YES
We show the termination of the TRS R:

  app(app(.(),|1|()),x) -> x
  app(app(.(),x),|1|()) -> x
  app(app(.(),app(i(),x)),x) -> |1|()
  app(app(.(),x),app(i(),x)) -> |1|()
  app(app(.(),app(i(),y)),app(app(.(),y),z)) -> z
  app(app(.(),y),app(app(.(),app(i(),y)),z)) -> z
  app(i(),|1|()) -> |1|()
  app(i(),app(i(),x)) -> x
  app(app(map(),f),nil()) -> nil()
  app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
  app(app(filter(),f),nil()) -> nil()
  app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(app(filter2(),app(f,x)),f),x),xs)
  app(app(app(app(filter2(),true()),f),x),xs) -> app(app(cons(),x),app(app(filter(),f),xs))
  app(app(app(app(filter2(),false()),f),x),xs) -> app(app(filter(),f),xs)

Instead we prove termination of the uncurried system:

  app(._0(),y) -> ._1(y)
  app(._1(x1),y) -> ._2(x1,y)
  app(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(filter_0(),y) -> filter_1(y)
  app(filter_1(x1),y) -> filter_2(x1,y)
  app(filter2_0(),y) -> filter2_1(y)
  app(filter2_1(x1),y) -> filter2_2(x1,y)
  app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
  app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
  app(i_0(),y) -> i_1(y)
  app(map_0(),y) -> map_1(y)
  app(map_1(x1),y) -> map_2(x1,y)
  ._2(|1|_0(),x) -> x
  ._2(x,|1|_0()) -> x
  ._2(i_1(x),x) -> |1|_0()
  ._2(x,i_1(x)) -> |1|_0()
  ._2(i_1(y),._2(y,z)) -> z
  ._2(y,._2(i_1(y),z)) -> z
  i_1(|1|_0()) -> |1|_0()
  i_1(i_1(x)) -> x
  map_2(f,nil_0()) -> nil_0()
  map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
  filter_2(f,nil_0()) -> nil_0()
  filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
  filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
  filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)


-- SCC decomposition.

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

p1: app#(._1(x1),y) -> ._2#(x1,y)
p2: app#(filter_1(x1),y) -> filter_2#(x1,y)
p3: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p4: app#(i_0(),y) -> i_1#(y)
p5: app#(map_1(x1),y) -> map_2#(x1,y)
p6: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p7: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p8: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p9: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p10: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p11: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(._0(),y) -> ._1(y)
r2: app(._1(x1),y) -> ._2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(filter_0(),y) -> filter_1(y)
r6: app(filter_1(x1),y) -> filter_2(x1,y)
r7: app(filter2_0(),y) -> filter2_1(y)
r8: app(filter2_1(x1),y) -> filter2_2(x1,y)
r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r11: app(i_0(),y) -> i_1(y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: ._2(|1|_0(),x) -> x
r15: ._2(x,|1|_0()) -> x
r16: ._2(i_1(x),x) -> |1|_0()
r17: ._2(x,i_1(x)) -> |1|_0()
r18: ._2(i_1(y),._2(y,z)) -> z
r19: ._2(y,._2(i_1(y),z)) -> z
r20: i_1(|1|_0()) -> |1|_0()
r21: i_1(i_1(x)) -> x
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p5, p6, p7, p8, p9, p10, p11}


-- Reduction pair.

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

p1: app#(filter_1(x1),y) -> filter_2#(x1,y)
p2: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p3: app#(map_1(x1),y) -> map_2#(x1,y)
p4: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p5: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p6: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p7: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p8: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p9: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(._0(),y) -> ._1(y)
r2: app(._1(x1),y) -> ._2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(filter_0(),y) -> filter_1(y)
r6: app(filter_1(x1),y) -> filter_2(x1,y)
r7: app(filter2_0(),y) -> filter2_1(y)
r8: app(filter2_1(x1),y) -> filter2_2(x1,y)
r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r11: app(i_0(),y) -> i_1(y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: ._2(|1|_0(),x) -> x
r15: ._2(x,|1|_0()) -> x
r16: ._2(i_1(x),x) -> |1|_0()
r17: ._2(x,i_1(x)) -> |1|_0()
r18: ._2(i_1(y),._2(y,z)) -> z
r19: ._2(y,._2(i_1(y),z)) -> z
r20: i_1(|1|_0()) -> |1|_0()
r21: i_1(i_1(x)) -> x
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          app#_A(x1,x2) = x1
          filter_1_A(x1) = x1 + 7
          filter_2#_A(x1,x2) = x1 + 6
          cons_2_A(x1,x2) = 1
          map_1_A(x1) = x1 + 3
          map_2#_A(x1,x2) = x1 + 2
          filter2_3_A(x1,x2,x3) = x1 + x2 + x3 + 7
          filter2_4#_A(x1,x2,x3,x4) = x2 + 6
          false_0_A() = 5
          app_A(x1,x2) = x1 + x2 + 3
          true_0_A() = 2
          ._0_A() = 1
          ._1_A(x1) = x1 + 2
          ._2_A(x1,x2) = x1 + x2 + 3
          cons_0_A() = 3
          cons_1_A(x1) = 2
          filter_0_A() = 5
          filter_2_A(x1,x2) = 4
          filter2_0_A() = 1
          filter2_1_A(x1) = x1 + 3
          filter2_2_A(x1,x2) = x1 + x2 + 5
          filter2_4_A(x1,x2,x3,x4) = 4
          i_0_A() = 1
          i_1_A(x1) = x1 + 2
          map_0_A() = 1
          map_2_A(x1,x2) = 2
          |1|_0_A() = 1
          nil_0_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{15, x1 - 58, x2 + 1}
          filter_1_A(x1) = x1 + 37
          filter_2#_A(x1,x2) = max{15, x1 - 28, x2 - 12}
          cons_2_A(x1,x2) = max{x1 + 13, x2}
          map_1_A(x1) = x1 + 31
          map_2#_A(x1,x2) = max{15, x1 - 27, x2 + 1}
          filter2_3_A(x1,x2,x3) = max{x1 - 57, x2 + 30, x3 + 57}
          filter2_4#_A(x1,x2,x3,x4) = max{15, x1 - 115, x2 - 28, x3 - 1, x4 - 12}
          false_0_A = 118
          app_A(x1,x2) = max{x1 - 44, x2 + 109}
          true_0_A = 114
          ._0_A = 35
          ._1_A(x1) = x1 + 45
          ._2_A(x1,x2) = max{109, x1 + 1, x2 + 1}
          cons_0_A = 31
          cons_1_A(x1) = x1 + 57
          filter_0_A = 31
          filter_2_A(x1,x2) = max{15, x1 - 14, x2 + 1}
          filter2_0_A = 31
          filter2_1_A(x1) = x1 + 31
          filter2_2_A(x1,x2) = max{x1 - 13, x2 + 74}
          filter2_4_A(x1,x2,x3,x4) = max{15, x1 - 119, x2 - 14, x3 + 13, x4 + 1}
          i_0_A = 31
          i_1_A(x1) = max{107, x1}
          map_0_A = 31
          map_2_A(x1,x2) = max{x1 - 13, x2 + 109}
          |1|_0_A = 109
          nil_0_A = 16
          app#!_A(x1,x2) = max{3, x1 - 58, x2 - 142}
          filter_1!_A(x1) = 0
          filter_2#!_A(x1,x2) = 0
          cons_2!_A(x1,x2) = max{2, x1 - 124, x2 - 141}
          map_1!_A(x1) = max{2, x1 - 34}
          map_2#!_A(x1,x2) = max{0, x2 - 142}
          filter2_3!_A(x1,x2,x3) = max{0, x1 - 114, x2 - 157, x3 - 126}
          filter2_4#!_A(x1,x2,x3,x4) = 0
          false_0!_A = 0
          app!_A(x1,x2) = max{4, x1 - 32, x2 - 3}
          true_0!_A = 6
          ._0!_A = 1
          ._1!_A(x1) = max{2, x1 - 107}
          ._2!_A(x1,x2) = max{13, x1 - 108, x2 - 108}
          cons_0!_A = 2
          cons_1!_A(x1) = 1
          filter_0!_A = 0
          filter_2!_A(x1,x2) = max{5, x1 - 156, x2 - 17}
          filter2_0!_A = 1
          filter2_1!_A(x1) = max{2, x1 - 116}
          filter2_2!_A(x1,x2) = max{4, x1 - 115, x2 - 158}
          filter2_4!_A(x1,x2,x3,x4) = max{1, x1 - 113, x2 - 156, x3 - 125, x4 - 17}
          i_0!_A = 1
          i_1!_A(x1) = max{2, x1 - 110}
          map_0!_A = 1
          map_2!_A(x1,x2) = max{3, x1 - 33, x2 - 17}
          |1|_0!_A = 0
          nil_0!_A = 0
  
    partial status:
  
      pi(app#) = [1]
      pi(filter_1) = [1]
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(map_1) = [1]
      pi(map_2#) = []
      pi(filter2_3) = [3]
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(._0) = []
      pi(._1) = [1]
      pi(._2) = [1, 2]
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_2) = []
      pi(filter2_0) = []
      pi(filter2_1) = [1]
      pi(filter2_2) = [1, 2]
      pi(filter2_4) = []
      pi(i_0) = []
      pi(i_1) = []
      pi(map_0) = []
      pi(map_2) = []
      pi(|1|_0) = []
      pi(nil_0) = []

The following rules are removed:

  p1, p2, p3, p5, p6



-- SCC decomposition.

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

p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p2: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p3: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p4: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(._0(),y) -> ._1(y)
r2: app(._1(x1),y) -> ._2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(filter_0(),y) -> filter_1(y)
r6: app(filter_1(x1),y) -> filter_2(x1,y)
r7: app(filter2_0(),y) -> filter2_1(y)
r8: app(filter2_1(x1),y) -> filter2_2(x1,y)
r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r11: app(i_0(),y) -> i_1(y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: ._2(|1|_0(),x) -> x
r15: ._2(x,|1|_0()) -> x
r16: ._2(i_1(x),x) -> |1|_0()
r17: ._2(x,i_1(x)) -> |1|_0()
r18: ._2(i_1(y),._2(y,z)) -> z
r19: ._2(y,._2(i_1(y),z)) -> z
r20: i_1(|1|_0()) -> |1|_0()
r21: i_1(i_1(x)) -> x
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

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: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(._0(),y) -> ._1(y)
r2: app(._1(x1),y) -> ._2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(filter_0(),y) -> filter_1(y)
r6: app(filter_1(x1),y) -> filter_2(x1,y)
r7: app(filter2_0(),y) -> filter2_1(y)
r8: app(filter2_1(x1),y) -> filter2_2(x1,y)
r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r11: app(i_0(),y) -> i_1(y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: ._2(|1|_0(),x) -> x
r15: ._2(x,|1|_0()) -> x
r16: ._2(i_1(x),x) -> |1|_0()
r17: ._2(x,i_1(x)) -> |1|_0()
r18: ._2(i_1(y),._2(y,z)) -> z
r19: ._2(y,._2(i_1(y),z)) -> z
r20: i_1(|1|_0()) -> |1|_0()
r21: i_1(i_1(x)) -> x
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          map_2#_A(x1,x2) = x2
          cons_2_A(x1,x2) = x1 + x2 + 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          map_2#_A(x1,x2) = 0
          cons_2_A(x1,x2) = max{4, x1 - 4, x2 - 4}
          map_2#!_A(x1,x2) = 6
          cons_2!_A(x1,x2) = max{5, x1 - 17, x2 - 12}
  
    partial status:
  
      pi(map_2#) = [2]
      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: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p2: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p3: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(._0(),y) -> ._1(y)
r2: app(._1(x1),y) -> ._2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(filter_0(),y) -> filter_1(y)
r6: app(filter_1(x1),y) -> filter_2(x1,y)
r7: app(filter2_0(),y) -> filter2_1(y)
r8: app(filter2_1(x1),y) -> filter2_2(x1,y)
r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r11: app(i_0(),y) -> i_1(y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: ._2(|1|_0(),x) -> x
r15: ._2(x,|1|_0()) -> x
r16: ._2(i_1(x),x) -> |1|_0()
r17: ._2(x,i_1(x)) -> |1|_0()
r18: ._2(i_1(y),._2(y,z)) -> z
r19: ._2(y,._2(i_1(y),z)) -> z
r20: i_1(|1|_0()) -> |1|_0()
r21: i_1(i_1(x)) -> x
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r7, r8, r11, r12, r14, r15, r16, r17, r18, r19, r20, r21, r22, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          filter2_4#_A(x1,x2,x3,x4) = x4 + 6
          false_0_A() = 1
          filter_2#_A(x1,x2) = x2 + 5
          cons_2_A(x1,x2) = x2 + 7
          app_A(x1,x2) = x1 + x2 + 13
          true_0_A() = 4
          ._0_A() = 1
          ._1_A(x1) = x1 + 2
          ._2_A(x1,x2) = x1 + x2 + 1
          cons_0_A() = 1
          cons_1_A(x1) = x1 + 2
          filter_0_A() = 1
          filter_1_A(x1) = x1 + 2
          filter2_0_A() = 3
          filter2_1_A(x1) = x1 + 2
          filter2_2_A(x1,x2) = x1 + x2 + 1
          i_0_A() = 1
          i_1_A(x1) = x1 + 2
          map_0_A() = 1
          map_1_A(x1) = x1 + 9
          |1|_0_A() = 1
          map_2_A(x1,x2) = x1 + 8
          nil_0_A() = 1
          filter_2_A(x1,x2) = x1 + x2 + 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 1, x3 - 29, x4 - 1}
          false_0_A = 17
          filter_2#_A(x1,x2) = max{1, x1 - 1, x2 - 4}
          cons_2_A(x1,x2) = max{x1 - 25, x2 + 4}
          app_A(x1,x2) = max{24, x1 - 8, x2 + 9}
          true_0_A = 22
          ._0_A = 0
          ._1_A(x1) = x1 + 9
          ._2_A(x1,x2) = max{24, x1 + 1, x2 + 1}
          cons_0_A = 0
          cons_1_A(x1) = x1 + 7
          filter_0_A = 0
          filter_1_A(x1) = max{0, x1 - 26}
          filter2_0_A = 0
          filter2_1_A(x1) = x1
          filter2_2_A(x1,x2) = max{x1 - 8, x2}
          i_0_A = 0
          i_1_A(x1) = max{22, x1}
          map_0_A = 0
          map_1_A(x1) = x1
          |1|_0_A = 24
          map_2_A(x1,x2) = max{11, x1 - 8, x2 - 2}
          nil_0_A = 11
          filter_2_A(x1,x2) = max{10, x1 - 34, x2}
          filter2_4#!_A(x1,x2,x3,x4) = max{6, x2 - 2, x3 - 30, x4 - 5}
          false_0!_A = 1
          filter_2#!_A(x1,x2) = max{6, x1 - 2, x2 - 5}
          cons_2!_A(x1,x2) = max{5, x1 - 31, x2 - 43}
          app!_A(x1,x2) = max{7, x1 - 1, x2 - 27}
          true_0!_A = 2
          ._0!_A = 2
          ._1!_A(x1) = max{1, x1 - 28}
          ._2!_A(x1,x2) = max{9, x1 - 23, x2 - 27}
          cons_0!_A = 1
          cons_1!_A(x1) = 2
          filter_0!_A = 1
          filter_1!_A(x1) = max{2, x1 - 27}
          filter2_0!_A = 9
          filter2_1!_A(x1) = max{8, x1 - 28}
          filter2_2!_A(x1,x2) = max{9, x1 + 1, x2 - 3}
          i_0!_A = 1
          i_1!_A(x1) = max{2, x1 - 28}
          map_0!_A = 1
          map_1!_A(x1) = max{0, x1 - 28}
          |1|_0!_A = 0
          map_2!_A(x1,x2) = max{0, x2 - 50}
          nil_0!_A = 4
          filter_2!_A(x1,x2) = max{3, x1 - 27, x2 - 47}
  
    partial status:
  
      pi(filter2_4#) = [4]
      pi(false_0) = []
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(app) = [2]
      pi(true_0) = []
      pi(._0) = []
      pi(._1) = [1]
      pi(._2) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_1) = [1]
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = [2]
      pi(i_0) = []
      pi(i_1) = []
      pi(map_0) = []
      pi(map_1) = [1]
      pi(|1|_0) = []
      pi(map_2) = [1]
      pi(nil_0) = []
      pi(filter_2) = []

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.