YES
We show the termination of the TRS R:

  app(app(*(),x),app(app(+(),y),z)) -> app(app(+(),app(app(*(),x),y)),app(app(*(),x),z))
  app(app(map(),f),nil()) -> nil()
  app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
  app(app(filter(),f),nil()) -> nil()
  app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(app(filter2(),app(f,x)),f),x),xs)
  app(app(app(app(filter2(),true()),f),x),xs) -> app(app(cons(),x),app(app(filter(),f),xs))
  app(app(app(app(filter2(),false()),f),x),xs) -> app(app(filter(),f),xs)

Instead we prove termination of the uncurried system:

  app(*_0(),y) -> *_1(y)
  app(*_1(x1),y) -> *_2(x1,y)
  app(+_0(),y) -> +_1(y)
  app(+_1(x1),y) -> +_2(x1,y)
  app(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_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(map_0(),y) -> map_1(y)
  app(map_1(x1),y) -> map_2(x1,y)
  *_2(x,+_2(y,z)) -> +_2(*_2(x,y),*_2(x,z))
  map_2(f,nil_0()) -> nil_0()
  map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
  filter_2(f,nil_0()) -> nil_0()
  filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
  filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
  filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)


-- SCC decomposition.

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

p1: app#(*_1(x1),y) -> *_2#(x1,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: *_2#(x,+_2(y,z)) -> *_2#(x,y)
p6: *_2#(x,+_2(y,z)) -> *_2#(x,z)
p7: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p8: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p9: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p10: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p11: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p12: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(*_0(),y) -> *_1(y)
r2: app(*_1(x1),y) -> *_2(x1,y)
r3: app(+_0(),y) -> +_1(y)
r4: app(+_1(x1),y) -> +_2(x1,y)
r5: app(cons_0(),y) -> cons_1(y)
r6: app(cons_1(x1),y) -> cons_2(x1,y)
r7: app(filter_0(),y) -> filter_1(y)
r8: app(filter_1(x1),y) -> filter_2(x1,y)
r9: app(filter2_0(),y) -> filter2_1(y)
r10: app(filter2_1(x1),y) -> filter2_2(x1,y)
r11: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r12: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r13: app(map_0(),y) -> map_1(y)
r14: app(map_1(x1),y) -> map_2(x1,y)
r15: *_2(x,+_2(y,z)) -> +_2(*_2(x,y),*_2(x,z))
r16: map_2(f,nil_0()) -> nil_0()
r17: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r18: filter_2(f,nil_0()) -> nil_0()
r19: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r20: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r21: filter2_4(false_0(),f,x,xs) -> filter_2(f,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#(f,cons_2(x,xs)) -> app#(f,x)
p2: app#(map_1(x1),y) -> map_2#(x1,y)
p3: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p4: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p5: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p6: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p7: app#(filter_1(x1),y) -> filter_2#(x1,y)
p8: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p9: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(*_0(),y) -> *_1(y)
r2: app(*_1(x1),y) -> *_2(x1,y)
r3: app(+_0(),y) -> +_1(y)
r4: app(+_1(x1),y) -> +_2(x1,y)
r5: app(cons_0(),y) -> cons_1(y)
r6: app(cons_1(x1),y) -> cons_2(x1,y)
r7: app(filter_0(),y) -> filter_1(y)
r8: app(filter_1(x1),y) -> filter_2(x1,y)
r9: app(filter2_0(),y) -> filter2_1(y)
r10: app(filter2_1(x1),y) -> filter2_2(x1,y)
r11: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r12: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r13: app(map_0(),y) -> map_1(y)
r14: app(map_1(x1),y) -> map_2(x1,y)
r15: *_2(x,+_2(y,z)) -> +_2(*_2(x,y),*_2(x,z))
r16: map_2(f,nil_0()) -> nil_0()
r17: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r18: filter_2(f,nil_0()) -> nil_0()
r19: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r20: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r21: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r7, r9, r10, r13, r15, 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) = 1
            cons_2_A(x1,x2) = max{x1 + 3, x2 + 3}
            app#_A(x1,x2) = 1
            map_1_A(x1) = 0
            filter2_3_A(x1,x2,x3) = max{x1 + 1, x3 + 1}
            filter2_4#_A(x1,x2,x3,x4) = 1
            false_0_A = 4
            filter_2#_A(x1,x2) = 1
            filter_1_A(x1) = 0
            app_A(x1,x2) = max{0, x1 - 5, x2 - 3}
            true_0_A = 7
            *_0_A = 10
            *_1_A(x1) = 5
            *_2_A(x1,x2) = 0
            +_0_A = 5
            +_1_A(x1) = 0
            +_2_A(x1,x2) = 0
            cons_0_A = 4
            cons_1_A(x1) = 0
            filter_0_A = 4
            filter2_0_A = 10
            filter2_1_A(x1) = 5
            filter2_2_A(x1,x2) = 0
            map_0_A = 5
            filter_2_A(x1,x2) = 0
            nil_0_A = 0
            map_2#!_A(x1,x2) = 13
            cons_2!_A(x1,x2) = 4
            app#!_A(x1,x2) = 13
            map_1!_A(x1) = 14
            filter2_3!_A(x1,x2,x3) = max{1, x1 - 1, x2 - 1, x3 - 1}
            filter2_4#!_A(x1,x2,x3,x4) = 13
            false_0!_A = 12
            filter_2#!_A(x1,x2) = 13
            filter_1!_A(x1) = 1
            app!_A(x1,x2) = 0
            true_0!_A = 1
            *_0!_A = 2
            *_1!_A(x1) = 1
            *_2!_A(x1,x2) = 0
            +_0!_A = 2
            +_1!_A(x1) = 1
            +_2!_A(x1,x2) = 0
            cons_0!_A = 6
            cons_1!_A(x1) = 5
            filter_0!_A = 2
            filter2_0!_A = 2
            filter2_1!_A(x1) = 1
            filter2_2!_A(x1,x2) = 0
            map_0!_A = 15
            filter_2!_A(x1,x2) = max{11, x1 - 2, x2 - 3}
            nil_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 - 30, x2 - 22}
            cons_2_A(x1,x2) = max{x1 + 21, x2 + 14}
            app#_A(x1,x2) = max{1, x1 - 51, x2 - 1}
            map_1_A(x1) = max{25, x1 + 21}
            filter2_3_A(x1,x2,x3) = max{25, x1 + 21, x2 + 21, x3 + 21}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 30, x3 - 30, x4 - 8}
            false_0_A = 48
            filter_2#_A(x1,x2) = max{1, x1 - 30, x2 - 22}
            filter_1_A(x1) = max{50, x1 + 21}
            app_A(x1,x2) = max{71, x1 - 26, x2 + 47}
            true_0_A = 48
            *_0_A = 25
            *_1_A(x1) = x1 + 47
            *_2_A(x1,x2) = max{4, x2 - 3}
            +_0_A = 0
            +_1_A(x1) = x1 + 26
            +_2_A(x1,x2) = max{1, x1, x2 - 5}
            cons_0_A = 50
            cons_1_A(x1) = 25
            filter_0_A = 25
            filter2_0_A = 0
            filter2_1_A(x1) = x1 + 25
            filter2_2_A(x1,x2) = max{48, x1 - 1, x2 + 47}
            map_0_A = 25
            filter_2_A(x1,x2) = max{48, x1 - 19, x2}
            nil_0_A = 48
            map_2#!_A(x1,x2) = max{8, x2 - 5}
            cons_2!_A(x1,x2) = 7
            app#!_A(x1,x2) = x2 + 15
            map_1!_A(x1) = 7
            filter2_3!_A(x1,x2,x3) = max{25, x1 + 15, x2 - 1, x3 - 1}
            filter2_4#!_A(x1,x2,x3,x4) = max{14, x4 + 1}
            false_0!_A = 12
            filter_2#!_A(x1,x2) = max{13, x2 - 1}
            filter_1!_A(x1) = max{12, x1 + 10}
            app!_A(x1,x2) = max{x1 - 49, x2 + 24}
            true_0!_A = 12
            *_0!_A = 23
            *_1!_A(x1) = 22
            *_2!_A(x1,x2) = max{23, x1 - 3, x2 - 2}
            +_0!_A = 26
            +_1!_A(x1) = 25
            +_2!_A(x1,x2) = max{22, x1 - 24, x2 - 21}
            cons_0!_A = 26
            cons_1!_A(x1) = 25
            filter_0!_A = 13
            filter2_0!_A = 26
            filter2_1!_A(x1) = 25
            filter2_2!_A(x1,x2) = max{20, x2 - 2}
            map_0!_A = 8
            filter_2!_A(x1,x2) = max{x1 + 22, x2 + 1}
            nil_0!_A = 19
        
  
    partial status:
  
      pi(map_2#) = []
      pi(cons_2) = []
      pi(app#) = []
      pi(map_1) = []
      pi(filter2_3) = [3]
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(filter_2#) = []
      pi(filter_1) = []
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(*_0) = []
      pi(*_1) = []
      pi(*_2) = []
      pi(+_0) = []
      pi(+_1) = []
      pi(+_2) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(map_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: *_2#(x,+_2(y,z)) -> *_2#(x,y)
p2: *_2#(x,+_2(y,z)) -> *_2#(x,z)

and R consists of:

r1: app(*_0(),y) -> *_1(y)
r2: app(*_1(x1),y) -> *_2(x1,y)
r3: app(+_0(),y) -> +_1(y)
r4: app(+_1(x1),y) -> +_2(x1,y)
r5: app(cons_0(),y) -> cons_1(y)
r6: app(cons_1(x1),y) -> cons_2(x1,y)
r7: app(filter_0(),y) -> filter_1(y)
r8: app(filter_1(x1),y) -> filter_2(x1,y)
r9: app(filter2_0(),y) -> filter2_1(y)
r10: app(filter2_1(x1),y) -> filter2_2(x1,y)
r11: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r12: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r13: app(map_0(),y) -> map_1(y)
r14: app(map_1(x1),y) -> map_2(x1,y)
r15: *_2(x,+_2(y,z)) -> +_2(*_2(x,y),*_2(x,z))
r16: map_2(f,nil_0()) -> nil_0()
r17: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r18: filter_2(f,nil_0()) -> nil_0()
r19: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r20: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r21: filter2_4(false_0(),f,x,xs) -> filter_2(f,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:
          
            *_2#_A(x1,x2) = 0
            +_2_A(x1,x2) = max{3, x1 - 3, x2 - 3}
            *_2#!_A(x1,x2) = max{1, x2 - 14}
            +_2!_A(x1,x2) = max{3, x1 - 10, x2 - 13}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            *_2#_A(x1,x2) = 0
            +_2_A(x1,x2) = max{4, x1 - 3, x2 - 3}
            *_2#!_A(x1,x2) = max{2, x1 - 1, x2 - 5}
            +_2!_A(x1,x2) = max{1, x1 - 13, x2 - 6}
        
  
    partial status:
  
      pi(*_2#) = [2]
      pi(+_2) = [1, 2]

The following rules are removed:

  p1, p2

Then no dependency pair remains.