YES
We show the termination of the TRS R:

  app(app(|:|(),app(app(|:|(),app(app(|:|(),app(app(|:|(),C()),x)),y)),z)),u) -> app(app(|:|(),app(app(|:|(),x),z)),app(app(|:|(),app(app(|:|(),app(app(|:|(),x),y)),z)),u))
  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(|:|_0(),y) -> |:|_1(y)
  app(|:|_1(x1),y) -> |:|_2(x1,y)
  |:|_2(|:|_2(|:|_2(|:|_2(C_0(),x),y),z),u) -> |:|_2(|:|_2(x,z),|:|_2(|:|_2(|:|_2(x,y),z),u))
  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#(|:|_1(x1),y) -> |:|_2#(x1,y)
p5: |:|_2#(|:|_2(|:|_2(|:|_2(C_0(),x),y),z),u) -> |:|_2#(|:|_2(x,z),|:|_2(|:|_2(|:|_2(x,y),z),u))
p6: |:|_2#(|:|_2(|:|_2(|:|_2(C_0(),x),y),z),u) -> |:|_2#(x,z)
p7: |:|_2#(|:|_2(|:|_2(|:|_2(C_0(),x),y),z),u) -> |:|_2#(|:|_2(|:|_2(x,y),z),u)
p8: |:|_2#(|:|_2(|:|_2(|:|_2(C_0(),x),y),z),u) -> |:|_2#(|:|_2(x,y),z)
p9: |:|_2#(|:|_2(|:|_2(|:|_2(C_0(),x),y),z),u) -> |:|_2#(x,y)
p10: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p11: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p12: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p13: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p14: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p15: 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(|:|_0(),y) -> |:|_1(y)
r12: app(|:|_1(x1),y) -> |:|_2(x1,y)
r13: |:|_2(|:|_2(|:|_2(|:|_2(C_0(),x),y),z),u) -> |:|_2(|:|_2(x,z),|:|_2(|:|_2(|:|_2(x,y),z),u))
r14: map_2(f,nil_0()) -> nil_0()
r15: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r16: filter_2(f,nil_0()) -> nil_0()
r17: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r18: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r19: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p10, p11, p12, p13, p14, p15}
  {p5, p6, p7, p8, 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(|:|_0(),y) -> |:|_1(y)
r12: app(|:|_1(x1),y) -> |:|_2(x1,y)
r13: |:|_2(|:|_2(|:|_2(|:|_2(C_0(),x),y),z),u) -> |:|_2(|:|_2(x,z),|:|_2(|:|_2(|:|_2(x,y),z),u))
r14: map_2(f,nil_0()) -> nil_0()
r15: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r16: filter_2(f,nil_0()) -> nil_0()
r17: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r18: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r19: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r3, r5, r6, r7, r11

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) = x1 + 7
              map_1_A(x1) = x1 + 5
              map_2#_A(x1,x2) = 0
              filter2_3_A(x1,x2,x3) = x1 + x2 + 1
              filter2_4#_A(x1,x2,x3,x4) = 0
              false_0_A() = 5
              app_A(x1,x2) = x1 + x2 + 4
              true_0_A() = 5
              cons_0_A() = 2
              cons_1_A(x1) = x1
              filter_0_A() = 0
              filter2_0_A() = 0
              filter2_1_A(x1) = x1
              filter2_2_A(x1,x2) = x1 + x2
              |:|_0_A() = 1
              |:|_1_A(x1) = 0
              app#!_A(x1,x2) = x1 + 1
              filter_1!_A(x1) = x1 + 2
              filter_2#!_A(x1,x2) = x1 + 1
              cons_2!_A(x1,x2) = 4
              map_1!_A(x1) = x1 + 3
              map_2#!_A(x1,x2) = x1 + 2
              filter2_3!_A(x1,x2,x3) = x1 + x2 + x3 + 1
              filter2_4#!_A(x1,x2,x3,x4) = x2 + 1
              false_0!_A() = 0
              app!_A(x1,x2) = x1 + x2 + 4
              true_0!_A() = 4
              cons_0!_A() = 6
              cons_1!_A(x1) = 5
              filter_0!_A() = 3
              filter2_0!_A() = 0
              filter2_1!_A(x1) = 1
              filter2_2!_A(x1,x2) = 0
              |:|_0!_A() = 0
              |:|_1!_A(x1) = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{1, x1 - 39}
            filter_1_A(x1) = x1 + 9
            filter_2#_A(x1,x2) = max{1, x1 - 39}
            cons_2_A(x1,x2) = max{40, x2}
            map_1_A(x1) = max{38, x1 + 2}
            map_2#_A(x1,x2) = max{1, x1 - 37}
            filter2_3_A(x1,x2,x3) = max{x1 + 13, x2 + 23, x3 + 11}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 39}
            false_0_A = 0
            app_A(x1,x2) = max{37, x1 - 10, x2 + 33}
            true_0_A = 0
            cons_0_A = 2
            cons_1_A(x1) = 2
            filter_0_A = 2
            filter2_0_A = 2
            filter2_1_A(x1) = x1 + 33
            filter2_2_A(x1,x2) = max{x1 + 23, x2 + 33}
            |:|_0_A = 2
            |:|_1_A(x1) = max{2, x1 - 4}
            app#!_A(x1,x2) = max{4, x1 - 2}
            filter_1!_A(x1) = x1 + 18
            filter_2#!_A(x1,x2) = x1 + 6
            cons_2!_A(x1,x2) = 5
            map_1!_A(x1) = max{7, x1 + 1}
            map_2#!_A(x1,x2) = max{6, x1 - 1}
            filter2_3!_A(x1,x2,x3) = 0
            filter2_4#!_A(x1,x2,x3,x4) = x2 + 6
            false_0!_A = 20
            app!_A(x1,x2) = max{2, x1 - 2}
            true_0!_A = 5
            cons_0!_A = 7
            cons_1!_A(x1) = max{6, x1 - 1}
            filter_0!_A = 1
            filter2_0!_A = 4
            filter2_1!_A(x1) = max{3, x1 - 1}
            filter2_2!_A(x1,x2) = 1
            |:|_0!_A = 1
            |:|_1!_A(x1) = max{3, x1 + 1}
        
  
    partial status:
  
      pi(app#) = []
      pi(filter_1) = [1]
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(map_1) = [1]
      pi(map_2#) = []
      pi(filter2_3) = [1]
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(cons_0) = []
      pi(cons_1) = [1]
      pi(filter_0) = []
      pi(filter2_0) = []
      pi(filter2_1) = [1]
      pi(filter2_2) = [1]
      pi(|:|_0) = []
      pi(|:|_1) = []

The following rules are removed:

  p1, p2, p3, p5, p6



-- 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)
p2: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p3: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p4: 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(|:|_0(),y) -> |:|_1(y)
r12: app(|:|_1(x1),y) -> |:|_2(x1,y)
r13: |:|_2(|:|_2(|:|_2(|:|_2(C_0(),x),y),z),u) -> |:|_2(|:|_2(x,z),|:|_2(|:|_2(|:|_2(x,y),z),u))
r14: map_2(f,nil_0()) -> nil_0()
r15: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r16: filter_2(f,nil_0()) -> nil_0()
r17: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r18: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r19: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1}
  {p2, p3, p4}


-- 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(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(|:|_0(),y) -> |:|_1(y)
r12: app(|:|_1(x1),y) -> |:|_2(x1,y)
r13: |:|_2(|:|_2(|:|_2(|:|_2(C_0(),x),y),z),u) -> |:|_2(|:|_2(x,z),|:|_2(|:|_2(|:|_2(x,y),z),u))
r14: map_2(f,nil_0()) -> nil_0()
r15: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r16: filter_2(f,nil_0()) -> nil_0()
r17: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r18: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r19: 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) = x1 + x2 + 5
              map_2#!_A(x1,x2) = x2 + 6
              cons_2!_A(x1,x2) = 5
        
        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 - 17, x2 - 12}
            map_2#!_A(x1,x2) = max{4, x2 - 11}
            cons_2!_A(x1,x2) = max{3, x1 - 23, x2 - 12}
        
  
    partial status:
  
      pi(map_2#) = []
      pi(cons_2) = [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: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p2: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p3: 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(|:|_0(),y) -> |:|_1(y)
r12: app(|:|_1(x1),y) -> |:|_2(x1,y)
r13: |:|_2(|:|_2(|:|_2(|:|_2(C_0(),x),y),z),u) -> |:|_2(|:|_2(x,z),|:|_2(|:|_2(|:|_2(x,y),z),u))
r14: map_2(f,nil_0()) -> nil_0()
r15: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r16: filter_2(f,nil_0()) -> nil_0()
r17: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r18: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r19: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r5, 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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              filter2_4#_A(x1,x2,x3,x4) = 0
              false_0_A() = 1
              filter_2#_A(x1,x2) = 0
              cons_2_A(x1,x2) = 8
              app_A(x1,x2) = x1 + 1
              true_0_A() = 1
              filter2_0_A() = 1
              filter2_1_A(x1) = 1
              filter_2_A(x1,x2) = x1 + 5
              nil_0_A() = 3
              filter2_4#!_A(x1,x2,x3,x4) = 3
              false_0!_A() = 4
              filter_2#!_A(x1,x2) = 3
              cons_2!_A(x1,x2) = 4
              app!_A(x1,x2) = 5
              true_0!_A() = 1
              filter2_0!_A() = 2
              filter2_1!_A(x1) = 1
              filter_2!_A(x1,x2) = 2
              nil_0!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 1, x3 - 12, x4 - 13}
            false_0_A = 1
            filter_2#_A(x1,x2) = max{1, x1 - 1, x2 - 13}
            cons_2_A(x1,x2) = max{12, x1 + 1, x2}
            app_A(x1,x2) = max{x1 + 10, x2 + 10}
            true_0_A = 18
            filter2_0_A = 1
            filter2_1_A(x1) = max{11, x1 - 22}
            filter_2_A(x1,x2) = max{15, x1 - 40, x2 + 1}
            nil_0_A = 23
            filter2_4#!_A(x1,x2,x3,x4) = 3
            false_0!_A = 2
            filter_2#!_A(x1,x2) = 2
            cons_2!_A(x1,x2) = max{0, x2 - 25}
            app!_A(x1,x2) = max{37, x1 + 23}
            true_0!_A = 3
            filter2_0!_A = 1
            filter2_1!_A(x1) = max{38, x1 - 1}
            filter_2!_A(x1,x2) = max{1, x1 - 3, x2 - 28}
            nil_0!_A = 0
        
  
    partial status:
  
      pi(filter2_4#) = [4]
      pi(false_0) = []
      pi(filter_2#) = [2]
      pi(cons_2) = [2]
      pi(app) = [1]
      pi(true_0) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter_2) = []
      pi(nil_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: |:|_2#(|:|_2(|:|_2(|:|_2(C_0(),x),y),z),u) -> |:|_2#(|:|_2(x,z),|:|_2(|:|_2(|:|_2(x,y),z),u))
p2: |:|_2#(|:|_2(|:|_2(|:|_2(C_0(),x),y),z),u) -> |:|_2#(x,y)
p3: |:|_2#(|:|_2(|:|_2(|:|_2(C_0(),x),y),z),u) -> |:|_2#(|:|_2(x,y),z)
p4: |:|_2#(|:|_2(|:|_2(|:|_2(C_0(),x),y),z),u) -> |:|_2#(|:|_2(|:|_2(x,y),z),u)
p5: |:|_2#(|:|_2(|:|_2(|:|_2(C_0(),x),y),z),u) -> |:|_2#(x,z)

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(|:|_0(),y) -> |:|_1(y)
r12: app(|:|_1(x1),y) -> |:|_2(x1,y)
r13: |:|_2(|:|_2(|:|_2(|:|_2(C_0(),x),y),z),u) -> |:|_2(|:|_2(x,z),|:|_2(|:|_2(|:|_2(x,y),z),u))
r14: map_2(f,nil_0()) -> nil_0()
r15: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r16: filter_2(f,nil_0()) -> nil_0()
r17: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r18: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r19: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r13

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:
              |:|_2#_A(x1,x2) = 0
              |:|_2_A(x1,x2) = 7
              C_0_A() = 8
              |:|_2#!_A(x1,x2) = 19
              |:|_2!_A(x1,x2) = 20
              C_0!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            |:|_2#_A(x1,x2) = 0
            |:|_2_A(x1,x2) = max{20, x1 + 4, x2}
            C_0_A = 0
            |:|_2#!_A(x1,x2) = max{x1 + 9, x2 + 2}
            |:|_2!_A(x1,x2) = max{32, x1 + 31}
            C_0!_A = 33
        
  
    partial status:
  
      pi(|:|_2#) = [1]
      pi(|:|_2) = [1, 2]
      pi(C_0) = []

The following rules are removed:

  p1, p2, p3, p4, p5

Then no dependency pair remains.