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:
  
      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!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{135, x2 + 89}
            cons_2_A(x1,x2) = max{32, x1 + 13, x2}
            app#_A(x1,x2) = max{135, x2 + 102}
            map_1_A(x1) = max{82, x1 + 56}
            filter2_3_A(x1,x2,x3) = max{x1 + 28, x2 + 102, x3 + 44}
            filter2_4#_A(x1,x2,x3,x4) = max{135, x4 + 102}
            false_0_A = 0
            filter_2#_A(x1,x2) = max{135, x2 + 102}
            filter_1_A(x1) = x1 + 64
            app_A(x1,x2) = max{120, x1 - 14, x2 + 116}
            true_0_A = 155
            ._0_A = 0
            ._1_A(x1) = x1 + 56
            ._2_A(x1,x2) = max{117, x1, x2}
            cons_0_A = 13
            cons_1_A(x1) = x1 + 45
            filter_0_A = 13
            filter_2_A(x1,x2) = x2 + 116
            filter2_0_A = 60
            filter2_1_A(x1) = x1 + 56
            filter2_2_A(x1,x2) = max{x1 + 42, x2 + 116}
            filter2_4_A(x1,x2,x3,x4) = max{x3 + 13, x4 + 116}
            i_0_A = 13
            i_1_A(x1) = max{120, x1 + 2}
            map_0_A = 0
            map_2_A(x1,x2) = max{x1 - 1, x2 + 116}
            |1|_0_A = 118
            nil_0_A = 52
            map_2#!_A(x1,x2) = max{7, x1 - 28, x2 - 3}
            cons_2!_A(x1,x2) = 6
            app#!_A(x1,x2) = max{9, x1 - 83, x2 - 2}
            map_1!_A(x1) = 4
            filter2_3!_A(x1,x2,x3) = 1
            filter2_4#!_A(x1,x2,x3,x4) = max{18, x1 - 156, x2 - 58, x3 - 40, x4 - 14}
            false_0!_A = 7
            filter_2#!_A(x1,x2) = max{8, x1 - 58, x2 - 14}
            filter_1!_A(x1) = 7
            app!_A(x1,x2) = max{121, x1 - 57, x2 + 3}
            true_0!_A = 9
            ._0!_A = 0
            ._1!_A(x1) = 0
            ._2!_A(x1,x2) = 119
            cons_0!_A = 1
            cons_1!_A(x1) = max{0, x1 - 1}
            filter_0!_A = 8
            filter_2!_A(x1,x2) = max{x1 + 6, x2 - 33}
            filter2_0!_A = 1
            filter2_1!_A(x1) = max{2, x1 - 1}
            filter2_2!_A(x1,x2) = 121
            filter2_4!_A(x1,x2,x3,x4) = max{x1 - 154, x2 + 6, x3 - 20, x4 - 33}
            i_0!_A = 121
            i_1!_A(x1) = max{120, x1 - 1}
            map_0!_A = 1
            map_2!_A(x1,x2) = max{26, x1 - 56, x2 - 53}
            |1|_0!_A = 118
            nil_0!_A = 7
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{0, x1 - 10, x2 - 19}
            cons_2_A(x1,x2) = max{18, x1, x2}
            app#_A(x1,x2) = max{0, x1 - 22, x2 - 19}
            map_1_A(x1) = x1 + 12
            filter2_3_A(x1,x2,x3) = max{x1 - 53, x2 + 2, x3 + 17}
            filter2_4#_A(x1,x2,x3,x4) = max{0, x1 - 75, x2 - 20, x3 - 19, x4 - 19}
            false_0_A = 58
            filter_2#_A(x1,x2) = max{0, x1 - 20, x2 - 19}
            filter_1_A(x1) = x1 + 4
            app_A(x1,x2) = max{56, x1 - 13, x2 + 17}
            true_0_A = 56
            ._0_A = 0
            ._1_A(x1) = x1 + 17
            ._2_A(x1,x2) = max{18, x1, x2}
            cons_0_A = 12
            cons_1_A(x1) = x1 + 17
            filter_0_A = 12
            filter_2_A(x1,x2) = max{20, x1 - 11, x2 + 1}
            filter2_0_A = 12
            filter2_1_A(x1) = x1 + 12
            filter2_2_A(x1,x2) = max{x1 - 1, x2 + 17}
            filter2_4_A(x1,x2,x3,x4) = max{20, x1 - 66, x2 - 11, x3 + 1, x4 + 1}
            i_0_A = 12
            i_1_A(x1) = max{21, x1 + 1}
            map_0_A = 12
            map_2_A(x1,x2) = max{56, x1 - 1, x2 + 17}
            |1|_0_A = 19
            nil_0_A = 21
            map_2#!_A(x1,x2) = max{6, x1 + 1, x2 - 12}
            cons_2!_A(x1,x2) = max{0, x2 - 55}
            app#!_A(x1,x2) = max{5, x1 - 5}
            map_1!_A(x1) = max{0, x1 - 34}
            filter2_3!_A(x1,x2,x3) = max{4, x1 - 57, x2 - 35, x3 - 20}
            filter2_4#!_A(x1,x2,x3,x4) = max{16, x1 - 57, x2 - 2, x4 - 2}
            false_0!_A = 1
            filter_2#!_A(x1,x2) = max{4, x1 - 2, x2 - 2}
            filter_1!_A(x1) = 3
            app!_A(x1,x2) = max{17, x1 - 35, x2 + 2}
            true_0!_A = 5
            ._0!_A = 4
            ._1!_A(x1) = max{3, x1 - 23}
            ._2!_A(x1,x2) = max{4, x1 - 22, x2 - 22}
            cons_0!_A = 1
            cons_1!_A(x1) = max{0, x1 - 1}
            filter_0!_A = 3
            filter_2!_A(x1,x2) = max{0, x1 - 34, x2 - 19}
            filter2_0!_A = 19
            filter2_1!_A(x1) = 18
            filter2_2!_A(x1,x2) = max{3, x1 - 58}
            filter2_4!_A(x1,x2,x3,x4) = max{0, x2 - 34, x3 - 19, x4 - 19}
            i_0!_A = 19
            i_1!_A(x1) = max{18, x1 - 20}
            map_0!_A = 1
            map_2!_A(x1,x2) = max{0, x1 - 34}
            |1|_0!_A = 0
            nil_0!_A = 1
        
  
    partial status:
  
      pi(map_2#) = []
      pi(cons_2) = []
      pi(app#) = []
      pi(map_1) = [1]
      pi(filter2_3) = [2]
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(filter_2#) = []
      pi(filter_1) = [1]
      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) = [1]
      pi(filter2_0) = []
      pi(filter2_1) = [1]
      pi(filter2_2) = [1, 2]
      pi(filter2_4) = [2]
      pi(i_0) = []
      pi(i_1) = [1]
      pi(map_0) = []
      pi(map_2) = [1]
      pi(|1|_0) = []
      pi(nil_0) = []

The following rules are removed:

  p1, p2, p4, p6, 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: 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: ._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}


-- 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:
  
      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!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = 0
            cons_2_A(x1,x2) = max{3, x1 - 11, x2 - 6}
            map_2#!_A(x1,x2) = max{1, x2 - 24}
            cons_2!_A(x1,x2) = max{2, x1 - 20, x2 - 23}
        
        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 - 9, x2 - 4}
            map_2#!_A(x1,x2) = max{6, x1 - 1, x2 - 37}
            cons_2!_A(x1,x2) = max{5, x1 - 16, x2 - 37}
        
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [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: ._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:
  
      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!):
        
          max/plus interpretation on natural numbers:
          
            filter2_4#_A(x1,x2,x3,x4) = max{13, x2 - 1, x4 - 1}
            false_0_A = 5
            filter_2#_A(x1,x2) = max{13, x1 - 1, x2 - 8}
            cons_2_A(x1,x2) = max{20, x1 + 9, x2 + 7}
            app_A(x1,x2) = 5
            true_0_A = 5
            filter2_4#!_A(x1,x2,x3,x4) = max{x3 + 5, x4 + 3}
            false_0!_A = 5
            filter_2#!_A(x1,x2) = max{4, x2 - 3}
            cons_2!_A(x1,x2) = 2
            app!_A(x1,x2) = max{0, x2 - 9}
            true_0!_A = 1
        
        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 - 1, x4 - 3}
            false_0_A = 18
            filter_2#_A(x1,x2) = max{1, x1 - 1, x2 - 6}
            cons_2_A(x1,x2) = max{x1 + 6, x2 + 3}
            app_A(x1,x2) = max{25, x1 - 4}
            true_0_A = 18
            filter2_4#!_A(x1,x2,x3,x4) = max{x2 + 1, x3 - 1, x4 - 4}
            false_0!_A = 6
            filter_2#!_A(x1,x2) = max{x1 + 2, x2 - 6}
            cons_2!_A(x1,x2) = 0
            app!_A(x1,x2) = max{5, x1 - 11, x2 + 2}
            true_0!_A = 6
        
  
    partial status:
  
      pi(filter2_4#) = [4]
      pi(false_0) = []
      pi(filter_2#) = [2]
      pi(cons_2) = [2]
      pi(app) = [1, 2]
      pi(true_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:
  
      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!):
        
          max/plus interpretation on natural numbers:
          
            i_1#_A(x1) = 0
            ._2_A(x1,x2) = max{3, x1 - 4, x2 - 5}
            i_1#!_A(x1) = max{1, x1 - 12}
            ._2!_A(x1,x2) = max{5, x1 - 12, x2 - 4}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            i_1#_A(x1) = 0
            ._2_A(x1,x2) = max{6, x1 - 7, x2 - 8}
            i_1#!_A(x1) = max{1, x1 - 38}
            ._2!_A(x1,x2) = max{6, x1 - 38, x2 - 34}
        
  
    partial status:
  
      pi(i_1#) = [1]
      pi(._2) = [1, 2]

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

  r14, r15, r16, r17, r18, r19, r20

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!):
        
          max/plus interpretation on natural numbers:
          
            ._2#_A(x1,x2) = max{0, x1 - 8, x2 - 11}
            ._2_A(x1,x2) = max{7, x1 + 3, x2}
            |1|_0_A = 6
            i_1_A(x1) = max{2, x1 - 8}
            ._2#!_A(x1,x2) = max{4, x1 - 61, x2 - 66}
            ._2!_A(x1,x2) = max{3, x1 - 51, x2 - 60}
            |1|_0!_A = 0
            i_1!_A(x1) = max{1, x1 - 72}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            ._2#_A(x1,x2) = max{0, x1 - 5, x2 - 6}
            ._2_A(x1,x2) = max{4, x1 + 1, x2}
            |1|_0_A = 4
            i_1_A(x1) = max{2, x1 - 5}
            ._2#!_A(x1,x2) = max{9, x1 - 5, x2 - 6}
            ._2!_A(x1,x2) = max{8, x1 - 5, x2 - 6}
            |1|_0!_A = 0
            i_1!_A(x1) = max{7, x1 - 13}
        
  
    partial status:
  
      pi(._2#) = [1]
      pi(._2) = [1, 2]
      pi(|1|_0) = []
      pi(i_1) = [1]

The following rules are removed:

  p1, p2

Then no dependency pair remains.