YES
We show the termination of the TRS R:

  app(app(app(f(),|0|()),|1|()),x) -> app(app(app(f(),app(s(),x)),x),x)
  app(app(app(f(),x),y),app(s(),z)) -> app(s(),app(app(app(f(),|0|()),|1|()),z))
  app(app(map(),|fun|),nil()) -> nil()
  app(app(map(),|fun|),app(app(cons(),x),xs)) -> app(app(cons(),app(|fun|,x)),app(app(map(),|fun|),xs))
  app(app(filter(),|fun|),nil()) -> nil()
  app(app(filter(),|fun|),app(app(cons(),x),xs)) -> app(app(app(app(filter2(),app(|fun|,x)),|fun|),x),xs)
  app(app(app(app(filter2(),true()),|fun|),x),xs) -> app(app(cons(),x),app(app(filter(),|fun|),xs))
  app(app(app(app(filter2(),false()),|fun|),x),xs) -> app(app(filter(),|fun|),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(f_0(),y) -> f_1(y)
  app(f_1(x1),y) -> f_2(x1,y)
  app(f_2(x1,x2),y) -> f_3(x1,x2,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(s_0(),y) -> s_1(y)
  f_3(|0|_0(),|1|_0(),x) -> f_3(s_1(x),x,x)
  f_3(x,y,s_1(z)) -> s_1(f_3(|0|_0(),|1|_0(),z))
  map_2(|fun|,nil_0()) -> nil_0()
  map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
  filter_2(|fun|,nil_0()) -> nil_0()
  filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
  filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
  filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)


-- SCC decomposition.

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

p1: app#(f_2(x1,x2),y) -> f_3#(x1,x2,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: f_3#(|0|_0(),|1|_0(),x) -> f_3#(s_1(x),x,x)
p6: f_3#(x,y,s_1(z)) -> f_3#(|0|_0(),|1|_0(),z)
p7: map_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p8: map_2#(|fun|,cons_2(x,xs)) -> map_2#(|fun|,xs)
p9: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs)
p10: filter_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p11: filter2_4#(true_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p12: filter2_4#(false_0(),|fun|,x,xs) -> filter_2#(|fun|,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(f_0(),y) -> f_1(y)
r4: app(f_1(x1),y) -> f_2(x1,y)
r5: app(f_2(x1,x2),y) -> f_3(x1,x2,y)
r6: app(filter_0(),y) -> filter_1(y)
r7: app(filter_1(x1),y) -> filter_2(x1,y)
r8: app(filter2_0(),y) -> filter2_1(y)
r9: app(filter2_1(x1),y) -> filter2_2(x1,y)
r10: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r11: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: app(s_0(),y) -> s_1(y)
r15: f_3(|0|_0(),|1|_0(),x) -> f_3(s_1(x),x,x)
r16: f_3(x,y,s_1(z)) -> s_1(f_3(|0|_0(),|1|_0(),z))
r17: map_2(|fun|,nil_0()) -> nil_0()
r18: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r19: filter_2(|fun|,nil_0()) -> nil_0()
r20: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r21: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r22: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: map_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p2: app#(map_1(x1),y) -> map_2#(x1,y)
p3: map_2#(|fun|,cons_2(x,xs)) -> map_2#(|fun|,xs)
p4: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p5: filter2_4#(false_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p6: filter_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p7: app#(filter_1(x1),y) -> filter_2#(x1,y)
p8: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs)
p9: filter2_4#(true_0(),|fun|,x,xs) -> filter_2#(|fun|,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(f_0(),y) -> f_1(y)
r4: app(f_1(x1),y) -> f_2(x1,y)
r5: app(f_2(x1,x2),y) -> f_3(x1,x2,y)
r6: app(filter_0(),y) -> filter_1(y)
r7: app(filter_1(x1),y) -> filter_2(x1,y)
r8: app(filter2_0(),y) -> filter2_1(y)
r9: app(filter2_1(x1),y) -> filter2_2(x1,y)
r10: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r11: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: app(s_0(),y) -> s_1(y)
r15: f_3(|0|_0(),|1|_0(),x) -> f_3(s_1(x),x,x)
r16: f_3(x,y,s_1(z)) -> s_1(f_3(|0|_0(),|1|_0(),z))
r17: map_2(|fun|,nil_0()) -> nil_0()
r18: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r19: filter_2(|fun|,nil_0()) -> nil_0()
r20: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r21: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r22: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,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:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = 3
        cons_2_A(x1,x2) = max{7, x2 + 4}
        app#_A(x1,x2) = 3
        map_1_A(x1) = max{2, x1 - 5}
        filter2_3_A(x1,x2,x3) = 5
        filter2_4#_A(x1,x2,x3,x4) = 3
        false_0_A = 6
        filter_2#_A(x1,x2) = 3
        filter_1_A(x1) = 9
        app_A(x1,x2) = max{11, x2 + 8}
        true_0_A = 6
        cons_0_A = 1
        cons_1_A(x1) = max{6, x1 - 5}
        f_0_A = 1
        f_1_A(x1) = max{10, x1 - 5}
        f_2_A(x1,x2) = max{0, x2 - 5}
        f_3_A(x1,x2,x3) = x3 + 2
        filter_0_A = 10
        filter_2_A(x1,x2) = x2 + 3
        filter2_0_A = 1
        filter2_1_A(x1) = max{0, x1 - 5}
        filter2_2_A(x1,x2) = max{10, x2 - 5}
        filter2_4_A(x1,x2,x3,x4) = x4 + 7
        map_0_A = 1
        map_2_A(x1,x2) = max{5, x2 + 2}
        s_0_A = 1
        s_1_A(x1) = max{0, x1 - 4}
        |0|_0_A = 0
        |1|_0_A = 1
        nil_0_A = 4
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          map_2#_A(x1,x2) = max{0, x1 - 3, x2 - 26}
          cons_2_A(x1,x2) = max{13, x1, x2}
          app#_A(x1,x2) = max{0, x1 - 13, x2 - 26}
          map_1_A(x1) = x1 + 10
          filter2_3_A(x1,x2,x3) = max{x1 - 12, x2, x3 + 12}
          filter2_4#_A(x1,x2,x3,x4) = max{0, x1 - 38, x2 - 13, x3 - 26, x4 - 26}
          false_0_A = 0
          filter_2#_A(x1,x2) = max{0, x1 - 13, x2 - 26}
          filter_1_A(x1) = max{10, x1 + 4}
          app_A(x1,x2) = max{26, x1 - 11, x2 + 12}
          true_0_A = 2
          cons_0_A = 10
          cons_1_A(x1) = max{13, x1 + 12}
          f_0_A = 17
          f_1_A(x1) = x1 + 10
          f_2_A(x1,x2) = max{x1 - 1, x2 + 10}
          f_3_A(x1,x2,x3) = max{13, x1 - 32, x2 - 45, x3 - 45}
          filter_0_A = 10
          filter_2_A(x1,x2) = max{21, x1 - 11, x2 + 7}
          filter2_0_A = 10
          filter2_1_A(x1) = x1 + 10
          filter2_2_A(x1,x2) = max{x1 - 1, x2 + 12}
          filter2_4_A(x1,x2,x3,x4) = max{21, x1 - 27, x2 - 11, x3 + 1, x4 + 7}
          map_0_A = 10
          map_2_A(x1,x2) = max{26, x1 - 11, x2 + 12}
          s_0_A = 10
          s_1_A(x1) = max{13, x1 - 14}
          |0|_0_A = 15
          |1|_0_A = 28
          nil_0_A = 27
          map_2#!_A(x1,x2) = x1 + 6
          cons_2!_A(x1,x2) = 0
          app#!_A(x1,x2) = max{6, x1 - 3}
          map_1!_A(x1) = max{4, x1 + 2}
          filter2_3!_A(x1,x2,x3) = max{5, x3 + 1}
          filter2_4#!_A(x1,x2,x3,x4) = max{6, x2 - 3}
          false_0!_A = 1
          filter_2#!_A(x1,x2) = max{6, x1 - 3}
          filter_1!_A(x1) = max{4, x1 - 3}
          app!_A(x1,x2) = max{26, x1 - 14, x2 + 3}
          true_0!_A = 0
          cons_0!_A = 0
          cons_1!_A(x1) = max{0, x1 - 41}
          f_0!_A = 1
          f_1!_A(x1) = 4
          f_2!_A(x1,x2) = max{25, x1 - 16, x2 - 1}
          f_3!_A(x1,x2,x3) = max{31, x1 - 17, x3 + 15}
          filter_0!_A = 5
          filter_2!_A(x1,x2) = max{25, x1 + 24, x2 + 11}
          filter2_0!_A = 1
          filter2_1!_A(x1) = max{2, x1 - 6}
          filter2_2!_A(x1,x2) = max{4, x1 - 5, x2 - 1}
          filter2_4!_A(x1,x2,x3,x4) = max{25, x1 - 1, x2 - 1, x3 - 1}
          map_0!_A = 1
          map_2!_A(x1,x2) = max{0, x1 - 5, x2 - 28}
          s_0!_A = 31
          s_1!_A(x1) = max{30, x1 - 1}
          |0|_0!_A = 32
          |1|_0!_A = 29
          nil_0!_A = 0
  
    partial status:
  
      pi(map_2#) = []
      pi(cons_2) = []
      pi(app#) = []
      pi(map_1) = []
      pi(filter2_3) = []
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(filter_2#) = []
      pi(filter_1) = []
      pi(app) = []
      pi(true_0) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(f_0) = []
      pi(f_1) = []
      pi(f_2) = []
      pi(f_3) = []
      pi(filter_0) = []
      pi(filter_2) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(filter2_4) = []
      pi(map_0) = []
      pi(map_2) = [2]
      pi(s_0) = []
      pi(s_1) = []
      pi(|0|_0) = []
      pi(|1|_0) = []
      pi(nil_0) = []

The following rules are removed:

  p2, p7



-- SCC decomposition.

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

p1: map_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p2: map_2#(|fun|,cons_2(x,xs)) -> map_2#(|fun|,xs)
p3: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p4: filter2_4#(false_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p5: filter_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p6: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs)
p7: filter2_4#(true_0(),|fun|,x,xs) -> filter_2#(|fun|,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(f_0(),y) -> f_1(y)
r4: app(f_1(x1),y) -> f_2(x1,y)
r5: app(f_2(x1,x2),y) -> f_3(x1,x2,y)
r6: app(filter_0(),y) -> filter_1(y)
r7: app(filter_1(x1),y) -> filter_2(x1,y)
r8: app(filter2_0(),y) -> filter2_1(y)
r9: app(filter2_1(x1),y) -> filter2_2(x1,y)
r10: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r11: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: app(s_0(),y) -> s_1(y)
r15: f_3(|0|_0(),|1|_0(),x) -> f_3(s_1(x),x,x)
r16: f_3(x,y,s_1(z)) -> s_1(f_3(|0|_0(),|1|_0(),z))
r17: map_2(|fun|,nil_0()) -> nil_0()
r18: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r19: filter_2(|fun|,nil_0()) -> nil_0()
r20: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r21: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r22: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: map_2#(|fun|,cons_2(x,xs)) -> map_2#(|fun|,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(f_0(),y) -> f_1(y)
r4: app(f_1(x1),y) -> f_2(x1,y)
r5: app(f_2(x1,x2),y) -> f_3(x1,x2,y)
r6: app(filter_0(),y) -> filter_1(y)
r7: app(filter_1(x1),y) -> filter_2(x1,y)
r8: app(filter2_0(),y) -> filter2_1(y)
r9: app(filter2_1(x1),y) -> filter2_2(x1,y)
r10: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r11: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: app(s_0(),y) -> s_1(y)
r15: f_3(|0|_0(),|1|_0(),x) -> f_3(s_1(x),x,x)
r16: f_3(x,y,s_1(z)) -> s_1(f_3(|0|_0(),|1|_0(),z))
r17: map_2(|fun|,nil_0()) -> nil_0()
r18: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r19: filter_2(|fun|,nil_0()) -> nil_0()
r20: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r21: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r22: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,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{13, x2}
        cons_2_A(x1,x2) = max{x1 + 5, x2 + 13}
  
    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{9, x1 - 4, x2 - 5}
          map_2#!_A(x1,x2) = 0
          cons_2!_A(x1,x2) = max{4, x1 - 33, x2 - 68}
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [1]

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: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p2: filter2_4#(true_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p3: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs)
p4: filter2_4#(false_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p5: filter_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(f_0(),y) -> f_1(y)
r4: app(f_1(x1),y) -> f_2(x1,y)
r5: app(f_2(x1,x2),y) -> f_3(x1,x2,y)
r6: app(filter_0(),y) -> filter_1(y)
r7: app(filter_1(x1),y) -> filter_2(x1,y)
r8: app(filter2_0(),y) -> filter2_1(y)
r9: app(filter2_1(x1),y) -> filter2_2(x1,y)
r10: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r11: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: app(s_0(),y) -> s_1(y)
r15: f_3(|0|_0(),|1|_0(),x) -> f_3(s_1(x),x,x)
r16: f_3(x,y,s_1(z)) -> s_1(f_3(|0|_0(),|1|_0(),z))
r17: map_2(|fun|,nil_0()) -> nil_0()
r18: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r19: filter_2(|fun|,nil_0()) -> nil_0()
r20: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r21: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r22: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,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:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = 0
        filter2_3_A(x1,x2,x3) = max{11, x3 - 7}
        filter2_4#_A(x1,x2,x3,x4) = 0
        true_0_A = 8
        filter_2#_A(x1,x2) = 0
        cons_2_A(x1,x2) = max{9, x2 + 4}
        app_A(x1,x2) = max{11, x2 + 7}
        false_0_A = 8
        cons_0_A = 1
        cons_1_A(x1) = max{0, x1 - 7}
        f_0_A = 1
        f_1_A(x1) = max{11, x1 - 7}
        f_2_A(x1,x2) = max{0, x2 - 7}
        f_3_A(x1,x2,x3) = 8
        filter_0_A = 1
        filter_1_A(x1) = x1 + 6
        filter_2_A(x1,x2) = max{5, x2}
        filter2_0_A = 12
        filter2_1_A(x1) = 11
        filter2_2_A(x1,x2) = max{0, x2 - 7}
        filter2_4_A(x1,x2,x3,x4) = max{9, x4 + 4}
        map_0_A = 1
        map_1_A(x1) = max{0, x1 - 7}
        map_2_A(x1,x2) = max{2, x2 + 1}
        s_0_A = 1
        s_1_A(x1) = max{0, x1 - 10}
        |0|_0_A = 0
        |1|_0_A = 9
        nil_0_A = 6
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 37, x2 - 34}
          filter2_3_A(x1,x2,x3) = max{x1 - 8, x2 + 4, x3 + 11}
          filter2_4#_A(x1,x2,x3,x4) = max{0, x1 - 45, x2 - 33, x3 - 34, x4 - 38}
          true_0_A = 44
          filter_2#_A(x1,x2) = max{0, x1 - 33, x2 - 38}
          cons_2_A(x1,x2) = max{17, x1 + 4, x2}
          app_A(x1,x2) = max{25, x1 - 7, x2 + 11}
          false_0_A = 44
          cons_0_A = 6
          cons_1_A(x1) = x1 + 11
          f_0_A = 6
          f_1_A(x1) = max{12, x1 + 6}
          f_2_A(x1,x2) = max{x1 - 1, x2 + 6}
          f_3_A(x1,x2,x3) = max{x1 - 13, x2 - 13, x3 + 10}
          filter_0_A = 6
          filter_1_A(x1) = x1 + 6
          filter_2_A(x1,x2) = max{25, x2}
          filter2_0_A = 6
          filter2_1_A(x1) = max{12, x1 + 6}
          filter2_2_A(x1,x2) = max{x1 - 1, x2 + 11}
          filter2_4_A(x1,x2,x3,x4) = max{25, x3 + 4, x4}
          map_0_A = 45
          map_1_A(x1) = max{38, x1 + 6}
          map_2_A(x1,x2) = max{31, x1 - 3, x2 + 11}
          s_0_A = 31
          s_1_A(x1) = max{25, x1 - 11}
          |0|_0_A = 12
          |1|_0_A = 26
          nil_0_A = 32
          app#!_A(x1,x2) = max{17, x1 - 3, x2 - 1}
          filter2_3!_A(x1,x2,x3) = max{0, x1 - 48, x2 - 36}
          filter2_4#!_A(x1,x2,x3,x4) = max{16, x1 - 26, x2 - 3, x3 + 1, x4 - 2}
          true_0!_A = 1
          filter_2#!_A(x1,x2) = max{18, x1 - 3, x2 - 2}
          cons_2!_A(x1,x2) = max{0, x1 - 26, x2 - 30}
          app!_A(x1,x2) = max{19, x1 - 46, x2 - 28}
          false_0!_A = 1
          cons_0!_A = 1
          cons_1!_A(x1) = max{0, x1 - 27}
          f_0!_A = 1
          f_1!_A(x1) = max{2, x1 - 41}
          f_2!_A(x1,x2) = max{1, x1 - 40, x2 - 80}
          f_3!_A(x1,x2,x3) = max{13, x1 - 53, x2 - 80, x3 - 29}
          filter_0!_A = 1
          filter_1!_A(x1) = 4
          filter_2!_A(x1,x2) = max{5, x1 - 41, x2 - 29}
          filter2_0!_A = 20
          filter2_1!_A(x1) = 19
          filter2_2!_A(x1,x2) = max{1, x1 - 49, x2 - 37}
          filter2_4!_A(x1,x2,x3,x4) = max{0, x3 - 26, x4 - 30}
          map_0!_A = 1
          map_1!_A(x1) = max{2, x1 - 34}
          map_2!_A(x1,x2) = max{1, x1 - 33, x2 - 19}
          s_0!_A = 1
          s_1!_A(x1) = max{12, x1 - 2}
          |0|_0!_A = 11
          |1|_0!_A = 14
          nil_0!_A = 4
  
    partial status:
  
      pi(app#) = []
      pi(filter2_3) = []
      pi(filter2_4#) = []
      pi(true_0) = []
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(app) = [2]
      pi(false_0) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(f_0) = []
      pi(f_1) = []
      pi(f_2) = []
      pi(f_3) = []
      pi(filter_0) = []
      pi(filter_1) = [1]
      pi(filter_2) = [2]
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(filter2_4) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(map_2) = [2]
      pi(s_0) = []
      pi(s_1) = []
      pi(|0|_0) = []
      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: filter2_4#(true_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p2: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs)
p3: filter2_4#(false_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p4: filter_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(f_0(),y) -> f_1(y)
r4: app(f_1(x1),y) -> f_2(x1,y)
r5: app(f_2(x1,x2),y) -> f_3(x1,x2,y)
r6: app(filter_0(),y) -> filter_1(y)
r7: app(filter_1(x1),y) -> filter_2(x1,y)
r8: app(filter2_0(),y) -> filter2_1(y)
r9: app(filter2_1(x1),y) -> filter2_2(x1,y)
r10: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r11: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: app(s_0(),y) -> s_1(y)
r15: f_3(|0|_0(),|1|_0(),x) -> f_3(s_1(x),x,x)
r16: f_3(x,y,s_1(z)) -> s_1(f_3(|0|_0(),|1|_0(),z))
r17: map_2(|fun|,nil_0()) -> nil_0()
r18: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r19: filter_2(|fun|,nil_0()) -> nil_0()
r20: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r21: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r22: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: filter2_4#(true_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p2: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs)
p3: filter2_4#(false_0(),|fun|,x,xs) -> filter_2#(|fun|,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(f_0(),y) -> f_1(y)
r4: app(f_1(x1),y) -> f_2(x1,y)
r5: app(f_2(x1,x2),y) -> f_3(x1,x2,y)
r6: app(filter_0(),y) -> filter_1(y)
r7: app(filter_1(x1),y) -> filter_2(x1,y)
r8: app(filter2_0(),y) -> filter2_1(y)
r9: app(filter2_1(x1),y) -> filter2_2(x1,y)
r10: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r11: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: app(s_0(),y) -> s_1(y)
r15: f_3(|0|_0(),|1|_0(),x) -> f_3(s_1(x),x,x)
r16: f_3(x,y,s_1(z)) -> s_1(f_3(|0|_0(),|1|_0(),z))
r17: map_2(|fun|,nil_0()) -> nil_0()
r18: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r19: filter_2(|fun|,nil_0()) -> nil_0()
r20: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r21: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r22: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The set of usable rules consists of

  r1, r3, r4, r6, r9, r10, r12, r17, r19

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
        true_0_A = 21
        filter_2#_A(x1,x2) = x2
        cons_2_A(x1,x2) = max{x1 - 9, x2}
        app_A(x1,x2) = max{12, x1 - 6, x2}
        false_0_A = 14
        cons_0_A = 1
        cons_1_A(x1) = max{5, x1 - 1}
        f_0_A = 3
        f_1_A(x1) = max{4, x1 - 1}
        f_2_A(x1,x2) = max{5, x1 - 8, x2 - 1}
        filter_0_A = 1
        filter_1_A(x1) = max{7, x1 - 1}
        filter2_1_A(x1) = max{32, x1 - 1}
        filter2_2_A(x1,x2) = max{25, x1 - 8, x2 - 1}
        filter2_3_A(x1,x2,x3) = max{19, x1 - 14, x2 - 7, x3 - 3}
        map_0_A = 1
        map_1_A(x1) = max{2, x1}
        map_2_A(x1,x2) = max{3, x1 - 5, x2}
        nil_0_A = 4
        filter_2_A(x1,x2) = max{22, x1 - 26, x2}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          filter2_4#_A(x1,x2,x3,x4) = max{15, x2 - 1, x3 - 13, x4 + 2}
          true_0_A = 38
          filter_2#_A(x1,x2) = max{15, x1 - 1, x2 + 2}
          cons_2_A(x1,x2) = max{12, x1 - 15, x2}
          app_A(x1,x2) = max{14, x1 - 12, x2 + 1}
          false_0_A = 38
          cons_0_A = 2
          cons_1_A(x1) = max{2, x1 - 35}
          f_0_A = 2
          f_1_A(x1) = max{2, x1 + 1}
          f_2_A(x1,x2) = max{2, x1 - 17, x2 - 33}
          filter_0_A = 2
          filter_1_A(x1) = max{2, x1 + 1}
          filter2_1_A(x1) = max{11, x1 + 9}
          filter2_2_A(x1,x2) = max{2, x1 - 3, x2 + 1}
          filter2_3_A(x1,x2,x3) = max{x1 - 15, x2 - 11, x3 + 1}
          map_0_A = 2
          map_1_A(x1) = max{2, x1 - 15}
          map_2_A(x1,x2) = max{14, x1 - 27, x2 + 1}
          nil_0_A = 15
          filter_2_A(x1,x2) = max{14, x1 - 39, x2 + 1}
          filter2_4#!_A(x1,x2,x3,x4) = x2 + 3
          true_0!_A = 1
          filter_2#!_A(x1,x2) = x1 + 3
          cons_2!_A(x1,x2) = max{4, x1 - 37, x2 - 50}
          app!_A(x1,x2) = max{6, x1 - 3, x2 - 20}
          false_0!_A = 1
          cons_0!_A = 8
          cons_1!_A(x1) = max{7, x1 - 38}
          f_0!_A = 1
          f_1!_A(x1) = max{2, x1 - 20}
          f_2!_A(x1,x2) = max{3, x1 - 20, x2 - 66}
          filter_0!_A = 1
          filter_1!_A(x1) = max{7, x1 - 21}
          filter2_1!_A(x1) = max{0, x1 - 21}
          filter2_2!_A(x1,x2) = max{7, x2 - 21}
          filter2_3!_A(x1,x2,x3) = max{0, x1 - 7, x2 - 3, x3 - 36}
          map_0!_A = 1
          map_1!_A(x1) = max{2, x1 - 21}
          map_2!_A(x1,x2) = max{5, x1 - 4, x2 - 20}
          nil_0!_A = 4
          filter_2!_A(x1,x2) = x2 + 16
  
    partial status:
  
      pi(filter2_4#) = [4]
      pi(true_0) = []
      pi(filter_2#) = [2]
      pi(cons_2) = [2]
      pi(app) = [2]
      pi(false_0) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(f_0) = []
      pi(f_1) = []
      pi(f_2) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(filter2_3) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(map_2) = []
      pi(nil_0) = []
      pi(filter_2) = [2]

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: filter2_4#(true_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p2: filter2_4#(false_0(),|fun|,x,xs) -> filter_2#(|fun|,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(f_0(),y) -> f_1(y)
r4: app(f_1(x1),y) -> f_2(x1,y)
r5: app(f_2(x1,x2),y) -> f_3(x1,x2,y)
r6: app(filter_0(),y) -> filter_1(y)
r7: app(filter_1(x1),y) -> filter_2(x1,y)
r8: app(filter2_0(),y) -> filter2_1(y)
r9: app(filter2_1(x1),y) -> filter2_2(x1,y)
r10: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r11: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: app(s_0(),y) -> s_1(y)
r15: f_3(|0|_0(),|1|_0(),x) -> f_3(s_1(x),x,x)
r16: f_3(x,y,s_1(z)) -> s_1(f_3(|0|_0(),|1|_0(),z))
r17: map_2(|fun|,nil_0()) -> nil_0()
r18: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r19: filter_2(|fun|,nil_0()) -> nil_0()
r20: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r21: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r22: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

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

p1: f_3#(|0|_0(),|1|_0(),x) -> f_3#(s_1(x),x,x)
p2: f_3#(x,y,s_1(z)) -> f_3#(|0|_0(),|1|_0(),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(f_0(),y) -> f_1(y)
r4: app(f_1(x1),y) -> f_2(x1,y)
r5: app(f_2(x1,x2),y) -> f_3(x1,x2,y)
r6: app(filter_0(),y) -> filter_1(y)
r7: app(filter_1(x1),y) -> filter_2(x1,y)
r8: app(filter2_0(),y) -> filter2_1(y)
r9: app(filter2_1(x1),y) -> filter2_2(x1,y)
r10: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r11: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: app(s_0(),y) -> s_1(y)
r15: f_3(|0|_0(),|1|_0(),x) -> f_3(s_1(x),x,x)
r16: f_3(x,y,s_1(z)) -> s_1(f_3(|0|_0(),|1|_0(),z))
r17: map_2(|fun|,nil_0()) -> nil_0()
r18: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r19: filter_2(|fun|,nil_0()) -> nil_0()
r20: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r21: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r22: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,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:
      
        f_3#_A(x1,x2,x3) = 23
        |0|_0_A = 22
        |1|_0_A = 22
        s_1_A(x1) = max{24, x1 - 33}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          f_3#_A(x1,x2,x3) = max{0, x1 - 51, x2 - 32, x3 - 32}
          |0|_0_A = 50
          |1|_0_A = 31
          s_1_A(x1) = max{30, x1 + 19}
          f_3#!_A(x1,x2,x3) = max{1, x3 - 28}
          |0|_0!_A = 3
          |1|_0!_A = 3
          s_1!_A(x1) = max{5, x1 - 27}
  
    partial status:
  
      pi(f_3#) = []
      pi(|0|_0) = []
      pi(|1|_0) = []
      pi(s_1) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: f_3#(|0|_0(),|1|_0(),x) -> f_3#(s_1(x),x,x)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(f_0(),y) -> f_1(y)
r4: app(f_1(x1),y) -> f_2(x1,y)
r5: app(f_2(x1,x2),y) -> f_3(x1,x2,y)
r6: app(filter_0(),y) -> filter_1(y)
r7: app(filter_1(x1),y) -> filter_2(x1,y)
r8: app(filter2_0(),y) -> filter2_1(y)
r9: app(filter2_1(x1),y) -> filter2_2(x1,y)
r10: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r11: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r12: app(map_0(),y) -> map_1(y)
r13: app(map_1(x1),y) -> map_2(x1,y)
r14: app(s_0(),y) -> s_1(y)
r15: f_3(|0|_0(),|1|_0(),x) -> f_3(s_1(x),x,x)
r16: f_3(x,y,s_1(z)) -> s_1(f_3(|0|_0(),|1|_0(),z))
r17: map_2(|fun|,nil_0()) -> nil_0()
r18: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r19: filter_2(|fun|,nil_0()) -> nil_0()
r20: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r21: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r22: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The estimated dependency graph contains the following SCCs:

  (no SCCs)