YES
We show the termination of the TRS R:

  app(f(),app(app(cons(),nil()),y)) -> y
  app(f(),app(app(cons(),app(f(),app(app(cons(),nil()),y))),z)) -> app(app(app(copy(),n()),y),z)
  app(app(app(copy(),|0|()),y),z) -> app(f(),z)
  app(app(app(copy(),app(s(),x)),y),z) -> app(app(app(copy(),x),y),app(app(cons(),app(f(),y)),z))
  app(app(map(),|fun|),nil()) -> nil()
  app(app(map(),|fun|),app(app(cons(),x),xs)) -> app(app(cons(),app(|fun|,x)),app(app(map(),|fun|),xs))
  app(app(filter(),|fun|),nil()) -> nil()
  app(app(filter(),|fun|),app(app(cons(),x),xs)) -> app(app(app(app(filter2(),app(|fun|,x)),|fun|),x),xs)
  app(app(app(app(filter2(),true()),|fun|),x),xs) -> app(app(cons(),x),app(app(filter(),|fun|),xs))
  app(app(app(app(filter2(),false()),|fun|),x),xs) -> app(app(filter(),|fun|),xs)

Instead we prove termination of the uncurried system:

  app(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(copy_0(),y) -> copy_1(y)
  app(copy_1(x1),y) -> copy_2(x1,y)
  app(copy_2(x1,x2),y) -> copy_3(x1,x2,y)
  app(f_0(),y) -> f_1(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(s_0(),y) -> s_1(y)
  f_1(cons_2(nil_0(),y)) -> y
  f_1(cons_2(f_1(cons_2(nil_0(),y)),z)) -> copy_3(n_0(),y,z)
  copy_3(|0|_0(),y,z) -> f_1(z)
  copy_3(s_1(x),y,z) -> copy_3(x,y,cons_2(f_1(y),z))
  map_2(|fun|,nil_0()) -> nil_0()
  map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
  filter_2(|fun|,nil_0()) -> nil_0()
  filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
  filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
  filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)


-- SCC decomposition.

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

p1: app#(copy_2(x1,x2),y) -> copy_3#(x1,x2,y)
p2: app#(f_0(),y) -> f_1#(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: f_1#(cons_2(f_1(cons_2(nil_0(),y)),z)) -> copy_3#(n_0(),y,z)
p7: copy_3#(|0|_0(),y,z) -> f_1#(z)
p8: copy_3#(s_1(x),y,z) -> copy_3#(x,y,cons_2(f_1(y),z))
p9: copy_3#(s_1(x),y,z) -> f_1#(y)
p10: map_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p11: map_2#(|fun|,cons_2(x,xs)) -> map_2#(|fun|,xs)
p12: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs)
p13: filter_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p14: filter2_4#(true_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p15: filter2_4#(false_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(copy_0(),y) -> copy_1(y)
r4: app(copy_1(x1),y) -> copy_2(x1,y)
r5: app(copy_2(x1,x2),y) -> copy_3(x1,x2,y)
r6: app(f_0(),y) -> f_1(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(s_0(),y) -> s_1(y)
r16: f_1(cons_2(nil_0(),y)) -> y
r17: f_1(cons_2(f_1(cons_2(nil_0(),y)),z)) -> copy_3(n_0(),y,z)
r18: copy_3(|0|_0(),y,z) -> f_1(z)
r19: copy_3(s_1(x),y,z) -> copy_3(x,y,cons_2(f_1(y),z))
r20: map_2(|fun|,nil_0()) -> nil_0()
r21: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r22: filter_2(|fun|,nil_0()) -> nil_0()
r23: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r24: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r25: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The estimated dependency graph contains the following SCCs:

  {p3, p4, p5, p10, p11, p12, p13, p14, p15}
  {p8}


-- Reduction pair.

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

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

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(copy_0(),y) -> copy_1(y)
r4: app(copy_1(x1),y) -> copy_2(x1,y)
r5: app(copy_2(x1,x2),y) -> copy_3(x1,x2,y)
r6: app(f_0(),y) -> f_1(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(s_0(),y) -> s_1(y)
r16: f_1(cons_2(nil_0(),y)) -> y
r17: f_1(cons_2(f_1(cons_2(nil_0(),y)),z)) -> copy_3(n_0(),y,z)
r18: copy_3(|0|_0(),y,z) -> f_1(z)
r19: copy_3(s_1(x),y,z) -> copy_3(x,y,cons_2(f_1(y),z))
r20: map_2(|fun|,nil_0()) -> nil_0()
r21: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r22: filter_2(|fun|,nil_0()) -> nil_0()
r23: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r24: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r25: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The set of usable rules consists of

  r16

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              map_2#_A(x1,x2) = x1
              cons_2_A(x1,x2) = x1 + x2 + 11
              app#_A(x1,x2) = x1
              map_1_A(x1) = x1 + 5
              filter2_3_A(x1,x2,x3) = x1 + x2 + 7
              filter2_4#_A(x1,x2,x3,x4) = x2 + 7
              false_0_A() = 1
              filter_2#_A(x1,x2) = x1 + 7
              filter_1_A(x1) = x1 + 7
              app_A(x1,x2) = x1 + x2 + 1
              true_0_A() = 1
              f_1_A(x1) = x1 + 9
              nil_0_A() = 1
              map_2#!_A(x1,x2) = x1 + x2 + 43
              cons_2!_A(x1,x2) = 39
              app#!_A(x1,x2) = x1 + x2 + 47
              map_1!_A(x1) = 42
              filter2_3!_A(x1,x2,x3) = 46
              filter2_4#!_A(x1,x2,x3,x4) = x2 + x4 + 47
              false_0!_A() = 34
              filter_2#!_A(x1,x2) = x1 + x2 + 37
              filter_1!_A(x1) = 36
              app!_A(x1,x2) = x1 + 40
              true_0!_A() = 38
              f_1!_A(x1) = 39
              nil_0!_A() = 37
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{1, x2 - 75}
            cons_2_A(x1,x2) = max{74, x1 + 12, x2}
            app#_A(x1,x2) = max{1, x2 - 63}
            map_1_A(x1) = max{15, x1 - 7}
            filter2_3_A(x1,x2,x3) = max{46, x1 - 6, x2 - 7, x3 + 5}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x4 - 75}
            false_0_A = 0
            filter_2#_A(x1,x2) = max{1, x2 - 75}
            filter_1_A(x1) = max{3, x1 - 4}
            app_A(x1,x2) = max{x1 + 2, x2 + 2}
            true_0_A = 1
            f_1_A(x1) = x1 + 13
            nil_0_A = 10
            map_2#!_A(x1,x2) = max{31, x2 - 75}
            cons_2!_A(x1,x2) = max{29, x1 - 11}
            app#!_A(x1,x2) = max{30, x1 - 14, x2 - 76}
            map_1!_A(x1) = 24
            filter2_3!_A(x1,x2,x3) = max{0, x2 - 2}
            filter2_4#!_A(x1,x2,x3,x4) = max{31, x3 - 10}
            false_0!_A = 33
            filter_2#!_A(x1,x2) = max{32, x1 - 13, x2 - 21}
            filter_1!_A(x1) = max{0, x1 - 12}
            app!_A(x1,x2) = max{33, x1 + 13, x2 - 11}
            true_0!_A = 33
            f_1!_A(x1) = max{19, x1 - 48}
            nil_0!_A = 1
        
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [1, 2]
      pi(app#) = [2]
      pi(map_1) = []
      pi(filter2_3) = []
      pi(filter2_4#) = [4]
      pi(false_0) = []
      pi(filter_2#) = [2]
      pi(filter_1) = [1]
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(f_1) = [1]
      pi(nil_0) = []

The following rules are removed:

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

Then no dependency pair remains.

-- Reduction pair.

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

p1: copy_3#(s_1(x),y,z) -> copy_3#(x,y,cons_2(f_1(y),z))

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(copy_0(),y) -> copy_1(y)
r4: app(copy_1(x1),y) -> copy_2(x1,y)
r5: app(copy_2(x1,x2),y) -> copy_3(x1,x2,y)
r6: app(f_0(),y) -> f_1(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(s_0(),y) -> s_1(y)
r16: f_1(cons_2(nil_0(),y)) -> y
r17: f_1(cons_2(f_1(cons_2(nil_0(),y)),z)) -> copy_3(n_0(),y,z)
r18: copy_3(|0|_0(),y,z) -> f_1(z)
r19: copy_3(s_1(x),y,z) -> copy_3(x,y,cons_2(f_1(y),z))
r20: map_2(|fun|,nil_0()) -> nil_0()
r21: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r22: filter_2(|fun|,nil_0()) -> nil_0()
r23: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r24: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r25: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The set of usable rules consists of

  r16, r17

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              copy_3#_A(x1,x2,x3) = x1 + x3
              s_1_A(x1) = x1 + 3
              cons_2_A(x1,x2) = x2 + 2
              f_1_A(x1) = x1 + 4
              nil_0_A() = 3
              copy_3_A(x1,x2,x3) = x1 + x3 + 2
              n_0_A() = 1
              copy_3#!_A(x1,x2,x3) = x1 + x3 + 3
              s_1!_A(x1) = 0
              cons_2!_A(x1,x2) = x2 + 5
              f_1!_A(x1) = 5
              nil_0!_A() = 0
              copy_3!_A(x1,x2,x3) = x1 + 1
              n_0!_A() = 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            copy_3#_A(x1,x2,x3) = max{0, x1 - 3, x2 - 1}
            s_1_A(x1) = max{2, x1}
            cons_2_A(x1,x2) = max{x1 - 8, x2 + 40}
            f_1_A(x1) = max{68, x1 + 24}
            nil_0_A = 55
            copy_3_A(x1,x2,x3) = max{65, x1 - 26, x2 - 46, x3 - 61}
            n_0_A = 12
            copy_3#!_A(x1,x2,x3) = max{20, x1 - 22, x2 - 2, x3 - 17}
            s_1!_A(x1) = max{2, x1 - 23}
            cons_2!_A(x1,x2) = max{19, x1 - 65, x2 + 14}
            f_1!_A(x1) = max{14, x1 - 48}
            nil_0!_A = 0
            copy_3!_A(x1,x2,x3) = max{13, x1 - 21, x3 - 60}
            n_0!_A = 15
        
  
    partial status:
  
      pi(copy_3#) = [2, 3]
      pi(s_1) = []
      pi(cons_2) = [1, 2]
      pi(f_1) = [1]
      pi(nil_0) = []
      pi(copy_3) = []
      pi(n_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.