YES
We show the termination of the TRS R:

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

Instead we prove termination of the uncurried system:

  app(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(filter_0(),y) -> filter_1(y)
  app(filter_1(x1),y) -> filter_2(x1,y)
  app(filter2_0(),y) -> filter2_1(y)
  app(filter2_1(x1),y) -> filter2_2(x1,y)
  app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
  app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
  app(map_0(),y) -> map_1(y)
  app(map_1(x1),y) -> map_2(x1,y)
  app(minus_0(),y) -> minus_1(y)
  app(minus_1(x1),y) -> minus_2(x1,y)
  app(pred_0(),y) -> pred_1(y)
  app(quot_0(),y) -> quot_1(y)
  app(quot_1(x1),y) -> quot_2(x1,y)
  app(s_0(),y) -> s_1(y)
  pred_1(s_1(x)) -> x
  minus_2(x,|0|_0()) -> x
  minus_2(x,s_1(y)) -> pred_1(minus_2(x,y))
  quot_2(|0|_0(),s_1(y)) -> |0|_0()
  quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
  map_2(f,nil_0()) -> nil_0()
  map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
  filter_2(f,nil_0()) -> nil_0()
  filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
  filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
  filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)


-- SCC decomposition.

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

p1: app#(filter_1(x1),y) -> filter_2#(x1,y)
p2: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p3: app#(map_1(x1),y) -> map_2#(x1,y)
p4: app#(minus_1(x1),y) -> minus_2#(x1,y)
p5: app#(pred_0(),y) -> pred_1#(y)
p6: app#(quot_1(x1),y) -> quot_2#(x1,y)
p7: minus_2#(x,s_1(y)) -> pred_1#(minus_2(x,y))
p8: minus_2#(x,s_1(y)) -> minus_2#(x,y)
p9: quot_2#(s_1(x),s_1(y)) -> quot_2#(minus_2(x,y),s_1(y))
p10: quot_2#(s_1(x),s_1(y)) -> minus_2#(x,y)
p11: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p12: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p13: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p14: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p15: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p16: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(minus_0(),y) -> minus_1(y)
r12: app(minus_1(x1),y) -> minus_2(x1,y)
r13: app(pred_0(),y) -> pred_1(y)
r14: app(quot_0(),y) -> quot_1(y)
r15: app(quot_1(x1),y) -> quot_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: pred_1(s_1(x)) -> x
r18: minus_2(x,|0|_0()) -> x
r19: minus_2(x,s_1(y)) -> pred_1(minus_2(x,y))
r20: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r21: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p11, p12, p13, p14, p15, p16}
  {p9}
  {p8}


-- Reduction pair.

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

p1: app#(filter_1(x1),y) -> filter_2#(x1,y)
p2: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p3: app#(map_1(x1),y) -> map_2#(x1,y)
p4: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p5: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p6: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p7: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p8: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p9: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(minus_0(),y) -> minus_1(y)
r12: app(minus_1(x1),y) -> minus_2(x1,y)
r13: app(pred_0(),y) -> pred_1(y)
r14: app(quot_0(),y) -> quot_1(y)
r15: app(quot_1(x1),y) -> quot_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: pred_1(s_1(x)) -> x
r18: minus_2(x,|0|_0()) -> x
r19: minus_2(x,s_1(y)) -> pred_1(minus_2(x,y))
r20: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r21: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r2, r5, r11, r14, r16, r20, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{5, x1}
        filter_1_A(x1) = x1 + 7
        filter_2#_A(x1,x2) = x1 + 6
        cons_2_A(x1,x2) = max{3, x1 - 21, x2 - 21}
        map_1_A(x1) = max{8, x1 + 7}
        map_2#_A(x1,x2) = max{8, x1 + 6}
        filter2_3_A(x1,x2,x3) = max{9, x1 + 5, x2 + 7, x3 + 5}
        filter2_4#_A(x1,x2,x3,x4) = x2 + 6
        false_0_A = 3
        app_A(x1,x2) = max{20, x1 - 10, x2 - 10}
        true_0_A = 1
        cons_0_A = 1
        cons_1_A(x1) = max{2, x1 - 11}
        filter2_0_A = 8
        filter2_1_A(x1) = max{9, x1 - 11}
        minus_0_A = 1
        minus_1_A(x1) = max{2, x1 - 11}
        quot_0_A = 8
        quot_1_A(x1) = max{9, x1 - 11}
        s_0_A = 6
        s_1_A(x1) = max{7, x1 - 36}
        quot_2_A(x1,x2) = max{19, x1 - 25, x2 - 11}
        |0|_0_A = 0
        filter_2_A(x1,x2) = max{8, x1 - 53, x2 - 26}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        app > filter2_0 > map_1 = filter2_3 = filter2_1 = minus_1 = filter_2 > app# = filter_1 = map_2# = false_0 = minus_0 = s_1 = quot_2 = nil_0 > cons_2 = cons_1 = quot_1 = s_0 = |0|_0 > true_0 = cons_0 = quot_0 > filter_2# = filter2_4#
      
  
    partial status:
  
      pi(app#) = [1]
      pi(filter_1) = []
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(map_1) = []
      pi(map_2#) = [1]
      pi(filter2_3) = [3]
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(app) = []
      pi(true_0) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(minus_0) = []
      pi(minus_1) = []
      pi(quot_0) = []
      pi(quot_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(quot_2) = []
      pi(|0|_0) = []
      pi(filter_2) = []
      pi(nil_0) = []

The following rules are removed:

  p3



-- SCC decomposition.

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

p1: app#(filter_1(x1),y) -> filter_2#(x1,y)
p2: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p3: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p4: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p5: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p6: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p7: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p8: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(minus_0(),y) -> minus_1(y)
r12: app(minus_1(x1),y) -> minus_2(x1,y)
r13: app(pred_0(),y) -> pred_1(y)
r14: app(quot_0(),y) -> quot_1(y)
r15: app(quot_1(x1),y) -> quot_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: pred_1(s_1(x)) -> x
r18: minus_2(x,|0|_0()) -> x
r19: minus_2(x,s_1(y)) -> pred_1(minus_2(x,y))
r20: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r21: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(minus_0(),y) -> minus_1(y)
r12: app(minus_1(x1),y) -> minus_2(x1,y)
r13: app(pred_0(),y) -> pred_1(y)
r14: app(quot_0(),y) -> quot_1(y)
r15: app(quot_1(x1),y) -> quot_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: pred_1(s_1(x)) -> x
r18: minus_2(x,|0|_0()) -> x
r19: minus_2(x,s_1(y)) -> pred_1(minus_2(x,y))
r20: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r21: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r3, r5, r6, r7, r11, r14, r20, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = max{x1 + 11, x2 + 17}
        cons_2_A(x1,x2) = max{9, x1 - 6, x2}
        app_A(x1,x2) = max{5, x1 - 9, x2 - 24}
        cons_0_A = 3
        cons_1_A(x1) = max{4, x1 - 25}
        filter_0_A = 18
        filter_1_A(x1) = max{8, x1 - 25}
        filter2_0_A = 1
        filter2_1_A(x1) = max{2, x1 - 24}
        filter2_2_A(x1,x2) = max{4, x1 - 34, x2 - 25}
        filter2_3_A(x1,x2,x3) = max{5, x1 - 43, x2 - 34, x3 - 24}
        minus_0_A = 2
        minus_1_A(x1) = max{3, x1 - 25}
        quot_0_A = 1
        quot_1_A(x1) = max{2, x1 - 25}
        quot_2_A(x1,x2) = max{3, x1 - 62, x2 - 136}
        |0|_0_A = 0
        s_1_A(x1) = x1 + 26
        filter_2_A(x1,x2) = max{9, x1 - 8, x2 - 17}
        nil_0_A = 1
  
    base order 2:
  
      precedence:
      
        quot_2 > map_2# = cons_0 = s_1 = filter_2 > app = |0|_0 > cons_2 = filter_1 = filter2_3 = minus_1 = quot_1 > cons_1 = filter_0 = filter2_2 = minus_0 = quot_0 = nil_0 > filter2_1 > filter2_0
      
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [2]
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(filter2_3) = []
      pi(minus_0) = []
      pi(minus_1) = []
      pi(quot_0) = []
      pi(quot_1) = []
      pi(quot_2) = []
      pi(|0|_0) = []
      pi(s_1) = [1]
      pi(filter_2) = []
      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: app#(filter_1(x1),y) -> filter_2#(x1,y)
p2: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p3: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p4: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p5: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p6: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(minus_0(),y) -> minus_1(y)
r12: app(minus_1(x1),y) -> minus_2(x1,y)
r13: app(pred_0(),y) -> pred_1(y)
r14: app(quot_0(),y) -> quot_1(y)
r15: app(quot_1(x1),y) -> quot_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: pred_1(s_1(x)) -> x
r18: minus_2(x,|0|_0()) -> x
r19: minus_2(x,s_1(y)) -> pred_1(minus_2(x,y))
r20: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r21: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r2, r5, r6, r11, r14, r16, r20

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{7, x1 + 1}
        filter_1_A(x1) = max{12, x1 + 8}
        filter_2#_A(x1,x2) = x1 + 8
        cons_2_A(x1,x2) = max{5, x1 - 43, x2 - 43}
        filter2_4#_A(x1,x2,x3,x4) = x2 + 8
        app_A(x1,x2) = max{12, x1 - 15, x2 - 15}
        true_0_A = 4
        filter2_3_A(x1,x2,x3) = max{x1 + 1, x2 + 9, x3 - 36}
        false_0_A = 1
        cons_0_A = 3
        cons_1_A(x1) = max{4, x1 - 16}
        filter2_0_A = 1
        filter2_1_A(x1) = max{2, x1 - 16}
        filter2_2_A(x1,x2) = max{3, x1 - 32, x2 - 15}
        minus_0_A = 9
        minus_1_A(x1) = max{10, x1 - 16}
        quot_0_A = 8
        quot_1_A(x1) = max{9, x1 - 16}
        s_0_A = 8
        s_1_A(x1) = max{9, x1 - 65}
        quot_2_A(x1,x2) = max{10, x1 - 32, x2 - 29}
        |0|_0_A = 0
  
    base order 2:
  
      precedence:
      
        app > filter2_3 = cons_0 > filter2_2 = minus_1 > filter2_1 = minus_0 = s_1 = quot_2 > app# = filter_1 = filter2_0 = quot_1 = s_0 = |0|_0 > filter_2# = filter2_4# = quot_0 > cons_2 = false_0 > true_0 = cons_1
      
  
    partial status:
  
      pi(app#) = [1]
      pi(filter_1) = []
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(filter2_4#) = []
      pi(app) = []
      pi(true_0) = []
      pi(filter2_3) = []
      pi(false_0) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(minus_0) = []
      pi(minus_1) = []
      pi(quot_0) = []
      pi(quot_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(quot_2) = []
      pi(|0|_0) = []

The following rules are removed:

  p4



-- SCC decomposition.

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

p1: app#(filter_1(x1),y) -> filter_2#(x1,y)
p2: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p3: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p4: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p5: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(minus_0(),y) -> minus_1(y)
r12: app(minus_1(x1),y) -> minus_2(x1,y)
r13: app(pred_0(),y) -> pred_1(y)
r14: app(quot_0(),y) -> quot_1(y)
r15: app(quot_1(x1),y) -> quot_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: pred_1(s_1(x)) -> x
r18: minus_2(x,|0|_0()) -> x
r19: minus_2(x,s_1(y)) -> pred_1(minus_2(x,y))
r20: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r21: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p5}


-- Reduction pair.

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

p1: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p2: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p3: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(minus_0(),y) -> minus_1(y)
r12: app(minus_1(x1),y) -> minus_2(x1,y)
r13: app(pred_0(),y) -> pred_1(y)
r14: app(quot_0(),y) -> quot_1(y)
r15: app(quot_1(x1),y) -> quot_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: pred_1(s_1(x)) -> x
r18: minus_2(x,|0|_0()) -> x
r19: minus_2(x,s_1(y)) -> pred_1(minus_2(x,y))
r20: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r21: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r2, r3, r5, r6, r7, r9, r11, r14, r16, r20, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        filter_2#_A(x1,x2) = x2 + 40
        cons_2_A(x1,x2) = max{x1 + 25, x2 + 42}
        filter2_4#_A(x1,x2,x3,x4) = max{x3 + 24, x4 + 41}
        app_A(x1,x2) = max{64, x1 - 5, x2 + 44}
        false_0_A = 1
        true_0_A = 1
        cons_0_A = 4
        cons_1_A(x1) = x1 + 31
        filter_0_A = 1
        filter_1_A(x1) = max{2, x1 - 50}
        filter2_0_A = 4
        filter2_1_A(x1) = max{2, x1 - 1}
        filter2_2_A(x1,x2) = max{3, x1 - 7, x2 - 1}
        filter2_3_A(x1,x2,x3) = max{4, x1 - 12, x2 - 6, x3 - 1}
        map_0_A = 1
        map_1_A(x1) = max{4, x1 - 37}
        minus_0_A = 3
        minus_1_A(x1) = max{4, x1 - 1}
        quot_0_A = 3
        quot_1_A(x1) = max{4, x1 - 1}
        s_0_A = 1
        s_1_A(x1) = max{2, x1 - 56}
        quot_2_A(x1,x2) = max{3, x1 - 48, x2 - 3}
        |0|_0_A = 0
        filter_2_A(x1,x2) = max{45, x1 - 55, x2 - 44}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        cons_2 > filter2_4# > filter_2# > app = cons_1 = filter2_1 = filter2_2 = filter2_3 = quot_0 > cons_0 = filter2_0 = map_1 = filter_2 > false_0 = true_0 = filter_1 = map_0 = minus_1 = nil_0 > filter_0 = minus_0 = s_1 = quot_2 > quot_1 = s_0 = |0|_0
      
  
    partial status:
  
      pi(filter_2#) = [2]
      pi(cons_2) = [1, 2]
      pi(filter2_4#) = []
      pi(app) = [2]
      pi(false_0) = []
      pi(true_0) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(filter2_3) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(minus_0) = []
      pi(minus_1) = []
      pi(quot_0) = []
      pi(quot_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(quot_2) = []
      pi(|0|_0) = []
      pi(filter_2) = []
      pi(nil_0) = []

The following rules are removed:

  p3



-- SCC decomposition.

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

p1: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p2: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(minus_0(),y) -> minus_1(y)
r12: app(minus_1(x1),y) -> minus_2(x1,y)
r13: app(pred_0(),y) -> pred_1(y)
r14: app(quot_0(),y) -> quot_1(y)
r15: app(quot_1(x1),y) -> quot_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: pred_1(s_1(x)) -> x
r18: minus_2(x,|0|_0()) -> x
r19: minus_2(x,s_1(y)) -> pred_1(minus_2(x,y))
r20: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r21: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p2: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(minus_0(),y) -> minus_1(y)
r12: app(minus_1(x1),y) -> minus_2(x1,y)
r13: app(pred_0(),y) -> pred_1(y)
r14: app(quot_0(),y) -> quot_1(y)
r15: app(quot_1(x1),y) -> quot_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: pred_1(s_1(x)) -> x
r18: minus_2(x,|0|_0()) -> x
r19: minus_2(x,s_1(y)) -> pred_1(minus_2(x,y))
r20: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r21: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r3, r5, r9, r14, r20, r24

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{2, x2 - 2}
        cons_2_A(x1,x2) = max{x1 + 8, x2 + 4}
        filter2_4#_A(x1,x2,x3,x4) = max{x3 + 5, x4 - 1}
        app_A(x1,x2) = max{33, x2 + 22}
        false_0_A = 23
        cons_0_A = 14
        cons_1_A(x1) = max{23, x1 + 21}
        filter_0_A = 12
        filter_1_A(x1) = x1 + 13
        filter2_0_A = 14
        filter2_1_A(x1) = max{33, x1 + 13}
        map_0_A = 13
        map_1_A(x1) = max{23, x1 + 13}
        quot_0_A = 13
        quot_1_A(x1) = max{14, x1 + 13}
        quot_2_A(x1,x2) = max{21, x1 - 36, x2 - 88}
        |0|_0_A = 0
        s_1_A(x1) = max{35, x1 - 73}
        filter_2_A(x1,x2) = max{36, x1 - 14, x2 - 7}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        filter_2 > app = nil_0 > cons_2 = filter_1 = map_1 > filter2_4# = cons_1 = filter_0 = map_0 > filter_2# = cons_0 = quot_2 = s_1 > false_0 = filter2_1 = quot_1 = |0|_0 > filter2_0 = quot_0
      
  
    partial status:
  
      pi(filter_2#) = []
      pi(cons_2) = [2]
      pi(filter2_4#) = []
      pi(app) = [2]
      pi(false_0) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(quot_0) = []
      pi(quot_1) = []
      pi(quot_2) = []
      pi(|0|_0) = []
      pi(s_1) = []
      pi(filter_2) = []
      pi(nil_0) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(minus_0(),y) -> minus_1(y)
r12: app(minus_1(x1),y) -> minus_2(x1,y)
r13: app(pred_0(),y) -> pred_1(y)
r14: app(quot_0(),y) -> quot_1(y)
r15: app(quot_1(x1),y) -> quot_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: pred_1(s_1(x)) -> x
r18: minus_2(x,|0|_0()) -> x
r19: minus_2(x,s_1(y)) -> pred_1(minus_2(x,y))
r20: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r21: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,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: quot_2#(s_1(x),s_1(y)) -> quot_2#(minus_2(x,y),s_1(y))

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(minus_0(),y) -> minus_1(y)
r12: app(minus_1(x1),y) -> minus_2(x1,y)
r13: app(pred_0(),y) -> pred_1(y)
r14: app(quot_0(),y) -> quot_1(y)
r15: app(quot_1(x1),y) -> quot_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: pred_1(s_1(x)) -> x
r18: minus_2(x,|0|_0()) -> x
r19: minus_2(x,s_1(y)) -> pred_1(minus_2(x,y))
r20: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r21: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r2, r3, r5, r6, r7, r9, r11, r13, r14, r16, r17, r18, r19, r20, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        quot_2#_A(x1,x2) = max{4, x1 + 2}
        s_1_A(x1) = x1 + 14
        minus_2_A(x1,x2) = x1 + 3
        app_A(x1,x2) = max{x1 - 20, x2 + 18}
        cons_0_A = 5
        cons_1_A(x1) = max{6, x1 - 1}
        cons_2_A(x1,x2) = max{7, x1 - 22, x2 - 14}
        filter_0_A = 19
        filter_1_A(x1) = max{18, x1 - 1}
        filter2_0_A = 1
        filter2_1_A(x1) = max{2, x1 - 1}
        filter2_2_A(x1,x2) = max{3, x1 - 22, x2 - 20}
        filter2_3_A(x1,x2,x3) = max{4, x1 - 42, x2 - 40, x3 - 24}
        map_0_A = 4
        map_1_A(x1) = max{5, x1 - 1}
        minus_0_A = 39
        minus_1_A(x1) = max{19, x1 - 1}
        pred_0_A = 1
        pred_1_A(x1) = max{2, x1 - 13}
        quot_0_A = 1
        quot_1_A(x1) = max{0, x1 - 1}
        s_0_A = 5
        |0|_0_A = 7
        quot_2_A(x1,x2) = max{1, x2 - 7}
        filter_2_A(x1,x2) = max{13, x1 - 24, x2 - 1}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        map_0 > app = cons_1 = cons_2 = filter2_0 = map_1 = minus_0 = pred_0 > s_1 = cons_0 = filter2_3 = quot_2 = filter_2 > quot_2# = filter_1 = filter2_2 = quot_1 = s_0 = |0|_0 = nil_0 > minus_2 = filter_0 = filter2_1 = quot_0 > minus_1 = pred_1
      
  
    partial status:
  
      pi(quot_2#) = [1]
      pi(s_1) = [1]
      pi(minus_2) = [1]
      pi(app) = [2]
      pi(cons_0) = []
      pi(cons_1) = []
      pi(cons_2) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(filter2_3) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(minus_0) = []
      pi(minus_1) = []
      pi(pred_0) = []
      pi(pred_1) = []
      pi(quot_0) = []
      pi(quot_1) = []
      pi(s_0) = []
      pi(|0|_0) = []
      pi(quot_2) = []
      pi(filter_2) = []
      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: minus_2#(x,s_1(y)) -> minus_2#(x,y)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(minus_0(),y) -> minus_1(y)
r12: app(minus_1(x1),y) -> minus_2(x1,y)
r13: app(pred_0(),y) -> pred_1(y)
r14: app(quot_0(),y) -> quot_1(y)
r15: app(quot_1(x1),y) -> quot_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: pred_1(s_1(x)) -> x
r18: minus_2(x,|0|_0()) -> x
r19: minus_2(x,s_1(y)) -> pred_1(minus_2(x,y))
r20: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r21: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r2, r3, r5, r6, r7, r9, r11, r14, r20, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        minus_2#_A(x1,x2) = max{16, x2}
        s_1_A(x1) = max{16, x1 + 15}
        app_A(x1,x2) = max{13, x1 - 5, x2 - 4}
        cons_0_A = 2
        cons_1_A(x1) = max{3, x1 - 9}
        cons_2_A(x1,x2) = max{2, x1 - 14, x2 - 23}
        filter_0_A = 3
        filter_1_A(x1) = max{4, x1 - 28}
        filter2_0_A = 4
        filter2_1_A(x1) = max{3, x1 - 5}
        filter2_2_A(x1,x2) = max{2, x1 - 11, x2 - 5}
        filter2_3_A(x1,x2,x3) = max{1, x1 - 16, x2 - 10, x3 - 4}
        map_0_A = 1
        map_1_A(x1) = max{2, x1 - 14}
        minus_0_A = 1
        minus_1_A(x1) = max{4, x1 - 5}
        quot_0_A = 1
        quot_1_A(x1) = max{3, x1 - 5}
        quot_2_A(x1,x2) = max{2, x1 - 18, x2 - 32}
        |0|_0_A = 0
        filter_2_A(x1,x2) = max{11, x1 - 34, x2 - 19}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        quot_2 > s_1 = filter_2 = nil_0 > app = filter_0 = quot_0 = |0|_0 > minus_2# = cons_0 = filter_1 = filter2_3 = map_1 = minus_1 = quot_1 > cons_1 = cons_2 = filter2_1 = filter2_2 = map_0 = minus_0 > filter2_0
      
  
    partial status:
  
      pi(minus_2#) = [2]
      pi(s_1) = [1]
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(cons_2) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(filter2_3) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(minus_0) = []
      pi(minus_1) = []
      pi(quot_0) = []
      pi(quot_1) = []
      pi(quot_2) = []
      pi(|0|_0) = []
      pi(filter_2) = []
      pi(nil_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.