YES
We show the termination of the TRS R:

  app(app(|:|(),x),x) -> e()
  app(app(|:|(),e()),x) -> x
  app(app(|:|(),x),app(app(.(),x),y)) -> y
  app(app(|:|(),app(app(/(),x),y)),x) -> y
  app(app(/(),x),x) -> e()
  app(app(/(),x),e()) -> x
  app(app(/(),app(app(.(),y),x)),x) -> y
  app(app(/(),x),app(app(|:|(),y),x)) -> y
  app(app(.(),e()),x) -> x
  app(app(.(),x),e()) -> x
  app(app(.(),x),app(app(|:|(),x),y)) -> y
  app(app(.(),app(app(/(),y),x)),x) -> y
  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(/_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(map_0(),y) -> map_1(y)
  app(map_1(x1),y) -> map_2(x1,y)
  app(|:|_0(),y) -> |:|_1(y)
  app(|:|_1(x1),y) -> |:|_2(x1,y)
  |:|_2(x,x) -> e_0()
  |:|_2(e_0(),x) -> x
  |:|_2(x,._2(x,y)) -> y
  |:|_2(/_2(x,y),x) -> y
  /_2(x,x) -> e_0()
  /_2(x,e_0()) -> x
  /_2(._2(y,x),x) -> y
  /_2(x,|:|_2(y,x)) -> y
  ._2(e_0(),x) -> x
  ._2(x,e_0()) -> x
  ._2(x,|:|_2(x,y)) -> y
  ._2(/_2(y,x),x) -> y
  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#(/_1(x1),y) -> /_2#(x1,y)
p3: app#(filter_1(x1),y) -> filter_2#(x1,y)
p4: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p5: app#(map_1(x1),y) -> map_2#(x1,y)
p6: app#(|:|_1(x1),y) -> |:|_2#(x1,y)
p7: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p8: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p9: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p10: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p11: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p12: 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(/_0(),y) -> /_1(y)
r4: app(/_1(x1),y) -> /_2(x1,y)
r5: app(cons_0(),y) -> cons_1(y)
r6: app(cons_1(x1),y) -> cons_2(x1,y)
r7: app(filter_0(),y) -> filter_1(y)
r8: app(filter_1(x1),y) -> filter_2(x1,y)
r9: app(filter2_0(),y) -> filter2_1(y)
r10: app(filter2_1(x1),y) -> filter2_2(x1,y)
r11: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r12: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r13: app(map_0(),y) -> map_1(y)
r14: app(map_1(x1),y) -> map_2(x1,y)
r15: app(|:|_0(),y) -> |:|_1(y)
r16: app(|:|_1(x1),y) -> |:|_2(x1,y)
r17: |:|_2(x,x) -> e_0()
r18: |:|_2(e_0(),x) -> x
r19: |:|_2(x,._2(x,y)) -> y
r20: |:|_2(/_2(x,y),x) -> y
r21: /_2(x,x) -> e_0()
r22: /_2(x,e_0()) -> x
r23: /_2(._2(y,x),x) -> y
r24: /_2(x,|:|_2(y,x)) -> y
r25: ._2(e_0(),x) -> x
r26: ._2(x,e_0()) -> x
r27: ._2(x,|:|_2(x,y)) -> y
r28: ._2(/_2(y,x),x) -> y
r29: map_2(f,nil_0()) -> nil_0()
r30: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r31: filter_2(f,nil_0()) -> nil_0()
r32: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r33: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r34: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p3, p4, p5, p7, p8, p9, p10, p11, p12}


-- 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(/_0(),y) -> /_1(y)
r4: app(/_1(x1),y) -> /_2(x1,y)
r5: app(cons_0(),y) -> cons_1(y)
r6: app(cons_1(x1),y) -> cons_2(x1,y)
r7: app(filter_0(),y) -> filter_1(y)
r8: app(filter_1(x1),y) -> filter_2(x1,y)
r9: app(filter2_0(),y) -> filter2_1(y)
r10: app(filter2_1(x1),y) -> filter2_2(x1,y)
r11: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r12: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r13: app(map_0(),y) -> map_1(y)
r14: app(map_1(x1),y) -> map_2(x1,y)
r15: app(|:|_0(),y) -> |:|_1(y)
r16: app(|:|_1(x1),y) -> |:|_2(x1,y)
r17: |:|_2(x,x) -> e_0()
r18: |:|_2(e_0(),x) -> x
r19: |:|_2(x,._2(x,y)) -> y
r20: |:|_2(/_2(x,y),x) -> y
r21: /_2(x,x) -> e_0()
r22: /_2(x,e_0()) -> x
r23: /_2(._2(y,x),x) -> y
r24: /_2(x,|:|_2(y,x)) -> y
r25: ._2(e_0(),x) -> x
r26: ._2(x,e_0()) -> x
r27: ._2(x,|:|_2(x,y)) -> y
r28: ._2(/_2(y,x),x) -> y
r29: map_2(f,nil_0()) -> nil_0()
r30: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r31: filter_2(f,nil_0()) -> nil_0()
r32: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r33: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r34: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r17, r18, r20, r22, r24, r29, r31

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{2, x1 - 30, x2 - 35}
            filter_1_A(x1) = max{2, x1 + 1}
            filter_2#_A(x1,x2) = max{2, x1 - 29, x2 - 35}
            cons_2_A(x1,x2) = max{x1 + 26, x2 + 36}
            map_1_A(x1) = max{2, x1 + 1}
            map_2#_A(x1,x2) = max{2, x1 - 29, x2 - 37}
            filter2_3_A(x1,x2,x3) = max{x1 + 1, x2 + 1, x3 + 29}
            filter2_4#_A(x1,x2,x3,x4) = max{2, x2 - 29, x3 - 9, x4 - 35}
            false_0_A = 97
            app_A(x1,x2) = max{97, x1 - 3, x2 + 60}
            true_0_A = 97
            |:|_2_A(x1,x2) = max{97, x1 + 1, x2 + 1}
            e_0_A = 2
            /_2_A(x1,x2) = max{61, x1 + 1, x2 - 1}
            map_2_A(x1,x2) = max{123, x2 - 26}
            nil_0_A = 27
            filter_2_A(x1,x2) = max{61, x1 - 38}
            app#!_A(x1,x2) = max{23, x2 + 13}
            filter_1!_A(x1) = max{10, x1 + 3}
            filter_2#!_A(x1,x2) = max{11, x2 - 2}
            cons_2!_A(x1,x2) = 4
            map_1!_A(x1) = max{1, x1 - 32}
            map_2#!_A(x1,x2) = max{12, x2 - 12}
            filter2_3!_A(x1,x2,x3) = max{4, x1 - 32, x2 - 29, x3 - 1}
            filter2_4#!_A(x1,x2,x3,x4) = max{12, x4 - 1}
            false_0!_A = 3
            app!_A(x1,x2) = max{x1 - 30, x2 + 25}
            true_0!_A = 12
            |:|_2!_A(x1,x2) = max{13, x1 - 2, x2 - 1}
            e_0!_A = 12
            /_2!_A(x1,x2) = max{11, x1 - 2, x2 - 1}
            map_2!_A(x1,x2) = max{8, x1 - 31, x2 - 86}
            nil_0!_A = 1
            filter_2!_A(x1,x2) = max{2, x2 - 28}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{1, x1 - 11, x2 - 17}
            filter_1_A(x1) = x1
            filter_2#_A(x1,x2) = max{1, x1 - 11, x2 - 17}
            cons_2_A(x1,x2) = max{16, x1 + 5, x2 + 5}
            map_1_A(x1) = max{10, x1 + 2}
            map_2#_A(x1,x2) = max{1, x1 - 9, x2 - 17}
            filter2_3_A(x1,x2,x3) = max{x1, x2, x3 + 5}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 11, x3 - 12, x4 - 17}
            false_0_A = 8
            app_A(x1,x2) = 6
            true_0_A = 8
            |:|_2_A(x1,x2) = max{2, x1 + 1, x2 + 1}
            e_0_A = 2
            /_2_A(x1,x2) = max{2, x1 + 1, x2 - 1}
            map_2_A(x1,x2) = max{7, x1 - 4, x2 - 4}
            nil_0_A = 6
            filter_2_A(x1,x2) = max{6, x1 - 4, x2 - 4}
            app#!_A(x1,x2) = max{2, x1 - 6}
            filter_1!_A(x1) = 5
            filter_2#!_A(x1,x2) = max{9, x1 - 12, x2 - 20}
            cons_2!_A(x1,x2) = 0
            map_1!_A(x1) = x1 + 4
            map_2#!_A(x1,x2) = max{3, x1 - 5}
            filter2_3!_A(x1,x2,x3) = max{0, x1 - 10, x2 - 10, x3 - 17}
            filter2_4#!_A(x1,x2,x3,x4) = max{1, x3 - 16}
            false_0!_A = 8
            app!_A(x1,x2) = max{5, x1 - 11, x2 - 18}
            true_0!_A = 10
            |:|_2!_A(x1,x2) = max{4, x1 - 17, x2 - 19}
            e_0!_A = 5
            /_2!_A(x1,x2) = max{4, x1 - 12, x2 - 19}
            map_2!_A(x1,x2) = max{0, x1 - 12, x2 - 17}
            nil_0!_A = 0
            filter_2!_A(x1,x2) = 7
        
  
    partial status:
  
      pi(app#) = []
      pi(filter_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(|:|_2) = [1, 2]
      pi(e_0) = []
      pi(/_2) = [1, 2]
      pi(map_2) = [2]
      pi(nil_0) = []
      pi(filter_2) = [1]

The following rules are removed:

  p1, p2, p3, p4, p5, p6, p7, p8, p9

Then no dependency pair remains.