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

  r3, r6, r9, r14, r16, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{64, x2 + 60}
        filter_1_A(x1) = x1 + 54
        filter_2#_A(x1,x2) = max{59, x2 + 1}
        cons_2_A(x1,x2) = max{65, x1 + 61, x2}
        map_1_A(x1) = max{61, x1 + 55}
        map_2#_A(x1,x2) = max{62, x2 + 2}
        filter2_3_A(x1,x2,x3) = max{x2 + 55, x3 + 75}
        filter2_4#_A(x1,x2,x3,x4) = max{59, x4 + 1}
        false_0_A = 1
        app_A(x1,x2) = max{x1 - 9, x2 + 57}
        true_0_A = 1
        filter_0_A = 44
        filter2_1_A(x1) = max{58, x1 + 20}
        filter2_2_A(x1,x2) = max{x1 + 8, x2 + 56}
        map_0_A = 71
        quot_0_A = 68
        quot_1_A(x1) = max{58, x1 + 10}
        s_0_A = 51
        s_1_A(x1) = max{41, x1 + 14}
        filter_2_A(x1,x2) = max{44, x1 - 1, x2 + 4}
        nil_0_A = 53
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{78, x1 - 43, x2 - 2}
          filter_1_A(x1) = x1 + 8
          filter_2#_A(x1,x2) = max{78, x1 - 35, x2 - 2}
          cons_2_A(x1,x2) = max{79, x1 + 2, x2}
          map_1_A(x1) = x1 + 2
          map_2#_A(x1,x2) = max{78, x1 - 41, x2 - 4}
          filter2_3_A(x1,x2,x3) = max{x1 + 9, x2 + 32, x3 + 42}
          filter2_4#_A(x1,x2,x3,x4) = max{78, x2 - 35, x3 - 1, x4 - 2}
          false_0_A = 86
          app_A(x1,x2) = max{86, x1 - 9, x2 + 41}
          true_0_A = 86
          filter_0_A = 8
          filter2_1_A(x1) = x1 + 27
          filter2_2_A(x1,x2) = max{x1 + 18, x2 + 41}
          map_0_A = 8
          quot_0_A = 8
          quot_1_A(x1) = max{8, x1 - 1}
          s_0_A = 8
          s_1_A(x1) = max{42, x1}
          filter_2_A(x1,x2) = max{86, x1 - 11, x2 + 6}
          nil_0_A = 87
          app#!_A(x1,x2) = max{3, x1 - 2}
          filter_1!_A(x1) = max{1, x1 - 1}
          filter_2#!_A(x1,x2) = max{x1 + 5, x2 - 77}
          cons_2!_A(x1,x2) = 0
          map_1!_A(x1) = x1 + 1
          map_2#!_A(x1,x2) = max{2, x1 + 1}
          filter2_3!_A(x1,x2,x3) = max{0, x1 - 12, x2 - 53, x3 - 87}
          filter2_4#!_A(x1,x2,x3,x4) = max{x2 + 5, x3 - 76, x4 - 77}
          false_0!_A = 2
          app!_A(x1,x2) = max{0, x1 - 29, x2 - 86}
          true_0!_A = 2
          filter_0!_A = 2
          filter2_1!_A(x1) = max{0, x1 - 1}
          filter2_2!_A(x1,x2) = max{0, x1 - 1}
          map_0!_A = 2
          quot_0!_A = 1
          quot_1!_A(x1) = max{0, x1 - 86}
          s_0!_A = 1
          s_1!_A(x1) = x1 + 46
          filter_2!_A(x1,x2) = 2
          nil_0!_A = 1
  
    partial status:
  
      pi(app#) = [2]
      pi(filter_1) = [1]
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(map_1) = [1]
      pi(map_2#) = []
      pi(filter2_3) = []
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(app) = [2]
      pi(true_0) = []
      pi(filter_0) = []
      pi(filter2_1) = [1]
      pi(filter2_2) = [1]
      pi(map_0) = []
      pi(quot_0) = []
      pi(quot_1) = [1]
      pi(s_0) = []
      pi(s_1) = []
      pi(filter_2) = []
      pi(nil_0) = []

The following rules are removed:

  p1, p2, p3, p5, p6



-- SCC decomposition.

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

p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p2: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p3: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p4: 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:

  {p1}
  {p2, p3, p4}


-- 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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = max{x1 + 6, x2 + 46}
        cons_2_A(x1,x2) = max{4, x1 - 40, x2 + 1}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          map_2#_A(x1,x2) = 0
          cons_2_A(x1,x2) = max{42, x1 - 6, x2 - 6}
          map_2#!_A(x1,x2) = 0
          cons_2!_A(x1,x2) = max{0, x1 - 11, x2 - 12}
  
    partial status:
  
      pi(map_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: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
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)

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, r17, r18, r19, r20, r21, r22, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        filter2_4#_A(x1,x2,x3,x4) = x2 + 2
        false_0_A = 13
        filter_2#_A(x1,x2) = x1 + 2
        cons_2_A(x1,x2) = max{0, x2 - 9}
        app_A(x1,x2) = x1 + 14
        true_0_A = 13
        cons_0_A = 1
        cons_1_A(x1) = 0
        filter_0_A = 2
        filter_1_A(x1) = 1
        pred_1_A(x1) = max{16, x1 - 18}
        s_1_A(x1) = max{47, x1 + 30}
        minus_2_A(x1,x2) = max{17, x1}
        |0|_0_A = 31
        quot_2_A(x1,x2) = x1 + 12
        map_2_A(x1,x2) = max{x1 + 8, x2 + 11}
        nil_0_A = 9
        filter_2_A(x1,x2) = x2 + 14
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          filter2_4#_A(x1,x2,x3,x4) = max{24, x2 - 1, x3 - 38, x4 - 1}
          false_0_A = 3
          filter_2#_A(x1,x2) = max{24, x1 - 1, x2 - 1}
          cons_2_A(x1,x2) = max{24, x1 - 37, x2 + 1}
          app_A(x1,x2) = max{29, x1 - 11, x2 + 22}
          true_0_A = 1
          cons_0_A = 0
          cons_1_A(x1) = max{0, x1 - 26}
          filter_0_A = 0
          filter_1_A(x1) = max{10, x1 - 26}
          pred_1_A(x1) = max{23, x1}
          s_1_A(x1) = max{30, x1}
          minus_2_A(x1,x2) = max{32, x1, x2 - 31}
          |0|_0_A = 0
          quot_2_A(x1,x2) = max{30, x1 - 33, x2 - 32}
          map_2_A(x1,x2) = max{23, x1 - 48, x2 + 22}
          nil_0_A = 24
          filter_2_A(x1,x2) = max{23, x1 - 36, x2}
          filter2_4#!_A(x1,x2,x3,x4) = x4 + 70
          false_0!_A = 3
          filter_2#!_A(x1,x2) = x2 + 70
          cons_2!_A(x1,x2) = 0
          app!_A(x1,x2) = max{99, x2 + 34}
          true_0!_A = 0
          cons_0!_A = 34
          cons_1!_A(x1) = max{35, x1 + 1}
          filter_0!_A = 36
          filter_1!_A(x1) = max{35, x1 - 1}
          pred_1!_A(x1) = 100
          s_1!_A(x1) = 0
          minus_2!_A(x1,x2) = x1 + 101
          |0|_0!_A = 0
          quot_2!_A(x1,x2) = max{98, x1 + 66, x2 + 35}
          map_2!_A(x1,x2) = 0
          nil_0!_A = 0
          filter_2!_A(x1,x2) = max{x1 + 26, x2 + 141}
  
    partial status:
  
      pi(filter2_4#) = [2]
      pi(false_0) = []
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(app) = [1]
      pi(true_0) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(pred_1) = []
      pi(s_1) = []
      pi(minus_2) = [1]
      pi(|0|_0) = []
      pi(quot_2) = []
      pi(map_2) = []
      pi(nil_0) = []
      pi(filter_2) = [2]

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.

-- 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

  r17, r18, r19

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        quot_2#_A(x1,x2) = max{1, x1 - 28}
        s_1_A(x1) = max{34, x1 + 32}
        minus_2_A(x1,x2) = max{7, x1 + 5}
        pred_1_A(x1) = max{6, x1 - 31}
        |0|_0_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          quot_2#_A(x1,x2) = max{0, x1 - 9, x2 - 17}
          s_1_A(x1) = max{12, x1 + 8}
          minus_2_A(x1,x2) = max{5, x1 + 4, x2 - 14}
          pred_1_A(x1) = max{5, x1 - 6}
          |0|_0_A = 3
          quot_2#!_A(x1,x2) = max{7, x2 + 2}
          s_1!_A(x1) = max{15, x1 - 4}
          minus_2!_A(x1,x2) = max{6, x1 - 5}
          pred_1!_A(x1) = max{16, x1 - 6}
          |0|_0!_A = 0
  
    partial status:
  
      pi(quot_2#) = []
      pi(s_1) = [1]
      pi(minus_2) = []
      pi(pred_1) = []
      pi(|0|_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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        minus_2#_A(x1,x2) = max{6, x2 + 2}
        s_1_A(x1) = x1 + 36
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          minus_2#_A(x1,x2) = 0
          s_1_A(x1) = max{3, x1 - 3}
          minus_2#!_A(x1,x2) = x1 + 2
          s_1!_A(x1) = max{6, x1 - 68}
  
    partial status:
  
      pi(minus_2#) = [2]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.