YES
We show the termination of the TRS R:

  app(app(ack(),|0|()),y) -> app(succ(),y)
  app(app(ack(),app(succ(),x)),y) -> app(app(ack(),x),app(succ(),|0|()))
  app(app(ack(),app(succ(),x)),app(succ(),y)) -> app(app(ack(),x),app(app(ack(),app(succ(),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(ack_0(),y) -> ack_1(y)
  app(ack_1(x1),y) -> ack_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)
  app(succ_0(),y) -> succ_1(y)
  ack_2(|0|_0(),y) -> succ_1(y)
  ack_2(succ_1(x),y) -> ack_2(x,succ_1(|0|_0()))
  ack_2(succ_1(x),succ_1(y)) -> ack_2(x,ack_2(succ_1(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#(ack_1(x1),y) -> ack_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: ack_2#(succ_1(x),y) -> ack_2#(x,succ_1(|0|_0()))
p6: ack_2#(succ_1(x),succ_1(y)) -> ack_2#(x,ack_2(succ_1(x),y))
p7: ack_2#(succ_1(x),succ_1(y)) -> ack_2#(succ_1(x),y)
p8: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p9: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p10: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p11: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p12: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p13: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(ack_0(),y) -> ack_1(y)
r2: app(ack_1(x1),y) -> ack_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(filter_0(),y) -> filter_1(y)
r6: app(filter_1(x1),y) -> filter_2(x1,y)
r7: app(filter2_0(),y) -> filter2_1(y)
r8: app(filter2_1(x1),y) -> filter2_2(x1,y)
r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r11: app(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: app(succ_0(),y) -> succ_1(y)
r14: ack_2(|0|_0(),y) -> succ_1(y)
r15: ack_2(succ_1(x),y) -> ack_2(x,succ_1(|0|_0()))
r16: ack_2(succ_1(x),succ_1(y)) -> ack_2(x,ack_2(succ_1(x),y))
r17: map_2(f,nil_0()) -> nil_0()
r18: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r19: filter_2(f,nil_0()) -> nil_0()
r20: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r21: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r22: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

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


-- 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(ack_0(),y) -> ack_1(y)
r2: app(ack_1(x1),y) -> ack_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(filter_0(),y) -> filter_1(y)
r6: app(filter_1(x1),y) -> filter_2(x1,y)
r7: app(filter2_0(),y) -> filter2_1(y)
r8: app(filter2_1(x1),y) -> filter2_2(x1,y)
r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r11: app(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: app(succ_0(),y) -> succ_1(y)
r14: ack_2(|0|_0(),y) -> succ_1(y)
r15: ack_2(succ_1(x),y) -> ack_2(x,succ_1(|0|_0()))
r16: ack_2(succ_1(x),succ_1(y)) -> ack_2(x,ack_2(succ_1(x),y))
r17: map_2(f,nil_0()) -> nil_0()
r18: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r19: filter_2(f,nil_0()) -> nil_0()
r20: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r21: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r22: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22

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{58, x1 + 28, x2}
            app#_A(x1,x2) = 1
            map_1_A(x1) = x1 + 28
            filter2_3_A(x1,x2,x3) = max{48, x1 + 28, x2 - 46, x3 + 28}
            filter2_4#_A(x1,x2,x3,x4) = 1
            false_0_A = 53
            filter_2#_A(x1,x2) = 1
            filter_1_A(x1) = max{49, x1 + 28}
            app_A(x1,x2) = max{58, x1 + 46, x2 + 28}
            true_0_A = 53
            ack_0_A = 13
            ack_1_A(x1) = x1 + 28
            ack_2_A(x1,x2) = max{74, x1 + 1, x2}
            cons_0_A = 0
            cons_1_A(x1) = max{0, x1 - 18}
            filter_0_A = 0
            filter_2_A(x1,x2) = max{18, x2 + 17}
            filter2_0_A = 0
            filter2_1_A(x1) = max{8, x1 - 46}
            filter2_2_A(x1,x2) = max{1, x1 - 18, x2 - 46}
            filter2_4_A(x1,x2,x3,x4) = max{75, x3 + 28, x4 + 17}
            map_0_A = 0
            map_2_A(x1,x2) = max{x1 + 74, x2 + 28}
            succ_0_A = 0
            succ_1_A(x1) = max{2, x1}
            |0|_0_A = 74
            nil_0_A = 58
            map_2#!_A(x1,x2) = x2 + 3
            cons_2!_A(x1,x2) = max{7, x1 + 5, x2 - 20}
            app#!_A(x1,x2) = x2 + 4
            map_1!_A(x1) = max{1, x1 - 1}
            filter2_3!_A(x1,x2,x3) = x1 + 1
            filter2_4#!_A(x1,x2,x3,x4) = max{3, x4 - 21}
            false_0!_A = 1
            filter_2#!_A(x1,x2) = max{2, x2 - 22}
            filter_1!_A(x1) = 70
            app!_A(x1,x2) = max{x1 + 58, x2 + 69}
            true_0!_A = 11
            ack_0!_A = 1
            ack_1!_A(x1) = max{70, x1 - 4}
            ack_2!_A(x1,x2) = max{0, x1 - 3}
            cons_0!_A = 1
            cons_1!_A(x1) = 7
            filter_0!_A = 71
            filter_2!_A(x1,x2) = max{10, x1 + 1, x2 - 1}
            filter2_0!_A = 1
            filter2_1!_A(x1) = x1 + 68
            filter2_2!_A(x1,x2) = max{67, x1 + 12}
            filter2_4!_A(x1,x2,x3,x4) = max{11, x1 - 47, x2 + 1, x3 + 6, x4 - 1}
            map_0!_A = 2
            map_2!_A(x1,x2) = max{x1 + 64, x2 + 66}
            succ_0!_A = 70
            succ_1!_A(x1) = 70
            |0|_0!_A = 0
            nil_0!_A = 9
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{1, x1 - 3}
            cons_2_A(x1,x2) = max{41, x2 + 36}
            app#_A(x1,x2) = max{1, x1 - 16}
            map_1_A(x1) = x1 + 13
            filter2_3_A(x1,x2,x3) = max{x1 + 11, x2 + 12, x3 + 15}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 7}
            false_0_A = 89
            filter_2#_A(x1,x2) = max{1, x1 - 7}
            filter_1_A(x1) = x1 + 9
            app_A(x1,x2) = max{x1 - 14, x2 + 41}
            true_0_A = 43
            ack_0_A = 0
            ack_1_A(x1) = max{4, x1 - 3}
            ack_2_A(x1,x2) = max{4, x2 - 5}
            cons_0_A = 0
            cons_1_A(x1) = x1 + 15
            filter_0_A = 0
            filter_2_A(x1,x2) = x2 + 5
            filter2_0_A = 0
            filter2_1_A(x1) = x1 + 39
            filter2_2_A(x1,x2) = max{x1 + 25, x2 + 26}
            filter2_4_A(x1,x2,x3,x4) = x4 + 41
            map_0_A = 0
            map_2_A(x1,x2) = max{5, x2}
            succ_0_A = 0
            succ_1_A(x1) = max{1, x1 - 5}
            |0|_0_A = 9
            nil_0_A = 5
            map_2#!_A(x1,x2) = x1 + 7
            cons_2!_A(x1,x2) = 0
            app#!_A(x1,x2) = max{6, x1 - 5}
            map_1!_A(x1) = max{6, x1 - 1}
            filter2_3!_A(x1,x2,x3) = 29
            filter2_4#!_A(x1,x2,x3,x4) = max{x1 - 39, x2 + 6, x3 + 6}
            false_0!_A = 15
            filter_2#!_A(x1,x2) = max{5, x1 + 2}
            filter_1!_A(x1) = max{47, x1 + 45}
            app!_A(x1,x2) = max{46, x1 + 2, x2 + 44}
            true_0!_A = 17
            ack_0!_A = 1
            ack_1!_A(x1) = max{0, x1 - 1}
            ack_2!_A(x1,x2) = max{48, x2 - 5}
            cons_0!_A = 1
            cons_1!_A(x1) = max{0, x1 - 1}
            filter_0!_A = 1
            filter_2!_A(x1,x2) = max{16, x2 - 6}
            filter2_0!_A = 46
            filter2_1!_A(x1) = max{45, x1 + 43}
            filter2_2!_A(x1,x2) = x1 + 42
            filter2_4!_A(x1,x2,x3,x4) = max{29, x4 + 17}
            map_0!_A = 1
            map_2!_A(x1,x2) = max{16, x1 + 1, x2 + 8}
            succ_0!_A = 47
            succ_1!_A(x1) = 47
            |0|_0!_A = 46
            nil_0!_A = 16
        
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [1, 2]
      pi(app#) = [2]
      pi(map_1) = [1]
      pi(filter2_3) = [1, 3]
      pi(filter2_4#) = [4]
      pi(false_0) = []
      pi(filter_2#) = [2]
      pi(filter_1) = []
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(ack_0) = []
      pi(ack_1) = []
      pi(ack_2) = []
      pi(cons_0) = []
      pi(cons_1) = [1]
      pi(filter_0) = []
      pi(filter_2) = [2]
      pi(filter2_0) = []
      pi(filter2_1) = [1]
      pi(filter2_2) = [1]
      pi(filter2_4) = [3, 4]
      pi(map_0) = []
      pi(map_2) = [1, 2]
      pi(succ_0) = []
      pi(succ_1) = []
      pi(|0|_0) = []
      pi(nil_0) = []

The following rules are removed:

  p1, p2, p4, p5, p6, p7, p8, p9



-- SCC decomposition.

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

p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(ack_0(),y) -> ack_1(y)
r2: app(ack_1(x1),y) -> ack_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(filter_0(),y) -> filter_1(y)
r6: app(filter_1(x1),y) -> filter_2(x1,y)
r7: app(filter2_0(),y) -> filter2_1(y)
r8: app(filter2_1(x1),y) -> filter2_2(x1,y)
r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r11: app(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: app(succ_0(),y) -> succ_1(y)
r14: ack_2(|0|_0(),y) -> succ_1(y)
r15: ack_2(succ_1(x),y) -> ack_2(x,succ_1(|0|_0()))
r16: ack_2(succ_1(x),succ_1(y)) -> ack_2(x,ack_2(succ_1(x),y))
r17: map_2(f,nil_0()) -> nil_0()
r18: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r19: filter_2(f,nil_0()) -> nil_0()
r20: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r21: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r22: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(ack_0(),y) -> ack_1(y)
r2: app(ack_1(x1),y) -> ack_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(filter_0(),y) -> filter_1(y)
r6: app(filter_1(x1),y) -> filter_2(x1,y)
r7: app(filter2_0(),y) -> filter2_1(y)
r8: app(filter2_1(x1),y) -> filter2_2(x1,y)
r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r11: app(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: app(succ_0(),y) -> succ_1(y)
r14: ack_2(|0|_0(),y) -> succ_1(y)
r15: ack_2(succ_1(x),y) -> ack_2(x,succ_1(|0|_0()))
r16: ack_2(succ_1(x),succ_1(y)) -> ack_2(x,ack_2(succ_1(x),y))
r17: map_2(f,nil_0()) -> nil_0()
r18: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r19: filter_2(f,nil_0()) -> nil_0()
r20: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r21: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r22: 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) = 0
            cons_2_A(x1,x2) = max{4, x1 - 9, x2 - 4}
            map_2#!_A(x1,x2) = max{0, x2 - 34}
            cons_2!_A(x1,x2) = max{7, x1 - 5, x2 - 33}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = 0
            cons_2_A(x1,x2) = max{3, x1 - 15, x2 - 4}
            map_2#!_A(x1,x2) = max{4, x1 - 1, x2 - 5}
            cons_2!_A(x1,x2) = max{3, x1 - 2, x2 - 5}
        
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [1, 2]

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: ack_2#(succ_1(x),y) -> ack_2#(x,succ_1(|0|_0()))
p2: ack_2#(succ_1(x),succ_1(y)) -> ack_2#(succ_1(x),y)
p3: ack_2#(succ_1(x),succ_1(y)) -> ack_2#(x,ack_2(succ_1(x),y))

and R consists of:

r1: app(ack_0(),y) -> ack_1(y)
r2: app(ack_1(x1),y) -> ack_2(x1,y)
r3: app(cons_0(),y) -> cons_1(y)
r4: app(cons_1(x1),y) -> cons_2(x1,y)
r5: app(filter_0(),y) -> filter_1(y)
r6: app(filter_1(x1),y) -> filter_2(x1,y)
r7: app(filter2_0(),y) -> filter2_1(y)
r8: app(filter2_1(x1),y) -> filter2_2(x1,y)
r9: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r10: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r11: app(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: app(succ_0(),y) -> succ_1(y)
r14: ack_2(|0|_0(),y) -> succ_1(y)
r15: ack_2(succ_1(x),y) -> ack_2(x,succ_1(|0|_0()))
r16: ack_2(succ_1(x),succ_1(y)) -> ack_2(x,ack_2(succ_1(x),y))
r17: map_2(f,nil_0()) -> nil_0()
r18: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r19: filter_2(f,nil_0()) -> nil_0()
r20: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r21: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r22: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r14, r15, r16

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:
          
            ack_2#_A(x1,x2) = max{0, x1 - 7, x2 - 4}
            succ_1_A(x1) = max{2, x1}
            |0|_0_A = 1
            ack_2_A(x1,x2) = max{3, x1 - 3, x2}
            ack_2#!_A(x1,x2) = max{7, x1 - 7, x2 - 4}
            succ_1!_A(x1) = 5
            |0|_0!_A = 1
            ack_2!_A(x1,x2) = max{6, x1 - 10, x2 - 7}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            ack_2#_A(x1,x2) = max{0, x1 - 10, x2 - 7}
            succ_1_A(x1) = max{5, x1}
            |0|_0_A = 3
            ack_2_A(x1,x2) = max{6, x1 - 7, x2}
            ack_2#!_A(x1,x2) = max{29, x1 - 13, x2 - 7}
            succ_1!_A(x1) = max{28, x1 - 15}
            |0|_0!_A = 1
            ack_2!_A(x1,x2) = max{28, x1 - 26, x2 - 19}
        
  
    partial status:
  
      pi(ack_2#) = [1, 2]
      pi(succ_1) = [1]
      pi(|0|_0) = []
      pi(ack_2) = [1, 2]

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.