YES
We show the termination of the TRS R:

  app(f(),app(g(),x)) -> app(g(),app(f(),app(f(),x)))
  app(f(),app(h(),x)) -> app(h(),app(g(),x))
  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(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(g_0(),y) -> g_1(y)
  app(h_0(),y) -> h_1(y)
  app(map_0(),y) -> map_1(y)
  app(map_1(x1),y) -> map_2(x1,y)
  f_1(g_1(x)) -> g_1(f_1(f_1(x)))
  f_1(h_1(x)) -> h_1(g_1(x))
  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#(f_0(),y) -> f_1#(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#(map_1(x1),y) -> map_2#(x1,y)
p5: f_1#(g_1(x)) -> f_1#(f_1(x))
p6: f_1#(g_1(x)) -> f_1#(x)
p7: map_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p8: map_2#(|fun|,cons_2(x,xs)) -> map_2#(|fun|,xs)
p9: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs)
p10: filter_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p11: filter2_4#(true_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p12: 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(f_0(),y) -> f_1(y)
r4: app(filter_0(),y) -> filter_1(y)
r5: app(filter_1(x1),y) -> filter_2(x1,y)
r6: app(filter2_0(),y) -> filter2_1(y)
r7: app(filter2_1(x1),y) -> filter2_2(x1,y)
r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r10: app(g_0(),y) -> g_1(y)
r11: app(h_0(),y) -> h_1(y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: f_1(g_1(x)) -> g_1(f_1(f_1(x)))
r15: f_1(h_1(x)) -> h_1(g_1(x))
r16: map_2(|fun|,nil_0()) -> nil_0()
r17: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r18: filter_2(|fun|,nil_0()) -> nil_0()
r19: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r20: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r21: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p7, p8, p9, p10, p11, p12}
  {p5, p6}


-- 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(f_0(),y) -> f_1(y)
r4: app(filter_0(),y) -> filter_1(y)
r5: app(filter_1(x1),y) -> filter_2(x1,y)
r6: app(filter2_0(),y) -> filter2_1(y)
r7: app(filter2_1(x1),y) -> filter2_2(x1,y)
r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r10: app(g_0(),y) -> g_1(y)
r11: app(h_0(),y) -> h_1(y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: f_1(g_1(x)) -> g_1(f_1(f_1(x)))
r15: f_1(h_1(x)) -> h_1(g_1(x))
r16: map_2(|fun|,nil_0()) -> nil_0()
r17: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r18: filter_2(|fun|,nil_0()) -> nil_0()
r19: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r20: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r21: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The set of usable rules consists of

  r4, r6, r11, r14, r15, r16, r18

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) = 0
              cons_2_A(x1,x2) = 1
              app#_A(x1,x2) = 0
              map_1_A(x1) = x1 + 2
              filter2_3_A(x1,x2,x3) = x1 + x2 + x3 + 6
              filter2_4#_A(x1,x2,x3,x4) = 0
              false_0_A() = 1
              filter_2#_A(x1,x2) = 0
              filter_1_A(x1) = x1 + 6
              app_A(x1,x2) = x1 + x2 + 1
              true_0_A() = 1
              filter_0_A() = 5
              filter2_0_A() = 3
              filter2_1_A(x1) = x1 + 4
              h_0_A() = 1
              h_1_A(x1) = 1
              f_1_A(x1) = 2
              g_1_A(x1) = 1
              map_2_A(x1,x2) = x2 + 1
              nil_0_A() = 1
              filter_2_A(x1,x2) = x1 + 1
              map_2#!_A(x1,x2) = x1 + 1
              cons_2!_A(x1,x2) = 6
              app#!_A(x1,x2) = x1
              map_1!_A(x1) = x1 + 2
              filter2_3!_A(x1,x2,x3) = x1 + x2 + x3 + 6
              filter2_4#!_A(x1,x2,x3,x4) = x2 + 5
              false_0!_A() = 1
              filter_2#!_A(x1,x2) = x1 + 5
              filter_1!_A(x1) = 0
              app!_A(x1,x2) = 4
              true_0!_A() = 1
              filter_0!_A() = 1
              filter2_0!_A() = 0
              filter2_1!_A(x1) = 1
              h_0!_A() = 0
              h_1!_A(x1) = 1
              f_1!_A(x1) = 6
              g_1!_A(x1) = 5
              map_2!_A(x1,x2) = 3
              nil_0!_A() = 0
              filter_2!_A(x1,x2) = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{1, x1 - 54, x2 - 40}
            cons_2_A(x1,x2) = max{39, x1 + 13, x2 + 7}
            app#_A(x1,x2) = max{1, x1 - 54, x2 - 27}
            map_1_A(x1) = max{16, x1 + 13}
            filter2_3_A(x1,x2,x3) = max{x1 + 53, x2 + 15, x3 + 53}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 54, x3 - 27, x4 - 33}
            false_0_A = 49
            filter_2#_A(x1,x2) = max{1, x1 - 54, x2 - 40}
            filter_1_A(x1) = max{23, x1 + 22}
            app_A(x1,x2) = max{49, x1 - 54, x2 + 22}
            true_0_A = 49
            filter_0_A = 0
            filter2_0_A = 0
            filter2_1_A(x1) = max{23, x1 - 8}
            h_0_A = 0
            h_1_A(x1) = max{23, x1 - 30}
            f_1_A(x1) = max{27, x1 - 28}
            g_1_A(x1) = max{26, x1 - 28}
            map_2_A(x1,x2) = max{30, x1 - 41, x2 + 23}
            nil_0_A = 0
            filter_2_A(x1,x2) = max{31, x1 - 46, x2 + 23}
            map_2#!_A(x1,x2) = max{3, x1 - 1, x2 - 9}
            cons_2!_A(x1,x2) = x1 + 2
            app#!_A(x1,x2) = max{5, x1 - 14, x2 + 1}
            map_1!_A(x1) = max{2, x1 - 54}
            filter2_3!_A(x1,x2,x3) = max{x1 + 1, x2 + 77, x3 + 13}
            filter2_4#!_A(x1,x2,x3,x4) = max{38, x2 + 6, x3 + 12, x4 + 6}
            false_0!_A = 8
            filter_2#!_A(x1,x2) = max{7, x1 + 6, x2 - 1}
            filter_1!_A(x1) = max{8, x1 - 2}
            app!_A(x1,x2) = x1 + 7
            true_0!_A = 63
            filter_0!_A = 9
            filter2_0!_A = 2
            filter2_1!_A(x1) = max{1, x1 - 2}
            h_0!_A = 9
            h_1!_A(x1) = x1 + 27
            f_1!_A(x1) = 54
            g_1!_A(x1) = max{28, x1 - 27}
            map_2!_A(x1,x2) = max{3, x1 - 53, x2 - 40}
            nil_0!_A = 0
            filter_2!_A(x1,x2) = max{62, x1 - 1, x2 - 8}
        
  
    partial status:
  
      pi(map_2#) = []
      pi(cons_2) = [1]
      pi(app#) = []
      pi(map_1) = []
      pi(filter2_3) = []
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(filter_2#) = []
      pi(filter_1) = []
      pi(app) = []
      pi(true_0) = []
      pi(filter_0) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(h_0) = []
      pi(h_1) = []
      pi(f_1) = []
      pi(g_1) = []
      pi(map_2) = [2]
      pi(nil_0) = []
      pi(filter_2) = []

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(f_0(),y) -> f_1(y)
r4: app(filter_0(),y) -> filter_1(y)
r5: app(filter_1(x1),y) -> filter_2(x1,y)
r6: app(filter2_0(),y) -> filter2_1(y)
r7: app(filter2_1(x1),y) -> filter2_2(x1,y)
r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r10: app(g_0(),y) -> g_1(y)
r11: app(h_0(),y) -> h_1(y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: f_1(g_1(x)) -> g_1(f_1(f_1(x)))
r15: f_1(h_1(x)) -> h_1(g_1(x))
r16: map_2(|fun|,nil_0()) -> nil_0()
r17: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r18: filter_2(|fun|,nil_0()) -> nil_0()
r19: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r20: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r21: 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(f_0(),y) -> f_1(y)
r4: app(filter_0(),y) -> filter_1(y)
r5: app(filter_1(x1),y) -> filter_2(x1,y)
r6: app(filter2_0(),y) -> filter2_1(y)
r7: app(filter2_1(x1),y) -> filter2_2(x1,y)
r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r10: app(g_0(),y) -> g_1(y)
r11: app(h_0(),y) -> h_1(y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: f_1(g_1(x)) -> g_1(f_1(f_1(x)))
r15: f_1(h_1(x)) -> h_1(g_1(x))
r16: map_2(|fun|,nil_0()) -> nil_0()
r17: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r18: filter_2(|fun|,nil_0()) -> nil_0()
r19: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r20: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r21: 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!):
        
          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 + 2
              map_2#!_A(x1,x2) = x2 + 4
              cons_2!_A(x1,x2) = 3
        
        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{1, x1 - 16, x2 - 16}
            map_2#!_A(x1,x2) = max{1, x1 - 1}
            cons_2!_A(x1,x2) = 0
        
  
    partial status:
  
      pi(map_2#) = [1]
      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(),|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(f_0(),y) -> f_1(y)
r4: app(filter_0(),y) -> filter_1(y)
r5: app(filter_1(x1),y) -> filter_2(x1,y)
r6: app(filter2_0(),y) -> filter2_1(y)
r7: app(filter2_1(x1),y) -> filter2_2(x1,y)
r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r10: app(g_0(),y) -> g_1(y)
r11: app(h_0(),y) -> h_1(y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: f_1(g_1(x)) -> g_1(f_1(f_1(x)))
r15: f_1(h_1(x)) -> h_1(g_1(x))
r16: map_2(|fun|,nil_0()) -> nil_0()
r17: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r18: filter_2(|fun|,nil_0()) -> nil_0()
r19: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r20: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r21: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The set of usable rules consists of

  r6, r7, r10, r11, r12, r16, r18

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:
              filter2_4#_A(x1,x2,x3,x4) = x4
              false_0_A() = 1
              filter_2#_A(x1,x2) = x2
              cons_2_A(x1,x2) = x2 + 3
              app_A(x1,x2) = x1 + x2 + 1
              true_0_A() = 1
              filter2_0_A() = 1
              filter2_1_A(x1) = x1 + 1
              filter2_2_A(x1,x2) = x1 + x2 + 2
              g_0_A() = 1
              g_1_A(x1) = 1
              h_0_A() = 1
              h_1_A(x1) = 1
              map_0_A() = 1
              map_1_A(x1) = x1 + 1
              map_2_A(x1,x2) = x1 + 1
              nil_0_A() = 1
              filter_2_A(x1,x2) = x1 + 1
              filter2_4#!_A(x1,x2,x3,x4) = 4
              false_0!_A() = 5
              filter_2#!_A(x1,x2) = 4
              cons_2!_A(x1,x2) = x1 + 2
              app!_A(x1,x2) = x2 + 8
              true_0!_A() = 3
              filter2_0!_A() = 0
              filter2_1!_A(x1) = 1
              filter2_2!_A(x1,x2) = 2
              g_0!_A() = 0
              g_1!_A(x1) = 2
              h_0!_A() = 0
              h_1!_A(x1) = 1
              map_0!_A() = 8
              map_1!_A(x1) = x1 + 7
              map_2!_A(x1,x2) = x2 + 6
              nil_0!_A() = 0
              filter_2!_A(x1,x2) = x1 + 6
        
        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 - 53, x4 - 14}
            false_0_A = 36
            filter_2#_A(x1,x2) = max{1, x1 - 1, x2 - 16}
            cons_2_A(x1,x2) = max{15, x1 - 37, x2 + 2}
            app_A(x1,x2) = max{36, x1 - 18, x2 + 2}
            true_0_A = 36
            filter2_0_A = 3
            filter2_1_A(x1) = max{3, x1 + 2}
            filter2_2_A(x1,x2) = max{5, x1 - 16, x2 - 11}
            g_0_A = 3
            g_1_A(x1) = max{6, x1 - 8}
            h_0_A = 3
            h_1_A(x1) = max{3, x1 - 22}
            map_0_A = 3
            map_1_A(x1) = max{5, x1 + 2}
            map_2_A(x1,x2) = max{3, x1 - 16, x2 + 2}
            nil_0_A = 4
            filter_2_A(x1,x2) = max{3, x1 - 17, x2}
            filter2_4#!_A(x1,x2,x3,x4) = max{43, x3 - 1, x4 + 41}
            false_0!_A = 2
            filter_2#!_A(x1,x2) = x2 + 40
            cons_2!_A(x1,x2) = max{x1 + 12, x2 + 5}
            app!_A(x1,x2) = max{4, x1 - 4, x2 + 2}
            true_0!_A = 12
            filter2_0!_A = 4
            filter2_1!_A(x1) = max{3, x1 + 1}
            filter2_2!_A(x1,x2) = max{4, x1 - 41, x2 + 2}
            g_0!_A = 1
            g_1!_A(x1) = max{4, x1 - 8}
            h_0!_A = 4
            h_1!_A(x1) = max{3, x1 - 9}
            map_0!_A = 4
            map_1!_A(x1) = 3
            map_2!_A(x1,x2) = max{9, x1 - 12, x2 - 5}
            nil_0!_A = 1
            filter_2!_A(x1,x2) = max{3, x2 - 19}
        
  
    partial status:
  
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(app) = []
      pi(true_0) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(g_0) = []
      pi(g_1) = []
      pi(h_0) = []
      pi(h_1) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(map_2) = []
      pi(nil_0) = []
      pi(filter_2) = []

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: f_1#(g_1(x)) -> f_1#(f_1(x))
p2: f_1#(g_1(x)) -> f_1#(x)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(f_0(),y) -> f_1(y)
r4: app(filter_0(),y) -> filter_1(y)
r5: app(filter_1(x1),y) -> filter_2(x1,y)
r6: app(filter2_0(),y) -> filter2_1(y)
r7: app(filter2_1(x1),y) -> filter2_2(x1,y)
r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r10: app(g_0(),y) -> g_1(y)
r11: app(h_0(),y) -> h_1(y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: f_1(g_1(x)) -> g_1(f_1(f_1(x)))
r15: f_1(h_1(x)) -> h_1(g_1(x))
r16: map_2(|fun|,nil_0()) -> nil_0()
r17: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r18: filter_2(|fun|,nil_0()) -> nil_0()
r19: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r20: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r21: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The set of usable rules consists of

  r14, r15

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:
              f_1#_A(x1) = 0
              g_1_A(x1) = 4
              f_1_A(x1) = x1 + 3
              h_1_A(x1) = 3
              f_1#!_A(x1) = 1
              g_1!_A(x1) = 2
              f_1!_A(x1) = 3
              h_1!_A(x1) = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            f_1#_A(x1) = max{0, x1 - 32}
            g_1_A(x1) = max{43, x1 + 31}
            f_1_A(x1) = max{12, x1}
            h_1_A(x1) = 2
            f_1#!_A(x1) = max{10, x1 - 1}
            g_1!_A(x1) = 8
            f_1!_A(x1) = 9
            h_1!_A(x1) = 10
        
  
    partial status:
  
      pi(f_1#) = []
      pi(g_1) = []
      pi(f_1) = []
      pi(h_1) = [1]

The following rules are removed:

  p1, p2

Then no dependency pair remains.