YES
We show the termination of the TRS R:

  app(f(),a()) -> app(f(),b())
  app(g(),b()) -> app(g(),a())
  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(a_0()) -> f_1(b_0())
  g_1(b_0()) -> g_1(a_0())
  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#(g_0(),y) -> g_1#(y)
p5: app#(map_1(x1),y) -> map_2#(x1,y)
p6: f_1#(a_0()) -> f_1#(b_0())
p7: g_1#(b_0()) -> g_1#(a_0())
p8: map_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p9: map_2#(|fun|,cons_2(x,xs)) -> map_2#(|fun|,xs)
p10: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs)
p11: filter_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p12: filter2_4#(true_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p13: 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(a_0()) -> f_1(b_0())
r14: g_1(b_0()) -> g_1(a_0())
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, p5, p8, p9, p10, p11, p12, p13}


-- 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(a_0()) -> f_1(b_0())
r14: g_1(b_0()) -> g_1(a_0())
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:
          
            map_2#_A(x1,x2) = max{1, x1 - 10, x2 - 70}
            cons_2_A(x1,x2) = max{x1 + 69, x2 + 38}
            app#_A(x1,x2) = max{1, x1 - 10, x2 - 1}
            map_1_A(x1) = x1 + 8
            filter2_3_A(x1,x2,x3) = max{x1 + 8, x2 + 8, x3 + 8}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 2, x3 - 2, x4 - 32}
            false_0_A = 2
            filter_2#_A(x1,x2) = max{1, x1 - 2, x2 - 70}
            filter_1_A(x1) = x1 + 8
            app_A(x1,x2) = max{7, x1 - 17, x2 - 17}
            true_0_A = 0
            map_2#!_A(x1,x2) = max{1, x1 - 9, x2 - 36}
            cons_2!_A(x1,x2) = max{3, x1 - 36, x2 - 10}
            app#!_A(x1,x2) = max{32, x1 - 9, x2 - 35}
            map_1!_A(x1) = max{2, x1 - 4}
            filter2_3!_A(x1,x2,x3) = max{1, x2 - 10, x3 - 1}
            filter2_4#!_A(x1,x2,x3,x4) = max{32, x2 - 9, x3 - 3, x4 - 35}
            false_0!_A = 33
            filter_2#!_A(x1,x2) = max{32, x1 - 9, x2 - 72}
            filter_1!_A(x1) = max{3, x1 - 37}
            app!_A(x1,x2) = max{33, x1 - 9, x2 - 3}
            true_0!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{1, x1 - 16, x2 - 36}
            cons_2_A(x1,x2) = max{35, x1 + 10, x2 + 27}
            app#_A(x1,x2) = max{1, x1 - 27, x2 - 36}
            map_1_A(x1) = max{21, x1 + 11}
            filter2_3_A(x1,x2,x3) = max{19, x1 + 11, x2 + 11, x3 + 11}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 16, x3 - 26, x4 - 36}
            false_0_A = 16
            filter_2#_A(x1,x2) = max{1, x1 - 16, x2 - 36}
            filter_1_A(x1) = max{12, x1 + 11}
            app_A(x1,x2) = max{4, x2 - 11}
            true_0_A = 16
            map_2#!_A(x1,x2) = max{7, x1 - 3, x2 - 36}
            cons_2!_A(x1,x2) = max{0, x2 - 37}
            app#!_A(x1,x2) = max{6, x1 - 13, x2 - 27}
            map_1!_A(x1) = max{0, x1 - 85}
            filter2_3!_A(x1,x2,x3) = max{4, x1 - 65, x2 - 4, x3 - 87}
            filter2_4#!_A(x1,x2,x3,x4) = max{5, x2 - 3, x3 - 19, x4 - 28}
            false_0!_A = 6
            filter_2#!_A(x1,x2) = max{5, x1 - 3, x2 - 28}
            filter_1!_A(x1) = max{8, x1 - 87}
            app!_A(x1,x2) = max{7, x1 - 76, x2 - 86}
            true_0!_A = 4
        
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [1, 2]
      pi(app#) = [2]
      pi(map_1) = [1]
      pi(filter2_3) = [3]
      pi(filter2_4#) = [4]
      pi(false_0) = []
      pi(filter_2#) = [2]
      pi(filter_1) = [1]
      pi(app) = [2]
      pi(true_0) = []

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(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: f_1(a_0()) -> f_1(b_0())
r14: g_1(b_0()) -> g_1(a_0())
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:

  {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(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: f_1(a_0()) -> f_1(b_0())
r14: g_1(b_0()) -> g_1(a_0())
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:
          
            map_2#_A(x1,x2) = 0
            cons_2_A(x1,x2) = max{3, x1 - 15, x2 - 4}
            map_2#!_A(x1,x2) = max{0, x2 - 8}
            cons_2!_A(x1,x2) = max{3, x1 - 2, x2 - 7}
        
        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{4, x1 - 9, x2 - 4}
            map_2#!_A(x1,x2) = max{5, x1 - 1, x2 - 35}
            cons_2!_A(x1,x2) = max{4, x1 - 29, x2 - 34}
        
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [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(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: f_1(a_0()) -> f_1(b_0())
r14: g_1(b_0()) -> g_1(a_0())
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

  r4, r6, r11, r15, 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:
          
            filter2_4#_A(x1,x2,x3,x4) = max{2, x2 - 1, x3 - 30, x4 - 4}
            false_0_A = 46
            filter_2#_A(x1,x2) = max{2, x1 - 1, x2 - 4}
            cons_2_A(x1,x2) = max{5, x1 - 26, x2 + 4}
            app_A(x1,x2) = max{22, x2 + 16}
            true_0_A = 46
            filter_0_A = 6
            filter_1_A(x1) = max{17, x1 + 6}
            filter2_0_A = 58
            filter2_1_A(x1) = x1 + 6
            map_0_A = 1
            map_1_A(x1) = max{17, x1}
            map_2_A(x1,x2) = max{18, x2 + 16}
            nil_0_A = 1
            filter_2_A(x1,x2) = max{23, x2}
            filter2_4#!_A(x1,x2,x3,x4) = max{30, x4 + 28}
            false_0!_A = 30
            filter_2#!_A(x1,x2) = max{29, x2 + 27}
            cons_2!_A(x1,x2) = max{34, x1 + 6, x2 + 29}
            app!_A(x1,x2) = max{54, x1 - 2, x2 + 35}
            true_0!_A = 1
            filter_0!_A = 1
            filter_1!_A(x1) = max{0, x1 - 1}
            filter2_0!_A = 1
            filter2_1!_A(x1) = max{56, x1 - 1}
            map_0!_A = 1
            map_1!_A(x1) = max{0, x1 - 4}
            map_2!_A(x1,x2) = max{48, x1 - 3}
            nil_0!_A = 36
            filter_2!_A(x1,x2) = max{53, x2 + 34}
        
        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 - 48, x4 - 37}
            false_0_A = 12
            filter_2#_A(x1,x2) = max{1, x1 - 1, x2 - 37}
            cons_2_A(x1,x2) = max{36, x1 - 11, x2}
            app_A(x1,x2) = max{x1 + 23, x2 + 23}
            true_0_A = 10
            filter_0_A = 15
            filter_1_A(x1) = x1 + 23
            filter2_0_A = 15
            filter2_1_A(x1) = max{0, x1 - 26}
            map_0_A = 15
            map_1_A(x1) = max{0, x1 - 24}
            map_2_A(x1,x2) = max{38, x1 + 12, x2 + 1}
            nil_0_A = 0
            filter_2_A(x1,x2) = max{47, x1 - 1, x2 + 12}
            filter2_4#!_A(x1,x2,x3,x4) = 0
            false_0!_A = 5
            filter_2#!_A(x1,x2) = x2 + 48
            cons_2!_A(x1,x2) = max{2, x1 - 23, x2 - 13}
            app!_A(x1,x2) = max{36, x1 - 16, x2 - 24}
            true_0!_A = 3
            filter_0!_A = 6
            filter_1!_A(x1) = max{5, x1 - 11}
            filter2_0!_A = 1
            filter2_1!_A(x1) = max{2, x1 - 25}
            map_0!_A = 1
            map_1!_A(x1) = max{37, x1 - 16}
            map_2!_A(x1,x2) = max{26, x1 - 15}
            nil_0!_A = 7
            filter_2!_A(x1,x2) = max{6, x1 - 12, x2 - 1}
        
  
    partial status:
  
      pi(filter2_4#) = [4]
      pi(false_0) = []
      pi(filter_2#) = [2]
      pi(cons_2) = [2]
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(filter_0) = []
      pi(filter_1) = [1]
      pi(filter2_0) = []
      pi(filter2_1) = [1]
      pi(map_0) = []
      pi(map_1) = [1]
      pi(map_2) = [2]
      pi(nil_0) = []
      pi(filter_2) = []

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.