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

  (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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              map_2#_A(x1,x2) = x1 + x2
              cons_2_A(x1,x2) = x1 + x2 + 5
              app#_A(x1,x2) = x1 + x2
              map_1_A(x1) = x1 + 3
              filter2_3_A(x1,x2,x3) = x1 + x2 + x3 + 7
              filter2_4#_A(x1,x2,x3,x4) = x2 + x4 + 7
              false_0_A() = 1
              filter_2#_A(x1,x2) = x1 + x2 + 7
              filter_1_A(x1) = x1 + 7
              app_A(x1,x2) = x1 + x2 + 1
              true_0_A() = 1
              map_2#!_A(x1,x2) = 6
              cons_2!_A(x1,x2) = 4
              app#!_A(x1,x2) = 5
              map_1!_A(x1) = 7
              filter2_3!_A(x1,x2,x3) = x1 + x2 + x3 + 7
              filter2_4#!_A(x1,x2,x3,x4) = 6
              false_0!_A() = 9
              filter_2#!_A(x1,x2) = 6
              filter_1!_A(x1) = x1 + 7
              app!_A(x1,x2) = 2
              true_0!_A() = 5
        
        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) = x2
              cons_2_A(x1,x2) = x1 + x2 + 5
              app#_A(x1,x2) = x2
              map_1_A(x1) = 3
              filter2_3_A(x1,x2,x3) = x1 + x2 + x3
              filter2_4#_A(x1,x2,x3,x4) = x4
              false_0_A() = 1
              filter_2#_A(x1,x2) = x2
              filter_1_A(x1) = 3
              app_A(x1,x2) = x1 + x2 + 1
              true_0_A() = 1
              map_2#!_A(x1,x2) = 1
              cons_2!_A(x1,x2) = x2 + 4
              app#!_A(x1,x2) = 0
              map_1!_A(x1) = 2
              filter2_3!_A(x1,x2,x3) = 16
              filter2_4#!_A(x1,x2,x3,x4) = x2 + x4 + 15
              false_0!_A() = 0
              filter_2#!_A(x1,x2) = x1 + x2 + 11
              filter_1!_A(x1) = 7
              app!_A(x1,x2) = 5
              true_0!_A() = 10
        
  
    partial status:
  
      pi(map_2#) = [1]
      pi(cons_2) = []
      pi(app#) = [1]
      pi(map_1) = [1]
      pi(filter2_3) = [2]
      pi(filter2_4#) = [2]
      pi(false_0) = []
      pi(filter_2#) = [1]
      pi(filter_1) = [1]
      pi(app) = []
      pi(true_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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              map_2#_A(x1,x2) = x2
              cons_2_A(x1,x2) = x2 + 2
              map_2#!_A(x1,x2) = x2 + 6
              cons_2!_A(x1,x2) = 7
        
        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) = x2
              cons_2_A(x1,x2) = x1 + x2 + 9
              map_2#!_A(x1,x2) = x2 + 30
              cons_2!_A(x1,x2) = 29
        
  
    partial status:
  
      pi(map_2#) = [1]
      pi(cons_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

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:
              ack_2#_A(x1,x2) = 0
              succ_1_A(x1) = 1
              |0|_0_A() = 2
              ack_2_A(x1,x2) = x1 + 2
              ack_2#!_A(x1,x2) = 6
              succ_1!_A(x1) = 6
              |0|_0!_A() = 5
              ack_2!_A(x1,x2) = x1 + 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              ack_2#_A(x1,x2) = 0
              succ_1_A(x1) = 1
              |0|_0_A() = 0
              ack_2_A(x1,x2) = x1 + 1
              ack_2#!_A(x1,x2) = 1
              succ_1!_A(x1) = 1
              |0|_0!_A() = 0
              ack_2!_A(x1,x2) = 1
        
  
    partial status:
  
      pi(ack_2#) = [1]
      pi(succ_1) = [1]
      pi(|0|_0) = []
      pi(ack_2) = [2]

The following rules are removed:

  p1, p3



-- SCC decomposition.

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

p1: ack_2#(succ_1(x),succ_1(y)) -> 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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: ack_2#(succ_1(x),succ_1(y)) -> 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

  (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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              ack_2#_A(x1,x2) = 0
              succ_1_A(x1) = 2
              ack_2#!_A(x1,x2) = 6
              succ_1!_A(x1) = 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              ack_2#_A(x1,x2) = x2
              succ_1_A(x1) = x1 + 1
              ack_2#!_A(x1,x2) = x1 + x2 + 2
              succ_1!_A(x1) = 2
        
  
    partial status:
  
      pi(ack_2#) = [1]
      pi(succ_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.