YES
We show the termination of the TRS R:

  app(app(minus(),x),|0|()) -> x
  app(app(minus(),app(s(),x)),app(s(),y)) -> app(app(minus(),x),y)
  app(f(),|0|()) -> app(s(),|0|())
  app(f(),app(s(),x)) -> app(app(minus(),app(s(),x)),app(g(),app(f(),x)))
  app(g(),|0|()) -> |0|()
  app(g(),app(s(),x)) -> app(app(minus(),app(s(),x)),app(f(),app(g(),x)))
  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(map_0(),y) -> map_1(y)
  app(map_1(x1),y) -> map_2(x1,y)
  app(minus_0(),y) -> minus_1(y)
  app(minus_1(x1),y) -> minus_2(x1,y)
  app(s_0(),y) -> s_1(y)
  minus_2(x,|0|_0()) -> x
  minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
  f_1(|0|_0()) -> s_1(|0|_0())
  f_1(s_1(x)) -> minus_2(s_1(x),g_1(f_1(x)))
  g_1(|0|_0()) -> |0|_0()
  g_1(s_1(x)) -> minus_2(s_1(x),f_1(g_1(x)))
  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#(map_1(x1),y) -> map_2#(x1,y)
p6: app#(minus_1(x1),y) -> minus_2#(x1,y)
p7: minus_2#(s_1(x),s_1(y)) -> minus_2#(x,y)
p8: f_1#(s_1(x)) -> minus_2#(s_1(x),g_1(f_1(x)))
p9: f_1#(s_1(x)) -> g_1#(f_1(x))
p10: f_1#(s_1(x)) -> f_1#(x)
p11: g_1#(s_1(x)) -> minus_2#(s_1(x),f_1(g_1(x)))
p12: g_1#(s_1(x)) -> f_1#(g_1(x))
p13: g_1#(s_1(x)) -> g_1#(x)
p14: map_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p15: map_2#(|fun|,cons_2(x,xs)) -> map_2#(|fun|,xs)
p16: filter_2#(|fun|,cons_2(x,xs)) -> filter2_4#(app(|fun|,x),|fun|,x,xs)
p17: filter_2#(|fun|,cons_2(x,xs)) -> app#(|fun|,x)
p18: filter2_4#(true_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p19: 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(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: app(minus_0(),y) -> minus_1(y)
r14: app(minus_1(x1),y) -> minus_2(x1,y)
r15: app(s_0(),y) -> s_1(y)
r16: minus_2(x,|0|_0()) -> x
r17: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r18: f_1(|0|_0()) -> s_1(|0|_0())
r19: f_1(s_1(x)) -> minus_2(s_1(x),g_1(f_1(x)))
r20: g_1(|0|_0()) -> |0|_0()
r21: g_1(s_1(x)) -> minus_2(s_1(x),f_1(g_1(x)))
r22: map_2(|fun|,nil_0()) -> nil_0()
r23: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r24: filter_2(|fun|,nil_0()) -> nil_0()
r25: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r26: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r27: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p5, p14, p15, p16, p17, p18, p19}
  {p9, p10, p12, p13}
  {p7}


-- 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(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: app(minus_0(),y) -> minus_1(y)
r14: app(minus_1(x1),y) -> minus_2(x1,y)
r15: app(s_0(),y) -> s_1(y)
r16: minus_2(x,|0|_0()) -> x
r17: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r18: f_1(|0|_0()) -> s_1(|0|_0())
r19: f_1(s_1(x)) -> minus_2(s_1(x),g_1(f_1(x)))
r20: g_1(|0|_0()) -> |0|_0()
r21: g_1(s_1(x)) -> minus_2(s_1(x),f_1(g_1(x)))
r22: map_2(|fun|,nil_0()) -> nil_0()
r23: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r24: filter_2(|fun|,nil_0()) -> nil_0()
r25: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r26: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r27: 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) = max{1, x1 - 10, x2 - 35}
            cons_2_A(x1,x2) = max{x1 + 34, x2 + 34}
            app#_A(x1,x2) = max{1, x1 - 27, x2 - 35}
            map_1_A(x1) = x1 + 17
            filter2_3_A(x1,x2,x3) = max{x1 + 14, x2 + 26, x3 + 9}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 20, x3 - 18, x4 - 35}
            false_0_A = 0
            filter_2#_A(x1,x2) = max{1, x1 - 20, x2 - 35}
            filter_1_A(x1) = max{25, x1 + 7}
            app_A(x1,x2) = max{1, x1 - 27, x2 - 27}
            true_0_A = 0
            map_2#!_A(x1,x2) = max{2, x1 - 1, x2 - 68}
            cons_2!_A(x1,x2) = max{4, x1 - 66, x2 - 36}
            app#!_A(x1,x2) = max{1, x1 - 15, x2 - 67}
            map_1!_A(x1) = max{10, x1 - 1}
            filter2_3!_A(x1,x2,x3) = max{12, x2 - 10, x3 - 68}
            filter2_4#!_A(x1,x2,x3,x4) = max{10, x2 - 9, x3 - 34, x4 - 68}
            false_0!_A = 11
            filter_2#!_A(x1,x2) = max{10, x1 - 9, x2 - 68}
            filter_1!_A(x1) = max{11, x1 - 1}
            app!_A(x1,x2) = max{9, x1 - 10, x2 - 35}
            true_0!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{1, x1 - 37, x2 - 34}
            cons_2_A(x1,x2) = max{x1 + 9, x2 + 33}
            app#_A(x1,x2) = max{1, x1 - 37, x2 - 25}
            map_1_A(x1) = x1 + 4
            filter2_3_A(x1,x2,x3) = max{x1 + 36, x2, x3 - 51}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 37, x3 - 88, x4 - 25}
            false_0_A = 0
            filter_2#_A(x1,x2) = max{1, x1 - 37, x2 - 34}
            filter_1_A(x1) = max{11, x1 + 1}
            app_A(x1,x2) = max{1, x1 - 52, x2 - 52}
            true_0_A = 0
            map_2#!_A(x1,x2) = max{6, x1 - 3, x2 - 26}
            cons_2!_A(x1,x2) = max{5, x1 - 88, x2 - 26}
            app#!_A(x1,x2) = max{7, x1 - 3, x2 - 26}
            map_1!_A(x1) = max{0, x1 - 1}
            filter2_3!_A(x1,x2,x3) = max{1, x1 - 36, x2 - 83, x3 - 90}
            filter2_4#!_A(x1,x2,x3,x4) = max{8, x2 - 2, x3 - 53, x4 - 25}
            false_0!_A = 2
            filter_2#!_A(x1,x2) = max{8, x1 - 2, x2 - 34}
            filter_1!_A(x1) = max{1, x1 - 34}
            app!_A(x1,x2) = max{7, x1 - 33, x2 - 38}
            true_0!_A = 4
        
  
    partial status:
  
      pi(map_2#) = [1, 2]
      pi(cons_2) = [1, 2]
      pi(app#) = [1, 2]
      pi(map_1) = [1]
      pi(filter2_3) = [1, 2, 3]
      pi(filter2_4#) = [2, 4]
      pi(false_0) = []
      pi(filter_2#) = [1, 2]
      pi(filter_1) = [1]
      pi(app) = [2]
      pi(true_0) = []

The following rules are removed:

  p1, p2, p3, p4, p6, p7, p8



-- SCC decomposition.

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

p1: filter2_4#(false_0(),|fun|,x,xs) -> filter_2#(|fun|,xs)
p2: 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(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: app(minus_0(),y) -> minus_1(y)
r14: app(minus_1(x1),y) -> minus_2(x1,y)
r15: app(s_0(),y) -> s_1(y)
r16: minus_2(x,|0|_0()) -> x
r17: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r18: f_1(|0|_0()) -> s_1(|0|_0())
r19: f_1(s_1(x)) -> minus_2(s_1(x),g_1(f_1(x)))
r20: g_1(|0|_0()) -> |0|_0()
r21: g_1(s_1(x)) -> minus_2(s_1(x),f_1(g_1(x)))
r22: map_2(|fun|,nil_0()) -> nil_0()
r23: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r24: filter_2(|fun|,nil_0()) -> nil_0()
r25: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r26: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r27: 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: g_1#(s_1(x)) -> f_1#(g_1(x))
p2: f_1#(s_1(x)) -> f_1#(x)
p3: f_1#(s_1(x)) -> g_1#(f_1(x))
p4: g_1#(s_1(x)) -> g_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(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: app(minus_0(),y) -> minus_1(y)
r14: app(minus_1(x1),y) -> minus_2(x1,y)
r15: app(s_0(),y) -> s_1(y)
r16: minus_2(x,|0|_0()) -> x
r17: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r18: f_1(|0|_0()) -> s_1(|0|_0())
r19: f_1(s_1(x)) -> minus_2(s_1(x),g_1(f_1(x)))
r20: g_1(|0|_0()) -> |0|_0()
r21: g_1(s_1(x)) -> minus_2(s_1(x),f_1(g_1(x)))
r22: map_2(|fun|,nil_0()) -> nil_0()
r23: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r24: filter_2(|fun|,nil_0()) -> nil_0()
r25: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r26: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r27: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The set of usable rules consists of

  r16, r17, r18, r19, r20, 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:
          
            g_1#_A(x1) = max{0, x1 - 123}
            s_1_A(x1) = x1 + 90
            f_1#_A(x1) = max{0, x1 - 92}
            g_1_A(x1) = max{60, x1 + 9}
            f_1_A(x1) = max{122, x1}
            minus_2_A(x1,x2) = max{25, x1, x2 - 61}
            |0|_0_A = 14
            g_1#!_A(x1) = max{2, x1 - 167}
            s_1!_A(x1) = max{91, x1 - 44}
            f_1#!_A(x1) = max{1, x1 - 87}
            g_1!_A(x1) = max{90, x1 - 132}
            f_1!_A(x1) = max{92, x1 - 1}
            minus_2!_A(x1,x2) = max{13, x1 - 133, x2 - 176}
            |0|_0!_A = 89
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            g_1#_A(x1) = max{0, x1 - 54}
            s_1_A(x1) = max{21, x1 + 19}
            f_1#_A(x1) = max{0, x1 - 49}
            g_1_A(x1) = max{48, x1 + 14}
            f_1_A(x1) = max{34, x1 + 24}
            minus_2_A(x1,x2) = max{44, x1 + 14, x2 - 73}
            |0|_0_A = 49
            g_1#!_A(x1) = 0
            s_1!_A(x1) = max{16, x1 - 14}
            f_1#!_A(x1) = 1
            g_1!_A(x1) = max{37, x1 - 35}
            f_1!_A(x1) = max{36, x1 - 63}
            minus_2!_A(x1,x2) = max{36, x1 - 64, x2 - 89}
            |0|_0!_A = 15
        
  
    partial status:
  
      pi(g_1#) = []
      pi(s_1) = [1]
      pi(f_1#) = []
      pi(g_1) = [1]
      pi(f_1) = [1]
      pi(minus_2) = [1]
      pi(|0|_0) = []

The following rules are removed:

  p1, p2, p3



-- SCC decomposition.

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

p1: g_1#(s_1(x)) -> g_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(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: app(minus_0(),y) -> minus_1(y)
r14: app(minus_1(x1),y) -> minus_2(x1,y)
r15: app(s_0(),y) -> s_1(y)
r16: minus_2(x,|0|_0()) -> x
r17: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r18: f_1(|0|_0()) -> s_1(|0|_0())
r19: f_1(s_1(x)) -> minus_2(s_1(x),g_1(f_1(x)))
r20: g_1(|0|_0()) -> |0|_0()
r21: g_1(s_1(x)) -> minus_2(s_1(x),f_1(g_1(x)))
r22: map_2(|fun|,nil_0()) -> nil_0()
r23: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r24: filter_2(|fun|,nil_0()) -> nil_0()
r25: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r26: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r27: filter2_4(false_0(),|fun|,x,xs) -> filter_2(|fun|,xs)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: g_1#(s_1(x)) -> g_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(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: app(minus_0(),y) -> minus_1(y)
r14: app(minus_1(x1),y) -> minus_2(x1,y)
r15: app(s_0(),y) -> s_1(y)
r16: minus_2(x,|0|_0()) -> x
r17: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r18: f_1(|0|_0()) -> s_1(|0|_0())
r19: f_1(s_1(x)) -> minus_2(s_1(x),g_1(f_1(x)))
r20: g_1(|0|_0()) -> |0|_0()
r21: g_1(s_1(x)) -> minus_2(s_1(x),f_1(g_1(x)))
r22: map_2(|fun|,nil_0()) -> nil_0()
r23: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r24: filter_2(|fun|,nil_0()) -> nil_0()
r25: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r26: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r27: 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
            s_1_A(x1) = max{9, x1 - 32}
            g_1#!_A(x1) = max{1, x1 - 23}
            s_1!_A(x1) = max{7, x1 - 22}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            g_1#_A(x1) = 0
            s_1_A(x1) = max{12, x1 - 36}
            g_1#!_A(x1) = max{1, x1 - 23}
            s_1!_A(x1) = max{5, x1 - 23}
        
  
    partial status:
  
      pi(g_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: minus_2#(s_1(x),s_1(y)) -> minus_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(map_0(),y) -> map_1(y)
r12: app(map_1(x1),y) -> map_2(x1,y)
r13: app(minus_0(),y) -> minus_1(y)
r14: app(minus_1(x1),y) -> minus_2(x1,y)
r15: app(s_0(),y) -> s_1(y)
r16: minus_2(x,|0|_0()) -> x
r17: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r18: f_1(|0|_0()) -> s_1(|0|_0())
r19: f_1(s_1(x)) -> minus_2(s_1(x),g_1(f_1(x)))
r20: g_1(|0|_0()) -> |0|_0()
r21: g_1(s_1(x)) -> minus_2(s_1(x),f_1(g_1(x)))
r22: map_2(|fun|,nil_0()) -> nil_0()
r23: map_2(|fun|,cons_2(x,xs)) -> cons_2(app(|fun|,x),map_2(|fun|,xs))
r24: filter_2(|fun|,nil_0()) -> nil_0()
r25: filter_2(|fun|,cons_2(x,xs)) -> filter2_4(app(|fun|,x),|fun|,x,xs)
r26: filter2_4(true_0(),|fun|,x,xs) -> cons_2(x,filter_2(|fun|,xs))
r27: 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:
          
            minus_2#_A(x1,x2) = 0
            s_1_A(x1) = max{11, x1 - 31}
            minus_2#!_A(x1,x2) = max{1, x1 - 31}
            s_1!_A(x1) = max{5, x1 - 30}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus_2#_A(x1,x2) = 0
            s_1_A(x1) = max{14, x1 - 49}
            minus_2#!_A(x1,x2) = max{65, x1 - 56, x2 - 15}
            s_1!_A(x1) = max{64, x1 - 56}
        
  
    partial status:
  
      pi(minus_2#) = [1]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.