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

  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

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = max{19, x1 + 18}
        cons_2_A(x1,x2) = max{7, x2}
        app#_A(x1,x2) = max{19, x1 + 17}
        map_1_A(x1) = max{11, x1 + 3}
        filter2_3_A(x1,x2,x3) = max{x1 - 9, x2, x3 + 8}
        filter2_4#_A(x1,x2,x3,x4) = max{24, x2 + 17}
        false_0_A = 25
        filter_2#_A(x1,x2) = max{24, x1 + 17}
        filter_1_A(x1) = max{23, x1}
        app_A(x1,x2) = max{22, x1 + 9, x2 + 9}
        true_0_A = 25
        cons_0_A = 1
        cons_1_A(x1) = max{11, x1 + 1}
        copy_0_A = 1
        copy_1_A(x1) = max{11, x1 - 9}
        copy_2_A(x1,x2) = max{0, x1 - 9, x2 - 9}
        copy_3_A(x1,x2,x3) = max{11, x3 + 4}
        f_0_A = 6
        f_1_A(x1) = max{5, x1 + 4}
        filter_0_A = 15
        filter_2_A(x1,x2) = 21
        filter2_0_A = 1
        filter2_1_A(x1) = max{11, x1 - 9}
        filter2_2_A(x1,x2) = max{21, x1 - 9, x2 - 9}
        filter2_4_A(x1,x2,x3,x4) = 21
        map_0_A = 1
        map_2_A(x1,x2) = max{21, x2 + 8}
        s_0_A = 1
        s_1_A(x1) = x1 + 4
        nil_0_A = 9
        n_0_A = 1
        |0|_0_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          map_2#_A(x1,x2) = max{0, x1 - 49, x2 - 23}
          cons_2_A(x1,x2) = max{x1 + 9, x2}
          app#_A(x1,x2) = max{0, x1 - 103, x2 - 23}
          map_1_A(x1) = max{102, x1 + 54}
          filter2_3_A(x1,x2,x3) = max{x1 + 1, x2 + 1, x3 + 22}
          filter2_4#_A(x1,x2,x3,x4) = max{0, x1 - 102, x2 - 103, x3 - 81, x4 - 32}
          false_0_A = 80
          filter_2#_A(x1,x2) = max{0, x1 - 103, x2 - 32}
          filter_1_A(x1) = max{12, x1 + 1}
          app_A(x1,x2) = max{80, x1 - 13, x2 + 54}
          true_0_A = 79
          cons_0_A = 12
          cons_1_A(x1) = x1 + 36
          copy_0_A = 12
          copy_1_A(x1) = max{12, x1 - 1}
          copy_2_A(x1,x2) = max{x1 - 14, x2 + 54}
          copy_3_A(x1,x2,x3) = max{80, x1 - 153, x2 + 41, x3 + 16}
          f_0_A = 0
          f_1_A(x1) = max{55, x1 + 16}
          filter_0_A = 24
          filter_2_A(x1,x2) = max{55, x2 + 1}
          filter2_0_A = 0
          filter2_1_A(x1) = x1 + 27
          filter2_2_A(x1,x2) = max{x1 + 14, x2 + 22}
          filter2_4_A(x1,x2,x3,x4) = max{55, x3 + 9, x4 + 1}
          map_0_A = 115
          map_2_A(x1,x2) = max{89, x1 - 1, x2 + 54}
          s_0_A = 94
          s_1_A(x1) = max{81, x1 + 54}
          nil_0_A = 90
          n_0_A = 81
          |0|_0_A = 152
          map_2#!_A(x1,x2) = x1 + 54
          cons_2!_A(x1,x2) = 81
          app#!_A(x1,x2) = 0
          map_1!_A(x1) = max{2, x1 - 1}
          filter2_3!_A(x1,x2,x3) = max{72, x1 - 10, x2 - 2, 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{71, x1 - 1}
          app!_A(x1,x2) = max{85, x1 - 23, x2 + 70}
          true_0!_A = 82
          cons_0!_A = 1
          cons_1!_A(x1) = max{2, x1 - 1}
          copy_0!_A = 1
          copy_1!_A(x1) = 2
          copy_2!_A(x1,x2) = max{68, x2 - 1}
          copy_3!_A(x1,x2,x3) = max{69, x1 - 82, x2 + 30, x3 + 5}
          f_0!_A = 1
          f_1!_A(x1) = max{69, x1 + 5}
          filter_0!_A = 72
          filter_2!_A(x1,x2) = max{86, x2 + 80}
          filter2_0!_A = 1
          filter2_1!_A(x1) = max{2, x1 - 12}
          filter2_2!_A(x1,x2) = max{71, x1 - 11, x2 - 1}
          filter2_4!_A(x1,x2,x3,x4) = max{86, x3 + 80, x4 + 80}
          map_0!_A = 1
          map_2!_A(x1,x2) = max{84, x1 + 82, x2 - 8}
          s_0!_A = 1
          s_1!_A(x1) = max{0, x1 - 83}
          nil_0!_A = 83
          n_0!_A = 70
          |0|_0!_A = 1
  
    partial status:
  
      pi(map_2#) = [1]
      pi(cons_2) = []
      pi(app#) = []
      pi(map_1) = []
      pi(filter2_3) = [3]
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(filter_2#) = []
      pi(filter_1) = [1]
      pi(app) = []
      pi(true_0) = []
      pi(cons_0) = []
      pi(cons_1) = [1]
      pi(copy_0) = []
      pi(copy_1) = []
      pi(copy_2) = []
      pi(copy_3) = []
      pi(f_0) = []
      pi(f_1) = []
      pi(filter_0) = []
      pi(filter_2) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(filter2_4) = []
      pi(map_0) = []
      pi(map_2) = []
      pi(s_0) = []
      pi(s_1) = [1]
      pi(nil_0) = []
      pi(n_0) = []
      pi(|0|_0) = []

The following rules are removed:

  p1, p2



-- 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: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p3: filter2_4#(false_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p4: filter_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p5: app#(filter_1(x1),y) -> filter_2#(x1,y)
p6: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs)
p7: 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, p5, p6, p7}


-- 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:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = max{7, x2}
        cons_2_A(x1,x2) = max{x1 + 1, x2 + 8}
  
    base order 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{3, x1 - 14, x2 - 34}
          map_2#!_A(x1,x2) = 0
          cons_2!_A(x1,x2) = max{6, x1 - 23, x2 - 21}
  
    partial status:
  
      pi(map_2#) = [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: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p2: filter2_4#(true_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#(false_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p5: filter_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p6: app#(filter_1(x1),y) -> filter_2#(x1,y)

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

  r17, r22

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{10, x2 + 1}
        filter2_3_A(x1,x2,x3) = max{0, x2 - 1}
        filter2_4#_A(x1,x2,x3,x4) = max{9, x4 - 5}
        true_0_A = 3
        filter_2#_A(x1,x2) = max{9, x2 - 5}
        cons_2_A(x1,x2) = max{22, x1 + 7, x2}
        app_A(x1,x2) = 0
        false_0_A = 3
        filter_1_A(x1) = 8
        f_1_A(x1) = 3
        nil_0_A = 30
        copy_3_A(x1,x2,x3) = max{1, x1 - 5}
        n_0_A = 2
        filter_2_A(x1,x2) = max{20, x1 + 8, x2 + 10}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{1, x1 - 11, x2}
          filter2_3_A(x1,x2,x3) = max{10, x1 - 3, x2 + 6, x3 + 6}
          filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 11, x3 - 5, x4}
          true_0_A = 1
          filter_2#_A(x1,x2) = max{1, x1 - 11, x2}
          cons_2_A(x1,x2) = max{x1, x2}
          app_A(x1,x2) = max{15, x1, x2}
          false_0_A = 1
          filter_1_A(x1) = max{10, x1 + 6}
          f_1_A(x1) = max{22, x1 - 3}
          nil_0_A = 0
          copy_3_A(x1,x2,x3) = max{18, x1 - 20, x2 - 7, x3 - 3}
          n_0_A = 0
          filter_2_A(x1,x2) = max{12, x1 - 3}
          app#!_A(x1,x2) = max{0, x1 - 11}
          filter2_3!_A(x1,x2,x3) = max{x1 + 1, x2 - 7}
          filter2_4#!_A(x1,x2,x3,x4) = max{3, x2 - 6, x3 - 53, x4 - 53}
          true_0!_A = 4
          filter_2#!_A(x1,x2) = max{3, x1 - 6, x2 - 53}
          cons_2!_A(x1,x2) = 0
          app!_A(x1,x2) = 0
          false_0!_A = 4
          filter_1!_A(x1) = max{2, x1 - 7}
          f_1!_A(x1) = max{2, x1 - 23}
          nil_0!_A = 4
          copy_3!_A(x1,x2,x3) = max{5, x1 - 27, x2 - 22, x3 - 51}
          n_0!_A = 0
          filter_2!_A(x1,x2) = max{5, x1 + 2, x2 + 2}
  
    partial status:
  
      pi(app#) = [2]
      pi(filter2_3) = []
      pi(filter2_4#) = []
      pi(true_0) = []
      pi(filter_2#) = []
      pi(cons_2) = [1, 2]
      pi(app) = []
      pi(false_0) = []
      pi(filter_1) = []
      pi(f_1) = []
      pi(nil_0) = []
      pi(copy_3) = []
      pi(n_0) = []
      pi(filter_2) = [2]

The following rules are removed:

  p1, p5, p6



-- SCC decomposition.

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

p1: filter2_4#(true_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#(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:

  {p1, p2, p3}


-- Reduction pair.

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

p1: filter2_4#(true_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#(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 set of usable rules consists of

  r2, r22

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        filter2_4#_A(x1,x2,x3,x4) = max{7, x4 + 4}
        true_0_A = 2
        filter_2#_A(x1,x2) = max{3, x2 + 1}
        cons_2_A(x1,x2) = max{16, x2 + 5}
        app_A(x1,x2) = max{35, x2 + 17}
        false_0_A = 2
        cons_1_A(x1) = 0
        filter_2_A(x1,x2) = x2 + 18
        nil_0_A = 9
  
    base order 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}
          true_0_A = 27
          filter_2#_A(x1,x2) = max{1, x1 - 1}
          cons_2_A(x1,x2) = max{18, x1 - 45}
          app_A(x1,x2) = max{38, x1 - 11}
          false_0_A = 27
          cons_1_A(x1) = x1
          filter_2_A(x1,x2) = max{62, x1 - 21, x2}
          nil_0_A = 14
          filter2_4#!_A(x1,x2,x3,x4) = 1
          true_0!_A = 48
          filter_2#!_A(x1,x2) = 0
          cons_2!_A(x1,x2) = 7
          app!_A(x1,x2) = max{4, x1 - 1, x2 - 61}
          false_0!_A = 1
          cons_1!_A(x1) = max{8, x1 - 13}
          filter_2!_A(x1,x2) = max{x1 + 3, x2 + 93}
          nil_0!_A = 46
  
    partial status:
  
      pi(filter2_4#) = [4]
      pi(true_0) = []
      pi(filter_2#) = [2]
      pi(cons_2) = [2]
      pi(app) = [2]
      pi(false_0) = []
      pi(cons_1) = []
      pi(filter_2) = []
      pi(nil_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: 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

  r22

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        copy_3#_A(x1,x2,x3) = max{14, x1, x3 + 6}
        s_1_A(x1) = x1 + 15
        cons_2_A(x1,x2) = max{7, x2 - 5}
        f_1_A(x1) = 0
        filter_2_A(x1,x2) = max{4, x1 - 15, x2 - 5}
        nil_0_A = 0
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          copy_3#_A(x1,x2,x3) = max{0, x1 - 4, x2 - 2, x3 - 22}
          s_1_A(x1) = max{2, x1}
          cons_2_A(x1,x2) = 21
          f_1_A(x1) = 0
          filter_2_A(x1,x2) = max{21, x1 - 24, x2 - 21}
          nil_0_A = 0
          copy_3#!_A(x1,x2,x3) = max{29, x1 - 4, x2 - 21, x3}
          s_1!_A(x1) = max{22, x1 - 4}
          cons_2!_A(x1,x2) = max{23, x2 - 23}
          f_1!_A(x1) = max{28, x1 - 21}
          filter_2!_A(x1,x2) = max{26, x1 - 47, x2 - 21}
          nil_0!_A = 25
  
    partial status:
  
      pi(copy_3#) = [1]
      pi(s_1) = []
      pi(cons_2) = []
      pi(f_1) = []
      pi(filter_2) = []
      pi(nil_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.