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(double(),|0|()) -> |0|()
  app(double(),app(s(),x)) -> app(s(),app(s(),app(double(),x)))
  app(app(plus(),|0|()),y) -> y
  app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),y))
  app(app(plus(),app(s(),x)),y) -> app(app(plus(),x),app(s(),y))
  app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),app(app(minus(),x),y)),app(double(),y)))
  app(app(map(),f),nil()) -> nil()
  app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
  app(app(filter(),f),nil()) -> nil()
  app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(app(filter2(),app(f,x)),f),x),xs)
  app(app(app(app(filter2(),true()),f),x),xs) -> app(app(cons(),x),app(app(filter(),f),xs))
  app(app(app(app(filter2(),false()),f),x),xs) -> app(app(filter(),f),xs)

Instead we prove termination of the uncurried system:

  app(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(double_0(),y) -> double_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(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(plus_0(),y) -> plus_1(y)
  app(plus_1(x1),y) -> plus_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)
  double_1(|0|_0()) -> |0|_0()
  double_1(s_1(x)) -> s_1(s_1(double_1(x)))
  plus_2(|0|_0(),y) -> y
  plus_2(s_1(x),y) -> s_1(plus_2(x,y))
  plus_2(s_1(x),y) -> plus_2(x,s_1(y))
  plus_2(s_1(x),y) -> s_1(plus_2(minus_2(x,y),double_1(y)))
  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#(double_0(),y) -> double_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#(map_1(x1),y) -> map_2#(x1,y)
p5: app#(minus_1(x1),y) -> minus_2#(x1,y)
p6: app#(plus_1(x1),y) -> plus_2#(x1,y)
p7: minus_2#(s_1(x),s_1(y)) -> minus_2#(x,y)
p8: double_1#(s_1(x)) -> double_1#(x)
p9: plus_2#(s_1(x),y) -> plus_2#(x,y)
p10: plus_2#(s_1(x),y) -> plus_2#(x,s_1(y))
p11: plus_2#(s_1(x),y) -> plus_2#(minus_2(x,y),double_1(y))
p12: plus_2#(s_1(x),y) -> minus_2#(x,y)
p13: plus_2#(s_1(x),y) -> double_1#(y)
p14: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p15: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p16: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p17: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p18: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p19: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(double_0(),y) -> double_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(map_0(),y) -> map_1(y)
r11: app(map_1(x1),y) -> map_2(x1,y)
r12: app(minus_0(),y) -> minus_1(y)
r13: app(minus_1(x1),y) -> minus_2(x1,y)
r14: app(plus_0(),y) -> plus_1(y)
r15: app(plus_1(x1),y) -> plus_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: minus_2(x,|0|_0()) -> x
r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r19: double_1(|0|_0()) -> |0|_0()
r20: double_1(s_1(x)) -> s_1(s_1(double_1(x)))
r21: plus_2(|0|_0(),y) -> y
r22: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r23: plus_2(s_1(x),y) -> plus_2(x,s_1(y))
r24: plus_2(s_1(x),y) -> s_1(plus_2(minus_2(x,y),double_1(y)))
r25: map_2(f,nil_0()) -> nil_0()
r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r27: filter_2(f,nil_0()) -> nil_0()
r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r30: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p14, p15, p16, p17, p18, p19}
  {p9, p10, p11}
  {p8}
  {p7}


-- Reduction pair.

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

p1: map_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: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p5: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p6: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p7: app#(filter_1(x1),y) -> filter_2#(x1,y)
p8: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p9: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(double_0(),y) -> double_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(map_0(),y) -> map_1(y)
r11: app(map_1(x1),y) -> map_2(x1,y)
r12: app(minus_0(),y) -> minus_1(y)
r13: app(minus_1(x1),y) -> minus_2(x1,y)
r14: app(plus_0(),y) -> plus_1(y)
r15: app(plus_1(x1),y) -> plus_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: minus_2(x,|0|_0()) -> x
r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r19: double_1(|0|_0()) -> |0|_0()
r20: double_1(s_1(x)) -> s_1(s_1(double_1(x)))
r21: plus_2(|0|_0(),y) -> y
r22: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r23: plus_2(s_1(x),y) -> plus_2(x,s_1(y))
r24: plus_2(s_1(x),y) -> s_1(plus_2(minus_2(x,y),double_1(y)))
r25: map_2(f,nil_0()) -> nil_0()
r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r27: filter_2(f,nil_0()) -> nil_0()
r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r30: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r3, r7, r16, r19, r20, r25

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              map_2#_A(x1,x2) = 0
              cons_2_A(x1,x2) = 1
              app#_A(x1,x2) = 0
              map_1_A(x1) = x1 + 1
              filter2_3_A(x1,x2,x3) = x1 + x2 + x3 + 1
              filter2_4#_A(x1,x2,x3,x4) = 0
              false_0_A() = 1
              filter_2#_A(x1,x2) = 0
              filter_1_A(x1) = x1 + 1
              app_A(x1,x2) = x1 + x2 + 1
              true_0_A() = 1
              cons_0_A() = 1
              cons_1_A(x1) = x1 + 1
              double_0_A() = 1
              double_1_A(x1) = 1
              filter2_1_A(x1) = x1 + 1
              filter2_2_A(x1,x2) = x1 + x2 + 1
              s_0_A() = 1
              s_1_A(x1) = 1
              |0|_0_A() = 1
              map_2_A(x1,x2) = x2 + 1
              nil_0_A() = 1
              map_2#!_A(x1,x2) = 1
              cons_2!_A(x1,x2) = 3
              app#!_A(x1,x2) = 1
              map_1!_A(x1) = x1 + 2
              filter2_3!_A(x1,x2,x3) = x1 + 4
              filter2_4#!_A(x1,x2,x3,x4) = 1
              false_0!_A() = 7
              filter_2#!_A(x1,x2) = 1
              filter_1!_A(x1) = 6
              app!_A(x1,x2) = x2 + 5
              true_0!_A() = 4
              cons_0!_A() = 5
              cons_1!_A(x1) = 4
              double_0!_A() = 0
              double_1!_A(x1) = 3
              filter2_1!_A(x1) = x1 + 6
              filter2_2!_A(x1,x2) = 3
              s_0!_A() = 1
              s_1!_A(x1) = 0
              |0|_0!_A() = 0
              map_2!_A(x1,x2) = 3
              nil_0!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{1, x1 - 13, x2 - 36}
            cons_2_A(x1,x2) = max{25, x1 + 10, x2 + 3}
            app#_A(x1,x2) = max{1, x1 - 13, x2 - 26}
            map_1_A(x1) = max{12, x1 + 10}
            filter2_3_A(x1,x2,x3) = max{x1 - 72, x2 + 7, x3 - 98}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 13, x3 - 111, x4 - 26}
            false_0_A = 85
            filter_2#_A(x1,x2) = max{1, x1 - 13, x2 - 26}
            filter_1_A(x1) = x1 + 7
            app_A(x1,x2) = max{85, x1 - 24, x2 - 24}
            true_0_A = 85
            cons_0_A = 16
            cons_1_A(x1) = max{16, x1 - 98}
            double_0_A = 23
            double_1_A(x1) = max{49, x1 - 26}
            filter2_1_A(x1) = max{16, x1 - 24}
            filter2_2_A(x1,x2) = max{16, x1 - 48, x2 - 24}
            s_0_A = 18
            s_1_A(x1) = max{25, x1 - 24}
            |0|_0_A = 0
            map_2_A(x1,x2) = max{95, x1 - 14, x2 - 24}
            nil_0_A = 3
            map_2#!_A(x1,x2) = max{3, x1 - 1, x2 - 4}
            cons_2!_A(x1,x2) = max{0, x1 - 114, x2 - 5}
            app#!_A(x1,x2) = max{x1 - 8, x2 + 5}
            map_1!_A(x1) = max{1, x1 - 2}
            filter2_3!_A(x1,x2,x3) = max{x1 + 2, x3 + 2}
            filter2_4#!_A(x1,x2,x3,x4) = max{x2 - 2, x3 - 114, x4 + 4}
            false_0!_A = 1
            filter_2#!_A(x1,x2) = max{3, x1 - 3, x2 - 4}
            filter_1!_A(x1) = max{0, x1 - 4}
            app!_A(x1,x2) = max{0, x1 - 17}
            true_0!_A = 4
            cons_0!_A = 2
            cons_1!_A(x1) = 1
            double_0!_A = 6
            double_1!_A(x1) = 5
            filter2_1!_A(x1) = max{1, x1 - 1}
            filter2_2!_A(x1,x2) = max{0, x1 - 42}
            s_0!_A = 1
            s_1!_A(x1) = max{2, x1 - 72}
            |0|_0!_A = 1
            map_2!_A(x1,x2) = max{0, x1 - 1, x2 - 55}
            nil_0!_A = 1
        
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [1, 2]
      pi(app#) = [2]
      pi(map_1) = []
      pi(filter2_3) = []
      pi(filter2_4#) = [4]
      pi(false_0) = []
      pi(filter_2#) = [2]
      pi(filter_1) = []
      pi(app) = [1]
      pi(true_0) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(double_0) = []
      pi(double_1) = []
      pi(filter2_1) = [1]
      pi(filter2_2) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(|0|_0) = []
      pi(map_2) = []
      pi(nil_0) = []

The following rules are removed:

  p1, p2, p3, p4, p5, p6, p7, p8, p9

Then no dependency pair remains.

-- Reduction pair.

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

p1: plus_2#(s_1(x),y) -> plus_2#(minus_2(x,y),double_1(y))
p2: plus_2#(s_1(x),y) -> plus_2#(x,s_1(y))
p3: plus_2#(s_1(x),y) -> plus_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(double_0(),y) -> double_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(map_0(),y) -> map_1(y)
r11: app(map_1(x1),y) -> map_2(x1,y)
r12: app(minus_0(),y) -> minus_1(y)
r13: app(minus_1(x1),y) -> minus_2(x1,y)
r14: app(plus_0(),y) -> plus_1(y)
r15: app(plus_1(x1),y) -> plus_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: minus_2(x,|0|_0()) -> x
r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r19: double_1(|0|_0()) -> |0|_0()
r20: double_1(s_1(x)) -> s_1(s_1(double_1(x)))
r21: plus_2(|0|_0(),y) -> y
r22: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r23: plus_2(s_1(x),y) -> plus_2(x,s_1(y))
r24: plus_2(s_1(x),y) -> s_1(plus_2(minus_2(x,y),double_1(y)))
r25: map_2(f,nil_0()) -> nil_0()
r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r27: filter_2(f,nil_0()) -> nil_0()
r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r30: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r17, r18, r19, r20

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus_2#_A(x1,x2) = 0
              s_1_A(x1) = x1
              minus_2_A(x1,x2) = x1 + 5
              double_1_A(x1) = 1
              |0|_0_A() = 1
              plus_2#!_A(x1,x2) = 16
              s_1!_A(x1) = x1 + 18
              minus_2!_A(x1,x2) = 17
              double_1!_A(x1) = x1 + 20
              |0|_0!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus_2#_A(x1,x2) = max{0, x2 - 14}
            s_1_A(x1) = max{11, x1}
            minus_2_A(x1,x2) = max{5, x1}
            double_1_A(x1) = max{13, x1 - 11}
            |0|_0_A = 11
            plus_2#!_A(x1,x2) = max{x1 + 38, x2 + 10}
            s_1!_A(x1) = x1 + 25
            minus_2!_A(x1,x2) = x2 + 12
            double_1!_A(x1) = max{37, x1 + 28}
            |0|_0!_A = 40
        
  
    partial status:
  
      pi(plus_2#) = [1]
      pi(s_1) = [1]
      pi(minus_2) = [1]
      pi(double_1) = [1]
      pi(|0|_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: double_1#(s_1(x)) -> double_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(double_0(),y) -> double_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(map_0(),y) -> map_1(y)
r11: app(map_1(x1),y) -> map_2(x1,y)
r12: app(minus_0(),y) -> minus_1(y)
r13: app(minus_1(x1),y) -> minus_2(x1,y)
r14: app(plus_0(),y) -> plus_1(y)
r15: app(plus_1(x1),y) -> plus_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: minus_2(x,|0|_0()) -> x
r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r19: double_1(|0|_0()) -> |0|_0()
r20: double_1(s_1(x)) -> s_1(s_1(double_1(x)))
r21: plus_2(|0|_0(),y) -> y
r22: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r23: plus_2(s_1(x),y) -> plus_2(x,s_1(y))
r24: plus_2(s_1(x),y) -> s_1(plus_2(minus_2(x,y),double_1(y)))
r25: map_2(f,nil_0()) -> nil_0()
r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r27: filter_2(f,nil_0()) -> nil_0()
r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r30: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              double_1#_A(x1) = 0
              s_1_A(x1) = 6
              double_1#!_A(x1) = 0
              s_1!_A(x1) = 7
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            double_1#_A(x1) = 0
            s_1_A(x1) = max{39, x1 + 33}
            double_1#!_A(x1) = max{1, x1 - 1}
            s_1!_A(x1) = 6
        
  
    partial status:
  
      pi(double_1#) = []
      pi(s_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(double_0(),y) -> double_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(map_0(),y) -> map_1(y)
r11: app(map_1(x1),y) -> map_2(x1,y)
r12: app(minus_0(),y) -> minus_1(y)
r13: app(minus_1(x1),y) -> minus_2(x1,y)
r14: app(plus_0(),y) -> plus_1(y)
r15: app(plus_1(x1),y) -> plus_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: minus_2(x,|0|_0()) -> x
r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r19: double_1(|0|_0()) -> |0|_0()
r20: double_1(s_1(x)) -> s_1(s_1(double_1(x)))
r21: plus_2(|0|_0(),y) -> y
r22: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r23: plus_2(s_1(x),y) -> plus_2(x,s_1(y))
r24: plus_2(s_1(x),y) -> s_1(plus_2(minus_2(x,y),double_1(y)))
r25: map_2(f,nil_0()) -> nil_0()
r26: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r27: filter_2(f,nil_0()) -> nil_0()
r28: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r29: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r30: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              minus_2#_A(x1,x2) = 0
              s_1_A(x1) = 4
              minus_2#!_A(x1,x2) = 12
              s_1!_A(x1) = 11
        
        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{0, x1 - 14}
            minus_2#!_A(x1,x2) = 9
            s_1!_A(x1) = x1 + 8
        
  
    partial status:
  
      pi(minus_2#) = [1, 2]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.