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

  r24

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        not_2#_A(x1,x2) = 7
        a#_A(x1,x2) = 7
        qs_0_A = 8
        qs_1#_A(x1) = 7
        cons_2_A(x1,x2) = 0
        filter_2#_A(x1,x2) = 7
        not_1_A(x1) = max{2, x1 - 41}
        le_1_A(x1) = max{17, x1 - 5}
        filter_1_A(x1) = 0
        filter_2_A(x1,x2) = max{8, x1 + 6}
        if_3_A(x1,x2,x3) = max{9, x1 + 3, x2 - 18, x3 - 5}
        true_0_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          not_2#_A(x1,x2) = max{2, x1 - 50}
          a#_A(x1,x2) = max{2, x1 - 50}
          qs_0_A = 48
          qs_1#_A(x1) = 2
          cons_2_A(x1,x2) = max{23, x1 + 1, x2 - 50}
          filter_2#_A(x1,x2) = max{2, x1 + 1}
          not_1_A(x1) = x1
          le_1_A(x1) = 0
          filter_1_A(x1) = x1 + 51
          filter_2_A(x1,x2) = max{x1 + 201, x2 + 26}
          if_3_A(x1,x2,x3) = max{202, x1 - 19, x2 + 28, x3 + 178}
          true_0_A = 18
          not_2#!_A(x1,x2) = max{2, x1 - 48}
          a#!_A(x1,x2) = max{2, x1 - 48}
          qs_0!_A = 12
          qs_1#!_A(x1) = 2
          cons_2!_A(x1,x2) = max{4, x1 + 1, x2 - 100}
          filter_2#!_A(x1,x2) = x1 + 2
          not_1!_A(x1) = 9
          le_1!_A(x1) = max{5, x1 - 200}
          filter_1!_A(x1) = max{8, x1 + 3}
          filter_2!_A(x1,x2) = max{10, x1 - 66, x2 - 22}
          if_3!_A(x1,x2,x3) = max{9, x1 - 67, x2 + 2, x3 - 99}
          true_0!_A = 3
  
    partial status:
  
      pi(not_2#) = []
      pi(a#) = []
      pi(qs_0) = []
      pi(qs_1#) = []
      pi(cons_2) = []
      pi(filter_2#) = []
      pi(not_1) = []
      pi(le_1) = []
      pi(filter_1) = []
      pi(filter_2) = []
      pi(if_3) = [1]
      pi(true_0) = []

The following rules are removed:

  p7



-- SCC decomposition.

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: 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:

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


-- 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)
p3: a#(qs_0(),y) -> qs_1#(y)
p4: qs_1#(cons_2(x,xs)) -> qs_1#(filter_2(le_1(x),xs))
p5: qs_1#(cons_2(x,xs)) -> filter_2#(le_1(x),xs)
p6: filter_2#(f,cons_2(x,xs)) -> a#(f,x)
p7: filter_2#(f,cons_2(x,xs)) -> filter_2#(f,xs)
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#(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

  r4, r19, r20, r24, r25

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        not_2#_A(x1,x2) = max{58, x1 - 1}
        a#_A(x1,x2) = max{1, x1 - 4}
        not_1_A(x1) = max{63, x1 + 4}
        qs_0_A = 71
        qs_1#_A(x1) = 67
        cons_2_A(x1,x2) = max{x1 + 13, x2 + 44}
        filter_2_A(x1,x2) = x2 + 3
        le_1_A(x1) = 40
        filter_2#_A(x1,x2) = x1 + 3
        a_A(x1,x2) = max{x1 - 72, x2 + 74}
        cons_1_A(x1) = x1 + 85
        nil_0_A = 20
        if_3_A(x1,x2,x3) = max{x2 + 13, x3 + 44}
        true_0_A = 76
        false_0_A = 76
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          not_2#_A(x1,x2) = max{0, x1 - 27, x2 - 33}
          a#_A(x1,x2) = max{0, x1 - 35, x2 - 33}
          not_1_A(x1) = max{26, x1 + 8}
          qs_0_A = 0
          qs_1#_A(x1) = max{0, x1 - 34}
          cons_2_A(x1,x2) = max{24, x1 + 1, x2}
          filter_2_A(x1,x2) = max{23, x1 - 32, x2}
          le_1_A(x1) = max{26, x1 - 6}
          filter_2#_A(x1,x2) = max{0, x1 - 35, x2 - 34}
          a_A(x1,x2) = max{17, x1 + 1, x2 + 1}
          cons_1_A(x1) = x1 + 26
          nil_0_A = 24
          if_3_A(x1,x2,x3) = max{24, x2 + 1, x3}
          true_0_A = 34
          false_0_A = 29
          not_2#!_A(x1,x2) = max{26, x1 - 28}
          a#!_A(x1,x2) = max{27, x1 - 1, x2 + 7}
          not_1!_A(x1) = max{34, x1 - 27}
          qs_0!_A = 1
          qs_1#!_A(x1) = 0
          cons_2!_A(x1,x2) = 0
          filter_2!_A(x1,x2) = 0
          le_1!_A(x1) = max{30, x1 - 1}
          filter_2#!_A(x1,x2) = 1
          a!_A(x1,x2) = max{33, x1 - 27}
          cons_1!_A(x1) = max{0, x1 - 1}
          nil_0!_A = 30
          if_3!_A(x1,x2,x3) = 0
          true_0!_A = 3
          false_0!_A = 3
  
    partial status:
  
      pi(not_2#) = []
      pi(a#) = []
      pi(not_1) = [1]
      pi(qs_0) = []
      pi(qs_1#) = []
      pi(cons_2) = []
      pi(filter_2) = []
      pi(le_1) = []
      pi(filter_2#) = []
      pi(a) = [2]
      pi(cons_1) = [1]
      pi(nil_0) = []
      pi(if_3) = []
      pi(true_0) = []
      pi(false_0) = []

The following rules are removed:

  p1, p2, p3, p5, p6, p9



-- SCC decomposition.

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: filter_2#(f,cons_2(x,xs)) -> filter_2#(f,xs)
p3: 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 estimated dependency graph contains the following SCCs:

  {p1, p3}
  {p2}


-- 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

  r19, r20, r24, r25

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        qs_1#_A(x1) = max{33, x1}
        cons_2_A(x1,x2) = max{33, x2}
        filter_2_A(x1,x2) = max{11, x1 - 1, x2}
        le_1_A(x1) = 34
        not_1_A(x1) = max{0, x1 - 35}
        nil_0_A = 14
        if_3_A(x1,x2,x3) = max{33, x3}
        a_A(x1,x2) = x1 + 4
        true_0_A = 3
        false_0_A = 5
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          qs_1#_A(x1) = x1
          cons_2_A(x1,x2) = max{x1, x2 + 13}
          filter_2_A(x1,x2) = x2
          le_1_A(x1) = 6
          not_1_A(x1) = max{5, x1 - 6}
          nil_0_A = 0
          if_3_A(x1,x2,x3) = max{x2, x3 + 13}
          a_A(x1,x2) = max{0, x1 - 8}
          true_0_A = 3
          false_0_A = 3
          qs_1#!_A(x1) = max{1, x1 - 4}
          cons_2!_A(x1,x2) = 9
          filter_2!_A(x1,x2) = max{x1 + 2, x2}
          le_1!_A(x1) = max{26, x1 - 5}
          not_1!_A(x1) = max{8, x1 - 6}
          nil_0!_A = 13
          if_3!_A(x1,x2,x3) = max{x2 - 17, x3 + 10}
          a!_A(x1,x2) = max{27, x1 - 6, x2 - 16}
          true_0!_A = 10
          false_0!_A = 11
  
    partial status:
  
      pi(qs_1#) = [1]
      pi(cons_2) = [2]
      pi(filter_2) = [2]
      pi(le_1) = []
      pi(not_1) = []
      pi(nil_0) = []
      pi(if_3) = [3]
      pi(a) = []
      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: 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:
  
      max/plus interpretation on natural numbers:
      
        filter_2#_A(x1,x2) = max{9, x2 + 4}
        cons_2_A(x1,x2) = max{x1 + 22, x2 + 8}
  
    base order 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{11, x1 - 34, x2 - 35}
          filter_2#!_A(x1,x2) = 0
          cons_2!_A(x1,x2) = max{4, x1 - 102, x2 - 7}
  
    partial status:
  
      pi(filter_2#) = [2]
      pi(cons_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:
  
      max/plus interpretation on natural numbers:
      
        append_2#_A(x1,x2) = x1 + 8
        cons_2_A(x1,x2) = max{57, x1 - 7, x2 + 9}
  
    base order 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{7, x1 - 21, x2 - 19}
          append_2#!_A(x1,x2) = max{x1 + 46, x2 + 25}
          cons_2!_A(x1,x2) = max{23, x1 - 107, x2 - 114}
  
    partial status:
  
      pi(append_2#) = [1]
      pi(cons_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:
  
      max/plus interpretation on natural numbers:
      
        le_2#_A(x1,x2) = 0
        s_1_A(x1) = max{4, x1 - 78}
  
    base order 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{7, x1 + 1}
          le_2#!_A(x1,x2) = max{4, x1 - 2, x2 - 8}
          s_1!_A(x1) = max{3, x1 - 82}
  
    partial status:
  
      pi(le_2#) = []
      pi(s_1) = []

The following rules are removed:

  p1

Then no dependency pair remains.