YES
We show the termination of the TRS R:

  app(g(),app(h(),app(g(),x))) -> app(g(),x)
  app(g(),app(g(),x)) -> app(g(),app(h(),app(g(),x)))
  app(h(),app(h(),x)) -> app(h(),app(app(f(),app(h(),x)),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(f_1(x1),y) -> f_2(x1,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)
  g_1(h_1(g_1(x))) -> g_1(x)
  g_1(g_1(x)) -> g_1(h_1(g_1(x)))
  h_1(h_1(x)) -> h_1(f_2(h_1(x),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#(filter_1(x1),y) -> filter_2#(x1,y)
p2: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p3: app#(g_0(),y) -> g_1#(y)
p4: app#(h_0(),y) -> h_1#(y)
p5: app#(map_1(x1),y) -> map_2#(x1,y)
p6: g_1#(g_1(x)) -> g_1#(h_1(g_1(x)))
p7: g_1#(g_1(x)) -> h_1#(g_1(x))
p8: h_1#(h_1(x)) -> h_1#(f_2(h_1(x),x))
p9: map_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p10: map_2#(|fun|,cons_2(x,xs)) -> map_2#(|fun|,xs)
p11: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs)
p12: filter_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p13: filter2_4#(true_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p14: 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(f_1(x1),y) -> f_2(x1,y)
r5: app(filter_0(),y) -> filter_1(y)
r6: app(filter_1(x1),y) -> filter_2(x1,y)
r7: app(filter2_0(),y) -> filter2_1(y)
r8: app(filter2_1(x1),y) -> filter2_2(x1,y)
r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r11: app(g_0(),y) -> g_1(y)
r12: app(h_0(),y) -> h_1(y)
r13: app(map_0(),y) -> map_1(y)
r14: app(map_1(x1),y) -> map_2(x1,y)
r15: g_1(h_1(g_1(x))) -> g_1(x)
r16: g_1(g_1(x)) -> g_1(h_1(g_1(x)))
r17: h_1(h_1(x)) -> h_1(f_2(h_1(x),x))
r18: map_2(|fun|,nil_0()) -> nil_0()
r19: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r20: filter_2(|fun|,nil_0()) -> nil_0()
r21: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r22: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r23: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p5, p9, p10, p11, p12, p13, p14}
  {p6}


-- Reduction pair.

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

p1: app#(filter_1(x1),y) -> filter_2#(x1,y)
p2: filter_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p3: app#(map_1(x1),y) -> map_2#(x1,y)
p4: map_2#(|fun|,cons_2(x,xs)) -> map_2#(|fun|,xs)
p5: map_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p6: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p7: filter2_4#(false_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
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(f_1(x1),y) -> f_2(x1,y)
r5: app(filter_0(),y) -> filter_1(y)
r6: app(filter_1(x1),y) -> filter_2(x1,y)
r7: app(filter2_0(),y) -> filter2_1(y)
r8: app(filter2_1(x1),y) -> filter2_2(x1,y)
r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r11: app(g_0(),y) -> g_1(y)
r12: app(h_0(),y) -> h_1(y)
r13: app(map_0(),y) -> map_1(y)
r14: app(map_1(x1),y) -> map_2(x1,y)
r15: g_1(h_1(g_1(x))) -> g_1(x)
r16: g_1(g_1(x)) -> g_1(h_1(g_1(x)))
r17: h_1(h_1(x)) -> h_1(f_2(h_1(x),x))
r18: map_2(|fun|,nil_0()) -> nil_0()
r19: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r20: filter_2(|fun|,nil_0()) -> nil_0()
r21: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r22: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r23: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The set of usable rules consists of

  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:
          
            app#_A(x1,x2) = max{1, x1 - 31, x2 - 10}
            filter_1_A(x1) = max{19, x1 + 15}
            filter_2#_A(x1,x2) = max{1, x1 - 16, x2 - 21}
            cons_2_A(x1,x2) = max{20, x1 + 11, x2 + 11}
            map_1_A(x1) = max{19, x1 + 2}
            map_2#_A(x1,x2) = max{1, x1 - 31, x2 - 21}
            filter2_3_A(x1,x2,x3) = max{x1 + 25, x2 + 27, x3 + 30}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 16, x3 - 10, x4 - 21}
            false_0_A = 11
            app_A(x1,x2) = max{13, x1 - 29, x2 + 1}
            true_0_A = 2
            filter_2_A(x1,x2) = max{29, x1 - 13, x2 + 2}
            nil_0_A = 30
            app#!_A(x1,x2) = 3
            filter_1!_A(x1) = x1 + 33
            filter_2#!_A(x1,x2) = 3
            cons_2!_A(x1,x2) = 4
            map_1!_A(x1) = 1
            map_2#!_A(x1,x2) = 3
            filter2_3!_A(x1,x2,x3) = max{x1 + 7, x2 + 7}
            filter2_4#!_A(x1,x2,x3,x4) = 3
            false_0!_A = 32
            app!_A(x1,x2) = max{x1 - 20, x2 + 30}
            true_0!_A = 5
            filter_2!_A(x1,x2) = x1 + 33
            nil_0!_A = 32
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{4, x1 - 53, x2 - 42}
            filter_1_A(x1) = max{52, x1 + 3}
            filter_2#_A(x1,x2) = max{4, x1 - 53, x2 - 42}
            cons_2_A(x1,x2) = max{41, x1 + 39, x2 + 24}
            map_1_A(x1) = max{52, x1 + 37}
            map_2#_A(x1,x2) = max{4, x1 - 16, x2 - 42}
            filter2_3_A(x1,x2,x3) = max{x1 + 48, x2, x3 + 56}
            filter2_4#_A(x1,x2,x3,x4) = max{4, x2 - 53, x3 - 3, x4 - 42}
            false_0_A = 5
            app_A(x1,x2) = max{51, x1 - 55, x2 + 2}
            true_0_A = 5
            filter_2_A(x1,x2) = max{29, x1 - 79, x2 + 3}
            nil_0_A = 20
            app#!_A(x1,x2) = x2 + 6
            filter_1!_A(x1) = max{3, x1 + 2}
            filter_2#!_A(x1,x2) = max{1, x2 - 32}
            cons_2!_A(x1,x2) = max{x1 + 41, x2 - 30}
            map_1!_A(x1) = max{3, x1 - 16}
            map_2#!_A(x1,x2) = max{5, x2 - 32}
            filter2_3!_A(x1,x2,x3) = 50
            filter2_4#!_A(x1,x2,x3,x4) = max{5, x4 - 31}
            false_0!_A = 54
            app!_A(x1,x2) = max{2, x1}
            true_0!_A = 54
            filter_2!_A(x1,x2) = max{53, x1 + 1, x2}
            nil_0!_A = 0
        
  
    partial status:
  
      pi(app#) = []
      pi(filter_1) = [1]
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(map_1) = [1]
      pi(map_2#) = []
      pi(filter2_3) = []
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(filter_2) = []
      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: g_1#(g_1(x)) -> g_1#(h_1(g_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(f_1(x1),y) -> f_2(x1,y)
r5: app(filter_0(),y) -> filter_1(y)
r6: app(filter_1(x1),y) -> filter_2(x1,y)
r7: app(filter2_0(),y) -> filter2_1(y)
r8: app(filter2_1(x1),y) -> filter2_2(x1,y)
r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r11: app(g_0(),y) -> g_1(y)
r12: app(h_0(),y) -> h_1(y)
r13: app(map_0(),y) -> map_1(y)
r14: app(map_1(x1),y) -> map_2(x1,y)
r15: g_1(h_1(g_1(x))) -> g_1(x)
r16: g_1(g_1(x)) -> g_1(h_1(g_1(x)))
r17: h_1(h_1(x)) -> h_1(f_2(h_1(x),x))
r18: map_2(|fun|,nil_0()) -> nil_0()
r19: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r20: filter_2(|fun|,nil_0()) -> nil_0()
r21: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r22: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r23: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The set of usable rules consists of

  r15, 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:
          
            g_1#_A(x1) = max{0, x1 - 21}
            g_1_A(x1) = max{20, x1 + 6}
            h_1_A(x1) = max{3, x1 - 5}
            f_2_A(x1,x2) = max{4, x1 - 4, x2 - 5}
            g_1#!_A(x1) = max{1, x1 - 30}
            g_1!_A(x1) = max{3, x1 - 28}
            h_1!_A(x1) = max{2, x1 - 49}
            f_2!_A(x1,x2) = max{3, x1 - 48, x2 - 27}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            g_1#_A(x1) = max{0, x1 - 26}
            g_1_A(x1) = max{25, x1 + 17}
            h_1_A(x1) = max{9, x1 - 3}
            f_2_A(x1,x2) = max{2, x1 - 11, x2 - 3}
            g_1#!_A(x1) = max{2, x1 - 45}
            g_1!_A(x1) = max{3, x1 - 18}
            h_1!_A(x1) = max{1, x1 - 44}
            f_2!_A(x1,x2) = max{2, x1 - 10, x2 - 44}
        
  
    partial status:
  
      pi(g_1#) = [1]
      pi(g_1) = []
      pi(h_1) = []
      pi(f_2) = []

The following rules are removed:

  p1

Then no dependency pair remains.