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

  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!):
        
          max/plus interpretation on natural numbers:
          
            not_2#_A(x1,x2) = max{0, x1 - 88, x2 - 94}
            a#_A(x1,x2) = max{0, x1 - 102, x2 - 94}
            qs_0_A = 101
            qs_1#_A(x1) = max{0, x1 - 94}
            cons_2_A(x1,x2) = max{3, x1 + 2, x2}
            filter_2#_A(x1,x2) = max{0, x1 - 101, x2 - 96}
            not_1_A(x1) = x1 + 14
            le_1_A(x1) = max{11, x1 - 5}
            filter_1_A(x1) = x1 + 1
            filter_2_A(x1,x2) = max{20, x1 - 29, x2}
            nil_0_A = 12
            if_3_A(x1,x2,x3) = max{4, x2 + 2, x3}
            a_A(x1,x2) = max{0, x1 - 102}
            true_0_A = 3
            false_0_A = 3
            not_2#!_A(x1,x2) = max{8, x1 - 92, x2 - 81}
            a#!_A(x1,x2) = max{8, x1 - 92, x2 - 81}
            qs_0!_A = 2
            qs_1#!_A(x1) = max{9, x1 - 82}
            cons_2!_A(x1,x2) = max{2, x2 - 61}
            filter_2#!_A(x1,x2) = max{8, x1 - 92, x2 - 82}
            not_1!_A(x1) = max{10, x1 - 50}
            le_1!_A(x1) = max{10, x1 - 2}
            filter_1!_A(x1) = max{8, x1 - 14}
            filter_2!_A(x1,x2) = max{10, x1 - 15, x2 - 59}
            nil_0!_A = 0
            if_3!_A(x1,x2,x3) = max{9, x2 - 58, x3 - 60}
            a!_A(x1,x2) = max{7, x2 - 13}
            true_0!_A = 3
            false_0!_A = 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            not_2#_A(x1,x2) = max{0, x1 - 100, x2 - 23}
            a#_A(x1,x2) = max{0, x1 - 100, x2 - 23}
            qs_0_A = 99
            qs_1#_A(x1) = max{0, x1 - 25}
            cons_2_A(x1,x2) = max{10, x1 + 2, x2}
            filter_2#_A(x1,x2) = max{0, x1 - 44, x2 - 25}
            not_1_A(x1) = max{33, x1}
            le_1_A(x1) = max{43, x1 - 1}
            filter_1_A(x1) = x1 + 98
            filter_2_A(x1,x2) = max{24, x1 - 58, x2}
            nil_0_A = 12
            if_3_A(x1,x2,x3) = max{11, x2 + 2, x3}
            a_A(x1,x2) = max{2, x1 - 157, x2 - 157}
            true_0_A = 5
            false_0_A = 5
            not_2#!_A(x1,x2) = max{40, x1 - 57, x2 - 34}
            a#!_A(x1,x2) = max{40, x1 - 57, x2 - 34}
            qs_0!_A = 37
            qs_1#!_A(x1) = max{41, x1 - 35}
            cons_2!_A(x1,x2) = max{5, x1 - 59}
            filter_2#!_A(x1,x2) = max{41, x1 - 57, x2 - 36}
            not_1!_A(x1) = max{39, x1 - 1}
            le_1!_A(x1) = max{42, x1 - 5}
            filter_1!_A(x1) = max{8, x1 - 59}
            filter_2!_A(x1,x2) = 0
            nil_0!_A = 0
            if_3!_A(x1,x2,x3) = max{6, x2 - 1, x3 - 25}
            a!_A(x1,x2) = max{7, x2 - 59}
            true_0!_A = 4
            false_0!_A = 3
        
  
    partial status:
  
      pi(not_2#) = []
      pi(a#) = []
      pi(qs_0) = []
      pi(qs_1#) = []
      pi(cons_2) = [2]
      pi(filter_2#) = []
      pi(not_1) = [1]
      pi(le_1) = [1]
      pi(filter_1) = [1]
      pi(filter_2) = [2]
      pi(nil_0) = []
      pi(if_3) = [3]
      pi(a) = [1]
      pi(true_0) = []
      pi(false_0) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

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

  {p7, p9}
  {p1, p3, p4, p5, p6}


-- Reduction pair.

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

p1: qs_1#(cons_2(x,xs)) -> qs_1#(filter_2(le_1(x),xs))
p2: qs_1#(cons_2(x,xs)) -> qs_1#(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 set of usable rules consists of

  r10, r12, r19, r20, r21, r22, r23, r24, r25, r29

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:
          
            qs_1#_A(x1) = max{94, x1 + 2}
            cons_2_A(x1,x2) = max{91, x1 + 89, x2}
            filter_2_A(x1,x2) = max{x1 + 65, x2}
            le_1_A(x1) = 27
            not_1_A(x1) = 0
            a_A(x1,x2) = max{0, x1 - 12}
            le_0_A = 43
            not_0_A = 131
            nil_0_A = 0
            if_3_A(x1,x2,x3) = max{91, x2 + 89, x3}
            le_2_A(x1,x2) = 16
            |0|_0_A = 13
            true_0_A = 1
            s_1_A(x1) = x1 + 10
            false_0_A = 1
            qs_1_A(x1) = max{31, x1 - 32}
            qs_1#!_A(x1) = max{15, x1 - 78}
            cons_2!_A(x1,x2) = max{95, x2 - 31}
            filter_2!_A(x1,x2) = max{x1 + 67, x2 + 7}
            le_1!_A(x1) = max{9, x1 - 2}
            not_1!_A(x1) = max{96, x1}
            a!_A(x1,x2) = max{8, x1 - 34, x2 + 5}
            le_0!_A = 10
            not_0!_A = 97
            nil_0!_A = 0
            if_3!_A(x1,x2,x3) = max{x2 + 95, x3 + 1}
            le_2!_A(x1,x2) = max{x1 - 1, x2 + 9}
            |0|_0!_A = 12
            true_0!_A = 11
            s_1!_A(x1) = max{11, x1 - 2}
            false_0!_A = 10
            qs_1!_A(x1) = max{0, x1 - 90}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            qs_1#_A(x1) = max{109, x1 + 102}
            cons_2_A(x1,x2) = max{36, x1 + 34, x2 + 31}
            filter_2_A(x1,x2) = max{7, x2 + 2}
            le_1_A(x1) = x1 + 9
            not_1_A(x1) = max{84, x1 - 94}
            a_A(x1,x2) = max{132, x1 - 85, x2 + 85}
            le_0_A = 174
            not_0_A = 168
            nil_0_A = 8
            if_3_A(x1,x2,x3) = max{x2 + 36, x3 + 31}
            le_2_A(x1,x2) = max{182, x1 - 164, x2 - 1}
            |0|_0_A = 0
            true_0_A = 91
            s_1_A(x1) = max{163, x1 + 112}
            false_0_A = 182
            qs_1_A(x1) = max{82, x1 + 36}
            qs_1#!_A(x1) = max{1, x1 - 8}
            cons_2!_A(x1,x2) = max{28, x2}
            filter_2!_A(x1,x2) = max{22, x1 - 1}
            le_1!_A(x1) = max{51, x1 + 43}
            not_1!_A(x1) = max{45, x1 - 218}
            a!_A(x1,x2) = max{50, x1 - 122, x2 + 44}
            le_0!_A = 52
            not_0!_A = 46
            nil_0!_A = 17
            if_3!_A(x1,x2,x3) = x3 + 29
            le_2!_A(x1,x2) = max{49, x1 - 113, x2 - 164}
            |0|_0!_A = 49
            true_0!_A = 48
            s_1!_A(x1) = max{45, x1 - 165}
            false_0!_A = 47
            qs_1!_A(x1) = max{28, x1 + 8}
        
  
    partial status:
  
      pi(qs_1#) = []
      pi(cons_2) = [1, 2]
      pi(filter_2) = [2]
      pi(le_1) = []
      pi(not_1) = []
      pi(a) = [1, 2]
      pi(le_0) = []
      pi(not_0) = []
      pi(nil_0) = []
      pi(if_3) = [2, 3]
      pi(le_2) = []
      pi(|0|_0) = []
      pi(true_0) = []
      pi(s_1) = [1]
      pi(false_0) = []
      pi(qs_1) = []

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: not_2#(f,b) -> a#(f,b)
p2: a#(filter_1(x1),y) -> filter_2#(x1,y)
p3: filter_2#(f,cons_2(x,xs)) -> a#(f,x)
p4: a#(not_1(x1),y) -> not_2#(x1,y)
p5: 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!):
        
          max/plus interpretation on natural numbers:
          
            not_2#_A(x1,x2) = max{0, x1 - 120, x2 - 12}
            a#_A(x1,x2) = max{0, x1 - 122, x2 - 12}
            filter_1_A(x1) = max{121, x1 + 116}
            filter_2#_A(x1,x2) = max{0, x1 - 6, x2 - 12}
            cons_2_A(x1,x2) = max{x1 + 5, x2 + 11}
            not_1_A(x1) = max{9, x1 + 2}
            not_2#!_A(x1,x2) = max{3, x1 - 128, x2 - 104}
            a#!_A(x1,x2) = max{3, x1 - 128, x2 - 104}
            filter_1!_A(x1) = max{5, x1 - 137}
            filter_2#!_A(x1,x2) = max{3, x1 - 128, x2 - 109}
            cons_2!_A(x1,x2) = max{3, x1 - 104, x2 - 100}
            not_1!_A(x1) = max{2, x1 - 128}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            not_2#_A(x1,x2) = max{0, x1 - 132, x2 - 21}
            a#_A(x1,x2) = max{0, x1 - 132, x2 - 21}
            filter_1_A(x1) = x1 + 131
            filter_2#_A(x1,x2) = max{0, x1 - 1, x2 - 21}
            cons_2_A(x1,x2) = max{20, x1 + 10, x2 + 12}
            not_1_A(x1) = max{126, x1 + 1}
            not_2#!_A(x1,x2) = max{1, x1 - 127, x2 - 200}
            a#!_A(x1,x2) = max{1, x1 - 127, x2 - 200}
            filter_1!_A(x1) = max{2, x1 - 1}
            filter_2#!_A(x1,x2) = max{3, x1 - 1, x2 - 201}
            cons_2!_A(x1,x2) = max{7, x1 - 199, x2 - 202}
            not_1!_A(x1) = max{10, x1 - 202}
        
  
    partial status:
  
      pi(not_2#) = []
      pi(a#) = []
      pi(filter_1) = [1]
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(not_1) = [1]

The following rules are removed:

  p2, p3



-- SCC decomposition.

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

p1: not_2#(f,b) -> a#(f,b)
p2: a#(not_1(x1),y) -> not_2#(x1,y)
p3: 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 estimated dependency graph contains the following SCCs:

  {p1, p2}
  {p3}


-- Reduction pair.

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

p1: not_2#(f,b) -> a#(f,b)
p2: a#(not_1(x1),y) -> not_2#(x1,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!):
        
          max/plus interpretation on natural numbers:
          
            not_2#_A(x1,x2) = max{0, x1 - 109, x2 - 1}
            a#_A(x1,x2) = max{0, x1 - 118, x2 - 1}
            not_1_A(x1) = max{117, x1 + 9}
            not_2#!_A(x1,x2) = max{1, x1 - 236, x2 - 1}
            a#!_A(x1,x2) = max{1, x1 - 236, x2 - 1}
            not_1!_A(x1) = max{5, x1 - 235}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            not_2#_A(x1,x2) = max{0, x1 - 15, x2 - 1}
            a#_A(x1,x2) = max{0, x1 - 15, x2 - 1}
            not_1_A(x1) = max{14, x1}
            not_2#!_A(x1,x2) = max{5, x1 - 15, x2 - 1}
            a#!_A(x1,x2) = max{5, x1 - 15, x2 - 1}
            not_1!_A(x1) = max{4, x1 - 17}
        
  
    partial status:
  
      pi(not_2#) = [1]
      pi(a#) = [1]
      pi(not_1) = [1]

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: not_2#(f,b) -> a#(f,b)

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:

  (no SCCs)

-- 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!):
        
          max/plus interpretation on natural numbers:
          
            filter_2#_A(x1,x2) = 0
            cons_2_A(x1,x2) = max{13, x1 + 11, x2 - 47}
            filter_2#!_A(x1,x2) = max{0, x2 - 98}
            cons_2!_A(x1,x2) = max{7, x1 - 243, x2 - 97}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            filter_2#_A(x1,x2) = 0
            cons_2_A(x1,x2) = max{2, x1 - 23, x2 - 4}
            filter_2#!_A(x1,x2) = max{6, x1 - 1, x2 - 147}
            cons_2!_A(x1,x2) = max{5, x1 - 60, x2 - 146}
        
  
    partial status:
  
      pi(filter_2#) = [2]
      pi(cons_2) = [1, 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: 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!):
        
          max/plus interpretation on natural numbers:
          
            append_2#_A(x1,x2) = 0
            cons_2_A(x1,x2) = max{6, x1 - 47, x2 - 58}
            append_2#!_A(x1,x2) = max{1, x1 - 362}
            cons_2!_A(x1,x2) = max{6, x1 - 183, x2 - 361}
        
        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{2, x1 - 41, x2 - 45}
            append_2#!_A(x1,x2) = max{7, x1 - 103, x2 - 1}
            cons_2!_A(x1,x2) = max{6, x1 - 112, x2 - 102}
        
  
    partial status:
  
      pi(append_2#) = [1]
      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: 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!):
        
          max/plus interpretation on natural numbers:
          
            le_2#_A(x1,x2) = max{x1, x2}
            s_1_A(x1) = x1 + 7
            le_2#!_A(x1,x2) = max{4, x1 - 140, x2 - 2}
            s_1!_A(x1) = max{3, x1 - 141}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            le_2#_A(x1,x2) = 0
            s_1_A(x1) = max{1, x1 - 3}
            le_2#!_A(x1,x2) = max{7, x1 - 264, x2 - 266}
            s_1!_A(x1) = max{6, x1 - 265}
        
  
    partial status:
  
      pi(le_2#) = [1]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.