YES
We show the termination of the TRS R:

  app(f(),app(g(),x)) -> app(g(),app(g(),app(f(),x)))
  app(f(),app(g(),x)) -> app(g(),app(g(),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(map_0(),y) -> map_1(y)
  app(map_1(x1),y) -> map_2(x1,y)
  f_1(g_1(x)) -> g_1(g_1(f_1(x)))
  f_1(g_1(x)) -> g_1(g_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#(x)
p6: map_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p7: map_2#(|fun|,cons_2(x,xs)) -> map_2#(|fun|,xs)
p8: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs)
p9: filter_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p10: filter2_4#(true_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p11: 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(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: f_1(g_1(x)) -> g_1(g_1(f_1(x)))
r14: f_1(g_1(x)) -> g_1(g_1(g_1(x)))
r15: map_2(|fun|,nil_0()) -> nil_0()
r16: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r17: filter_2(|fun|,nil_0()) -> nil_0()
r18: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r19: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r20: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The estimated dependency graph contains the following SCCs:

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


-- 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(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: f_1(g_1(x)) -> g_1(g_1(f_1(x)))
r14: f_1(g_1(x)) -> g_1(g_1(g_1(x)))
r15: map_2(|fun|,nil_0()) -> nil_0()
r16: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r17: filter_2(|fun|,nil_0()) -> nil_0()
r18: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r19: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r20: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The set of usable rules consists of

  r8, r11

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{2, x1 - 27, x2 - 25}
            cons_2_A(x1,x2) = max{x1 + 24, x2 + 10}
            app#_A(x1,x2) = max{2, x1 - 27, x2 - 25}
            map_1_A(x1) = max{1, x1}
            filter2_3_A(x1,x2,x3) = max{x1 - 1, x2, x3}
            filter2_4#_A(x1,x2,x3,x4) = max{2, x2 - 27, x3 - 27, x4 - 25}
            false_0_A = 29
            filter_2#_A(x1,x2) = max{2, x1 - 27, x2 - 25}
            filter_1_A(x1) = max{28, x1 + 26}
            app_A(x1,x2) = max{3, x1 + 1, x2 + 2}
            true_0_A = 29
            filter2_2_A(x1,x2) = max{x1 + 2, x2 + 24}
            map_0_A = 14
            map_2#!_A(x1,x2) = 0
            cons_2!_A(x1,x2) = max{x1 + 12, x2 + 11}
            app#!_A(x1,x2) = max{11, x2 - 1}
            map_1!_A(x1) = max{13, x1 + 1}
            filter2_3!_A(x1,x2,x3) = max{23, x1 + 2, x2 + 1, x3 + 18}
            filter2_4#!_A(x1,x2,x3,x4) = max{0, x4 - 2}
            false_0!_A = 0
            filter_2#!_A(x1,x2) = 0
            filter_1!_A(x1) = max{22, x1 + 1}
            app!_A(x1,x2) = max{21, x1, x2 + 18}
            true_0!_A = 0
            filter2_2!_A(x1,x2) = max{22, x2 + 1}
            map_0!_A = 1
        
        2. 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) = x1
              cons_2_A(x1,x2) = 14
              app#_A(x1,x2) = x1
              map_1_A(x1) = x1 + 1
              filter2_3_A(x1,x2,x3) = x1 + x2 + x3 + 1
              filter2_4#_A(x1,x2,x3,x4) = x2 + 1
              false_0_A() = 1
              filter_2#_A(x1,x2) = x1 + 1
              filter_1_A(x1) = x1 + 13
              app_A(x1,x2) = x1 + x2 + 5
              true_0_A() = 5
              filter2_2_A(x1,x2) = x1 + x2 + 13
              map_0_A() = 6
              map_2#!_A(x1,x2) = 7
              cons_2!_A(x1,x2) = 9
              app#!_A(x1,x2) = 8
              map_1!_A(x1) = x1 + 8
              filter2_3!_A(x1,x2,x3) = x2 + 6
              filter2_4#!_A(x1,x2,x3,x4) = 7
              false_0!_A() = 0
              filter_2#!_A(x1,x2) = 7
              filter_1!_A(x1) = x1 + 8
              app!_A(x1,x2) = x1 + x2 + 2
              true_0!_A() = 0
              filter2_2!_A(x1,x2) = 5
              map_0!_A() = 0
        
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [1, 2]
      pi(app#) = [2]
      pi(map_1) = []
      pi(filter2_3) = [1, 2]
      pi(filter2_4#) = [4]
      pi(false_0) = []
      pi(filter_2#) = [2]
      pi(filter_1) = [1]
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(filter2_2) = [1, 2]
      pi(map_0) = []

The following rules are removed:

  p1, p2, p3, p4, p6, p7, p8



-- 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(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(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: f_1(g_1(x)) -> g_1(g_1(f_1(x)))
r14: f_1(g_1(x)) -> g_1(g_1(g_1(x)))
r15: map_2(|fun|,nil_0()) -> nil_0()
r16: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r17: filter_2(|fun|,nil_0()) -> nil_0()
r18: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r19: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r20: 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: 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(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: f_1(g_1(x)) -> g_1(g_1(f_1(x)))
r14: f_1(g_1(x)) -> g_1(g_1(g_1(x)))
r15: map_2(|fun|,nil_0()) -> nil_0()
r16: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r17: filter_2(|fun|,nil_0()) -> nil_0()
r18: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r19: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r20: 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:
          
            f_1#_A(x1) = 0
            g_1_A(x1) = max{2, x1 - 3}
            f_1#!_A(x1) = 8
            g_1!_A(x1) = max{7, x1 - 5}
        
        2. 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) = x1
              g_1_A(x1) = x1 + 3
              f_1#!_A(x1) = x1 + 5
              g_1!_A(x1) = 4
        
  
    partial status:
  
      pi(f_1#) = []
      pi(g_1) = []

The following rules are removed:

  p1

Then no dependency pair remains.