YES
We show the termination of the TRS R:

  app(app(app(f(),|0|()),|1|()),x) -> app(app(app(f(),app(s(),x)),x),x)
  app(app(app(f(),x),y),app(s(),z)) -> app(s(),app(app(app(f(),|0|()),|1|()),z))
  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(f_2(x1,x2),y) -> f_3(x1,x2,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(map_0(),y) -> map_1(y)
  app(map_1(x1),y) -> map_2(x1,y)
  app(s_0(),y) -> s_1(y)
  f_3(|0|_0(),|1|_0(),x) -> f_3(s_1(x),x,x)
  f_3(x,y,s_1(z)) -> s_1(f_3(|0|_0(),|1|_0(),z))
  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_2(x1,x2),y) -> f_3#(x1,x2,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_3#(|0|_0(),|1|_0(),x) -> f_3#(s_1(x),x,x)
p6: f_3#(x,y,s_1(z)) -> f_3#(|0|_0(),|1|_0(),z)
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(f_1(x1),y) -> f_2(x1,y)
r5: app(f_2(x1,x2),y) -> f_3(x1,x2,y)
r6: app(filter_0(),y) -> filter_1(y)
r7: app(filter_1(x1),y) -> filter_2(x1,y)
r8: app(filter2_0(),y) -> filter2_1(y)
r9: app(filter2_1(x1),y) -> filter2_2(x1,y)
r10: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r11: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: app(s_0(),y) -> s_1(y)
r15: f_3(|0|_0(),|1|_0(),x) -> f_3(s_1(x),x,x)
r16: f_3(x,y,s_1(z)) -> s_1(f_3(|0|_0(),|1|_0(),z))
r17: map_2(|fun|,nil_0()) -> nil_0()
r18: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r19: filter_2(|fun|,nil_0()) -> nil_0()
r20: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r21: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r22: 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(f_1(x1),y) -> f_2(x1,y)
r5: app(f_2(x1,x2),y) -> f_3(x1,x2,y)
r6: app(filter_0(),y) -> filter_1(y)
r7: app(filter_1(x1),y) -> filter_2(x1,y)
r8: app(filter2_0(),y) -> filter2_1(y)
r9: app(filter2_1(x1),y) -> filter2_2(x1,y)
r10: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r11: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: app(s_0(),y) -> s_1(y)
r15: f_3(|0|_0(),|1|_0(),x) -> f_3(s_1(x),x,x)
r16: f_3(x,y,s_1(z)) -> s_1(f_3(|0|_0(),|1|_0(),z))
r17: map_2(|fun|,nil_0()) -> nil_0()
r18: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r19: filter_2(|fun|,nil_0()) -> nil_0()
r20: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r21: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r22: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The set of usable rules consists of

  r17, r19

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{36, x1 + 34}
            cons_2_A(x1,x2) = x1 + 23
            app#_A(x1,x2) = max{36, x1}
            map_1_A(x1) = max{35, x1 + 34}
            filter2_3_A(x1,x2,x3) = max{3, x1 - 53, x2, x3}
            filter2_4#_A(x1,x2,x3,x4) = max{36, x2}
            false_0_A = 0
            filter_2#_A(x1,x2) = max{36, x1}
            filter_1_A(x1) = max{35, x1 + 33}
            app_A(x1,x2) = x2 + 13
            true_0_A = 59
            map_2_A(x1,x2) = max{49, x1 - 10, x2 - 24}
            nil_0_A = 5
            filter_2_A(x1,x2) = max{4, x1 - 119, x2 + 1}
            map_2#!_A(x1,x2) = 5
            cons_2!_A(x1,x2) = max{55, x1 + 9, x2 + 7}
            app#!_A(x1,x2) = x2 + 8
            map_1!_A(x1) = max{6, x1 - 1}
            filter2_3!_A(x1,x2,x3) = max{4, x2 - 1, x3 - 2}
            filter2_4#!_A(x1,x2,x3,x4) = x4 + 7
            false_0!_A = 11
            filter_2#!_A(x1,x2) = max{7, x2 + 1}
            filter_1!_A(x1) = max{7, x1 + 1}
            app!_A(x1,x2) = 3
            true_0!_A = 1
            map_2!_A(x1,x2) = max{57, x1 - 2, x2}
            nil_0!_A = 15
            filter_2!_A(x1,x2) = max{x1 + 16, x2 + 9}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{1, x1 - 20, x2 - 5}
            cons_2_A(x1,x2) = max{4, x1 + 3, x2}
            app#_A(x1,x2) = max{1, x1 - 27, x2 - 2}
            map_1_A(x1) = max{8, x1 + 7}
            filter2_3_A(x1,x2,x3) = max{x1 - 68, x2 + 7, x3 + 26}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 20, x3 - 2, x4 - 5}
            false_0_A = 0
            filter_2#_A(x1,x2) = max{1, x1 - 20, x2 - 5}
            filter_1_A(x1) = max{9, x1 + 7}
            app_A(x1,x2) = max{0, x1 - 20}
            true_0_A = 0
            map_2_A(x1,x2) = max{30, x1 - 17, x2 - 28}
            nil_0_A = 30
            filter_2_A(x1,x2) = max{6, x1 - 89, x2 + 1}
            map_2#!_A(x1,x2) = max{16, x1 - 8, x2 - 5}
            cons_2!_A(x1,x2) = max{15, x1 + 7, x2 + 5}
            app#!_A(x1,x2) = x2 + 5
            map_1!_A(x1) = x1 + 17
            filter2_3!_A(x1,x2,x3) = max{0, x1 - 2, x2 - 3, x3 - 1}
            filter2_4#!_A(x1,x2,x3,x4) = x4 + 4
            false_0!_A = 5
            filter_2#!_A(x1,x2) = x2 + 3
            filter_1!_A(x1) = 1
            app!_A(x1,x2) = max{0, x1 - 10, x2 - 19}
            true_0!_A = 16
            map_2!_A(x1,x2) = max{0, x1 - 11, x2 - 22}
            nil_0!_A = 7
            filter_2!_A(x1,x2) = max{6, x1 - 2, x2 - 31}
        
  
    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) = []
      pi(app) = [2]
      pi(true_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_3#(|0|_0(),|1|_0(),x) -> f_3#(s_1(x),x,x)
p2: f_3#(x,y,s_1(z)) -> f_3#(|0|_0(),|1|_0(),z)

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(f_2(x1,x2),y) -> f_3(x1,x2,y)
r6: app(filter_0(),y) -> filter_1(y)
r7: app(filter_1(x1),y) -> filter_2(x1,y)
r8: app(filter2_0(),y) -> filter2_1(y)
r9: app(filter2_1(x1),y) -> filter2_2(x1,y)
r10: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r11: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: app(s_0(),y) -> s_1(y)
r15: f_3(|0|_0(),|1|_0(),x) -> f_3(s_1(x),x,x)
r16: f_3(x,y,s_1(z)) -> s_1(f_3(|0|_0(),|1|_0(),z))
r17: map_2(|fun|,nil_0()) -> nil_0()
r18: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r19: filter_2(|fun|,nil_0()) -> nil_0()
r20: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r21: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r22: 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_3#_A(x1,x2,x3) = max{0, x1 - 22, x2 - 22, x3 - 15}
            |0|_0_A = 21
            |1|_0_A = 21
            s_1_A(x1) = max{8, x1 + 4}
            f_3#!_A(x1,x2,x3) = max{2, x1 - 18, x2 - 18, x3 - 9}
            |0|_0!_A = 1
            |1|_0!_A = 1
            s_1!_A(x1) = max{4, x1 - 9}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            f_3#_A(x1,x2,x3) = max{0, x1 - 7, x2 - 8, x3 - 6}
            |0|_0_A = 6
            |1|_0_A = 7
            s_1_A(x1) = max{4, x1 + 1}
            f_3#!_A(x1,x2,x3) = max{4, x1 - 13, x2 - 21, x3 - 10}
            |0|_0!_A = 1
            |1|_0!_A = 3
            s_1!_A(x1) = max{4, x1 - 9}
        
  
    partial status:
  
      pi(f_3#) = [3]
      pi(|0|_0) = []
      pi(|1|_0) = []
      pi(s_1) = [1]

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: f_3#(|0|_0(),|1|_0(),x) -> f_3#(s_1(x),x,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(f_2(x1,x2),y) -> f_3(x1,x2,y)
r6: app(filter_0(),y) -> filter_1(y)
r7: app(filter_1(x1),y) -> filter_2(x1,y)
r8: app(filter2_0(),y) -> filter2_1(y)
r9: app(filter2_1(x1),y) -> filter2_2(x1,y)
r10: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r11: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: app(s_0(),y) -> s_1(y)
r15: f_3(|0|_0(),|1|_0(),x) -> f_3(s_1(x),x,x)
r16: f_3(x,y,s_1(z)) -> s_1(f_3(|0|_0(),|1|_0(),z))
r17: map_2(|fun|,nil_0()) -> nil_0()
r18: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r19: filter_2(|fun|,nil_0()) -> nil_0()
r20: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r21: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r22: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The estimated dependency graph contains the following SCCs:

  (no SCCs)