YES
We show the termination of the TRS R:

  a(a(append(),nil()),ys) -> ys
  a(a(append(),a(a(cons(),x),xs)),ys) -> a(a(cons(),x),a(a(append(),xs),ys))
  a(a(filter(),f),nil()) -> nil()
  a(a(filter(),f),a(a(cons(),x),xs)) -> a(a(a(if(),a(f,x)),x),a(a(filter(),f),xs))
  a(a(le(),|0|()),y) -> true()
  a(a(le(),a(s(),x)),|0|()) -> false()
  a(a(le(),a(s(),x)),a(s(),y)) -> a(a(le(),x),y)
  a(a(a(if(),true()),x),xs) -> a(a(cons(),x),xs)
  a(a(a(if(),false()),x),xs) -> xs
  a(a(not(),f),b) -> a(not2(),a(f,b))
  a(not2(),true()) -> false()
  a(not2(),false()) -> true()
  a(qs(),nil()) -> nil()
  a(qs(),a(a(cons(),x),xs)) -> a(a(append(),a(qs(),a(a(filter(),a(le(),x)),xs))),a(a(cons(),x),a(qs(),a(a(filter(),a(not(),a(le(),x))),xs))))

Instead we prove termination of the uncurried system:

  a(append_0(),y) -> append_1(y)
  a(append_1(x1),y) -> append_2(x1,y)
  a(cons_0(),y) -> cons_1(y)
  a(cons_1(x1),y) -> cons_2(x1,y)
  a(filter_0(),y) -> filter_1(y)
  a(filter_1(x1),y) -> filter_2(x1,y)
  a(if_0(),y) -> if_1(y)
  a(if_1(x1),y) -> if_2(x1,y)
  a(if_2(x1,x2),y) -> if_3(x1,x2,y)
  a(le_0(),y) -> le_1(y)
  a(le_1(x1),y) -> le_2(x1,y)
  a(not_0(),y) -> not_1(y)
  a(not_1(x1),y) -> not_2(x1,y)
  a(not2_0(),y) -> not2_1(y)
  a(qs_0(),y) -> qs_1(y)
  a(s_0(),y) -> s_1(y)
  append_2(nil_0(),ys) -> ys
  append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))
  filter_2(f,nil_0()) -> nil_0()
  filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs))
  le_2(|0|_0(),y) -> true_0()
  le_2(s_1(x),|0|_0()) -> false_0()
  le_2(s_1(x),s_1(y)) -> le_2(x,y)
  if_3(true_0(),x,xs) -> cons_2(x,xs)
  if_3(false_0(),x,xs) -> xs
  not_2(f,b) -> not2_1(a(f,b))
  not2_1(true_0()) -> false_0()
  not2_1(false_0()) -> true_0()
  qs_1(nil_0()) -> nil_0()
  qs_1(cons_2(x,xs)) -> append_2(qs_1(filter_2(le_1(x),xs)),cons_2(x,qs_1(filter_2(not_1(le_1(x)),xs))))


-- SCC decomposition.

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

p1: a#(append_1(x1),y) -> append_2#(x1,y)
p2: a#(filter_1(x1),y) -> filter_2#(x1,y)
p3: a#(if_2(x1,x2),y) -> if_3#(x1,x2,y)
p4: a#(le_1(x1),y) -> le_2#(x1,y)
p5: a#(not_1(x1),y) -> not_2#(x1,y)
p6: a#(not2_0(),y) -> not2_1#(y)
p7: a#(qs_0(),y) -> qs_1#(y)
p8: append_2#(cons_2(x,xs),ys) -> append_2#(xs,ys)
p9: filter_2#(f,cons_2(x,xs)) -> if_3#(a(f,x),x,filter_2(f,xs))
p10: filter_2#(f,cons_2(x,xs)) -> a#(f,x)
p11: filter_2#(f,cons_2(x,xs)) -> filter_2#(f,xs)
p12: le_2#(s_1(x),s_1(y)) -> le_2#(x,y)
p13: not_2#(f,b) -> not2_1#(a(f,b))
p14: not_2#(f,b) -> a#(f,b)
p15: qs_1#(cons_2(x,xs)) -> append_2#(qs_1(filter_2(le_1(x),xs)),cons_2(x,qs_1(filter_2(not_1(le_1(x)),xs))))
p16: qs_1#(cons_2(x,xs)) -> qs_1#(filter_2(le_1(x),xs))
p17: qs_1#(cons_2(x,xs)) -> filter_2#(le_1(x),xs)
p18: qs_1#(cons_2(x,xs)) -> qs_1#(filter_2(not_1(le_1(x)),xs))
p19: qs_1#(cons_2(x,xs)) -> filter_2#(not_1(le_1(x)),xs)

and R consists of:

r1: a(append_0(),y) -> append_1(y)
r2: a(append_1(x1),y) -> append_2(x1,y)
r3: a(cons_0(),y) -> cons_1(y)
r4: a(cons_1(x1),y) -> cons_2(x1,y)
r5: a(filter_0(),y) -> filter_1(y)
r6: a(filter_1(x1),y) -> filter_2(x1,y)
r7: a(if_0(),y) -> if_1(y)
r8: a(if_1(x1),y) -> if_2(x1,y)
r9: a(if_2(x1,x2),y) -> if_3(x1,x2,y)
r10: a(le_0(),y) -> le_1(y)
r11: a(le_1(x1),y) -> le_2(x1,y)
r12: a(not_0(),y) -> not_1(y)
r13: a(not_1(x1),y) -> not_2(x1,y)
r14: a(not2_0(),y) -> not2_1(y)
r15: a(qs_0(),y) -> qs_1(y)
r16: a(s_0(),y) -> s_1(y)
r17: append_2(nil_0(),ys) -> ys
r18: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))
r19: filter_2(f,nil_0()) -> nil_0()
r20: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs))
r21: le_2(|0|_0(),y) -> true_0()
r22: le_2(s_1(x),|0|_0()) -> false_0()
r23: le_2(s_1(x),s_1(y)) -> le_2(x,y)
r24: if_3(true_0(),x,xs) -> cons_2(x,xs)
r25: if_3(false_0(),x,xs) -> xs
r26: not_2(f,b) -> not2_1(a(f,b))
r27: not2_1(true_0()) -> false_0()
r28: not2_1(false_0()) -> true_0()
r29: qs_1(nil_0()) -> nil_0()
r30: qs_1(cons_2(x,xs)) -> append_2(qs_1(filter_2(le_1(x),xs)),cons_2(x,qs_1(filter_2(not_1(le_1(x)),xs))))

The estimated dependency graph contains the following SCCs:

  {p2, p5, p7, p10, p11, p14, p16, p17, p18, p19}
  {p8}
  {p12}


-- Reduction pair.

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

p1: not_2#(f,b) -> a#(f,b)
p2: a#(qs_0(),y) -> qs_1#(y)
p3: qs_1#(cons_2(x,xs)) -> filter_2#(not_1(le_1(x)),xs)
p4: filter_2#(f,cons_2(x,xs)) -> filter_2#(f,xs)
p5: filter_2#(f,cons_2(x,xs)) -> a#(f,x)
p6: a#(not_1(x1),y) -> not_2#(x1,y)
p7: a#(filter_1(x1),y) -> filter_2#(x1,y)
p8: qs_1#(cons_2(x,xs)) -> qs_1#(filter_2(not_1(le_1(x)),xs))
p9: qs_1#(cons_2(x,xs)) -> filter_2#(le_1(x),xs)
p10: qs_1#(cons_2(x,xs)) -> qs_1#(filter_2(le_1(x),xs))

and R consists of:

r1: a(append_0(),y) -> append_1(y)
r2: a(append_1(x1),y) -> append_2(x1,y)
r3: a(cons_0(),y) -> cons_1(y)
r4: a(cons_1(x1),y) -> cons_2(x1,y)
r5: a(filter_0(),y) -> filter_1(y)
r6: a(filter_1(x1),y) -> filter_2(x1,y)
r7: a(if_0(),y) -> if_1(y)
r8: a(if_1(x1),y) -> if_2(x1,y)
r9: a(if_2(x1,x2),y) -> if_3(x1,x2,y)
r10: a(le_0(),y) -> le_1(y)
r11: a(le_1(x1),y) -> le_2(x1,y)
r12: a(not_0(),y) -> not_1(y)
r13: a(not_1(x1),y) -> not_2(x1,y)
r14: a(not2_0(),y) -> not2_1(y)
r15: a(qs_0(),y) -> qs_1(y)
r16: a(s_0(),y) -> s_1(y)
r17: append_2(nil_0(),ys) -> ys
r18: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))
r19: filter_2(f,nil_0()) -> nil_0()
r20: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs))
r21: le_2(|0|_0(),y) -> true_0()
r22: le_2(s_1(x),|0|_0()) -> false_0()
r23: le_2(s_1(x),s_1(y)) -> le_2(x,y)
r24: if_3(true_0(),x,xs) -> cons_2(x,xs)
r25: if_3(false_0(),x,xs) -> xs
r26: not_2(f,b) -> not2_1(a(f,b))
r27: not2_1(true_0()) -> false_0()
r28: not2_1(false_0()) -> true_0()
r29: qs_1(nil_0()) -> nil_0()
r30: qs_1(cons_2(x,xs)) -> append_2(qs_1(filter_2(le_1(x),xs)),cons_2(x,qs_1(filter_2(not_1(le_1(x)),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:
              not_2#_A(x1,x2) = x1
              a#_A(x1,x2) = x1
              qs_0_A() = 7
              qs_1#_A(x1) = 5
              cons_2_A(x1,x2) = x2
              filter_2#_A(x1,x2) = x1
              not_1_A(x1) = x1 + 2
              le_1_A(x1) = 3
              filter_1_A(x1) = x1 + 2
              filter_2_A(x1,x2) = 2
              not_2#!_A(x1,x2) = x1 + 3
              a#!_A(x1,x2) = x1 + 1
              qs_0!_A() = 8
              qs_1#!_A(x1) = 8
              cons_2!_A(x1,x2) = 9
              filter_2#!_A(x1,x2) = x1 + 2
              not_1!_A(x1) = x1 + 4
              le_1!_A(x1) = 3
              filter_1!_A(x1) = 6
              filter_2!_A(x1,x2) = 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            not_2#_A(x1,x2) = max{1, x1 - 120}
            a#_A(x1,x2) = max{1, x1 - 226}
            qs_0_A = 107
            qs_1#_A(x1) = 1
            cons_2_A(x1,x2) = max{28, x1 - 16, x2 - 16}
            filter_2#_A(x1,x2) = max{1, x1 - 226}
            not_1_A(x1) = max{173, x1 + 106}
            le_1_A(x1) = 119
            filter_1_A(x1) = x1 + 162
            filter_2_A(x1,x2) = 14
            not_2#!_A(x1,x2) = max{49, x2 + 3}
            a#!_A(x1,x2) = max{50, x1 - 108}
            qs_0!_A = 50
            qs_1#!_A(x1) = 50
            cons_2!_A(x1,x2) = 52
            filter_2#!_A(x1,x2) = max{53, x1 - 68}
            not_1!_A(x1) = 48
            le_1!_A(x1) = max{99, x1 + 55}
            filter_1!_A(x1) = 47
            filter_2!_A(x1,x2) = max{99, x1 + 44, x2 + 17}
        
  
    partial status:
  
      pi(not_2#) = [1]
      pi(a#) = [1]
      pi(qs_0) = []
      pi(qs_1#) = []
      pi(cons_2) = [1]
      pi(filter_2#) = [1]
      pi(not_1) = [1]
      pi(le_1) = []
      pi(filter_1) = [1]
      pi(filter_2) = [2]

The following rules are removed:

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



-- SCC decomposition.

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

p1: filter_2#(f,cons_2(x,xs)) -> filter_2#(f,xs)
p2: qs_1#(cons_2(x,xs)) -> qs_1#(filter_2(not_1(le_1(x)),xs))
p3: qs_1#(cons_2(x,xs)) -> qs_1#(filter_2(le_1(x),xs))

and R consists of:

r1: a(append_0(),y) -> append_1(y)
r2: a(append_1(x1),y) -> append_2(x1,y)
r3: a(cons_0(),y) -> cons_1(y)
r4: a(cons_1(x1),y) -> cons_2(x1,y)
r5: a(filter_0(),y) -> filter_1(y)
r6: a(filter_1(x1),y) -> filter_2(x1,y)
r7: a(if_0(),y) -> if_1(y)
r8: a(if_1(x1),y) -> if_2(x1,y)
r9: a(if_2(x1,x2),y) -> if_3(x1,x2,y)
r10: a(le_0(),y) -> le_1(y)
r11: a(le_1(x1),y) -> le_2(x1,y)
r12: a(not_0(),y) -> not_1(y)
r13: a(not_1(x1),y) -> not_2(x1,y)
r14: a(not2_0(),y) -> not2_1(y)
r15: a(qs_0(),y) -> qs_1(y)
r16: a(s_0(),y) -> s_1(y)
r17: append_2(nil_0(),ys) -> ys
r18: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))
r19: filter_2(f,nil_0()) -> nil_0()
r20: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs))
r21: le_2(|0|_0(),y) -> true_0()
r22: le_2(s_1(x),|0|_0()) -> false_0()
r23: le_2(s_1(x),s_1(y)) -> le_2(x,y)
r24: if_3(true_0(),x,xs) -> cons_2(x,xs)
r25: if_3(false_0(),x,xs) -> xs
r26: not_2(f,b) -> not2_1(a(f,b))
r27: not2_1(true_0()) -> false_0()
r28: not2_1(false_0()) -> true_0()
r29: qs_1(nil_0()) -> nil_0()
r30: qs_1(cons_2(x,xs)) -> append_2(qs_1(filter_2(le_1(x),xs)),cons_2(x,qs_1(filter_2(not_1(le_1(x)),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: filter_2#(f,cons_2(x,xs)) -> filter_2#(f,xs)

and R consists of:

r1: a(append_0(),y) -> append_1(y)
r2: a(append_1(x1),y) -> append_2(x1,y)
r3: a(cons_0(),y) -> cons_1(y)
r4: a(cons_1(x1),y) -> cons_2(x1,y)
r5: a(filter_0(),y) -> filter_1(y)
r6: a(filter_1(x1),y) -> filter_2(x1,y)
r7: a(if_0(),y) -> if_1(y)
r8: a(if_1(x1),y) -> if_2(x1,y)
r9: a(if_2(x1,x2),y) -> if_3(x1,x2,y)
r10: a(le_0(),y) -> le_1(y)
r11: a(le_1(x1),y) -> le_2(x1,y)
r12: a(not_0(),y) -> not_1(y)
r13: a(not_1(x1),y) -> not_2(x1,y)
r14: a(not2_0(),y) -> not2_1(y)
r15: a(qs_0(),y) -> qs_1(y)
r16: a(s_0(),y) -> s_1(y)
r17: append_2(nil_0(),ys) -> ys
r18: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))
r19: filter_2(f,nil_0()) -> nil_0()
r20: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs))
r21: le_2(|0|_0(),y) -> true_0()
r22: le_2(s_1(x),|0|_0()) -> false_0()
r23: le_2(s_1(x),s_1(y)) -> le_2(x,y)
r24: if_3(true_0(),x,xs) -> cons_2(x,xs)
r25: if_3(false_0(),x,xs) -> xs
r26: not_2(f,b) -> not2_1(a(f,b))
r27: not2_1(true_0()) -> false_0()
r28: not2_1(false_0()) -> true_0()
r29: qs_1(nil_0()) -> nil_0()
r30: qs_1(cons_2(x,xs)) -> append_2(qs_1(filter_2(le_1(x),xs)),cons_2(x,qs_1(filter_2(not_1(le_1(x)),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:
              filter_2#_A(x1,x2) = x2
              cons_2_A(x1,x2) = x1 + x2 + 4
              filter_2#!_A(x1,x2) = x2 + 7
              cons_2!_A(x1,x2) = x1 + x2 + 8
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            filter_2#_A(x1,x2) = x2
            cons_2_A(x1,x2) = max{x1 + 130, x2 + 67}
            filter_2#!_A(x1,x2) = max{4, x2 - 131}
            cons_2!_A(x1,x2) = max{129, x1 + 3, x2 - 38}
        
  
    partial status:
  
      pi(filter_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: qs_1#(cons_2(x,xs)) -> qs_1#(filter_2(not_1(le_1(x)),xs))
p2: qs_1#(cons_2(x,xs)) -> qs_1#(filter_2(le_1(x),xs))

and R consists of:

r1: a(append_0(),y) -> append_1(y)
r2: a(append_1(x1),y) -> append_2(x1,y)
r3: a(cons_0(),y) -> cons_1(y)
r4: a(cons_1(x1),y) -> cons_2(x1,y)
r5: a(filter_0(),y) -> filter_1(y)
r6: a(filter_1(x1),y) -> filter_2(x1,y)
r7: a(if_0(),y) -> if_1(y)
r8: a(if_1(x1),y) -> if_2(x1,y)
r9: a(if_2(x1,x2),y) -> if_3(x1,x2,y)
r10: a(le_0(),y) -> le_1(y)
r11: a(le_1(x1),y) -> le_2(x1,y)
r12: a(not_0(),y) -> not_1(y)
r13: a(not_1(x1),y) -> not_2(x1,y)
r14: a(not2_0(),y) -> not2_1(y)
r15: a(qs_0(),y) -> qs_1(y)
r16: a(s_0(),y) -> s_1(y)
r17: append_2(nil_0(),ys) -> ys
r18: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))
r19: filter_2(f,nil_0()) -> nil_0()
r20: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs))
r21: le_2(|0|_0(),y) -> true_0()
r22: le_2(s_1(x),|0|_0()) -> false_0()
r23: le_2(s_1(x),s_1(y)) -> le_2(x,y)
r24: if_3(true_0(),x,xs) -> cons_2(x,xs)
r25: if_3(false_0(),x,xs) -> xs
r26: not_2(f,b) -> not2_1(a(f,b))
r27: not2_1(true_0()) -> false_0()
r28: not2_1(false_0()) -> true_0()
r29: qs_1(nil_0()) -> nil_0()
r30: qs_1(cons_2(x,xs)) -> append_2(qs_1(filter_2(le_1(x),xs)),cons_2(x,qs_1(filter_2(not_1(le_1(x)),xs))))

The set of usable rules consists of

  r6, r19, r20, r24, 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:
              qs_1#_A(x1) = 0
              cons_2_A(x1,x2) = x2
              filter_2_A(x1,x2) = 2
              not_1_A(x1) = x1
              le_1_A(x1) = 0
              a_A(x1,x2) = 2
              filter_1_A(x1) = 3
              nil_0_A() = 2
              if_3_A(x1,x2,x3) = x3
              true_0_A() = 5
              false_0_A() = 8
              qs_1#!_A(x1) = 3
              cons_2!_A(x1,x2) = 3
              filter_2!_A(x1,x2) = 5
              not_1!_A(x1) = 2
              le_1!_A(x1) = 2
              a!_A(x1,x2) = 6
              filter_1!_A(x1) = 7
              nil_0!_A() = 0
              if_3!_A(x1,x2,x3) = 4
              true_0!_A() = 1
              false_0!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            qs_1#_A(x1) = max{0, x1 - 8}
            cons_2_A(x1,x2) = max{x1 - 12, x2 + 28}
            filter_2_A(x1,x2) = x2 + 11
            not_1_A(x1) = max{109, x1 + 15}
            le_1_A(x1) = x1 + 94
            a_A(x1,x2) = max{22, x2 + 20}
            filter_1_A(x1) = max{0, x1 - 26}
            nil_0_A = 25
            if_3_A(x1,x2,x3) = max{x2 - 1, x3 + 28}
            true_0_A = 25
            false_0_A = 25
            qs_1#!_A(x1) = x1 + 60
            cons_2!_A(x1,x2) = x2 + 71
            filter_2!_A(x1,x2) = x2 + 89
            not_1!_A(x1) = max{92, x1 - 2}
            le_1!_A(x1) = 71
            a!_A(x1,x2) = max{93, x2 + 90}
            filter_1!_A(x1) = max{94, x1 + 91}
            nil_0!_A = 1
            if_3!_A(x1,x2,x3) = max{116, x3 + 85}
            true_0!_A = 93
            false_0!_A = 92
        
  
    partial status:
  
      pi(qs_1#) = []
      pi(cons_2) = [2]
      pi(filter_2) = []
      pi(not_1) = []
      pi(le_1) = []
      pi(a) = [1]
      pi(filter_1) = []
      pi(nil_0) = []
      pi(if_3) = [3]
      pi(true_0) = []
      pi(false_0) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.

-- Reduction pair.

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

p1: append_2#(cons_2(x,xs),ys) -> append_2#(xs,ys)

and R consists of:

r1: a(append_0(),y) -> append_1(y)
r2: a(append_1(x1),y) -> append_2(x1,y)
r3: a(cons_0(),y) -> cons_1(y)
r4: a(cons_1(x1),y) -> cons_2(x1,y)
r5: a(filter_0(),y) -> filter_1(y)
r6: a(filter_1(x1),y) -> filter_2(x1,y)
r7: a(if_0(),y) -> if_1(y)
r8: a(if_1(x1),y) -> if_2(x1,y)
r9: a(if_2(x1,x2),y) -> if_3(x1,x2,y)
r10: a(le_0(),y) -> le_1(y)
r11: a(le_1(x1),y) -> le_2(x1,y)
r12: a(not_0(),y) -> not_1(y)
r13: a(not_1(x1),y) -> not_2(x1,y)
r14: a(not2_0(),y) -> not2_1(y)
r15: a(qs_0(),y) -> qs_1(y)
r16: a(s_0(),y) -> s_1(y)
r17: append_2(nil_0(),ys) -> ys
r18: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))
r19: filter_2(f,nil_0()) -> nil_0()
r20: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs))
r21: le_2(|0|_0(),y) -> true_0()
r22: le_2(s_1(x),|0|_0()) -> false_0()
r23: le_2(s_1(x),s_1(y)) -> le_2(x,y)
r24: if_3(true_0(),x,xs) -> cons_2(x,xs)
r25: if_3(false_0(),x,xs) -> xs
r26: not_2(f,b) -> not2_1(a(f,b))
r27: not2_1(true_0()) -> false_0()
r28: not2_1(false_0()) -> true_0()
r29: qs_1(nil_0()) -> nil_0()
r30: qs_1(cons_2(x,xs)) -> append_2(qs_1(filter_2(le_1(x),xs)),cons_2(x,qs_1(filter_2(not_1(le_1(x)),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:
              append_2#_A(x1,x2) = x1
              cons_2_A(x1,x2) = x1 + x2 + 8
              append_2#!_A(x1,x2) = x1 + 19
              cons_2!_A(x1,x2) = x1 + x2 + 18
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            append_2#_A(x1,x2) = 0
            cons_2_A(x1,x2) = max{20, x1 + 17, x2 - 3}
            append_2#!_A(x1,x2) = max{x1 + 16, x2 - 1}
            cons_2!_A(x1,x2) = x2 + 15
        
  
    partial status:
  
      pi(append_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: le_2#(s_1(x),s_1(y)) -> le_2#(x,y)

and R consists of:

r1: a(append_0(),y) -> append_1(y)
r2: a(append_1(x1),y) -> append_2(x1,y)
r3: a(cons_0(),y) -> cons_1(y)
r4: a(cons_1(x1),y) -> cons_2(x1,y)
r5: a(filter_0(),y) -> filter_1(y)
r6: a(filter_1(x1),y) -> filter_2(x1,y)
r7: a(if_0(),y) -> if_1(y)
r8: a(if_1(x1),y) -> if_2(x1,y)
r9: a(if_2(x1,x2),y) -> if_3(x1,x2,y)
r10: a(le_0(),y) -> le_1(y)
r11: a(le_1(x1),y) -> le_2(x1,y)
r12: a(not_0(),y) -> not_1(y)
r13: a(not_1(x1),y) -> not_2(x1,y)
r14: a(not2_0(),y) -> not2_1(y)
r15: a(qs_0(),y) -> qs_1(y)
r16: a(s_0(),y) -> s_1(y)
r17: append_2(nil_0(),ys) -> ys
r18: append_2(cons_2(x,xs),ys) -> cons_2(x,append_2(xs,ys))
r19: filter_2(f,nil_0()) -> nil_0()
r20: filter_2(f,cons_2(x,xs)) -> if_3(a(f,x),x,filter_2(f,xs))
r21: le_2(|0|_0(),y) -> true_0()
r22: le_2(s_1(x),|0|_0()) -> false_0()
r23: le_2(s_1(x),s_1(y)) -> le_2(x,y)
r24: if_3(true_0(),x,xs) -> cons_2(x,xs)
r25: if_3(false_0(),x,xs) -> xs
r26: not_2(f,b) -> not2_1(a(f,b))
r27: not2_1(true_0()) -> false_0()
r28: not2_1(false_0()) -> true_0()
r29: qs_1(nil_0()) -> nil_0()
r30: qs_1(cons_2(x,xs)) -> append_2(qs_1(filter_2(le_1(x),xs)),cons_2(x,qs_1(filter_2(not_1(le_1(x)),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:
              le_2#_A(x1,x2) = x1 + x2
              s_1_A(x1) = x1 + 2
              le_2#!_A(x1,x2) = x1 + x2 + 1
              s_1!_A(x1) = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            le_2#_A(x1,x2) = max{x1, x2}
            s_1_A(x1) = max{111, x1 + 105}
            le_2#!_A(x1,x2) = max{4, x1 - 106, x2 - 112}
            s_1!_A(x1) = max{3, x1 - 106}
        
  
    partial status:
  
      pi(le_2#) = []
      pi(s_1) = []

The following rules are removed:

  p1

Then no dependency pair remains.