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

  r1, r4, r6, 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!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{1, x1 - 47, x2 - 28}
            cons_2_A(x1,x2) = max{27, x1 + 23, x2 + 23}
            app#_A(x1,x2) = max{1, x1 - 47, x2 - 5}
            map_1_A(x1) = x1 + 11
            filter2_3_A(x1,x2,x3) = max{x1 - 13, x2 + 4, x3 + 39}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 47, x3 - 8, x4 - 5}
            false_0_A = 1
            filter_2#_A(x1,x2) = max{1, x1 - 47, x2 - 28}
            filter_1_A(x1) = max{2, x1 + 1}
            app_A(x1,x2) = max{x1 - 16, x2 + 54}
            true_0_A = 21
            cons_0_A = 0
            cons_1_A(x1) = x1 + 39
            filter_0_A = 13
            filter2_0_A = 0
            filter2_1_A(x1) = max{15, x1 - 13}
            g_0_A = 15
            g_1_A(x1) = max{26, x1 - 57}
            h_0_A = 0
            h_1_A(x1) = max{54, x1 - 27}
            map_0_A = 0
            map_2_A(x1,x2) = max{48, x1 - 22, x2 - 22}
            nil_0_A = 27
            filter_2_A(x1,x2) = max{26, x2}
            map_2#!_A(x1,x2) = max{2, x2 - 1}
            cons_2!_A(x1,x2) = max{x1 + 16, x2 + 22}
            app#!_A(x1,x2) = max{21, x2 + 3}
            map_1!_A(x1) = 1
            filter2_3!_A(x1,x2,x3) = max{x1 + 18, x2 + 25, x3 + 18}
            filter2_4#!_A(x1,x2,x3,x4) = max{19, x4 - 1}
            false_0!_A = 5
            filter_2#!_A(x1,x2) = max{4, x2 - 5}
            filter_1!_A(x1) = 24
            app!_A(x1,x2) = max{25, x1 + 12, x2 + 23}
            true_0!_A = 0
            cons_0!_A = 1
            cons_1!_A(x1) = max{19, x1 - 1}
            filter_0!_A = 25
            filter2_0!_A = 1
            filter2_1!_A(x1) = 0
            g_0!_A = 27
            g_1!_A(x1) = max{26, x1 - 115}
            h_0!_A = 1
            h_1!_A(x1) = 24
            map_0!_A = 1
            map_2!_A(x1,x2) = max{22, x2 - 28}
            nil_0!_A = 0
            filter_2!_A(x1,x2) = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{1, x1 - 50, x2 - 16}
            cons_2_A(x1,x2) = max{x1 + 15, x2 + 7}
            app#_A(x1,x2) = max{1, x1 - 50, x2 - 1}
            map_1_A(x1) = max{16, x1 + 15}
            filter2_3_A(x1,x2,x3) = max{x1 + 15, x2 + 1, x3 + 49}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 49, x3 - 1, x4 - 9}
            false_0_A = 49
            filter_2#_A(x1,x2) = max{1, x1 - 49, x2 - 16}
            filter_1_A(x1) = x1 + 16
            app_A(x1,x2) = max{64, x1 - 34, x2 + 49}
            true_0_A = 49
            cons_0_A = 16
            cons_1_A(x1) = x1 + 35
            filter_0_A = 16
            filter2_0_A = 0
            filter2_1_A(x1) = x1 + 31
            g_0_A = 0
            g_1_A(x1) = max{50, x1 - 54}
            h_0_A = 32
            h_1_A(x1) = max{50, x1 - 51}
            map_0_A = 16
            map_2_A(x1,x2) = max{x1 - 19, x2 + 64}
            nil_0_A = 65
            filter_2_A(x1,x2) = max{50, x1 - 40, x2 + 42}
            map_2#!_A(x1,x2) = max{1, x2 - 9}
            cons_2!_A(x1,x2) = max{x1 + 9, x2 - 64}
            app#!_A(x1,x2) = x2 + 5
            map_1!_A(x1) = max{14, x1 - 17}
            filter2_3!_A(x1,x2,x3) = max{15, x1 - 50, x2 + 8}
            filter2_4#!_A(x1,x2,x3,x4) = max{7, x2 - 1, x3 - 1, x4 + 6}
            false_0!_A = 3
            filter_2#!_A(x1,x2) = max{4, x2 - 9}
            filter_1!_A(x1) = 1
            app!_A(x1,x2) = max{13, x1 - 17, x2 + 11}
            true_0!_A = 9
            cons_0!_A = 1
            cons_1!_A(x1) = max{14, x1 - 1}
            filter_0!_A = 2
            filter2_0!_A = 16
            filter2_1!_A(x1) = 15
            g_0!_A = 2
            g_1!_A(x1) = max{1, x1 - 162}
            h_0!_A = 1
            h_1!_A(x1) = max{14, x1 - 52}
            map_0!_A = 1
            map_2!_A(x1,x2) = max{15, x2 - 66}
            nil_0!_A = 1
            filter_2!_A(x1,x2) = max{2, x1 - 86, x2 - 3}
        
  
    partial status:
  
      pi(map_2#) = []
      pi(cons_2) = []
      pi(app#) = []
      pi(map_1) = [1]
      pi(filter2_3) = []
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(filter_2#) = []
      pi(filter_1) = [1]
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(cons_0) = []
      pi(cons_1) = [1]
      pi(filter_0) = []
      pi(filter2_0) = []
      pi(filter2_1) = [1]
      pi(g_0) = []
      pi(g_1) = []
      pi(h_0) = []
      pi(h_1) = []
      pi(map_0) = []
      pi(map_2) = [2]
      pi(nil_0) = []
      pi(filter_2) = []

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: 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!):
        
          max/plus interpretation on natural numbers:
          
            f_1#_A(x1) = max{0, x1 - 33}
            g_1_A(x1) = x1 + 32
            f_1_A(x1) = x1
            h_1_A(x1) = 2
            f_1#!_A(x1) = 0
            g_1!_A(x1) = max{2, x1 - 28}
            f_1!_A(x1) = max{0, x1 - 29}
            h_1!_A(x1) = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            f_1#_A(x1) = max{0, x1 - 4}
            g_1_A(x1) = max{2, x1}
            f_1_A(x1) = max{3, x1}
            h_1_A(x1) = max{4, x1 - 3}
            f_1#!_A(x1) = 1
            g_1!_A(x1) = 6
            f_1!_A(x1) = 2
            h_1!_A(x1) = 1
        
  
    partial status:
  
      pi(f_1#) = [1]
      pi(g_1) = [1]
      pi(f_1) = []
      pi(h_1) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: f_1#(g_1(x)) -> f_1#(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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: f_1#(g_1(x)) -> f_1#(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!):
        
          max/plus interpretation on natural numbers:
          
            f_1#_A(x1) = max{28, x1 + 15}
            g_1_A(x1) = max{8, x1 - 14}
            f_1_A(x1) = max{13, x1 - 14}
            h_1_A(x1) = max{2, x1 - 9}
            f_1#!_A(x1) = 1
            g_1!_A(x1) = 4
            f_1!_A(x1) = max{3, x1 - 3}
            h_1!_A(x1) = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            f_1#_A(x1) = max{0, x1 - 13}
            g_1_A(x1) = max{15, x1 + 1}
            f_1_A(x1) = max{14, x1}
            h_1_A(x1) = 4
            f_1#!_A(x1) = max{9, x1 - 4}
            g_1!_A(x1) = 8
            f_1!_A(x1) = 12
            h_1!_A(x1) = max{7, x1 - 2}
        
  
    partial status:
  
      pi(f_1#) = []
      pi(g_1) = []
      pi(f_1) = []
      pi(h_1) = []

The following rules are removed:

  p1

Then no dependency pair remains.