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

  r1, r3, r5, r6, r7, r8, r10, r12, r16, r19, r20, r21, r22, r24, r25, r27, r28

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        not_2#_A(x1,x2) = max{96, x1 + 89, x2 + 58}
        a#_A(x1,x2) = max{x1 + 89, x2 + 58}
        qs_0_A = 9
        qs_1#_A(x1) = x1 + 57
        cons_2_A(x1,x2) = max{71, x1 + 43, x2}
        filter_2#_A(x1,x2) = max{x1 + 90, x2 + 16}
        not_1_A(x1) = max{7, x1}
        le_1_A(x1) = max{9, x1 + 1}
        filter_1_A(x1) = max{67, x1 + 2}
        filter_2_A(x1,x2) = max{56, x1 - 59, x2}
        a_A(x1,x2) = max{68, x1 - 10, x2 + 8}
        append_0_A = 11
        append_1_A(x1) = max{12, x1 - 1}
        cons_0_A = 8
        cons_1_A(x1) = max{9, x1 - 1}
        filter_0_A = 9
        if_0_A = 9
        if_1_A(x1) = max{11, x1 - 1}
        if_2_A(x1,x2) = max{0, x1 - 70, x2 - 1}
        le_0_A = 8
        not_0_A = 6
        s_0_A = 9
        s_1_A(x1) = 68
        nil_0_A = 0
        if_3_A(x1,x2,x3) = max{71, x2 + 43, x3}
        le_2_A(x1,x2) = max{71, x1 - 72, x2 - 72}
        |0|_0_A = 71
        true_0_A = 70
        false_0_A = 67
        not2_1_A(x1) = max{71, x1 - 80}
  
    base order 2:
  
      precedence:
      
        qs_1# > not_1 > a > cons_2 = filter_2 = cons_0 = filter_0 = if_0 = not_0 = if_3 = |0|_0 = not2_1 > not_2# = a# = filter_1 = cons_1 = if_1 = if_2 = le_0 = s_1 = le_2 = true_0 > filter_2# = le_1 = append_1 = s_0 = false_0 > append_0 > qs_0 = nil_0
      
  
    partial status:
  
      pi(not_2#) = [2]
      pi(a#) = [2]
      pi(qs_0) = []
      pi(qs_1#) = []
      pi(cons_2) = [1, 2]
      pi(filter_2#) = []
      pi(not_1) = []
      pi(le_1) = [1]
      pi(filter_1) = []
      pi(filter_2) = [2]
      pi(a) = []
      pi(append_0) = []
      pi(append_1) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(if_0) = []
      pi(if_1) = []
      pi(if_2) = []
      pi(le_0) = []
      pi(not_0) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(nil_0) = []
      pi(if_3) = [2, 3]
      pi(le_2) = []
      pi(|0|_0) = []
      pi(true_0) = []
      pi(false_0) = []
      pi(not2_1) = []

The following rules are removed:

  p5



-- 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: 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:

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


-- 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

  r1, r3, r5, r7, r8, r10, r12, r14, r16, r21, r22, r27, r28, r29

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        not_2#_A(x1,x2) = x1 + 33
        a#_A(x1,x2) = x1 + 32
        not_1_A(x1) = x1 + 34
        a_A(x1,x2) = max{x1 - 40, x2 + 47}
        append_0_A = 38
        append_1_A(x1) = max{39, x1 - 61}
        cons_0_A = 41
        cons_1_A(x1) = max{46, x1 - 1}
        filter_0_A = 39
        filter_1_A(x1) = max{4, x1 - 1}
        if_0_A = 1
        if_1_A(x1) = max{2, x1 - 1}
        if_2_A(x1,x2) = max{3, x1 - 42, x2 - 1}
        le_0_A = 28
        le_1_A(x1) = max{29, x1 - 35}
        not_0_A = 39
        not2_0_A = 41
        not2_1_A(x1) = max{3, x1 - 80}
        s_0_A = 38
        s_1_A(x1) = max{39, x1 - 77}
        le_2_A(x1,x2) = max{40, x1 - 76, x2 - 40}
        |0|_0_A = 3
        true_0_A = 1
        false_0_A = 2
        qs_1_A(x1) = max{30, x1 - 36}
        nil_0_A = 2
  
    base order 2:
  
      precedence:
      
        if_0 > not_2# = a = append_0 = cons_0 = filter_1 = if_1 = if_2 = not2_0 = s_0 = |0|_0 = qs_1 > a# = cons_1 = filter_0 = le_2 > not_1 = append_1 = le_1 = not2_1 = s_1 = nil_0 > le_0 = not_0 = false_0 > true_0
      
  
    partial status:
  
      pi(not_2#) = [1]
      pi(a#) = []
      pi(not_1) = [1]
      pi(a) = [2]
      pi(append_0) = []
      pi(append_1) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(if_0) = []
      pi(if_1) = []
      pi(if_2) = []
      pi(le_0) = []
      pi(le_1) = []
      pi(not_0) = []
      pi(not2_0) = []
      pi(not2_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(le_2) = []
      pi(|0|_0) = []
      pi(true_0) = []
      pi(false_0) = []
      pi(qs_1) = []
      pi(nil_0) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  (no SCCs)

-- 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

  r3, r5, r7, r8, r14, r16, r19, r20, r21, r22, r24, r25, r27, r28

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        qs_1#_A(x1) = x1 + 14
        cons_2_A(x1,x2) = max{x1 - 12, x2 + 20}
        filter_2_A(x1,x2) = x2 + 5
        le_1_A(x1) = x1 + 35
        not_1_A(x1) = max{26, x1 - 9}
        a_A(x1,x2) = max{0, x1 - 40, x2 - 40}
        cons_0_A = 39
        cons_1_A(x1) = max{0, x1 - 40}
        filter_0_A = 39
        filter_1_A(x1) = max{0, x1 - 40}
        if_0_A = 39
        if_1_A(x1) = max{0, x1 - 40}
        if_2_A(x1,x2) = max{0, x1 - 81, x2 - 40}
        not2_0_A = 44
        not2_1_A(x1) = max{3, x1 - 41}
        s_0_A = 44
        s_1_A(x1) = max{3, x1 - 42}
        nil_0_A = 0
        if_3_A(x1,x2,x3) = max{25, x2 - 12, x3 + 20}
        le_2_A(x1,x2) = max{36, x1 - 4, x2 - 41}
        |0|_0_A = 3
        true_0_A = 1
        false_0_A = 2
  
    base order 2:
  
      precedence:
      
        le_2 > le_1 = not_1 > a > filter_2 = not2_1 = s_1 = |0|_0 > cons_0 = filter_1 = not2_0 = s_0 = if_3 = false_0 > cons_2 = filter_0 = if_2 > qs_1# = cons_1 = if_1 = true_0 > if_0 = nil_0
      
  
    partial status:
  
      pi(qs_1#) = []
      pi(cons_2) = []
      pi(filter_2) = [2]
      pi(le_1) = [1]
      pi(not_1) = []
      pi(a) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(if_0) = []
      pi(if_1) = []
      pi(if_2) = []
      pi(not2_0) = []
      pi(not2_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(nil_0) = []
      pi(if_3) = [3]
      pi(le_2) = []
      pi(|0|_0) = []
      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: 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}


-- 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))

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

  r1, r3, r5, r7, r8, r10, r16, r19, r20, r21, r22, r24, r25, r27, r28

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        qs_1#_A(x1) = x1 + 39
        cons_2_A(x1,x2) = max{12, x1 - 25, x2 + 5}
        filter_2_A(x1,x2) = max{8, x2 + 4}
        le_1_A(x1) = 6
        a_A(x1,x2) = max{x1 + 27, x2 - 20}
        append_0_A = 1
        append_1_A(x1) = max{0, x1 - 36}
        cons_0_A = 1
        cons_1_A(x1) = max{11, x1 - 26}
        filter_0_A = 1
        filter_1_A(x1) = max{0, x1 - 28}
        if_0_A = 1
        if_1_A(x1) = max{16, x1 - 29}
        if_2_A(x1,x2) = max{18, x1 - 28, x2 - 51}
        le_0_A = 7
        s_0_A = 1
        s_1_A(x1) = max{18, x1 - 21}
        nil_0_A = 7
        if_3_A(x1,x2,x3) = max{12, x2 - 25, x3 + 5}
        le_2_A(x1,x2) = max{34, x1 - 19, x2 - 19}
        |0|_0_A = 3
        true_0_A = 1
        false_0_A = 2
        not2_1_A(x1) = max{18, x1 - 18}
  
    base order 2:
  
      precedence:
      
        qs_1# = le_2 = not2_1 > filter_2 = a > s_1 = if_3 = |0|_0 > cons_2 = filter_1 = if_2 = le_0 = s_0 = false_0 > append_1 = cons_1 = filter_0 = if_1 = true_0 > le_1 = append_0 = cons_0 = if_0 = nil_0
      
  
    partial status:
  
      pi(qs_1#) = []
      pi(cons_2) = []
      pi(filter_2) = []
      pi(le_1) = []
      pi(a) = [1]
      pi(append_0) = []
      pi(append_1) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(if_0) = []
      pi(if_1) = []
      pi(if_2) = []
      pi(le_0) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(nil_0) = []
      pi(if_3) = [3]
      pi(le_2) = []
      pi(|0|_0) = []
      pi(true_0) = []
      pi(false_0) = []
      pi(not2_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: 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

  r1, r3, r5, r7, r8, r10, r12, r14, r16, r21, r22, r27, r28, r29

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        filter_2#_A(x1,x2) = max{x1 + 23, x2 + 75}
        cons_2_A(x1,x2) = max{21, x1 - 52, x2 + 1}
        a_A(x1,x2) = max{48, x1 - 18, x2 - 5}
        append_0_A = 16
        append_1_A(x1) = max{17, x1 - 5}
        cons_0_A = 17
        cons_1_A(x1) = max{16, x1 - 5}
        filter_0_A = 1
        filter_1_A(x1) = max{2, x1 - 6}
        if_0_A = 16
        if_1_A(x1) = max{17, x1 - 6}
        if_2_A(x1,x2) = max{1, x1 - 24, x2 - 5}
        le_0_A = 16
        le_1_A(x1) = max{17, x1 - 69}
        not_0_A = 17
        not_1_A(x1) = max{19, x1 - 6}
        not2_0_A = 19
        not2_1_A(x1) = max{20, x1 - 49}
        s_0_A = 2
        s_1_A(x1) = max{3, x1 - 6}
        le_2_A(x1,x2) = max{49, x1 - 4, x2 - 4}
        |0|_0_A = 3
        true_0_A = 1
        false_0_A = 2
        qs_1_A(x1) = max{20, x1 - 16}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        filter_2# = le_2 > cons_2 = a = cons_0 = nil_0 > cons_1 = not2_1 = s_1 = |0|_0 = qs_1 > le_1 = not_1 = not2_0 = s_0 = false_0 > append_1 = filter_1 = le_0 = not_0 = true_0 > append_0 = filter_0 = if_2 > if_1 > if_0
      
  
    partial status:
  
      pi(filter_2#) = [2]
      pi(cons_2) = []
      pi(a) = []
      pi(append_0) = []
      pi(append_1) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(if_0) = []
      pi(if_1) = []
      pi(if_2) = []
      pi(le_0) = []
      pi(le_1) = []
      pi(not_0) = []
      pi(not_1) = []
      pi(not2_0) = []
      pi(not2_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(le_2) = []
      pi(|0|_0) = []
      pi(true_0) = []
      pi(false_0) = []
      pi(qs_1) = []
      pi(nil_0) = []

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

  r1, r3, r5, r7, r8, r10, r12, r14, r16, r21, r22, r27, r28, r29

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        append_2#_A(x1,x2) = max{x1 + 12, x2 + 18}
        cons_2_A(x1,x2) = max{37, x1 + 18, x2 + 6}
        a_A(x1,x2) = max{33, x1 - 34, x2 - 3}
        append_0_A = 33
        append_1_A(x1) = max{1, x1 - 4}
        cons_0_A = 31
        cons_1_A(x1) = max{32, x1 - 4}
        filter_0_A = 32
        filter_1_A(x1) = max{33, x1 - 3}
        if_0_A = 1
        if_1_A(x1) = max{32, x1 - 3}
        if_2_A(x1,x2) = max{1, x1 - 38, x2 - 4}
        le_0_A = 1
        le_1_A(x1) = max{31, x1 - 4}
        not_0_A = 1
        not_1_A(x1) = max{32, x1 - 12}
        not2_0_A = 2
        not2_1_A(x1) = max{3, x1 - 34}
        s_0_A = 2
        s_1_A(x1) = max{3, x1 - 40}
        le_2_A(x1,x2) = max{32, x1 - 39, x2 - 4}
        |0|_0_A = 3
        true_0_A = 1
        false_0_A = 2
        qs_1_A(x1) = max{36, x1 - 17}
        nil_0_A = 35
  
    base order 2:
  
      precedence:
      
        a > append_2# = cons_2 = if_0 = le_2 > cons_1 = if_2 = not2_1 = s_1 = |0|_0 = qs_1 > cons_0 = if_1 = le_1 = not_1 = not2_0 = s_0 = false_0 = nil_0 > append_1 = le_0 = not_0 = true_0 > append_0 = filter_1 > filter_0
      
  
    partial status:
  
      pi(append_2#) = [1]
      pi(cons_2) = [2]
      pi(a) = []
      pi(append_0) = []
      pi(append_1) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(if_0) = []
      pi(if_1) = []
      pi(if_2) = []
      pi(le_0) = []
      pi(le_1) = []
      pi(not_0) = []
      pi(not_1) = []
      pi(not2_0) = []
      pi(not2_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(le_2) = []
      pi(|0|_0) = []
      pi(true_0) = []
      pi(false_0) = []
      pi(qs_1) = []
      pi(nil_0) = []

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

  r1, r3, r4, r5, r7, r8, r10, r11, r12, r14, r21, r22, r23, r24, r27, r28, r29

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        le_2#_A(x1,x2) = max{x1 - 1, x2 + 4}
        s_1_A(x1) = max{5, x1 + 3}
        a_A(x1,x2) = max{12, x1 - 13, x2 - 3}
        append_0_A = 10
        append_1_A(x1) = max{11, x1 - 113}
        cons_0_A = 11
        cons_1_A(x1) = max{12, x1 - 3}
        cons_2_A(x1,x2) = max{3, x1 - 55, x2 - 171}
        filter_0_A = 8
        filter_1_A(x1) = max{9, x1 - 4}
        if_0_A = 8
        if_1_A(x1) = max{9, x1 - 4}
        if_2_A(x1,x2) = max{10, x1 - 18, x2 - 4}
        le_0_A = 1
        le_1_A(x1) = max{2, x1 - 56}
        le_2_A(x1,x2) = max{6, x1 - 70, x2 - 6}
        not_0_A = 1
        not_1_A(x1) = max{2, x1 - 4}
        not2_0_A = 27
        not2_1_A(x1) = max{13, x1 - 13}
        |0|_0_A = 5
        true_0_A = 1
        false_0_A = 4
        if_3_A(x1,x2,x3) = max{11, x1 - 32, x2 - 18, x3 - 118}
        qs_1_A(x1) = x1 + 6
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        le_2 > le_2# = s_1 = qs_1 > a > filter_0 = if_2 = not2_1 = |0|_0 = if_3 > append_0 = cons_2 = filter_1 = if_1 = not2_0 = false_0 = nil_0 > cons_1 = if_0 = le_1 = not_1 = true_0 > append_1 = cons_0 = le_0 = not_0
      
  
    partial status:
  
      pi(le_2#) = [2]
      pi(s_1) = [1]
      pi(a) = []
      pi(append_0) = []
      pi(append_1) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(cons_2) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(if_0) = []
      pi(if_1) = []
      pi(if_2) = []
      pi(le_0) = []
      pi(le_1) = []
      pi(le_2) = []
      pi(not_0) = []
      pi(not_1) = []
      pi(not2_0) = []
      pi(not2_1) = []
      pi(|0|_0) = []
      pi(true_0) = []
      pi(false_0) = []
      pi(if_3) = []
      pi(qs_1) = [1]
      pi(nil_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.