YES
We show the termination of the TRS R:

  app(D(),t()) -> |1|()
  app(D(),constant()) -> |0|()
  app(D(),app(app(+(),x),y)) -> app(app(+(),app(D(),x)),app(D(),y))
  app(D(),app(app(*(),x),y)) -> app(app(+(),app(app(*(),y),app(D(),x))),app(app(*(),x),app(D(),y)))
  app(D(),app(app(-(),x),y)) -> app(app(-(),app(D(),x)),app(D(),y))
  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(-_0(),y) -> -_1(y)
  app(-_1(x1),y) -> -_2(x1,y)
  app(D_0(),y) -> D_1(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)
  D_1(t_0()) -> |1|_0()
  D_1(constant_0()) -> |0|_0()
  D_1(+_2(x,y)) -> +_2(D_1(x),D_1(y))
  D_1(*_2(x,y)) -> +_2(*_2(y,D_1(x)),*_2(x,D_1(y)))
  D_1(-_2(x,y)) -> -_2(D_1(x),D_1(y))
  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#(D_0(),y) -> D_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: D_1#(+_2(x,y)) -> D_1#(x)
p6: D_1#(+_2(x,y)) -> D_1#(y)
p7: D_1#(*_2(x,y)) -> D_1#(x)
p8: D_1#(*_2(x,y)) -> D_1#(y)
p9: D_1#(-_2(x,y)) -> D_1#(x)
p10: D_1#(-_2(x,y)) -> D_1#(y)
p11: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p12: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p13: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p14: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p15: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p16: 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(-_0(),y) -> -_1(y)
r6: app(-_1(x1),y) -> -_2(x1,y)
r7: app(D_0(),y) -> D_1(y)
r8: app(cons_0(),y) -> cons_1(y)
r9: app(cons_1(x1),y) -> cons_2(x1,y)
r10: app(filter_0(),y) -> filter_1(y)
r11: app(filter_1(x1),y) -> filter_2(x1,y)
r12: app(filter2_0(),y) -> filter2_1(y)
r13: app(filter2_1(x1),y) -> filter2_2(x1,y)
r14: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r15: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r16: app(map_0(),y) -> map_1(y)
r17: app(map_1(x1),y) -> map_2(x1,y)
r18: D_1(t_0()) -> |1|_0()
r19: D_1(constant_0()) -> |0|_0()
r20: D_1(+_2(x,y)) -> +_2(D_1(x),D_1(y))
r21: D_1(*_2(x,y)) -> +_2(*_2(y,D_1(x)),*_2(x,D_1(y)))
r22: D_1(-_2(x,y)) -> -_2(D_1(x),D_1(y))
r23: map_2(f,nil_0()) -> nil_0()
r24: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r25: filter_2(f,nil_0()) -> nil_0()
r26: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r27: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r28: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p11, p12, p13, p14, p15, p16}
  {p5, p6, p7, p8, p9, p10}


-- 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(-_0(),y) -> -_1(y)
r6: app(-_1(x1),y) -> -_2(x1,y)
r7: app(D_0(),y) -> D_1(y)
r8: app(cons_0(),y) -> cons_1(y)
r9: app(cons_1(x1),y) -> cons_2(x1,y)
r10: app(filter_0(),y) -> filter_1(y)
r11: app(filter_1(x1),y) -> filter_2(x1,y)
r12: app(filter2_0(),y) -> filter2_1(y)
r13: app(filter2_1(x1),y) -> filter2_2(x1,y)
r14: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r15: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r16: app(map_0(),y) -> map_1(y)
r17: app(map_1(x1),y) -> map_2(x1,y)
r18: D_1(t_0()) -> |1|_0()
r19: D_1(constant_0()) -> |0|_0()
r20: D_1(+_2(x,y)) -> +_2(D_1(x),D_1(y))
r21: D_1(*_2(x,y)) -> +_2(*_2(y,D_1(x)),*_2(x,D_1(y)))
r22: D_1(-_2(x,y)) -> -_2(D_1(x),D_1(y))
r23: map_2(f,nil_0()) -> nil_0()
r24: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r25: filter_2(f,nil_0()) -> nil_0()
r26: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r27: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r28: 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:
          
            map_2#_A(x1,x2) = max{2, x1 - 32, x2 - 23}
            cons_2_A(x1,x2) = max{x1 + 15, x2 + 22}
            app#_A(x1,x2) = max{2, x1 - 45, x2 - 8}
            map_1_A(x1) = x1 + 13
            filter2_3_A(x1,x2,x3) = max{44, x1 + 10, x2 + 10, x3 + 10}
            filter2_4#_A(x1,x2,x3,x4) = max{2, x2 - 45, x3 - 35, x4 - 8}
            false_0_A = 12
            filter_2#_A(x1,x2) = max{2, x1 - 45, x2 - 21}
            filter_1_A(x1) = x1 + 46
            app_A(x1,x2) = max{10, x1 - 209, x2 + 2}
            true_0_A = 3
            map_2#!_A(x1,x2) = 39
            cons_2!_A(x1,x2) = max{41, x1 + 32, x2 + 4}
            app#!_A(x1,x2) = max{40, x2 + 21}
            map_1!_A(x1) = x1 + 36
            filter2_3!_A(x1,x2,x3) = max{38, x1 + 32, x2 + 29, x3 + 34}
            filter2_4#!_A(x1,x2,x3,x4) = max{39, x4 + 1}
            false_0!_A = 1
            filter_2#!_A(x1,x2) = max{0, x2 - 22}
            filter_1!_A(x1) = max{38, x1 - 14}
            app!_A(x1,x2) = max{37, x1 - 12}
            true_0!_A = 32
        
        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) = 0
              cons_2_A(x1,x2) = 10
              app#_A(x1,x2) = 0
              map_1_A(x1) = x1 + 3
              filter2_3_A(x1,x2,x3) = x1 + x2 + x3 + 7
              filter2_4#_A(x1,x2,x3,x4) = 0
              false_0_A() = 1
              filter_2#_A(x1,x2) = 0
              filter_1_A(x1) = x1 + 3
              app_A(x1,x2) = x1 + x2 + 1
              true_0_A() = 1
              map_2#!_A(x1,x2) = 5
              cons_2!_A(x1,x2) = 7
              app#!_A(x1,x2) = 6
              map_1!_A(x1) = 4
              filter2_3!_A(x1,x2,x3) = 4
              filter2_4#!_A(x1,x2,x3,x4) = 1
              false_0!_A() = 3
              filter_2#!_A(x1,x2) = 0
              filter_1!_A(x1) = 1
              app!_A(x1,x2) = 3
              true_0!_A() = 1
        
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [1, 2]
      pi(app#) = [2]
      pi(map_1) = []
      pi(filter2_3) = [2, 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, 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: D_1#(+_2(x,y)) -> D_1#(x)
p2: D_1#(-_2(x,y)) -> D_1#(y)
p3: D_1#(-_2(x,y)) -> D_1#(x)
p4: D_1#(*_2(x,y)) -> D_1#(y)
p5: D_1#(*_2(x,y)) -> D_1#(x)
p6: D_1#(+_2(x,y)) -> D_1#(y)

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(-_0(),y) -> -_1(y)
r6: app(-_1(x1),y) -> -_2(x1,y)
r7: app(D_0(),y) -> D_1(y)
r8: app(cons_0(),y) -> cons_1(y)
r9: app(cons_1(x1),y) -> cons_2(x1,y)
r10: app(filter_0(),y) -> filter_1(y)
r11: app(filter_1(x1),y) -> filter_2(x1,y)
r12: app(filter2_0(),y) -> filter2_1(y)
r13: app(filter2_1(x1),y) -> filter2_2(x1,y)
r14: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r15: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r16: app(map_0(),y) -> map_1(y)
r17: app(map_1(x1),y) -> map_2(x1,y)
r18: D_1(t_0()) -> |1|_0()
r19: D_1(constant_0()) -> |0|_0()
r20: D_1(+_2(x,y)) -> +_2(D_1(x),D_1(y))
r21: D_1(*_2(x,y)) -> +_2(*_2(y,D_1(x)),*_2(x,D_1(y)))
r22: D_1(-_2(x,y)) -> -_2(D_1(x),D_1(y))
r23: map_2(f,nil_0()) -> nil_0()
r24: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r25: filter_2(f,nil_0()) -> nil_0()
r26: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r27: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r28: 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:
          
            D_1#_A(x1) = 0
            +_2_A(x1,x2) = max{2, x1 - 23, x2 - 23}
            -_2_A(x1,x2) = max{2, x1 - 31, x2 - 31}
            *_2_A(x1,x2) = max{8, x1 - 12, x2 - 34}
            D_1#!_A(x1) = 11
            +_2!_A(x1,x2) = x1 + 11
            -_2!_A(x1,x2) = max{2, x1 - 26, x2 - 49}
            *_2!_A(x1,x2) = max{0, x1 - 39, x2 - 27}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              D_1#_A(x1) = x1
              +_2_A(x1,x2) = x1 + x2 + 5
              -_2_A(x1,x2) = x1 + x2 + 5
              *_2_A(x1,x2) = x1 + x2 + 5
              D_1#!_A(x1) = x1
              +_2!_A(x1,x2) = 5
              -_2!_A(x1,x2) = 2
              *_2!_A(x1,x2) = 3
        
  
    partial status:
  
      pi(D_1#) = []
      pi(+_2) = [1]
      pi(-_2) = []
      pi(*_2) = []

The following rules are removed:

  p1, p2, p3, p4, p5, p6

Then no dependency pair remains.