YES
We show the termination of the TRS R:

  app(f(),app(s(),x)) -> app(f(),x)
  app(g(),app(app(cons(),|0|()),y)) -> app(g(),y)
  app(g(),app(app(cons(),app(s(),x)),y)) -> app(s(),x)
  app(h(),app(app(cons(),x),y)) -> app(h(),app(g(),app(app(cons(),x),y)))
  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(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(g_0(),y) -> g_1(y)
  app(h_0(),y) -> h_1(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_1(s_1(x)) -> f_1(x)
  g_1(cons_2(|0|_0(),y)) -> g_1(y)
  g_1(cons_2(s_1(x),y)) -> s_1(x)
  h_1(cons_2(x,y)) -> h_1(g_1(cons_2(x,y)))
  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_0(),y) -> f_1#(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#(g_0(),y) -> g_1#(y)
p5: app#(h_0(),y) -> h_1#(y)
p6: app#(map_1(x1),y) -> map_2#(x1,y)
p7: f_1#(s_1(x)) -> f_1#(x)
p8: g_1#(cons_2(|0|_0(),y)) -> g_1#(y)
p9: h_1#(cons_2(x,y)) -> h_1#(g_1(cons_2(x,y)))
p10: h_1#(cons_2(x,y)) -> g_1#(cons_2(x,y))
p11: map_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p12: map_2#(|fun|,cons_2(x,xs)) -> map_2#(|fun|,xs)
p13: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs)
p14: filter_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p15: filter2_4#(true_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p16: 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(filter_0(),y) -> filter_1(y)
r5: app(filter_1(x1),y) -> filter_2(x1,y)
r6: app(filter2_0(),y) -> filter2_1(y)
r7: app(filter2_1(x1),y) -> filter2_2(x1,y)
r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r10: app(g_0(),y) -> g_1(y)
r11: app(h_0(),y) -> h_1(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_1(s_1(x)) -> f_1(x)
r16: g_1(cons_2(|0|_0(),y)) -> g_1(y)
r17: g_1(cons_2(s_1(x),y)) -> s_1(x)
r18: h_1(cons_2(x,y)) -> h_1(g_1(cons_2(x,y)))
r19: map_2(|fun|,nil_0()) -> nil_0()
r20: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r21: filter_2(|fun|,nil_0()) -> nil_0()
r22: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r23: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r24: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p6, p11, p12, p13, p14, p15, p16}
  {p7}
  {p9}
  {p8}


-- 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(filter_0(),y) -> filter_1(y)
r5: app(filter_1(x1),y) -> filter_2(x1,y)
r6: app(filter2_0(),y) -> filter2_1(y)
r7: app(filter2_1(x1),y) -> filter2_2(x1,y)
r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r10: app(g_0(),y) -> g_1(y)
r11: app(h_0(),y) -> h_1(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_1(s_1(x)) -> f_1(x)
r16: g_1(cons_2(|0|_0(),y)) -> g_1(y)
r17: g_1(cons_2(s_1(x),y)) -> s_1(x)
r18: h_1(cons_2(x,y)) -> h_1(g_1(cons_2(x,y)))
r19: map_2(|fun|,nil_0()) -> nil_0()
r20: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r21: filter_2(|fun|,nil_0()) -> nil_0()
r22: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r23: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r24: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The set of usable rules consists of

  r2, r3, r6, r8, r12, r15, r19, r21

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{53, x1 - 1, x2 + 1}
            cons_2_A(x1,x2) = max{19, x1 + 5, x2}
            app#_A(x1,x2) = max{53, x1 - 13, x2 + 1}
            map_1_A(x1) = x1 + 12
            filter2_3_A(x1,x2,x3) = max{65, x1 - 1, x2 + 13, x3 - 55}
            filter2_4#_A(x1,x2,x3,x4) = max{53, x2, x3 - 68, x4 - 1}
            false_0_A = 59
            filter_2#_A(x1,x2) = max{53, x1, x2 - 1}
            filter_1_A(x1) = max{65, x1 + 13}
            app_A(x1,x2) = max{x1 + 64, x2 + 13}
            true_0_A = 64
            cons_1_A(x1) = max{65, x1 - 46}
            f_0_A = 0
            f_1_A(x1) = 64
            filter2_0_A = 0
            filter2_1_A(x1) = max{30, x1 - 46}
            filter2_2_A(x1,x2) = max{94, x1 - 65, x2 - 48}
            map_0_A = 0
            s_1_A(x1) = max{30, x1 - 5}
            map_2_A(x1,x2) = max{70, x1 + 69, x2 + 13}
            nil_0_A = 130
            filter_2_A(x1,x2) = max{129, x1 - 1, x2 + 14}
            map_2#!_A(x1,x2) = max{6, x1 - 65, x2 - 6}
            cons_2!_A(x1,x2) = max{5, x1 - 30}
            app#!_A(x1,x2) = max{12, x1 - 66, x2 - 2}
            map_1!_A(x1) = max{1, x1}
            filter2_3!_A(x1,x2,x3) = max{10, x1 - 61, x2 - 55, x3 - 123}
            filter2_4#!_A(x1,x2,x3,x4) = max{11, x2 - 54, x3 - 122, x4 - 4}
            false_0!_A = 1
            filter_2#!_A(x1,x2) = max{2, x1 - 54, x2 - 4}
            filter_1!_A(x1) = max{5, x1 - 1}
            app!_A(x1,x2) = max{x1 + 47, x2 + 6}
            true_0!_A = 1
            cons_1!_A(x1) = max{0, x1 - 30}
            f_0!_A = 8
            f_1!_A(x1) = 7
            filter2_0!_A = 9
            filter2_1!_A(x1) = max{8, x1 + 5}
            filter2_2!_A(x1,x2) = max{7, x1 - 60}
            map_0!_A = 2
            s_1!_A(x1) = 48
            map_2!_A(x1,x2) = max{x1 + 56, x2 - 21}
            nil_0!_A = 3
            filter_2!_A(x1,x2) = max{8, x1 + 4, x2 - 12}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{30, x2 - 4}
            cons_2_A(x1,x2) = max{28, x1 + 5, x2}
            app#_A(x1,x2) = max{30, x2 + 1}
            map_1_A(x1) = max{17, x1 - 2}
            filter2_3_A(x1,x2,x3) = max{17, x1 + 9, x2 - 2, x3 + 9}
            filter2_4#_A(x1,x2,x3,x4) = max{30, x4 + 1}
            false_0_A = 37
            filter_2#_A(x1,x2) = max{30, x2 + 1}
            filter_1_A(x1) = x1 + 17
            app_A(x1,x2) = max{35, x1 - 4, x2 + 17}
            true_0_A = 45
            cons_1_A(x1) = max{18, x1 + 17}
            f_0_A = 3
            f_1_A(x1) = max{18, x1 + 17}
            filter2_0_A = 3
            filter2_1_A(x1) = max{20, x1 + 17}
            filter2_2_A(x1,x2) = max{x1 + 13, x2 + 20}
            map_0_A = 3
            s_1_A(x1) = max{29, x1}
            map_2_A(x1,x2) = max{22, x1 - 36, x2 + 18}
            nil_0_A = 32
            filter_2_A(x1,x2) = max{31, x1 - 34, x2 + 2}
            map_2#!_A(x1,x2) = 0
            cons_2!_A(x1,x2) = max{30, x1 + 2, x2 + 13}
            app#!_A(x1,x2) = max{31, x2 + 3}
            map_1!_A(x1) = max{0, x1 - 21}
            filter2_3!_A(x1,x2,x3) = max{49, x1 - 15, x3 - 1}
            filter2_4#!_A(x1,x2,x3,x4) = 0
            false_0!_A = 1
            filter_2#!_A(x1,x2) = 0
            filter_1!_A(x1) = max{0, x1 - 2}
            app!_A(x1,x2) = max{x1 - 19, x2 + 47}
            true_0!_A = 31
            cons_1!_A(x1) = max{29, x1 - 1}
            f_0!_A = 39
            f_1!_A(x1) = max{38, x1 - 30}
            filter2_0!_A = 37
            filter2_1!_A(x1) = max{36, x1 - 1}
            filter2_2!_A(x1,x2) = max{48, x2 + 2}
            map_0!_A = 1
            s_1!_A(x1) = max{47, x1 - 1}
            map_2!_A(x1,x2) = max{x1 - 22, x2 + 31}
            nil_0!_A = 28
            filter_2!_A(x1,x2) = max{0, x1 - 2, x2 - 1}
        
  
    partial status:
  
      pi(map_2#) = []
      pi(cons_2) = []
      pi(app#) = []
      pi(map_1) = [1]
      pi(filter2_3) = [2, 3]
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(filter_2#) = []
      pi(filter_1) = [1]
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(cons_1) = [1]
      pi(f_0) = []
      pi(f_1) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = [2]
      pi(map_0) = []
      pi(s_1) = []
      pi(map_2) = [1, 2]
      pi(nil_0) = []
      pi(filter_2) = [1]

The following rules are removed:

  p1, p2, p4, p6, p7



-- SCC decomposition.

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

p1: map_2#(|fun|,cons_2(x,xs)) -> map_2#(|fun|,xs)
p2: filter2_4#(false_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#(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(filter_0(),y) -> filter_1(y)
r5: app(filter_1(x1),y) -> filter_2(x1,y)
r6: app(filter2_0(),y) -> filter2_1(y)
r7: app(filter2_1(x1),y) -> filter2_2(x1,y)
r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r10: app(g_0(),y) -> g_1(y)
r11: app(h_0(),y) -> h_1(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_1(s_1(x)) -> f_1(x)
r16: g_1(cons_2(|0|_0(),y)) -> g_1(y)
r17: g_1(cons_2(s_1(x),y)) -> s_1(x)
r18: h_1(cons_2(x,y)) -> h_1(g_1(cons_2(x,y)))
r19: map_2(|fun|,nil_0()) -> nil_0()
r20: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r21: filter_2(|fun|,nil_0()) -> nil_0()
r22: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r23: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r24: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,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#(|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(filter_0(),y) -> filter_1(y)
r5: app(filter_1(x1),y) -> filter_2(x1,y)
r6: app(filter2_0(),y) -> filter2_1(y)
r7: app(filter2_1(x1),y) -> filter2_2(x1,y)
r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r10: app(g_0(),y) -> g_1(y)
r11: app(h_0(),y) -> h_1(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_1(s_1(x)) -> f_1(x)
r16: g_1(cons_2(|0|_0(),y)) -> g_1(y)
r17: g_1(cons_2(s_1(x),y)) -> s_1(x)
r18: h_1(cons_2(x,y)) -> h_1(g_1(cons_2(x,y)))
r19: map_2(|fun|,nil_0()) -> nil_0()
r20: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r21: filter_2(|fun|,nil_0()) -> nil_0()
r22: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r23: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r24: 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:
  
      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) = 0
            cons_2_A(x1,x2) = max{12, x1 - 4, x2 - 12}
            map_2#!_A(x1,x2) = max{0, x2 - 56}
            cons_2!_A(x1,x2) = max{0, x1 - 23, x2 - 56}
        
        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 - 9, x2 - 4}
            map_2#!_A(x1,x2) = max{1, x2 - 24}
            cons_2!_A(x1,x2) = max{3, x1 - 54, x2 - 23}
        
  
    partial status:
  
      pi(map_2#) = [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(),|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#(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(filter_0(),y) -> filter_1(y)
r5: app(filter_1(x1),y) -> filter_2(x1,y)
r6: app(filter2_0(),y) -> filter2_1(y)
r7: app(filter2_1(x1),y) -> filter2_2(x1,y)
r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r10: app(g_0(),y) -> g_1(y)
r11: app(h_0(),y) -> h_1(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_1(s_1(x)) -> f_1(x)
r16: g_1(cons_2(|0|_0(),y)) -> g_1(y)
r17: g_1(cons_2(s_1(x),y)) -> s_1(x)
r18: h_1(cons_2(x,y)) -> h_1(g_1(cons_2(x,y)))
r19: map_2(|fun|,nil_0()) -> nil_0()
r20: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r21: filter_2(|fun|,nil_0()) -> nil_0()
r22: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r23: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r24: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The set of usable rules consists of

  r3, r4, r11, r14, r15, r16, r17, r18, r19, r21

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, x3 - 23, x4}
            false_0_A = 12
            filter_2#_A(x1,x2) = max{1, x1 - 1, x2}
            cons_2_A(x1,x2) = max{x1 - 23, x2}
            app_A(x1,x2) = max{7, x1}
            true_0_A = 11
            f_0_A = 3
            f_1_A(x1) = 4
            filter_0_A = 4
            filter_1_A(x1) = 5
            h_0_A = 3
            h_1_A(x1) = 4
            s_0_A = 0
            s_1_A(x1) = 0
            g_1_A(x1) = max{2, x1 - 23}
            |0|_0_A = 0
            map_2_A(x1,x2) = max{7, x1 - 11, x2 - 11}
            nil_0_A = 7
            filter_2_A(x1,x2) = max{6, x2 + 1}
            filter2_4#!_A(x1,x2,x3,x4) = 10
            false_0!_A = 9
            filter_2#!_A(x1,x2) = 10
            cons_2!_A(x1,x2) = max{x1 + 1, x2 + 11}
            app!_A(x1,x2) = max{23, x1 - 2}
            true_0!_A = 18
            f_0!_A = 3
            f_1!_A(x1) = 2
            filter_0!_A = 1
            filter_1!_A(x1) = 1
            h_0!_A = 1
            h_1!_A(x1) = 12
            s_0!_A = 1
            s_1!_A(x1) = 11
            g_1!_A(x1) = x1 + 11
            |0|_0!_A = 1
            map_2!_A(x1,x2) = max{21, x1 + 19}
            nil_0!_A = 20
            filter_2!_A(x1,x2) = max{24, x1 + 1, x2}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            filter2_4#_A(x1,x2,x3,x4) = max{111, x3 - 1, x4 - 1}
            false_0_A = 55
            filter_2#_A(x1,x2) = max{111, x2 - 1}
            cons_2_A(x1,x2) = max{111, x1 + 85, x2 + 72}
            app_A(x1,x2) = max{x1 + 55, x2 + 136}
            true_0_A = 55
            f_0_A = 79
            f_1_A(x1) = x1 + 136
            filter_0_A = 79
            filter_1_A(x1) = max{56, x1 - 56}
            h_0_A = 80
            h_1_A(x1) = max{136, x1 - 112}
            s_0_A = 0
            s_1_A(x1) = max{55, x1 + 35}
            g_1_A(x1) = max{93, x1 - 18}
            |0|_0_A = 0
            map_2_A(x1,x2) = max{176, x1 + 140, x2 + 136}
            nil_0_A = 89
            filter_2_A(x1,x2) = max{114, x2 - 4}
            filter2_4#!_A(x1,x2,x3,x4) = max{5, x4 + 3}
            false_0!_A = 1
            filter_2#!_A(x1,x2) = max{2, x2 - 66}
            cons_2!_A(x1,x2) = max{21, x2 + 4}
            app!_A(x1,x2) = max{20, x1 - 57, x2 - 2}
            true_0!_A = 119
            f_0!_A = 22
            f_1!_A(x1) = max{21, x1 - 34}
            filter_0!_A = 1
            filter_1!_A(x1) = max{21, x1 - 3}
            h_0!_A = 1
            h_1!_A(x1) = max{22, x1 - 92}
            s_0!_A = 20
            s_1!_A(x1) = 19
            g_1!_A(x1) = 20
            |0|_0!_A = 21
            map_2!_A(x1,x2) = max{x1 + 75, x2 - 88}
            nil_0!_A = 0
            filter_2!_A(x1,x2) = 0
        
  
    partial status:
  
      pi(filter2_4#) = [4]
      pi(false_0) = []
      pi(filter_2#) = [2]
      pi(cons_2) = [2]
      pi(app) = []
      pi(true_0) = []
      pi(f_0) = []
      pi(f_1) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(h_0) = []
      pi(h_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(g_1) = []
      pi(|0|_0) = []
      pi(map_2) = []
      pi(nil_0) = []
      pi(filter_2) = []

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: f_1#(s_1(x)) -> f_1#(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(filter_0(),y) -> filter_1(y)
r5: app(filter_1(x1),y) -> filter_2(x1,y)
r6: app(filter2_0(),y) -> filter2_1(y)
r7: app(filter2_1(x1),y) -> filter2_2(x1,y)
r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r10: app(g_0(),y) -> g_1(y)
r11: app(h_0(),y) -> h_1(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_1(s_1(x)) -> f_1(x)
r16: g_1(cons_2(|0|_0(),y)) -> g_1(y)
r17: g_1(cons_2(s_1(x),y)) -> s_1(x)
r18: h_1(cons_2(x,y)) -> h_1(g_1(cons_2(x,y)))
r19: map_2(|fun|,nil_0()) -> nil_0()
r20: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r21: filter_2(|fun|,nil_0()) -> nil_0()
r22: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r23: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r24: 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:
  
      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:
          
            f_1#_A(x1) = x1
            s_1_A(x1) = max{3, x1}
            f_1#!_A(x1) = max{3, x1 - 15}
            s_1!_A(x1) = max{3, x1 - 14}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            f_1#_A(x1) = x1
            s_1_A(x1) = max{31, x1}
            f_1#!_A(x1) = max{4, x1 - 98}
            s_1!_A(x1) = max{4, x1 - 98}
        
  
    partial status:
  
      pi(f_1#) = [1]
      pi(s_1) = [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: h_1#(cons_2(x,y)) -> h_1#(g_1(cons_2(x,y)))

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(filter_0(),y) -> filter_1(y)
r5: app(filter_1(x1),y) -> filter_2(x1,y)
r6: app(filter2_0(),y) -> filter2_1(y)
r7: app(filter2_1(x1),y) -> filter2_2(x1,y)
r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r10: app(g_0(),y) -> g_1(y)
r11: app(h_0(),y) -> h_1(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_1(s_1(x)) -> f_1(x)
r16: g_1(cons_2(|0|_0(),y)) -> g_1(y)
r17: g_1(cons_2(s_1(x),y)) -> s_1(x)
r18: h_1(cons_2(x,y)) -> h_1(g_1(cons_2(x,y)))
r19: map_2(|fun|,nil_0()) -> nil_0()
r20: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r21: filter_2(|fun|,nil_0()) -> nil_0()
r22: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r23: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r24: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The set of usable rules consists of

  r16, r17

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:
          
            h_1#_A(x1) = max{0, x1 - 75}
            cons_2_A(x1,x2) = max{x1 - 35, x2 + 74}
            g_1_A(x1) = max{57, x1 - 83}
            |0|_0_A = 117
            s_1_A(x1) = 10
            h_1#!_A(x1) = max{5, x1 - 68}
            cons_2!_A(x1,x2) = max{4, x1 - 118, x2 - 76}
            g_1!_A(x1) = max{3, x1 - 83}
            |0|_0!_A = 4
            s_1!_A(x1) = max{8, x1 - 117}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            h_1#_A(x1) = max{0, x1 - 16}
            cons_2_A(x1,x2) = max{x1 - 18, x2 + 15}
            g_1_A(x1) = max{5, x1 - 20}
            |0|_0_A = 36
            s_1_A(x1) = 5
            h_1#!_A(x1) = max{5, x1 - 19}
            cons_2!_A(x1,x2) = max{5, x1 - 37, x2 - 9}
            g_1!_A(x1) = max{3, x1 - 20}
            |0|_0!_A = 1
            s_1!_A(x1) = max{6, x1 - 38}
        
  
    partial status:
  
      pi(h_1#) = [1]
      pi(cons_2) = [1, 2]
      pi(g_1) = [1]
      pi(|0|_0) = []
      pi(s_1) = [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: g_1#(cons_2(|0|_0(),y)) -> g_1#(y)

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(filter_0(),y) -> filter_1(y)
r5: app(filter_1(x1),y) -> filter_2(x1,y)
r6: app(filter2_0(),y) -> filter2_1(y)
r7: app(filter2_1(x1),y) -> filter2_2(x1,y)
r8: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r9: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r10: app(g_0(),y) -> g_1(y)
r11: app(h_0(),y) -> h_1(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_1(s_1(x)) -> f_1(x)
r16: g_1(cons_2(|0|_0(),y)) -> g_1(y)
r17: g_1(cons_2(s_1(x),y)) -> s_1(x)
r18: h_1(cons_2(x,y)) -> h_1(g_1(cons_2(x,y)))
r19: map_2(|fun|,nil_0()) -> nil_0()
r20: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r21: filter_2(|fun|,nil_0()) -> nil_0()
r22: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r23: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r24: 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:
  
      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:
          
            g_1#_A(x1) = 0
            cons_2_A(x1,x2) = max{3, x1 - 13, x2 - 4}
            |0|_0_A = 18
            g_1#!_A(x1) = max{2, x1 - 18}
            cons_2!_A(x1,x2) = max{3, x1 - 19, x2 - 17}
            |0|_0!_A = 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            g_1#_A(x1) = 0
            cons_2_A(x1,x2) = max{9, x1 - 21, x2 - 10}
            |0|_0_A = 69
            g_1#!_A(x1) = max{2, x1 - 45}
            cons_2!_A(x1,x2) = max{5, x1 - 70, x2 - 44}
            |0|_0!_A = 1
        
  
    partial status:
  
      pi(g_1#) = [1]
      pi(cons_2) = [2]
      pi(|0|_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.