YES
We show the termination of the TRS R:

  app(app(times(),x),app(app(plus(),y),app(s(),z))) -> app(app(plus(),app(app(times(),x),app(app(plus(),y),app(app(times(),app(s(),z)),|0|())))),app(app(times(),x),app(s(),z)))
  app(app(times(),x),|0|()) -> |0|()
  app(app(times(),x),app(s(),y)) -> app(app(plus(),app(app(times(),x),y)),x)
  app(app(plus(),x),|0|()) -> x
  app(app(plus(),x),app(s(),y)) -> app(s(),app(app(plus(),x),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(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(plus_0(),y) -> plus_1(y)
  app(plus_1(x1),y) -> plus_2(x1,y)
  app(s_0(),y) -> s_1(y)
  app(times_0(),y) -> times_1(y)
  app(times_1(x1),y) -> times_2(x1,y)
  times_2(x,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
  times_2(x,|0|_0()) -> |0|_0()
  times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
  plus_2(x,|0|_0()) -> x
  plus_2(x,s_1(y)) -> s_1(plus_2(x,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#(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#(plus_1(x1),y) -> plus_2#(x1,y)
p5: app#(times_1(x1),y) -> times_2#(x1,y)
p6: times_2#(x,plus_2(y,s_1(z))) -> plus_2#(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
p7: times_2#(x,plus_2(y,s_1(z))) -> times_2#(x,plus_2(y,times_2(s_1(z),|0|_0())))
p8: times_2#(x,plus_2(y,s_1(z))) -> plus_2#(y,times_2(s_1(z),|0|_0()))
p9: times_2#(x,plus_2(y,s_1(z))) -> times_2#(s_1(z),|0|_0())
p10: times_2#(x,plus_2(y,s_1(z))) -> times_2#(x,s_1(z))
p11: times_2#(x,s_1(y)) -> plus_2#(times_2(x,y),x)
p12: times_2#(x,s_1(y)) -> times_2#(x,y)
p13: plus_2#(x,s_1(y)) -> plus_2#(x,y)
p14: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p15: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p16: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p17: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p18: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p19: 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(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p14, p15, p16, p17, p18, p19}
  {p7, p10, p12}
  {p13}


-- 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(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r5, r13, r14, r17, r21, 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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              app#_A(x1,x2) = 0
              filter_1_A(x1) = x1
              filter_2#_A(x1,x2) = 0
              cons_2_A(x1,x2) = 1
              map_1_A(x1) = x1
              map_2#_A(x1,x2) = 0
              filter2_3_A(x1,x2,x3) = x1 + x2 + x3 + 1
              filter2_4#_A(x1,x2,x3,x4) = 0
              false_0_A() = 1
              app_A(x1,x2) = x1 + x2 + 1
              true_0_A() = 1
              filter2_0_A() = 1
              filter2_1_A(x1) = x1 + 1
              s_0_A() = 2
              s_1_A(x1) = x1 + 3
              times_0_A() = 1
              times_1_A(x1) = x1 + 1
              times_2_A(x1,x2) = 1
              |0|_0_A() = 1
              map_2_A(x1,x2) = x2 + 1
              nil_0_A() = 1
              filter_2_A(x1,x2) = x1 + 1
              app#!_A(x1,x2) = 8
              filter_1!_A(x1) = 6
              filter_2#!_A(x1,x2) = 7
              cons_2!_A(x1,x2) = 9
              map_1!_A(x1) = 4
              map_2#!_A(x1,x2) = 7
              filter2_3!_A(x1,x2,x3) = 0
              filter2_4#!_A(x1,x2,x3,x4) = 7
              false_0!_A() = 0
              app!_A(x1,x2) = 5
              true_0!_A() = 12
              filter2_0!_A() = 0
              filter2_1!_A(x1) = 1
              s_0!_A() = 0
              s_1!_A(x1) = 4
              times_0!_A() = 1
              times_1!_A(x1) = 0
              times_2!_A(x1,x2) = 3
              |0|_0!_A() = 2
              map_2!_A(x1,x2) = x1 + 10
              nil_0!_A() = 0
              filter_2!_A(x1,x2) = 11
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{1, x1 - 35, x2 - 213}
            filter_1_A(x1) = max{32, x1 + 31}
            filter_2#_A(x1,x2) = max{1, x1 - 4, x2 - 213}
            cons_2_A(x1,x2) = max{212, x1 + 36, x2 + 72}
            map_1_A(x1) = max{33, x1 + 31}
            map_2#_A(x1,x2) = max{1, x1 - 4, x2 - 213}
            filter2_3_A(x1,x2,x3) = max{x1 + 31, x2 + 34, x3 + 31}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 4, x3 - 177, x4 - 213}
            false_0_A = 146
            app_A(x1,x2) = max{152, x2 - 31}
            true_0_A = 181
            filter2_0_A = 32
            filter2_1_A(x1) = 32
            s_0_A = 32
            s_1_A(x1) = max{1, x1 - 31}
            times_0_A = 32
            times_1_A(x1) = 32
            times_2_A(x1,x2) = max{143, x2 + 73}
            |0|_0_A = 74
            map_2_A(x1,x2) = max{212, x1 - 103, x2 - 213}
            nil_0_A = 3
            filter_2_A(x1,x2) = max{37, x1 - 103, x2 - 103}
            app#!_A(x1,x2) = max{176, x1 - 32, x2 + 30}
            filter_1!_A(x1) = x1 + 6
            filter_2#!_A(x1,x2) = max{28, x2 - 32}
            cons_2!_A(x1,x2) = max{33, x1 + 31}
            map_1!_A(x1) = max{2, x1 - 1}
            map_2#!_A(x1,x2) = max{34, x2 - 37}
            filter2_3!_A(x1,x2,x3) = max{1, x1 - 147, x2 - 2}
            filter2_4#!_A(x1,x2,x3,x4) = x4 + 29
            false_0!_A = 320
            app!_A(x1,x2) = 0
            true_0!_A = 320
            filter2_0!_A = 2
            filter2_1!_A(x1) = max{1, x1 - 2}
            s_0!_A = 1
            s_1!_A(x1) = x1 + 26
            times_0!_A = 167
            times_1!_A(x1) = 166
            times_2!_A(x1,x2) = x2 + 228
            |0|_0!_A = 202
            map_2!_A(x1,x2) = max{34, x2 - 1}
            nil_0!_A = 1
            filter_2!_A(x1,x2) = max{319, x1 + 35, x2 + 105}
        
  
    partial status:
  
      pi(app#) = [2]
      pi(filter_1) = []
      pi(filter_2#) = [2]
      pi(cons_2) = [1, 2]
      pi(map_1) = [1]
      pi(map_2#) = [2]
      pi(filter2_3) = []
      pi(filter2_4#) = [4]
      pi(false_0) = []
      pi(app) = [2]
      pi(true_0) = []
      pi(filter2_0) = []
      pi(filter2_1) = [1]
      pi(s_0) = []
      pi(s_1) = [1]
      pi(times_0) = []
      pi(times_1) = []
      pi(times_2) = [2]
      pi(|0|_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: times_2#(x,s_1(y)) -> times_2#(x,y)
p2: times_2#(x,plus_2(y,s_1(z))) -> times_2#(x,s_1(z))
p3: times_2#(x,plus_2(y,s_1(z))) -> times_2#(x,plus_2(y,times_2(s_1(z),|0|_0())))

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(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r17, r19, r20

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              times_2#_A(x1,x2) = x2
              s_1_A(x1) = x1 + 3
              plus_2_A(x1,x2) = x1 + x2 + 5
              times_2_A(x1,x2) = 0
              |0|_0_A() = 0
              times_2#!_A(x1,x2) = x2
              s_1!_A(x1) = x1 + 9
              plus_2!_A(x1,x2) = x1 + x2 + 13
              times_2!_A(x1,x2) = 10
              |0|_0!_A() = 7
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            times_2#_A(x1,x2) = max{0, x1 - 1}
            s_1_A(x1) = 10
            plus_2_A(x1,x2) = max{13, x1 + 5}
            times_2_A(x1,x2) = max{28, x1 - 5, x2 + 2}
            |0|_0_A = 3
            times_2#!_A(x1,x2) = 4
            s_1!_A(x1) = max{27, x1 + 12}
            plus_2!_A(x1,x2) = max{26, x1 - 18, x2}
            times_2!_A(x1,x2) = 2
            |0|_0!_A = 3
        
  
    partial status:
  
      pi(times_2#) = [2]
      pi(s_1) = [1]
      pi(plus_2) = [1, 2]
      pi(times_2) = []
      pi(|0|_0) = []

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: plus_2#(x,s_1(y)) -> plus_2#(x,y)

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(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: 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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus_2#_A(x1,x2) = x2
              s_1_A(x1) = x1 + 3
              times_2_A(x1,x2) = x1 + x2 + 5
              |0|_0_A() = 1
              plus_2#!_A(x1,x2) = x2 + 6
              s_1!_A(x1) = 5
              times_2!_A(x1,x2) = x1 + 6
              |0|_0!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus_2#_A(x1,x2) = 0
            s_1_A(x1) = 0
            times_2_A(x1,x2) = max{75, x1 - 49, x2}
            |0|_0_A = 75
            plus_2#!_A(x1,x2) = max{103, x1 - 1, x2 - 24}
            s_1!_A(x1) = max{102, x1 + 24}
            times_2!_A(x1,x2) = max{101, x1 - 24, x2 + 49}
            |0|_0!_A = 50
        
  
    partial status:
  
      pi(plus_2#) = []
      pi(s_1) = [1]
      pi(times_2) = [1]
      pi(|0|_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.