YES
We show the termination of the TRS R:

  app(app(.(),|1|()),x) -> x
  app(app(.(),x),|1|()) -> x
  app(app(.(),app(i(),x)),x) -> |1|()
  app(app(.(),x),app(i(),x)) -> |1|()
  app(app(.(),app(i(),y)),app(app(.(),y),z)) -> z
  app(app(.(),y),app(app(.(),app(i(),y)),z)) -> z
  app(i(),|1|()) -> |1|()
  app(i(),app(i(),x)) -> x
  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(._0(),y) -> ._1(y)
  app(._1(x1),y) -> ._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(i_0(),y) -> i_1(y)
  app(map_0(),y) -> map_1(y)
  app(map_1(x1),y) -> map_2(x1,y)
  ._2(|1|_0(),x) -> x
  ._2(x,|1|_0()) -> x
  ._2(i_1(x),x) -> |1|_0()
  ._2(x,i_1(x)) -> |1|_0()
  ._2(i_1(y),._2(y,z)) -> z
  ._2(y,._2(i_1(y),z)) -> z
  i_1(|1|_0()) -> |1|_0()
  i_1(i_1(x)) -> x
  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#(._1(x1),y) -> ._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#(i_0(),y) -> i_1#(y)
p5: app#(map_1(x1),y) -> map_2#(x1,y)
p6: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p7: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p8: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p9: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p10: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p11: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(._0(),y) -> ._1(y)
r2: app(._1(x1),y) -> ._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(i_0(),y) -> i_1(y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: ._2(|1|_0(),x) -> x
r15: ._2(x,|1|_0()) -> x
r16: ._2(i_1(x),x) -> |1|_0()
r17: ._2(x,i_1(x)) -> |1|_0()
r18: ._2(i_1(y),._2(y,z)) -> z
r19: ._2(y,._2(i_1(y),z)) -> z
r20: i_1(|1|_0()) -> |1|_0()
r21: i_1(i_1(x)) -> x
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

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


-- 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(._0(),y) -> ._1(y)
r2: app(._1(x1),y) -> ._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(i_0(),y) -> i_1(y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: ._2(|1|_0(),x) -> x
r15: ._2(x,|1|_0()) -> x
r16: ._2(i_1(x),x) -> |1|_0()
r17: ._2(x,i_1(x)) -> |1|_0()
r18: ._2(i_1(y),._2(y,z)) -> z
r19: ._2(y,._2(i_1(y),z)) -> z
r20: i_1(|1|_0()) -> |1|_0()
r21: i_1(i_1(x)) -> x
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: 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, r23, r24, r25, r26, r27

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{5, x1 - 10}
        filter_1_A(x1) = max{18, x1 + 16}
        filter_2#_A(x1,x2) = max{5, x1 - 10}
        cons_2_A(x1,x2) = 5
        map_1_A(x1) = max{9, x1}
        map_2#_A(x1,x2) = max{5, x1 - 10}
        filter2_3_A(x1,x2,x3) = max{x1 - 7, x2, x3 - 8}
        filter2_4#_A(x1,x2,x3,x4) = max{5, x2 - 10}
        false_0_A = 18
        app_A(x1,x2) = max{27, x1 + 7, x2 + 17}
        true_0_A = 18
        ._0_A = 1
        ._1_A(x1) = max{18, x1 + 1}
        ._2_A(x1,x2) = max{26, x1 + 3, x2 + 1}
        cons_0_A = 29
        cons_1_A(x1) = 28
        filter_0_A = 1
        filter_2_A(x1,x2) = max{26, x1 + 3}
        filter2_0_A = 22
        filter2_1_A(x1) = max{28, x1 - 9}
        filter2_2_A(x1,x2) = max{29, x1 - 8, x2 - 7}
        filter2_4_A(x1,x2,x3,x4) = max{26, x2 + 3}
        i_0_A = 1
        i_1_A(x1) = x1 + 12
        map_0_A = 1
        map_2_A(x1,x2) = 18
        |1|_0_A = 11
        nil_0_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 62, x2 - 34}
          filter_1_A(x1) = max{13, x1 + 3}
          filter_2#_A(x1,x2) = max{0, x1 - 59, x2 - 34}
          cons_2_A(x1,x2) = max{11, x1, x2}
          map_1_A(x1) = max{14, x1 + 5}
          map_2#_A(x1,x2) = max{0, x1 - 57, x2 - 34}
          filter2_3_A(x1,x2,x3) = max{14, x1 + 5, x2 + 5, x3 + 5}
          filter2_4#_A(x1,x2,x3,x4) = max{0, x1 - 57, x2 - 59, x3 - 57, x4 - 34}
          false_0_A = 56
          app_A(x1,x2) = max{35, x1 - 4, x2 + 13}
          true_0_A = 12
          ._0_A = 3
          ._1_A(x1) = x1 + 5
          ._2_A(x1,x2) = max{35, x1 + 1, x2 + 13}
          cons_0_A = 3
          cons_1_A(x1) = max{14, x1 + 13}
          filter_0_A = 16
          filter_2_A(x1,x2) = max{35, x1 - 2, x2 + 1}
          filter2_0_A = 3
          filter2_1_A(x1) = x1 + 13
          filter2_2_A(x1,x2) = max{x1 + 9, x2 + 13}
          filter2_4_A(x1,x2,x3,x4) = max{35, x1 - 22, x2 - 2, x3 + 1, x4 + 1}
          i_0_A = 3
          i_1_A(x1) = max{14, x1}
          map_0_A = 3
          map_2_A(x1,x2) = max{35, x1 - 4, x2 + 13}
          |1|_0_A = 35
          nil_0_A = 11
          app#!_A(x1,x2) = 0
          filter_1!_A(x1) = max{0, x1 - 16}
          filter_2#!_A(x1,x2) = 0
          cons_2!_A(x1,x2) = 0
          map_1!_A(x1) = max{2, x1 - 15}
          map_2#!_A(x1,x2) = 0
          filter2_3!_A(x1,x2,x3) = max{0, x1 - 7, x2 - 16, x3 - 13}
          filter2_4#!_A(x1,x2,x3,x4) = 0
          false_0!_A = 1
          app!_A(x1,x2) = max{6, x1 - 15, x2 - 12}
          true_0!_A = 1
          ._0!_A = 1
          ._1!_A(x1) = 2
          ._2!_A(x1,x2) = max{3, x1 - 15, x2 - 15}
          cons_0!_A = 6
          cons_1!_A(x1) = max{7, x1 - 13}
          filter_0!_A = 1
          filter_2!_A(x1,x2) = max{5, x1 - 15, x2 - 38}
          filter2_0!_A = 1
          filter2_1!_A(x1) = max{0, x1 - 13}
          filter2_2!_A(x1,x2) = max{0, x2 - 16}
          filter2_4!_A(x1,x2,x3,x4) = max{1, x1 - 51, x2 - 15, x4 - 38}
          i_0!_A = 1
          i_1!_A(x1) = max{2, x1 - 31}
          map_0!_A = 1
          map_2!_A(x1,x2) = max{3, x1 - 16, x2 - 12}
          |1|_0!_A = 3
          nil_0!_A = 4
  
    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) = []
      pi(true_0) = []
      pi(._0) = []
      pi(._1) = [1]
      pi(._2) = [1]
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_2) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(filter2_4) = []
      pi(i_0) = []
      pi(i_1) = []
      pi(map_0) = []
      pi(map_2) = []
      pi(|1|_0) = []
      pi(nil_0) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: filter_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: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p5: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p6: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p7: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p8: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(._0(),y) -> ._1(y)
r2: app(._1(x1),y) -> ._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(i_0(),y) -> i_1(y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: ._2(|1|_0(),x) -> x
r15: ._2(x,|1|_0()) -> x
r16: ._2(i_1(x),x) -> |1|_0()
r17: ._2(x,i_1(x)) -> |1|_0()
r18: ._2(i_1(y),._2(y,z)) -> z
r19: ._2(y,._2(i_1(y),z)) -> z
r20: i_1(|1|_0()) -> |1|_0()
r21: i_1(i_1(x)) -> x
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8}


-- Reduction pair.

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

p1: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p2: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p3: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p4: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p5: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p6: app#(map_1(x1),y) -> map_2#(x1,y)
p7: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p8: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(._0(),y) -> ._1(y)
r2: app(._1(x1),y) -> ._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(i_0(),y) -> i_1(y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: ._2(|1|_0(),x) -> x
r15: ._2(x,|1|_0()) -> x
r16: ._2(i_1(x),x) -> |1|_0()
r17: ._2(x,i_1(x)) -> |1|_0()
r18: ._2(i_1(y),._2(y,z)) -> z
r19: ._2(y,._2(i_1(y),z)) -> z
r20: i_1(|1|_0()) -> |1|_0()
r21: i_1(i_1(x)) -> x
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: 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, r23, r24, r25, r26, r27

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        filter_2#_A(x1,x2) = max{57, x2 + 46}
        cons_2_A(x1,x2) = max{x1 + 14, x2}
        app#_A(x1,x2) = max{57, x2 + 46}
        filter2_3_A(x1,x2,x3) = max{20, x1 - 35, x2 - 19, x3 - 1}
        filter2_4#_A(x1,x2,x3,x4) = max{57, x4 + 46}
        true_0_A = 20
        app_A(x1,x2) = max{x1 + 16, x2 + 57}
        false_0_A = 20
        map_1_A(x1) = max{40, x1 + 19}
        map_2#_A(x1,x2) = max{57, x2 + 44}
        ._0_A = 23
        ._1_A(x1) = max{38, x1 - 14}
        ._2_A(x1,x2) = max{x1 + 2, x2 + 57}
        cons_0_A = 1
        cons_1_A(x1) = max{18, x1 + 14}
        filter_0_A = 1
        filter_1_A(x1) = max{2, x1 - 17}
        filter_2_A(x1,x2) = max{21, x1 - 18, x2 + 11}
        filter2_0_A = 1
        filter2_1_A(x1) = max{2, x1 - 37}
        filter2_2_A(x1,x2) = max{3, x1 - 36, x2 - 20}
        filter2_4_A(x1,x2,x3,x4) = max{21, x1 - 34, x2 - 18, x3 + 14, x4 + 11}
        i_0_A = 40
        i_1_A(x1) = x1 + 55
        map_0_A = 25
        map_2_A(x1,x2) = max{x1 + 35, x2 + 57}
        |1|_0_A = 56
        nil_0_A = 58
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          filter_2#_A(x1,x2) = max{0, x1 - 52, x2 - 54}
          cons_2_A(x1,x2) = max{x1 + 10, x2}
          app#_A(x1,x2) = max{0, x1 - 52, x2 - 44}
          filter2_3_A(x1,x2,x3) = max{x1 + 12, x2 + 10, x3 + 25}
          filter2_4#_A(x1,x2,x3,x4) = max{0, x1 - 92, x2 - 52, x3 - 44, x4 - 54}
          true_0_A = 91
          app_A(x1,x2) = max{x1 - 15, x2 + 48}
          false_0_A = 0
          map_1_A(x1) = x1 + 10
          map_2#_A(x1,x2) = max{0, x1 - 42, x2 - 44}
          ._0_A = 44
          ._1_A(x1) = max{42, x1 + 16}
          ._2_A(x1,x2) = max{48, x1 + 1, x2 + 1}
          cons_0_A = 14
          cons_1_A(x1) = max{44, x1 + 42}
          filter_0_A = 14
          filter_1_A(x1) = x1 + 14
          filter_2_A(x1,x2) = max{48, x1 - 52, x2 + 1}
          filter2_0_A = 14
          filter2_1_A(x1) = x1 + 42
          filter2_2_A(x1,x2) = max{x1 + 27, x2 + 28}
          filter2_4_A(x1,x2,x3,x4) = max{48, x1 - 92, x2 - 52, x3 + 10, x4 + 1}
          i_0_A = 42
          i_1_A(x1) = max{42, x1}
          map_0_A = 14
          map_2_A(x1,x2) = max{x1 - 5, x2 + 48}
          |1|_0_A = 44
          nil_0_A = 64
          filter_2#!_A(x1,x2) = max{14, x1 - 10, x2 - 50}
          cons_2!_A(x1,x2) = max{0, x2 - 46}
          app#!_A(x1,x2) = max{x1 - 11, x2 + 6}
          filter2_3!_A(x1,x2,x3) = max{0, x3 - 41}
          filter2_4#!_A(x1,x2,x3,x4) = max{14, x1 - 92, x2 - 10, x3 - 40, x4 - 50}
          true_0!_A = 1
          app!_A(x1,x2) = max{x1 - 43, x2 + 17}
          false_0!_A = 1
          map_1!_A(x1) = max{18, x1 - 45}
          map_2#!_A(x1,x2) = max{5, x1 - 11, x2 - 3}
          ._0!_A = 1
          ._1!_A(x1) = max{16, x1 - 44}
          ._2!_A(x1,x2) = max{17, x1 - 43, x2 - 43}
          cons_0!_A = 1
          cons_1!_A(x1) = max{17, x1 - 1}
          filter_0!_A = 15
          filter_1!_A(x1) = max{16, x1 - 1}
          filter_2!_A(x1,x2) = max{x1 + 19, x2 - 44}
          filter2_0!_A = 1
          filter2_1!_A(x1) = max{17, x1 - 3}
          filter2_2!_A(x1,x2) = max{0, x1 - 2, x2 - 1}
          filter2_4!_A(x1,x2,x3,x4) = max{x1 - 88, x2 + 19, x4 - 44}
          i_0!_A = 19
          i_1!_A(x1) = max{18, x1 - 45}
          map_0!_A = 17
          map_2!_A(x1,x2) = max{x1 - 44, x2 + 2}
          |1|_0!_A = 19
          nil_0!_A = 20
  
    partial status:
  
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(app#) = [2]
      pi(filter2_3) = []
      pi(filter2_4#) = []
      pi(true_0) = []
      pi(app) = [1, 2]
      pi(false_0) = []
      pi(map_1) = [1]
      pi(map_2#) = []
      pi(._0) = []
      pi(._1) = []
      pi(._2) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(filter_2) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(filter2_4) = []
      pi(i_0) = []
      pi(i_1) = [1]
      pi(map_0) = []
      pi(map_2) = []
      pi(|1|_0) = []
      pi(nil_0) = []

The following rules are removed:

  p1, p2, p6, p7



-- SCC decomposition.

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

p1: filter2_4#(true_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#(false_0(),f,x,xs) -> filter_2#(f,xs)
p4: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(._0(),y) -> ._1(y)
r2: app(._1(x1),y) -> ._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(i_0(),y) -> i_1(y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: ._2(|1|_0(),x) -> x
r15: ._2(x,|1|_0()) -> x
r16: ._2(i_1(x),x) -> |1|_0()
r17: ._2(x,i_1(x)) -> |1|_0()
r18: ._2(i_1(y),._2(y,z)) -> z
r19: ._2(y,._2(i_1(y),z)) -> z
r20: i_1(|1|_0()) -> |1|_0()
r21: i_1(i_1(x)) -> x
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: 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: filter2_4#(true_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#(false_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(._0(),y) -> ._1(y)
r2: app(._1(x1),y) -> ._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(i_0(),y) -> i_1(y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: ._2(|1|_0(),x) -> x
r15: ._2(x,|1|_0()) -> x
r16: ._2(i_1(x),x) -> |1|_0()
r17: ._2(x,i_1(x)) -> |1|_0()
r18: ._2(i_1(y),._2(y,z)) -> z
r19: ._2(y,._2(i_1(y),z)) -> z
r20: i_1(|1|_0()) -> |1|_0()
r21: i_1(i_1(x)) -> x
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r3, r4, r5, r7, r12, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        filter2_4#_A(x1,x2,x3,x4) = x4 + 3
        true_0_A = 15
        filter_2#_A(x1,x2) = x2 + 2
        cons_2_A(x1,x2) = x2 + 4
        app_A(x1,x2) = max{x1 + 14, x2 + 14}
        false_0_A = 15
        ._0_A = 1
        ._1_A(x1) = max{6, x1 - 15}
        cons_0_A = 2
        cons_1_A(x1) = x1 + 1
        filter_0_A = 1
        filter_1_A(x1) = max{8, x1 - 16}
        filter2_0_A = 2
        filter2_1_A(x1) = max{15, x1 + 5}
        map_0_A = 2
        map_1_A(x1) = max{15, x1 - 15}
        filter_2_A(x1,x2) = max{x1 - 3, x2 + 9}
        nil_0_A = 31
  
    base order 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 - 10, x4 - 10}
          true_0_A = 12
          filter_2#_A(x1,x2) = max{1, x1 - 1, x2 - 10}
          cons_2_A(x1,x2) = max{9, x1, x2}
          app_A(x1,x2) = max{x1 - 9, x2 + 11}
          false_0_A = 12
          ._0_A = 8
          ._1_A(x1) = max{8, x1 - 2}
          cons_0_A = 8
          cons_1_A(x1) = x1 + 9
          filter_0_A = 0
          filter_1_A(x1) = max{7, x1 - 15}
          filter2_0_A = 11
          filter2_1_A(x1) = max{8, x1 + 3}
          map_0_A = 8
          map_1_A(x1) = x1 + 11
          filter_2_A(x1,x2) = max{12, x1 - 23, x2}
          nil_0_A = 14
          filter2_4#!_A(x1,x2,x3,x4) = max{x3 + 2, x4 + 2}
          true_0!_A = 3
          filter_2#!_A(x1,x2) = max{2, x2 - 1}
          cons_2!_A(x1,x2) = max{0, x2 - 5}
          app!_A(x1,x2) = max{12, x1 - 10}
          false_0!_A = 3
          ._0!_A = 1
          ._1!_A(x1) = max{2, x1 - 14}
          cons_0!_A = 1
          cons_1!_A(x1) = max{0, x1 - 1}
          filter_0!_A = 1
          filter_1!_A(x1) = max{13, x1 - 1}
          filter2_0!_A = 14
          filter2_1!_A(x1) = max{13, x1 - 1}
          map_0!_A = 1
          map_1!_A(x1) = max{0, x1 - 1}
          filter_2!_A(x1,x2) = max{9, x1 - 9, x2 - 3}
          nil_0!_A = 10
  
    partial status:
  
      pi(filter2_4#) = [4]
      pi(true_0) = []
      pi(filter_2#) = [2]
      pi(cons_2) = []
      pi(app) = [1, 2]
      pi(false_0) = []
      pi(._0) = []
      pi(._1) = []
      pi(cons_0) = []
      pi(cons_1) = [1]
      pi(filter_0) = []
      pi(filter_1) = []
      pi(filter2_0) = []
      pi(filter2_1) = [1]
      pi(map_0) = []
      pi(map_1) = []
      pi(filter_2) = [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: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(._0(),y) -> ._1(y)
r2: app(._1(x1),y) -> ._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(i_0(),y) -> i_1(y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: ._2(|1|_0(),x) -> x
r15: ._2(x,|1|_0()) -> x
r16: ._2(i_1(x),x) -> |1|_0()
r17: ._2(x,i_1(x)) -> |1|_0()
r18: ._2(i_1(y),._2(y,z)) -> z
r19: ._2(y,._2(i_1(y),z)) -> z
r20: i_1(|1|_0()) -> |1|_0()
r21: i_1(i_1(x)) -> x
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: 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:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = max{9, x2}
        cons_2_A(x1,x2) = max{x1 + 9, x2 + 1}
  
    base order 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{16, x1 - 4, x2 - 4}
          map_2#!_A(x1,x2) = 0
          cons_2!_A(x1,x2) = max{2, x1 - 22, x2 - 15}
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.