YES
We show the termination of the TRS R:

  app(rev(),nil()) -> nil()
  app(rev(),app(app(cons(),x),l)) -> app(app(cons(),app(app(rev1(),x),l)),app(app(rev2(),x),l))
  app(app(rev1(),|0|()),nil()) -> |0|()
  app(app(rev1(),app(s(),x)),nil()) -> app(s(),x)
  app(app(rev1(),x),app(app(cons(),y),l)) -> app(app(rev1(),y),l)
  app(app(rev2(),x),nil()) -> nil()
  app(app(rev2(),x),app(app(cons(),y),l)) -> app(rev(),app(app(cons(),x),app(app(rev2(),y),l)))
  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(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)
  app(rev_0(),y) -> rev_1(y)
  app(rev1_0(),y) -> rev1_1(y)
  app(rev1_1(x1),y) -> rev1_2(x1,y)
  app(rev2_0(),y) -> rev2_1(y)
  app(rev2_1(x1),y) -> rev2_2(x1,y)
  app(s_0(),y) -> s_1(y)
  rev_1(nil_0()) -> nil_0()
  rev_1(cons_2(x,l)) -> cons_2(rev1_2(x,l),rev2_2(x,l))
  rev1_2(|0|_0(),nil_0()) -> |0|_0()
  rev1_2(s_1(x),nil_0()) -> s_1(x)
  rev1_2(x,cons_2(y,l)) -> rev1_2(y,l)
  rev2_2(x,nil_0()) -> nil_0()
  rev2_2(x,cons_2(y,l)) -> rev_1(cons_2(x,rev2_2(y,l)))
  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#(filter_1(x1),y) -> filter_2#(x1,y)
p2: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p3: app#(map_1(x1),y) -> map_2#(x1,y)
p4: app#(rev_0(),y) -> rev_1#(y)
p5: app#(rev1_1(x1),y) -> rev1_2#(x1,y)
p6: app#(rev2_1(x1),y) -> rev2_2#(x1,y)
p7: rev_1#(cons_2(x,l)) -> rev1_2#(x,l)
p8: rev_1#(cons_2(x,l)) -> rev2_2#(x,l)
p9: rev1_2#(x,cons_2(y,l)) -> rev1_2#(y,l)
p10: rev2_2#(x,cons_2(y,l)) -> rev_1#(cons_2(x,rev2_2(y,l)))
p11: rev2_2#(x,cons_2(y,l)) -> rev2_2#(y,l)
p12: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p13: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p14: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p15: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p16: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p17: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,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(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(rev_0(),y) -> rev_1(y)
r12: app(rev1_0(),y) -> rev1_1(y)
r13: app(rev1_1(x1),y) -> rev1_2(x1,y)
r14: app(rev2_0(),y) -> rev2_1(y)
r15: app(rev2_1(x1),y) -> rev2_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: rev_1(nil_0()) -> nil_0()
r18: rev_1(cons_2(x,l)) -> cons_2(rev1_2(x,l),rev2_2(x,l))
r19: rev1_2(|0|_0(),nil_0()) -> |0|_0()
r20: rev1_2(s_1(x),nil_0()) -> s_1(x)
r21: rev1_2(x,cons_2(y,l)) -> rev1_2(y,l)
r22: rev2_2(x,nil_0()) -> nil_0()
r23: rev2_2(x,cons_2(y,l)) -> rev_1(cons_2(x,rev2_2(y,l)))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p12, p13, p14, p15, p16, p17}
  {p8, p10, p11}
  {p9}


-- Reduction pair.

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

p1: app#(filter_1(x1),y) -> filter_2#(x1,y)
p2: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p3: app#(map_1(x1),y) -> map_2#(x1,y)
p4: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p5: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p6: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p7: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
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(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(rev_0(),y) -> rev_1(y)
r12: app(rev1_0(),y) -> rev1_1(y)
r13: app(rev1_1(x1),y) -> rev1_2(x1,y)
r14: app(rev2_0(),y) -> rev2_1(y)
r15: app(rev2_1(x1),y) -> rev2_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: rev_1(nil_0()) -> nil_0()
r18: rev_1(cons_2(x,l)) -> cons_2(rev1_2(x,l),rev2_2(x,l))
r19: rev1_2(|0|_0(),nil_0()) -> |0|_0()
r20: rev1_2(s_1(x),nil_0()) -> s_1(x)
r21: rev1_2(x,cons_2(y,l)) -> rev1_2(y,l)
r22: rev2_2(x,nil_0()) -> nil_0()
r23: rev2_2(x,cons_2(y,l)) -> rev_1(cons_2(x,rev2_2(y,l)))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: 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:
          
            app#_A(x1,x2) = 1
            filter_1_A(x1) = max{14, x1 + 13}
            filter_2#_A(x1,x2) = 1
            cons_2_A(x1,x2) = max{54, x1 + 40, x2 + 40}
            map_1_A(x1) = max{14, x1 + 13}
            map_2#_A(x1,x2) = 1
            filter2_3_A(x1,x2,x3) = max{14, x2 + 13, x3 + 13}
            filter2_4#_A(x1,x2,x3,x4) = 1
            false_0_A = 1
            app_A(x1,x2) = max{1, x1 - 13}
            true_0_A = 0
            app#!_A(x1,x2) = max{11, x2 + 9}
            filter_1!_A(x1) = 1
            filter_2#!_A(x1,x2) = max{2, x2 - 16}
            cons_2!_A(x1,x2) = max{7, x1 - 11, x2 - 15}
            map_1!_A(x1) = max{7, x1 + 1}
            map_2#!_A(x1,x2) = max{8, x2 - 30}
            filter2_3!_A(x1,x2,x3) = max{4, x1 - 10, x2 - 1, x3 - 1}
            filter2_4#!_A(x1,x2,x3,x4) = max{7, x4 - 15}
            false_0!_A = 1
            app!_A(x1,x2) = max{5, x1 - 3}
            true_0!_A = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              app#_A(x1,x2) = 0
              filter_1_A(x1) = x1 + 2
              filter_2#_A(x1,x2) = 0
              cons_2_A(x1,x2) = 4
              map_1_A(x1) = 2
              map_2#_A(x1,x2) = 0
              filter2_3_A(x1,x2,x3) = 2
              filter2_4#_A(x1,x2,x3,x4) = 0
              false_0_A() = 1
              app_A(x1,x2) = 1
              true_0_A() = 1
              app#!_A(x1,x2) = x1 + 3
              filter_1!_A(x1) = 5
              filter_2#!_A(x1,x2) = x1 + 4
              cons_2!_A(x1,x2) = 6
              map_1!_A(x1) = x1 + 8
              map_2#!_A(x1,x2) = x1 + 7
              filter2_3!_A(x1,x2,x3) = 6
              filter2_4#!_A(x1,x2,x3,x4) = 3
              false_0!_A() = 0
              app!_A(x1,x2) = x2 + 3
              true_0!_A() = 0
        
  
    partial status:
  
      pi(app#) = []
      pi(filter_1) = []
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(map_1) = [1]
      pi(map_2#) = []
      pi(filter2_3) = []
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(app) = []
      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: rev2_2#(x,cons_2(y,l)) -> rev_1#(cons_2(x,rev2_2(y,l)))
p2: rev_1#(cons_2(x,l)) -> rev2_2#(x,l)
p3: rev2_2#(x,cons_2(y,l)) -> rev2_2#(y,l)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(rev_0(),y) -> rev_1(y)
r12: app(rev1_0(),y) -> rev1_1(y)
r13: app(rev1_1(x1),y) -> rev1_2(x1,y)
r14: app(rev2_0(),y) -> rev2_1(y)
r15: app(rev2_1(x1),y) -> rev2_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: rev_1(nil_0()) -> nil_0()
r18: rev_1(cons_2(x,l)) -> cons_2(rev1_2(x,l),rev2_2(x,l))
r19: rev1_2(|0|_0(),nil_0()) -> |0|_0()
r20: rev1_2(s_1(x),nil_0()) -> s_1(x)
r21: rev1_2(x,cons_2(y,l)) -> rev1_2(y,l)
r22: rev2_2(x,nil_0()) -> nil_0()
r23: rev2_2(x,cons_2(y,l)) -> rev_1(cons_2(x,rev2_2(y,l)))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r17, r18, r19, r20, r21, r22, r23

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:
          
            rev2_2#_A(x1,x2) = max{11, x1 + 1, x2 - 12}
            cons_2_A(x1,x2) = max{22, x1 + 13, x2}
            rev_1#_A(x1) = max{11, x1 - 12}
            rev2_2_A(x1,x2) = max{x1 + 13, x2}
            rev_1_A(x1) = max{14, x1}
            nil_0_A = 19
            rev1_2_A(x1,x2) = 7
            |0|_0_A = 1
            s_1_A(x1) = 7
            rev2_2#!_A(x1,x2) = 1
            cons_2!_A(x1,x2) = max{3, x2 - 14}
            rev_1#!_A(x1) = 1
            rev2_2!_A(x1,x2) = max{x1 + 36, x2 + 14}
            rev_1!_A(x1) = max{35, x1 + 4}
            nil_0!_A = 34
            rev1_2!_A(x1,x2) = 3
            |0|_0!_A = 1
            s_1!_A(x1) = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              rev2_2#_A(x1,x2) = 0
              cons_2_A(x1,x2) = x2 + 3
              rev_1#_A(x1) = 0
              rev2_2_A(x1,x2) = x2
              rev_1_A(x1) = x1
              nil_0_A() = 0
              rev1_2_A(x1,x2) = 4
              |0|_0_A() = 1
              s_1_A(x1) = 4
              rev2_2#!_A(x1,x2) = x2 + 2
              cons_2!_A(x1,x2) = x2 + 3
              rev_1#!_A(x1) = x1 + 1
              rev2_2!_A(x1,x2) = 6
              rev_1!_A(x1) = 4
              nil_0!_A() = 1
              rev1_2!_A(x1,x2) = 3
              |0|_0!_A() = 0
              s_1!_A(x1) = 0
        
  
    partial status:
  
      pi(rev2_2#) = []
      pi(cons_2) = []
      pi(rev_1#) = []
      pi(rev2_2) = []
      pi(rev_1) = [1]
      pi(nil_0) = []
      pi(rev1_2) = []
      pi(|0|_0) = []
      pi(s_1) = []

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.

-- Reduction pair.

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

p1: rev1_2#(x,cons_2(y,l)) -> rev1_2#(y,l)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(rev_0(),y) -> rev_1(y)
r12: app(rev1_0(),y) -> rev1_1(y)
r13: app(rev1_1(x1),y) -> rev1_2(x1,y)
r14: app(rev2_0(),y) -> rev2_1(y)
r15: app(rev2_1(x1),y) -> rev2_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: rev_1(nil_0()) -> nil_0()
r18: rev_1(cons_2(x,l)) -> cons_2(rev1_2(x,l),rev2_2(x,l))
r19: rev1_2(|0|_0(),nil_0()) -> |0|_0()
r20: rev1_2(s_1(x),nil_0()) -> s_1(x)
r21: rev1_2(x,cons_2(y,l)) -> rev1_2(y,l)
r22: rev2_2(x,nil_0()) -> nil_0()
r23: rev2_2(x,cons_2(y,l)) -> rev_1(cons_2(x,rev2_2(y,l)))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  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:
          
            rev1_2#_A(x1,x2) = 0
            cons_2_A(x1,x2) = max{3, x1 - 22, x2 - 22}
            rev_1_A(x1) = max{22, x1 - 7}
            nil_0_A = 6
            rev1_2#!_A(x1,x2) = x1 + 1
            cons_2!_A(x1,x2) = max{40, x1 + 36, x2 + 20}
            rev_1!_A(x1) = max{39, x1 + 31}
            nil_0!_A = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              rev1_2#_A(x1,x2) = x2
              cons_2_A(x1,x2) = x1 + x2 + 5
              rev_1_A(x1) = x1 + 3
              nil_0_A() = 8
              rev1_2#!_A(x1,x2) = 0
              cons_2!_A(x1,x2) = 13
              rev_1!_A(x1) = x1 + 2
              nil_0!_A() = 1
        
  
    partial status:
  
      pi(rev1_2#) = [2]
      pi(cons_2) = [2]
      pi(rev_1) = [1]
      pi(nil_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.