_1!_A(x1) = 0
              filter_2!_A(x1,x2) = x1 + 16
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            not_2#_A(x1,x2) = max{1, x1 - 96}
            a#_A(x1,x2) = max{1, x1 - 96}
            sieve_0_A = 32
            sieve_1#_A(x1) = 1
            cons_2_A(x1,x2) = max{x1 + 45, x2}
            filter_2#_A(x1,x2) = max{1, x1 - 96}
            not_1_A(x1) = max{45, x1 + 31}
            divides_1_A(x1) = 64
            filter_1_A(x1) = max{32, x1 + 31}
            filter_2_A(x1,x2) = max{45, x1, x2}
            not_2#!_A(x1,x2) = 1
            a#!_A(x1,x2) = max{2, x1 - 46, x2 - 1}
            sieve_0!_A = 4
            sieve_1#!_A(x1) = 3
            cons_2!_A(x1,x2) = max{x1 + 32, x2 + 80}
            filter_2#!_A(x1,x2) = 3
            not_1!_A(x1) = 81
            divides_1!_A(x1) = x1 + 83
            filter_1!_A(x1) = max{2, x1 - 1}
            filter_2!_A(x1,x2) = max{4, x1 - 16, x2 - 11}
        
  
    partial status:
  
      pi(not_2#) = []
      pi(a#) = []
      pi(sieve_0) = []
      pi(sieve_1#) = []
      pi(cons_2) = [1]
      pi(filter_2#) = []
      pi(not_1) = [1]
      pi(divides_1) = []
      pi(filter_1) = [1]
      pi(filter_2) = [1, 2]

The following rules are removed:

  p1, p2, p3, p5, p6, p7



-- SCC decomposition.

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

p1: filter_2#(f,cons_2(x,xs)) -> filter_2#(f,xs)
p2: 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}
  {p2}


-- 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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              filter_2#_A(x1,x2) = x2
              cons_2_A(x1,x2) = x1 + x2 + 7
              filter_2#!_A(x1,x2) = x2
              cons_2!_A(x1,x2) = 3
        
        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{13, x1 - 24, x2 - 22}
            filter_2#!_A(x1,x2) = max{1, x1 - 1, x2 - 14}
            cons_2!_A(x1,x2) = max{x1 + 9, x2 + 2}
        
  
    partial status:
  
      pi(filter_2#) = [1]
      pi(cons_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: 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, r30

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              sieve_1#_A(x1) = x1
              cons_2_A(x1,x2) = x1 + x2 + 9
              filter_2_A(x1,x2) = x1 + x2 + 2
              not_1_A(x1) = x1 + 2
              divides_1_A(x1) = x1 + 2
              nil_0_A() = 1
              if_3_A(x1,x2,x3) = x2 + x3 + 9
              a_A(x1,x2) = 1
              true_0_A() = 8
              false_0_A() = 6
              sieve_1_A(x1) = x1 + 2
              sieve_1#!_A(x1) = x1
              cons_2!_A(x1,x2) = x1 + x2 + 10
              filter_2!_A(x1,x2) = x1 + x2 + 5
              not_1!_A(x1) = 18
              divides_1!_A(x1) = 18
              nil_0!_A() = 0
              if_3!_A(x1,x2,x3) = x2 + x3 + 11
              a!_A(x1,x2) = 20
              true_0!_A() = 14
              false_0!_A() = 15
              sieve_1!_A(x1) = 19
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            sieve_1#_A(x1) = max{0, x1 - 38}
            cons_2_A(x1,x2) = max{35, x1 - 9, x2}
            filter_2_A(x1,x2) = max{37, x1 - 49, x2}
            not_1_A(x1) = max{39, x1 + 15}
            divides_1_A(x1) = max{15, x1}
            nil_0_A = 38
            if_3_A(x1,x2,x3) = max{36, x2 - 9, x3}
            a_A(x1,x2) = max{14, x1 - 40, x2 - 40}
            true_0_A = 43
            false_0_A = 42
            sieve_1_A(x1) = max{47, x1 - 39}
            sieve_1#!_A(x1) = max{58, x1 + 22}
            cons_2!_A(x1,x2) = max{0, x2 - 47}
            filter_2!_A(x1,x2) = max{58, x1 - 35, x2 + 7}
            not_1!_A(x1) = max{2, x1 - 1}
            divides_1!_A(x1) = x1 + 40
            nil_0!_A = 4
            if_3!_A(x1,x2,x3) = max{41, x2 - 3, x3 - 38}
            a!_A(x1,x2) = max{58, x1 - 16, x2 + 8}
            true_0!_A = 10
            false_0!_A = 20
            sieve_1!_A(x1) = max{3, x1 - 39}
        
  
    partial status:
  
      pi(sieve_1#) = []
      pi(cons_2) = []
      pi(filter_2) = []
      pi(not_1) = [1]
      pi(divides_1) = []
      pi(nil_0) = []
      pi(if_3) = [3]
      pi(a) = [1]
      pi(true_0) = []
      pi(false_0) = []
      pi(sieve_1) = []

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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              div2_3#_A(x1,x2,x3) = 0
              |0|_0_A() = 1
              divides_2#_A(x1,x2) = 0
              s_1_A(x1) = 4
              div2_3#!_A(x1,x2,x3) = x1 + 2
              |0|_0!_A() = 1
              divides_2#!_A(x1,x2) = x1
              s_1!_A(x1) = x1 + 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            div2_3#_A(x1,x2,x3) = max{1, x1 - 13, x2 - 15}
            |0|_0_A = 15
            divides_2#_A(x1,x2) = max{1, x1 - 27, x2 - 15}
            s_1_A(x1) = x1 + 14
            div2_3#!_A(x1,x2,x3) = max{18, x1 - 2}
            |0|_0!_A = 16
            divides_2#!_A(x1,x2) = max{17, x1 - 15}
            s_1!_A(x1) = x1 + 23
        
  
    partial status:
  
      pi(div2_3#) = [1]
      pi(|0|_0) = []
      pi(divides_2#) = [1]
      pi(s_1) = [1]

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.