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

  r14, r15, r16, r17, r18, r19, r20, r21, r24

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:
          
            app#_A(x1,x2) = max{1, x1 - 18, x2 - 29}
            filter_1_A(x1) = max{15, x1 + 13}
            filter_2#_A(x1,x2) = max{1, x1 - 7, x2 - 29}
            cons_2_A(x1,x2) = max{28, x1 + 8, x2}
            map_1_A(x1) = x1 + 13
            map_2#_A(x1,x2) = max{1, x1 - 5, x2 - 29}
            filter2_3_A(x1,x2,x3) = max{x1 + 17, x2 + 11, x3 + 17}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 7, x3 - 21, x4 - 29}
            false_0_A = 0
            app_A(x1,x2) = max{17, x2 - 7}
            true_0_A = 14
            ._2_A(x1,x2) = max{3, x1 + 1, x2 + 1}
            |1|_0_A = 3
            i_1_A(x1) = max{1, x1}
            filter_2_A(x1,x2) = max{3, x1 - 2, x2 + 1}
            nil_0_A = 9
            app#!_A(x1,x2) = max{5, x1 - 2, x2 - 28}
            filter_1!_A(x1) = max{11, x1 + 8}
            filter_2#!_A(x1,x2) = max{12, x1 + 9, x2 - 29}
            cons_2!_A(x1,x2) = max{x1 + 4, x2 - 5}
            map_1!_A(x1) = max{7, x1 - 1}
            map_2#!_A(x1,x2) = max{x1 + 6, x2 - 29}
            filter2_3!_A(x1,x2,x3) = max{x1 + 13, x3 + 5}
            filter2_4#!_A(x1,x2,x3,x4) = max{12, x2 + 9, x4 - 29}
            false_0!_A = 0
            app!_A(x1,x2) = max{10, x1 - 16, x2 + 8}
            true_0!_A = 4
            ._2!_A(x1,x2) = max{9, x1 - 4, x2 - 2}
            |1|_0!_A = 0
            i_1!_A(x1) = max{11, x1 - 4}
            filter_2!_A(x1,x2) = 0
            nil_0!_A = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              app#_A(x1,x2) = x1
              filter_1_A(x1) = x1 + 2
              filter_2#_A(x1,x2) = x1 + 2
              cons_2_A(x1,x2) = 1
              map_1_A(x1) = x1 + 2
              map_2#_A(x1,x2) = x1
              filter2_3_A(x1,x2,x3) = x1 + x2 + x3 + 5
              filter2_4#_A(x1,x2,x3,x4) = x2 + 2
              false_0_A() = 1
              app_A(x1,x2) = x1 + x2 + 1
              true_0_A() = 1
              ._2_A(x1,x2) = x1 + x2
              |1|_0_A() = 1
              i_1_A(x1) = x1 + 1
              filter_2_A(x1,x2) = x1 + 1
              nil_0_A() = 1
              app#!_A(x1,x2) = x1 + 6
              filter_1!_A(x1) = 2
              filter_2#!_A(x1,x2) = 7
              cons_2!_A(x1,x2) = x2 + 5
              map_1!_A(x1) = 8
              map_2#!_A(x1,x2) = x1 + 7
              filter2_3!_A(x1,x2,x3) = x1 + x2 + x3 + 8
              filter2_4#!_A(x1,x2,x3,x4) = 7
              false_0!_A() = 0
              app!_A(x1,x2) = x1
              true_0!_A() = 4
              ._2!_A(x1,x2) = x1 + 1
              |1|_0!_A() = 0
              i_1!_A(x1) = 2
              filter_2!_A(x1,x2) = 3
              nil_0!_A() = 0
        
  
    partial status:
  
      pi(app#) = []
      pi(filter_1) = []
      pi(filter_2#) = []
      pi(cons_2) = [2]
      pi(map_1) = [1]
      pi(map_2#) = []
      pi(filter2_3) = [1, 2]
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(._2) = [1, 2]
      pi(|1|_0) = []
      pi(i_1) = [1]
      pi(filter_2) = []
      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:
  
      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{4, x1 - 9, x2 - 4}
            map_2#!_A(x1,x2) = 0
            cons_2!_A(x1,x2) = max{0, x1 - 29, x2 - 14}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          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 + 5
              map_2#!_A(x1,x2) = x2
              cons_2!_A(x1,x2) = x1 + x2 + 9
        
  
    partial status:
  
      pi(map_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

  r24

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{1, x2 - 1, x3 - 28, x4 - 6}
            false_0_A = 0
            filter_2#_A(x1,x2) = max{1, x1 - 1, x2 - 25}
            cons_2_A(x1,x2) = max{24, x1 - 3, x2 + 19}
            app_A(x1,x2) = 2
            true_0_A = 0
            filter_2_A(x1,x2) = max{21, x1 + 2, x2}
            nil_0_A = 22
            filter2_4#!_A(x1,x2,x3,x4) = 2
            false_0!_A = 23
            filter_2#!_A(x1,x2) = 2
            cons_2!_A(x1,x2) = max{x1 + 3, x2 + 22}
            app!_A(x1,x2) = max{26, x1 + 16, x2 + 23}
            true_0!_A = 1
            filter_2!_A(x1,x2) = max{22, x1 - 8, x2 - 25}
            nil_0!_A = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              filter2_4#_A(x1,x2,x3,x4) = x4
              false_0_A() = 1
              filter_2#_A(x1,x2) = x2
              cons_2_A(x1,x2) = x2 + 2
              app_A(x1,x2) = 0
              true_0_A() = 1
              filter_2_A(x1,x2) = x1 + x2 + 2
              nil_0_A() = 1
              filter2_4#!_A(x1,x2,x3,x4) = x4 + 5
              false_0!_A() = 0
              filter_2#!_A(x1,x2) = x2 + 4
              cons_2!_A(x1,x2) = x1 + x2 + 8
              app!_A(x1,x2) = x1 + x2 + 5
              true_0!_A() = 5
              filter_2!_A(x1,x2) = x2 + 8
              nil_0!_A() = 0
        
  
    partial status:
  
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(filter_2#) = []
      pi(cons_2) = [1]
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(filter_2) = [2]
      pi(nil_0) = []

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.