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

  r2, r3, r9, r15, r22

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{1, x1 - 97, x2 - 154}
            cons_2_A(x1,x2) = max{66, x1 + 4, x2}
            app#_A(x1,x2) = max{1, x1 - 110, x2 - 150}
            map_1_A(x1) = max{109, x1 + 13}
            filter2_3_A(x1,x2,x3) = max{x1 - 53, x2 + 109, x3 + 55}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 78, x3 - 150, x4 - 154}
            false_0_A = 73
            filter_2#_A(x1,x2) = max{1, x1 - 78, x2 - 154}
            filter_1_A(x1) = x1 + 32
            app_A(x1,x2) = max{75, x1 - 33, x2 + 13}
            true_0_A = 163
            cons_1_A(x1) = x1 + 53
            copy_0_A = 19
            copy_1_A(x1) = max{19, x1 - 6}
            filter2_0_A = 21
            filter2_1_A(x1) = x1 + 13
            s_0_A = 0
            s_1_A(x1) = max{75, x1 + 11}
            filter_2_A(x1,x2) = max{14, x2 + 10}
            nil_0_A = 64
            map_2#!_A(x1,x2) = max{67, x1 - 15, x2}
            cons_2!_A(x1,x2) = 8
            app#!_A(x1,x2) = max{67, x1 - 15, x2 + 4}
            map_1!_A(x1) = 2
            filter2_3!_A(x1,x2,x3) = max{123, x1 - 75, x2 - 1, x3 - 1}
            filter2_4#!_A(x1,x2,x3,x4) = max{x2 + 8, x3 - 2, x4 + 2}
            false_0!_A = 12
            filter_2#!_A(x1,x2) = max{x1 + 8, x2 + 2}
            filter_1!_A(x1) = max{4, x1 - 1}
            app!_A(x1,x2) = max{124, x1 - 20, x2 + 36}
            true_0!_A = 1
            cons_1!_A(x1) = max{8, x1 - 1}
            copy_0!_A = 2
            copy_1!_A(x1) = max{1, x1 - 81}
            filter2_0!_A = 1
            filter2_1!_A(x1) = max{0, x1 - 1}
            s_0!_A = 1
            s_1!_A(x1) = max{0, x1 - 78}
            filter_2!_A(x1,x2) = max{x1 + 2, x2 + 31}
            nil_0!_A = 10
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{49, x1 - 2, x2 - 7}
            cons_2_A(x1,x2) = max{x1 + 6, x2}
            app#_A(x1,x2) = max{49, x1 - 19, x2 - 1}
            map_1_A(x1) = max{65, x1 + 17}
            filter2_3_A(x1,x2,x3) = max{x1 - 21, x2 + 67, x3 + 17}
            filter2_4#_A(x1,x2,x3,x4) = max{49, x2 - 19, x3 - 2, x4 - 7}
            false_0_A = 40
            filter_2#_A(x1,x2) = max{49, x1 - 19, x2 - 7}
            filter_1_A(x1) = max{18, x1 + 17}
            app_A(x1,x2) = max{40, x1 - 19, x2 + 17}
            true_0_A = 40
            cons_1_A(x1) = x1 + 25
            copy_0_A = 18
            copy_1_A(x1) = x1 + 17
            filter2_0_A = 18
            filter2_1_A(x1) = x1 + 17
            s_0_A = 18
            s_1_A(x1) = max{40, x1}
            filter_2_A(x1,x2) = max{41, x1 - 2, x2 + 17}
            nil_0_A = 6
            map_2#!_A(x1,x2) = max{1, x1 - 50, x2 - 2}
            cons_2!_A(x1,x2) = 0
            app#!_A(x1,x2) = max{2, x1 - 66, x2 - 1}
            map_1!_A(x1) = max{0, x1 - 49}
            filter2_3!_A(x1,x2,x3) = max{24, x1 + 19, x2 - 1, x3 - 1}
            filter2_4#!_A(x1,x2,x3,x4) = 0
            false_0!_A = 1
            filter_2#!_A(x1,x2) = 0
            filter_1!_A(x1) = max{24, x1 - 1}
            app!_A(x1,x2) = max{23, x1 - 88, x2 + 21}
            true_0!_A = 1
            cons_1!_A(x1) = max{24, x1 - 1}
            copy_0!_A = 25
            copy_1!_A(x1) = max{24, x1 - 93}
            filter2_0!_A = 23
            filter2_1!_A(x1) = max{22, x1 - 1}
            s_0!_A = 26
            s_1!_A(x1) = max{25, x1 - 92}
            filter_2!_A(x1,x2) = 0
            nil_0!_A = 2
        
  
    partial status:
  
      pi(map_2#) = [1]
      pi(cons_2) = []
      pi(app#) = [1]
      pi(map_1) = [1]
      pi(filter2_3) = [2, 3]
      pi(filter2_4#) = [2]
      pi(false_0) = []
      pi(filter_2#) = [1]
      pi(filter_1) = [1]
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(cons_1) = [1]
      pi(copy_0) = []
      pi(copy_1) = [1]
      pi(filter2_0) = []
      pi(filter2_1) = [1]
      pi(s_0) = []
      pi(s_1) = [1]
      pi(filter_2) = [1]
      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#(|fun|,cons_2(x,xs)) -> map_2#(|fun|,xs)
p2: filter2_4#(false_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p3: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs)
p4: 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 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#(|fun|,cons_2(x,xs)) -> map_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

  (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{49, x1 - 42, x2 - 22}
            map_2#!_A(x1,x2) = max{1, x2 - 50}
            cons_2!_A(x1,x2) = max{5, x1 - 23, x2 - 10}
        
        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{10, x1 - 9, x2 - 4}
            map_2#!_A(x1,x2) = max{7, x1 - 1, x2 - 51}
            cons_2!_A(x1,x2) = max{6, x1 - 48, x2 - 51}
        
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [1, 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(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p2: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs)
p3: 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

  r1, r3, r7, r9, r13, r15, 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:
          
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 1, x4 - 9}
            false_0_A = 25
            filter_2#_A(x1,x2) = max{1, x1 - 1, x2 - 9}
            cons_2_A(x1,x2) = max{8, x2}
            app_A(x1,x2) = max{25, x2 + 9}
            true_0_A = 25
            cons_0_A = 10
            cons_1_A(x1) = max{10, x1 - 9}
            copy_0_A = 19
            copy_1_A(x1) = max{10, x1 - 9}
            filter_0_A = 10
            filter_1_A(x1) = max{10, x1 - 9}
            filter2_0_A = 10
            filter2_1_A(x1) = max{10, x1 - 9}
            map_0_A = 10
            map_1_A(x1) = max{10, x1 - 26}
            s_0_A = 0
            s_1_A(x1) = max{10, x1}
            map_2_A(x1,x2) = max{10, x1 - 16, x2 - 23}
            nil_0_A = 7
            filter2_4#!_A(x1,x2,x3,x4) = max{5, x4 - 9}
            false_0!_A = 4
            filter_2#!_A(x1,x2) = max{5, x2 - 9}
            cons_2!_A(x1,x2) = x2 + 15
            app!_A(x1,x2) = max{x1 - 11, x2 + 17}
            true_0!_A = 2
            cons_0!_A = 1
            cons_1!_A(x1) = max{0, x1 - 1}
            copy_0!_A = 8
            copy_1!_A(x1) = max{7, x1 - 18}
            filter_0!_A = 1
            filter_1!_A(x1) = max{2, x1 - 22}
            filter2_0!_A = 1
            filter2_1!_A(x1) = max{2, x1 - 1}
            map_0!_A = 6
            map_1!_A(x1) = max{5, x1 - 39}
            s_0!_A = 6
            s_1!_A(x1) = max{5, x1 - 9}
            map_2!_A(x1,x2) = max{16, x1 - 38, x2 - 9}
            nil_0!_A = 13
        
        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 - 11, x4 - 1}
            false_0_A = 37
            filter_2#_A(x1,x2) = max{1, x1 - 1, x2 - 1}
            cons_2_A(x1,x2) = max{x1 - 10, x2}
            app_A(x1,x2) = max{37, x1 - 5, x2 - 5}
            true_0_A = 37
            cons_0_A = 4
            cons_1_A(x1) = max{0, x1 - 5}
            copy_0_A = 4
            copy_1_A(x1) = 4
            filter_0_A = 4
            filter_1_A(x1) = max{4, x1 - 5}
            filter2_0_A = 4
            filter2_1_A(x1) = max{4, x1 - 5}
            map_0_A = 4
            map_1_A(x1) = max{0, x1 - 5}
            s_0_A = 4
            s_1_A(x1) = 0
            map_2_A(x1,x2) = max{27, x2}
            nil_0_A = 28
            filter2_4#!_A(x1,x2,x3,x4) = max{18, x3 + 6, x4 + 17}
            false_0!_A = 22
            filter_2#!_A(x1,x2) = max{18, x2 + 17}
            cons_2!_A(x1,x2) = 0
            app!_A(x1,x2) = max{19, x1 - 5}
            true_0!_A = 22
            cons_0!_A = 2
            cons_1!_A(x1) = max{1, x1 - 1}
            copy_0!_A = 1
            copy_1!_A(x1) = x1 + 24
            filter_0!_A = 21
            filter_1!_A(x1) = max{20, x1 - 1}
            filter2_0!_A = 19
            filter2_1!_A(x1) = max{20, x1 - 1}
            map_0!_A = 1
            map_1!_A(x1) = max{20, x1 - 7}
            s_0!_A = 1
            s_1!_A(x1) = max{0, x1 - 1}
            map_2!_A(x1,x2) = max{0, x1 - 6, x2 - 29}
            nil_0!_A = 22
        
  
    partial status:
  
      pi(filter2_4#) = [4]
      pi(false_0) = []
      pi(filter_2#) = [2]
      pi(cons_2) = [2]
      pi(app) = [2]
      pi(true_0) = []
      pi(cons_0) = []
      pi(cons_1) = [1]
      pi(copy_0) = []
      pi(copy_1) = []
      pi(filter_0) = []
      pi(filter_1) = [1]
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(map_0) = []
      pi(map_1) = [1]
      pi(s_0) = []
      pi(s_1) = [1]
      pi(map_2) = []
      pi(nil_0) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: filter2_4#(false_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p2: 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 estimated dependency graph contains the following SCCs:

  (no SCCs)

-- 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!):
        
          max/plus interpretation on natural numbers:
          
            copy_3#_A(x1,x2,x3) = max{0, x1 - 42, x2 - 40, x3 - 4}
            s_1_A(x1) = max{2, x1}
            cons_2_A(x1,x2) = max{3, x1 - 45, x2 - 4}
            f_1_A(x1) = max{10, x1 + 5}
            nil_0_A = 41
            copy_3_A(x1,x2,x3) = max{9, x1 - 42, x2 - 45, x3 - 4}
            n_0_A = 13
            copy_3#!_A(x1,x2,x3) = max{1, x1 - 111, x2 - 114, x3 - 118}
            s_1!_A(x1) = max{2, x1 - 105}
            cons_2!_A(x1,x2) = max{4, x1 - 120, x2 - 90}
            f_1!_A(x1) = 6
            nil_0!_A = 2
            copy_3!_A(x1,x2,x3) = max{3, x1 - 104, x2 - 149, x3 - 109}
            n_0!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            copy_3#_A(x1,x2,x3) = max{0, x1 - 72, x2 - 24, x3 - 4}
            s_1_A(x1) = x1 + 52
            cons_2_A(x1,x2) = max{3, x1 - 25, x2 - 4}
            f_1_A(x1) = max{10, x1 + 5}
            nil_0_A = 38
            copy_3_A(x1,x2,x3) = max{9, x1 - 72, x2 - 25, x3 - 4}
            n_0_A = 71
            copy_3#!_A(x1,x2,x3) = max{1, x1 - 76, x2 - 22, x3 - 6}
            s_1!_A(x1) = max{2, x1 - 73}
            cons_2!_A(x1,x2) = max{6, x1 - 33, x2 - 5}
            f_1!_A(x1) = max{5, x1 - 30}
            nil_0!_A = 3
            copy_3!_A(x1,x2,x3) = max{4, x1 - 72, x2 - 33, x3 - 13}
            n_0!_A = 0
        
  
    partial status:
  
      pi(copy_3#) = [1]
      pi(s_1) = [1]
      pi(cons_2) = [2]
      pi(f_1) = [1]
      pi(nil_0) = []
      pi(copy_3) = [1]
      pi(n_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.