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

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) = max{3, x1 - 30, x2 - 2}
            filter_1_A(x1) = x1 + 25
            filter_2#_A(x1,x2) = max{3, x1 - 12, x2 - 14}
            cons_2_A(x1,x2) = max{x1 + 12, x2}
            map_1_A(x1) = max{28, x1 + 25}
            map_2#_A(x1,x2) = max{3, x1 - 5, x2 - 13}
            filter2_3_A(x1,x2,x3) = max{x1 + 20, x2 + 18, x3 + 21}
            filter2_4#_A(x1,x2,x3,x4) = max{3, x1 - 28, x2 - 12, x3 - 9, x4 - 14}
            false_0_A = 29
            app_A(x1,x2) = max{27, x1 + 16, x2 + 25}
            true_0_A = 30
            cons_0_A = 21
            cons_1_A(x1) = max{25, x1 + 20}
            filter_0_A = 20
            filter_2_A(x1,x2) = max{x1 + 34, x2 + 21}
            filter2_0_A = 20
            filter2_1_A(x1) = max{28, x1 - 1}
            filter2_2_A(x1,x2) = max{x1 + 15, x2 + 20}
            filter2_4_A(x1,x2,x3,x4) = max{x1 - 31, x2 + 34, x3 + 12, x4 + 21}
            map_0_A = 18
            map_2_A(x1,x2) = max{x1 + 41, x2 + 25}
            |:|_0_A = 0
            |:|_1_A(x1) = x1 + 21
            |:|_2_A(x1,x2) = max{37, x1 + 3, x2}
            C_0_A = 2
            nil_0_A = 42
            app#!_A(x1,x2) = max{x1 - 10, x2 + 8}
            filter_1!_A(x1) = 0
            filter_2#!_A(x1,x2) = max{14, x1 + 2, x2 - 2}
            cons_2!_A(x1,x2) = 0
            map_1!_A(x1) = 0
            map_2#!_A(x1,x2) = max{x1 + 14, x2 + 7}
            filter2_3!_A(x1,x2,x3) = max{1, x1 - 26, x2 - 22}
            filter2_4#!_A(x1,x2,x3,x4) = max{9, x1 - 15, x2 + 2, x4 - 2}
            false_0!_A = 15
            app!_A(x1,x2) = max{0, x1 - 19}
            true_0!_A = 15
            cons_0!_A = 1
            cons_1!_A(x1) = 2
            filter_0!_A = 0
            filter_2!_A(x1,x2) = 0
            filter2_0!_A = 0
            filter2_1!_A(x1) = 0
            filter2_2!_A(x1,x2) = max{2, x1 - 26}
            filter2_4!_A(x1,x2,x3,x4) = 0
            map_0!_A = 0
            map_2!_A(x1,x2) = 7
            |:|_0!_A = 1
            |:|_1!_A(x1) = 0
            |:|_2!_A(x1,x2) = max{1, x1 - 1}
            C_0!_A = 37
            nil_0!_A = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 267, x2 - 187}
            filter_1_A(x1) = max{266, x1 + 60}
            filter_2#_A(x1,x2) = max{0, x1 - 267, x2 - 204}
            cons_2_A(x1,x2) = max{x1 + 17, x2}
            map_1_A(x1) = x1 + 60
            map_2#_A(x1,x2) = max{0, x1 - 267, x2 - 187}
            filter2_3_A(x1,x2,x3) = max{x1 - 27, x2 + 34, x3 + 107}
            filter2_4#_A(x1,x2,x3,x4) = max{0, x1 - 294, x2 - 267, x3 - 187, x4 - 204}
            false_0_A = 293
            app_A(x1,x2) = max{x1 - 61, x2 + 107}
            true_0_A = 293
            cons_0_A = 120
            cons_1_A(x1) = x1 + 95
            filter_0_A = 327
            filter_2_A(x1,x2) = max{205, x1 - 27, x2 + 2}
            filter2_0_A = 60
            filter2_1_A(x1) = x1 + 95
            filter2_2_A(x1,x2) = max{x1 + 34, x2 + 95}
            filter2_4_A(x1,x2,x3,x4) = max{95, x1 - 88, x2 - 27, x3 + 19, x4 + 2}
            map_0_A = 120
            map_2_A(x1,x2) = max{x1 - 1, x2 + 107}
            |:|_0_A = 107
            |:|_1_A(x1) = max{60, x1 - 1}
            |:|_2_A(x1,x2) = 0
            C_0_A = 96
            nil_0_A = 96
            app#!_A(x1,x2) = max{x1 - 267, x2 + 61}
            filter_1!_A(x1) = max{232, x1 + 12}
            filter_2#!_A(x1,x2) = max{x1, x2 + 4}
            cons_2!_A(x1,x2) = 9
            map_1!_A(x1) = 23
            map_2#!_A(x1,x2) = max{x1 - 208, x2 + 45}
            filter2_3!_A(x1,x2,x3) = x3 + 10
            filter2_4#!_A(x1,x2,x3,x4) = max{x1 - 91, x2, x3 + 9, x4 + 4}
            false_0!_A = 1
            app!_A(x1,x2) = max{22, x1 - 96, x2 + 11}
            true_0!_A = 10
            cons_0!_A = 1
            cons_1!_A(x1) = max{23, x1 + 1}
            filter_0!_A = 233
            filter_2!_A(x1,x2) = 9
            filter2_0!_A = 1
            filter2_1!_A(x1) = x1 + 13
            filter2_2!_A(x1,x2) = max{12, x2 - 2}
            filter2_4!_A(x1,x2,x3,x4) = 9
            map_0!_A = 1
            map_2!_A(x1,x2) = max{12, x2 - 97}
            |:|_0!_A = 11
            |:|_1!_A(x1) = 10
            |:|_2!_A(x1,x2) = 0
            C_0!_A = 1
            nil_0!_A = 1
        
  
    partial status:
  
      pi(app#) = []
      pi(filter_1) = [1]
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(map_1) = []
      pi(map_2#) = []
      pi(filter2_3) = []
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(app) = [2]
      pi(true_0) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_2) = []
      pi(filter2_0) = []
      pi(filter2_1) = [1]
      pi(filter2_2) = []
      pi(filter2_4) = []
      pi(map_0) = []
      pi(map_2) = []
      pi(|:|_0) = []
      pi(|:|_1) = []
      pi(|:|_2) = []
      pi(C_0) = []
      pi(nil_0) = []

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!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{0, x2 - 2}
            cons_2_A(x1,x2) = max{4, x1 - 5, x2 + 1}
            map_2#!_A(x1,x2) = max{1, x1 - 1, x2 - 7}
            cons_2!_A(x1,x2) = max{6, x1 - 31, x2 - 6}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{0, x2 - 2}
            cons_2_A(x1,x2) = max{9, x1 - 5, x2 + 1}
            map_2#!_A(x1,x2) = max{7, x1 - 1, x2 - 10}
            cons_2!_A(x1,x2) = max{6, x1 - 16, x2 - 21}
        
  
    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: 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

  (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:
          
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 1}
            false_0_A = 20
            filter_2#_A(x1,x2) = max{1, x1 - 1}
            cons_2_A(x1,x2) = max{22, x1 + 6, x2 + 2}
            app_A(x1,x2) = 9
            true_0_A = 1
            filter2_4#!_A(x1,x2,x3,x4) = max{18, x3 + 3, x4 - 2}
            false_0!_A = 21
            filter_2#!_A(x1,x2) = max{18, x2 - 2}
            cons_2!_A(x1,x2) = max{22, x2 + 7}
            app!_A(x1,x2) = max{21, x2 + 2}
            true_0!_A = 19
        
        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}
            false_0_A = 4
            filter_2#_A(x1,x2) = max{1, x1 - 1}
            cons_2_A(x1,x2) = max{18, x2 - 18}
            app_A(x1,x2) = max{16, x1 + 3}
            true_0_A = 0
            filter2_4#!_A(x1,x2,x3,x4) = max{23, x3 + 3, x4 + 19}
            false_0!_A = 21
            filter_2#!_A(x1,x2) = 20
            cons_2!_A(x1,x2) = 22
            app!_A(x1,x2) = 21
            true_0!_A = 23
        
  
    partial status:
  
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(filter_2#) = []
      pi(cons_2) = [2]
      pi(app) = []
      pi(true_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!):
        
          max/plus interpretation on natural numbers:
          
            |:|_2#_A(x1,x2) = 0
            |:|_2_A(x1,x2) = 0
            C_0_A = 0
            |:|_2#!_A(x1,x2) = 1
            |:|_2!_A(x1,x2) = 2
            C_0!_A = 3
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            |:|_2#_A(x1,x2) = 0
            |:|_2_A(x1,x2) = max{x1 + 1, x2}
            C_0_A = 3
            |:|_2#!_A(x1,x2) = x1 + 6
            |:|_2!_A(x1,x2) = 4
            C_0!_A = 8
        
  
    partial status:
  
      pi(|:|_2#) = []
      pi(|:|_2) = []
      pi(C_0) = []

The following rules are removed:

  p2, p3, p5



-- SCC decomposition.

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#(|:|_2(|:|_2(x,y),z),u)

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}


-- 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#(|:|_2(|:|_2(x,y),z),u)

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!):
        
          max/plus interpretation on natural numbers:
          
            |:|_2#_A(x1,x2) = max{0, x1 - 29, x2 - 38}
            |:|_2_A(x1,x2) = max{x1 + 9, x2}
            C_0_A = 1
            |:|_2#!_A(x1,x2) = 0
            |:|_2!_A(x1,x2) = max{0, x1 - 11}
            C_0!_A = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            |:|_2#_A(x1,x2) = max{0, x1 - 26, x2 - 30}
            |:|_2_A(x1,x2) = max{x1 + 4, x2}
            C_0_A = 13
            |:|_2#!_A(x1,x2) = max{32, x1 + 7, x2 + 30}
            |:|_2!_A(x1,x2) = max{8, x1 - 12}
            C_0!_A = 42
        
  
    partial status:
  
      pi(|:|_2#) = [1]
      pi(|:|_2) = []
      pi(C_0) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: |:|_2#(|:|_2(|:|_2(|:|_2(C_0(),x),y),z),u) -> |:|_2#(|:|_2(|:|_2(x,y),z),u)

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}


-- 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(|:|_2(x,y),z),u)

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!):
        
          max/plus interpretation on natural numbers:
          
            |:|_2#_A(x1,x2) = max{0, x1 - 32, x2 - 1}
            |:|_2_A(x1,x2) = max{9, x2 - 2}
            C_0_A = 10
            |:|_2#!_A(x1,x2) = 0
            |:|_2!_A(x1,x2) = 4
            C_0!_A = 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            |:|_2#_A(x1,x2) = x1
            |:|_2_A(x1,x2) = 0
            C_0_A = 34
            |:|_2#!_A(x1,x2) = 0
            |:|_2!_A(x1,x2) = 0
            C_0!_A = 1
        
  
    partial status:
  
      pi(|:|_2#) = [1]
      pi(|:|_2) = [1, 2]
      pi(C_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.