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(app(.(),app(app(.(),x),y)),z) -> app(app(.(),x),app(app(.(),y),z))
  app(i(),|1|()) -> |1|()
  app(i(),app(i(),x)) -> x
  app(i(),app(app(.(),x),y)) -> app(app(.(),app(i(),y)),app(i(),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
  ._2(._2(x,y),z) -> ._2(x,._2(y,z))
  i_1(|1|_0()) -> |1|_0()
  i_1(i_1(x)) -> x
  i_1(._2(x,y)) -> ._2(i_1(y),i_1(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: ._2#(._2(x,y),z) -> ._2#(x,._2(y,z))
p7: ._2#(._2(x,y),z) -> ._2#(y,z)
p8: i_1#(._2(x,y)) -> ._2#(i_1(y),i_1(x))
p9: i_1#(._2(x,y)) -> i_1#(y)
p10: i_1#(._2(x,y)) -> i_1#(x)
p11: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p12: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p13: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p14: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p15: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p16: 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: ._2(._2(x,y),z) -> ._2(x,._2(y,z))
r21: i_1(|1|_0()) -> |1|_0()
r22: i_1(i_1(x)) -> x
r23: i_1(._2(x,y)) -> ._2(i_1(y),i_1(x))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p5, p11, p12, p13, p14, p15, p16}
  {p9, p10}
  {p6, p7}


-- Reduction pair.

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

p1: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p2: app#(map_1(x1),y) -> map_2#(x1,y)
p3: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p4: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p5: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p6: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p7: app#(filter_1(x1),y) -> filter_2#(x1,y)
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: ._2(._2(x,y),z) -> ._2(x,._2(y,z))
r21: i_1(|1|_0()) -> |1|_0()
r22: i_1(i_1(x)) -> x
r23: i_1(._2(x,y)) -> ._2(i_1(y),i_1(x))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: 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, r28, r29

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = 2
        cons_2_A(x1,x2) = x2
        app#_A(x1,x2) = 2
        map_1_A(x1) = max{12, x1 - 2}
        filter2_3_A(x1,x2,x3) = max{5, x1 + 1}
        filter2_4#_A(x1,x2,x3,x4) = 2
        false_0_A = 3
        filter_2#_A(x1,x2) = 2
        filter_1_A(x1) = max{5, x1 - 3}
        app_A(x1,x2) = max{x1 - 6, x2 + 16}
        true_0_A = 3
        ._0_A = 8
        ._1_A(x1) = x1 + 7
        ._2_A(x1,x2) = max{8, x1, x2}
        cons_0_A = 1
        cons_1_A(x1) = max{2, x1 - 1}
        filter_0_A = 4
        filter_2_A(x1,x2) = max{0, x2 - 2}
        filter2_0_A = 4
        filter2_1_A(x1) = x1 + 15
        filter2_2_A(x1,x2) = max{10, x1 + 8, x2 + 4}
        filter2_4_A(x1,x2,x3,x4) = max{0, x4 - 2}
        i_0_A = 5
        i_1_A(x1) = max{6, x1 + 1}
        map_0_A = 5
        map_2_A(x1,x2) = max{5, x2 - 6}
        |1|_0_A = 7
        nil_0_A = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          map_2#_A(x1,x2) = max{1, x1 - 143, x2 - 16}
          cons_2_A(x1,x2) = max{x1 + 15, x2}
          app#_A(x1,x2) = max{1, x1 - 162, x2 - 15}
          map_1_A(x1) = max{161, x1 + 19}
          filter2_3_A(x1,x2,x3) = max{128, x1 + 76, x2 + 17, x3 + 112}
          filter2_4#_A(x1,x2,x3,x4) = max{1, x1 - 127, x2 - 145, x3 - 50, x4 - 30}
          false_0_A = 113
          filter_2#_A(x1,x2) = max{1, x1 - 145, x2 - 30}
          filter_1_A(x1) = max{128, x1 + 19}
          app_A(x1,x2) = max{128, x1 - 18, x2 + 112}
          true_0_A = 115
          ._0_A = 188
          ._1_A(x1) = max{129, x1 + 112}
          ._2_A(x1,x2) = max{128, x1, x2}
          cons_0_A = 188
          cons_1_A(x1) = x1 + 112
          filter_0_A = 19
          filter_2_A(x1,x2) = max{128, x1 - 2, x2 + 1}
          filter2_0_A = 145
          filter2_1_A(x1) = max{128, x1 + 112}
          filter2_2_A(x1,x2) = max{128, x1 + 94, x2 + 112}
          filter2_4_A(x1,x2,x3,x4) = max{128, x1 - 129, x2 - 2, x3 + 15, x4 + 1}
          i_0_A = 188
          i_1_A(x1) = max{170, x1 + 1}
          map_0_A = 195
          map_2_A(x1,x2) = max{143, x1 + 1, x2 + 112}
          |1|_0_A = 129
          nil_0_A = 144
          map_2#!_A(x1,x2) = max{x1 - 170, x2 + 9}
          cons_2!_A(x1,x2) = 5
          app#!_A(x1,x2) = max{11, x1 - 188, x2 + 10}
          map_1!_A(x1) = max{7, x1 - 170}
          filter2_3!_A(x1,x2,x3) = max{4, x1 - 112, x2 - 188, x3 - 147}
          filter2_4#!_A(x1,x2,x3,x4) = max{11, x2 - 187, x3 - 77, x4 - 4}
          false_0!_A = 4
          filter_2#!_A(x1,x2) = max{3, x1 - 187, x2 - 4}
          filter_1!_A(x1) = 0
          app!_A(x1,x2) = max{5, x1 - 187, x2 - 146}
          true_0!_A = 1
          ._0!_A = 1
          ._1!_A(x1) = max{0, x1 - 147}
          ._2!_A(x1,x2) = 0
          cons_0!_A = 1
          cons_1!_A(x1) = max{5, x1 - 147}
          filter_0!_A = 1
          filter_2!_A(x1,x2) = max{5, x1 + 1, x2 + 3}
          filter2_0!_A = 1
          filter2_1!_A(x1) = max{2, x1 - 147}
          filter2_2!_A(x1,x2) = max{3, x1 - 112}
          filter2_4!_A(x1,x2,x3,x4) = max{5, x1 - 110, x2 + 1, x4 + 3}
          i_0!_A = 1
          i_1!_A(x1) = 6
          map_0!_A = 1
          map_2!_A(x1,x2) = max{6, x1 - 167, x2 - 145}
          |1|_0!_A = 0
          nil_0!_A = 0
  
    partial status:
  
      pi(map_2#) = []
      pi(cons_2) = []
      pi(app#) = []
      pi(map_1) = []
      pi(filter2_3) = []
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(filter_2#) = []
      pi(filter_1) = []
      pi(app) = [2]
      pi(true_0) = []
      pi(._0) = []
      pi(._1) = []
      pi(._2) = [1, 2]
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_2) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(filter2_4) = []
      pi(i_0) = []
      pi(i_1) = [1]
      pi(map_0) = []
      pi(map_2) = []
      pi(|1|_0) = []
      pi(nil_0) = []

The following rules are removed:

  p1, p2, p7



-- 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: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p3: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p4: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p5: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p6: 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: ._2(._2(x,y),z) -> ._2(x,._2(y,z))
r21: i_1(|1|_0()) -> |1|_0()
r22: i_1(i_1(x)) -> x
r23: i_1(._2(x,y)) -> ._2(i_1(y),i_1(x))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1}
  {p2, p3, p4, p5, p6}


-- 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: ._2(._2(x,y),z) -> ._2(x,._2(y,z))
r21: i_1(|1|_0()) -> |1|_0()
r22: i_1(i_1(x)) -> x
r23: i_1(._2(x,y)) -> ._2(i_1(y),i_1(x))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: 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:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = max{15, x2}
        cons_2_A(x1,x2) = max{x1 + 9, x2 + 15}
  
    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{7, x2 - 3}
          map_2#!_A(x1,x2) = 0
          cons_2!_A(x1,x2) = max{3, x1 - 19, x2 - 28}
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [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: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p2: filter2_4#(true_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#(false_0(),f,x,xs) -> filter_2#(f,xs)
p5: filter_2#(f,cons_2(x,xs)) -> app#(f,x)

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: ._2(._2(x,y),z) -> ._2(x,._2(y,z))
r21: i_1(|1|_0()) -> |1|_0()
r22: i_1(i_1(x)) -> x
r23: i_1(._2(x,y)) -> ._2(i_1(y),i_1(x))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: 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, r28, r29

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = 8
        filter2_3_A(x1,x2,x3) = max{14, x1 - 10, x2 - 10, x3 - 10}
        filter2_4#_A(x1,x2,x3,x4) = 8
        true_0_A = 37
        filter_2#_A(x1,x2) = 8
        cons_2_A(x1,x2) = max{7, x1 - 25, x2}
        app_A(x1,x2) = max{14, x1 - 4, x2 + 6}
        false_0_A = 9
        ._0_A = 1
        ._1_A(x1) = x1 + 5
        ._2_A(x1,x2) = max{12, x1, x2}
        cons_0_A = 3
        cons_1_A(x1) = max{1, x1 - 1}
        filter_0_A = 1
        filter_1_A(x1) = max{3, x1 - 1}
        filter_2_A(x1,x2) = max{2, x1 - 15, x2 + 1}
        filter2_0_A = 6
        filter2_1_A(x1) = max{1, x1 - 1}
        filter2_2_A(x1,x2) = max{2, x1 - 6, x2 - 6}
        filter2_4_A(x1,x2,x3,x4) = max{2, x1 - 30, x2 - 15, x3 - 25, x4 + 1}
        i_0_A = 1
        i_1_A(x1) = max{13, x1 + 2}
        map_0_A = 1
        map_1_A(x1) = max{2, x1 - 25}
        map_2_A(x1,x2) = max{x1 - 29, x2 + 6}
        |1|_0_A = 1
        nil_0_A = 6
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 39, x2 - 73}
          filter2_3_A(x1,x2,x3) = max{x1 - 70, x2 + 8, x3 + 38}
          filter2_4#_A(x1,x2,x3,x4) = max{0, x1 - 109, x2 - 32, x3 - 71, x4 - 73}
          true_0_A = 78
          filter_2#_A(x1,x2) = max{0, x1 - 32, x2 - 73}
          cons_2_A(x1,x2) = max{42, x1 + 2, x2}
          app_A(x1,x2) = max{78, x1 - 8, x2 + 38}
          false_0_A = 78
          ._0_A = 0
          ._1_A(x1) = max{39, x1 + 9}
          ._2_A(x1,x2) = max{39, x1, x2}
          cons_0_A = 0
          cons_1_A(x1) = x1 + 10
          filter_0_A = 0
          filter_1_A(x1) = 7
          filter_2_A(x1,x2) = max{48, x2 + 1}
          filter2_0_A = 9
          filter2_1_A(x1) = max{2, x1 - 54}
          filter2_2_A(x1,x2) = max{x1 - 62, x2 + 38}
          filter2_4_A(x1,x2,x3,x4) = max{48, x3 + 2, x4 + 1}
          i_0_A = 7
          i_1_A(x1) = max{41, x1 + 1}
          map_0_A = 0
          map_1_A(x1) = x1 + 2
          map_2_A(x1,x2) = max{41, x1 - 6, x2 + 38}
          |1|_0_A = 40
          nil_0_A = 49
          app#!_A(x1,x2) = max{4, x1 - 32, x2 - 44}
          filter2_3!_A(x1,x2,x3) = max{2, x1 - 168, x2 - 134, x3 - 42}
          filter2_4#!_A(x1,x2,x3,x4) = max{5, x1 - 167, x2 - 25, x3 - 43, x4 - 45}
          true_0!_A = 1
          filter_2#!_A(x1,x2) = max{5, x1 - 25, x2 - 45}
          cons_2!_A(x1,x2) = max{55, x1 - 88, x2 - 46}
          app!_A(x1,x2) = max{57, x1 + 53, x2 + 5}
          false_0!_A = 1
          ._0!_A = 1
          ._1!_A(x1) = max{2, x1 - 84}
          ._2!_A(x1,x2) = max{3, x1 - 83, x2 - 83}
          cons_0!_A = 1
          cons_1!_A(x1) = max{54, x1 - 89}
          filter_0!_A = 1
          filter_1!_A(x1) = max{58, x1 - 1}
          filter_2!_A(x1,x2) = max{59, x1 - 133, x2 - 43}
          filter2_0!_A = 64
          filter2_1!_A(x1) = 63
          filter2_2!_A(x1,x2) = max{1, x1 - 169, x2 - 135}
          filter2_4!_A(x1,x2,x3,x4) = max{59, x1 - 125, x2 - 133, x3 - 41, x4 - 43}
          i_0!_A = 1
          i_1!_A(x1) = max{4, x1 - 41}
          map_0!_A = 1
          map_1!_A(x1) = max{2, x1 - 1}
          map_2!_A(x1,x2) = max{56, x1 + 54, x2 + 3}
          |1|_0!_A = 0
          nil_0!_A = 53
  
    partial status:
  
      pi(app#) = []
      pi(filter2_3) = []
      pi(filter2_4#) = []
      pi(true_0) = []
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(app) = []
      pi(false_0) = []
      pi(._0) = []
      pi(._1) = [1]
      pi(._2) = [1, 2]
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(filter_2) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(filter2_4) = []
      pi(i_0) = []
      pi(i_1) = [1]
      pi(map_0) = []
      pi(map_1) = []
      pi(map_2) = []
      pi(|1|_0) = []
      pi(nil_0) = []

The following rules are removed:

  p1, p5



-- SCC decomposition.

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

p1: filter2_4#(true_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#(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: ._2(._2(x,y),z) -> ._2(x,._2(y,z))
r21: i_1(|1|_0()) -> |1|_0()
r22: i_1(i_1(x)) -> x
r23: i_1(._2(x,y)) -> ._2(i_1(y),i_1(x))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

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: filter2_4#(true_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#(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: ._2(._2(x,y),z) -> ._2(x,._2(y,z))
r21: i_1(|1|_0()) -> |1|_0()
r22: i_1(i_1(x)) -> x
r23: i_1(._2(x,y)) -> ._2(i_1(y),i_1(x))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: 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, r28, r29

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        filter2_4#_A(x1,x2,x3,x4) = max{6, x4 + 1}
        true_0_A = 2
        filter_2#_A(x1,x2) = max{3, x2}
        cons_2_A(x1,x2) = x2 + 6
        app_A(x1,x2) = max{x1 - 7, x2 + 21}
        false_0_A = 2
        ._0_A = 1
        ._1_A(x1) = x1 + 8
        ._2_A(x1,x2) = max{9, x1, x2}
        cons_0_A = 6
        cons_1_A(x1) = 3
        filter_0_A = 1
        filter_1_A(x1) = max{6, x1 - 1}
        filter_2_A(x1,x2) = x2 + 5
        filter2_0_A = 1
        filter2_1_A(x1) = x1 + 17
        filter2_2_A(x1,x2) = max{x1 + 9, x2 - 1}
        filter2_3_A(x1,x2,x3) = max{x1 - 1, x2 - 9, x3 + 5}
        filter2_4_A(x1,x2,x3,x4) = x4 + 11
        i_0_A = 6
        i_1_A(x1) = max{20, x1 + 10}
        map_0_A = 5
        map_1_A(x1) = x1 + 6
        map_2_A(x1,x2) = x2 + 8
        |1|_0_A = 11
        nil_0_A = 8
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          filter2_4#_A(x1,x2,x3,x4) = max{0, x1 - 26, x2 - 26, x3 - 17, x4 - 7}
          true_0_A = 9
          filter_2#_A(x1,x2) = max{0, x1 - 26, x2 - 7}
          cons_2_A(x1,x2) = max{6, x1 - 10, x2}
          app_A(x1,x2) = max{x1 - 6, x2 + 9}
          false_0_A = 9
          ._0_A = 0
          ._1_A(x1) = x1 + 7
          ._2_A(x1,x2) = max{4, x1, x2}
          cons_0_A = 2
          cons_1_A(x1) = x1 + 2
          filter_0_A = 4
          filter_1_A(x1) = x1 + 2
          filter_2_A(x1,x2) = max{8, x1 - 37, x2 + 1}
          filter2_0_A = 2
          filter2_1_A(x1) = max{2, x1 - 13}
          filter2_2_A(x1,x2) = max{4, x1 - 19, x2 - 25}
          filter2_3_A(x1,x2,x3) = max{x1 - 25, x2 - 31, x3 + 2}
          filter2_4_A(x1,x2,x3,x4) = max{8, x1 - 31, x2 - 37, x3 - 9, x4 + 1}
          i_0_A = 2
          i_1_A(x1) = max{9, x1 + 3}
          map_0_A = 2
          map_1_A(x1) = x1 + 5
          map_2_A(x1,x2) = max{x1 - 10, x2 + 9}
          |1|_0_A = 5
          nil_0_A = 0
          filter2_4#!_A(x1,x2,x3,x4) = max{0, x4 - 4}
          true_0!_A = 6
          filter_2#!_A(x1,x2) = 0
          cons_2!_A(x1,x2) = max{2, x2 - 3}
          app!_A(x1,x2) = max{35, x1 - 3, x2 + 30}
          false_0!_A = 0
          ._0!_A = 1
          ._1!_A(x1) = max{32, x1 + 31}
          ._2!_A(x1,x2) = max{33, x1 - 10, x2 - 12}
          cons_0!_A = 1
          cons_1!_A(x1) = max{35, x1 - 1}
          filter_0!_A = 1
          filter_1!_A(x1) = max{34, x1 - 26}
          filter_2!_A(x1,x2) = max{33, x1 - 26, x2 - 1}
          filter2_0!_A = 1
          filter2_1!_A(x1) = max{35, x1 - 23}
          filter2_2!_A(x1,x2) = max{34, x1 - 22, x2 - 1}
          filter2_3!_A(x1,x2,x3) = max{36, x1 - 21, x2 - 27, x3 - 1}
          filter2_4!_A(x1,x2,x3,x4) = max{33, x1 - 20, x2 - 26, x4 - 1}
          i_0!_A = 1
          i_1!_A(x1) = max{34, x1 - 6}
          map_0!_A = 1
          map_1!_A(x1) = max{35, x1 - 1}
          map_2!_A(x1,x2) = max{31, x1 + 1, x2 + 25}
          |1|_0!_A = 0
          nil_0!_A = 32
  
    partial status:
  
      pi(filter2_4#) = [4]
      pi(true_0) = []
      pi(filter_2#) = [2]
      pi(cons_2) = []
      pi(app) = [2]
      pi(false_0) = []
      pi(._0) = []
      pi(._1) = [1]
      pi(._2) = [1, 2]
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(filter_2) = [2]
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = [1]
      pi(filter2_3) = []
      pi(filter2_4) = []
      pi(i_0) = []
      pi(i_1) = [1]
      pi(map_0) = []
      pi(map_1) = []
      pi(map_2) = []
      pi(|1|_0) = []
      pi(nil_0) = []

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.

-- Reduction pair.

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

p1: i_1#(._2(x,y)) -> i_1#(x)
p2: i_1#(._2(x,y)) -> i_1#(y)

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: ._2(._2(x,y),z) -> ._2(x,._2(y,z))
r21: i_1(|1|_0()) -> |1|_0()
r22: i_1(i_1(x)) -> x
r23: i_1(._2(x,y)) -> ._2(i_1(y),i_1(x))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: 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:
  
      max/plus interpretation on natural numbers:
      
        i_1#_A(x1) = max{13, x1 - 1}
        ._2_A(x1,x2) = max{x1 + 19, x2 + 36}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          i_1#_A(x1) = 0
          ._2_A(x1,x2) = max{33, x1 - 35, x2 - 34}
          i_1#!_A(x1) = 1
          ._2!_A(x1,x2) = max{6, x1 - 72, x2 - 34}
  
    partial status:
  
      pi(i_1#) = []
      pi(._2) = [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: ._2#(._2(x,y),z) -> ._2#(x,._2(y,z))
p2: ._2#(._2(x,y),z) -> ._2#(y,z)

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: ._2(._2(x,y),z) -> ._2(x,._2(y,z))
r21: i_1(|1|_0()) -> |1|_0()
r22: i_1(i_1(x)) -> x
r23: i_1(._2(x,y)) -> ._2(i_1(y),i_1(x))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: 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:
  
      max/plus interpretation on natural numbers:
      
        ._2#_A(x1,x2) = 0
        ._2_A(x1,x2) = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          ._2#_A(x1,x2) = max{0, x1 - 8}
          ._2_A(x1,x2) = max{7, x1 + 1, x2 + 1}
          ._2#!_A(x1,x2) = max{3, x1 - 3}
          ._2!_A(x1,x2) = max{5, x2 + 4}
  
    partial status:
  
      pi(._2#) = []
      pi(._2) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.