YES
We show the termination of the TRS R:

  app(rev(),nil()) -> nil()
  app(rev(),app(app(cons(),x),l)) -> app(app(cons(),app(app(rev1(),x),l)),app(app(rev2(),x),l))
  app(app(rev1(),|0|()),nil()) -> |0|()
  app(app(rev1(),app(s(),x)),nil()) -> app(s(),x)
  app(app(rev1(),x),app(app(cons(),y),l)) -> app(app(rev1(),y),l)
  app(app(rev2(),x),nil()) -> nil()
  app(app(rev2(),x),app(app(cons(),y),l)) -> app(rev(),app(app(cons(),x),app(app(rev2(),y),l)))
  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(rev_0(),y) -> rev_1(y)
  app(rev1_0(),y) -> rev1_1(y)
  app(rev1_1(x1),y) -> rev1_2(x1,y)
  app(rev2_0(),y) -> rev2_1(y)
  app(rev2_1(x1),y) -> rev2_2(x1,y)
  app(s_0(),y) -> s_1(y)
  rev_1(nil_0()) -> nil_0()
  rev_1(cons_2(x,l)) -> cons_2(rev1_2(x,l),rev2_2(x,l))
  rev1_2(|0|_0(),nil_0()) -> |0|_0()
  rev1_2(s_1(x),nil_0()) -> s_1(x)
  rev1_2(x,cons_2(y,l)) -> rev1_2(y,l)
  rev2_2(x,nil_0()) -> nil_0()
  rev2_2(x,cons_2(y,l)) -> rev_1(cons_2(x,rev2_2(y,l)))
  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#(rev_0(),y) -> rev_1#(y)
p5: app#(rev1_1(x1),y) -> rev1_2#(x1,y)
p6: app#(rev2_1(x1),y) -> rev2_2#(x1,y)
p7: rev_1#(cons_2(x,l)) -> rev1_2#(x,l)
p8: rev_1#(cons_2(x,l)) -> rev2_2#(x,l)
p9: rev1_2#(x,cons_2(y,l)) -> rev1_2#(y,l)
p10: rev2_2#(x,cons_2(y,l)) -> rev_1#(cons_2(x,rev2_2(y,l)))
p11: rev2_2#(x,cons_2(y,l)) -> rev2_2#(y,l)
p12: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p13: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p14: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p15: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p16: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p17: 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(rev_0(),y) -> rev_1(y)
r12: app(rev1_0(),y) -> rev1_1(y)
r13: app(rev1_1(x1),y) -> rev1_2(x1,y)
r14: app(rev2_0(),y) -> rev2_1(y)
r15: app(rev2_1(x1),y) -> rev2_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: rev_1(nil_0()) -> nil_0()
r18: rev_1(cons_2(x,l)) -> cons_2(rev1_2(x,l),rev2_2(x,l))
r19: rev1_2(|0|_0(),nil_0()) -> |0|_0()
r20: rev1_2(s_1(x),nil_0()) -> s_1(x)
r21: rev1_2(x,cons_2(y,l)) -> rev1_2(y,l)
r22: rev2_2(x,nil_0()) -> nil_0()
r23: rev2_2(x,cons_2(y,l)) -> rev_1(cons_2(x,rev2_2(y,l)))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p12, p13, p14, p15, p16, p17}
  {p8, p10, p11}
  {p9}


-- 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(rev_0(),y) -> rev_1(y)
r12: app(rev1_0(),y) -> rev1_1(y)
r13: app(rev1_1(x1),y) -> rev1_2(x1,y)
r14: app(rev2_0(),y) -> rev2_1(y)
r15: app(rev2_1(x1),y) -> rev2_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: rev_1(nil_0()) -> nil_0()
r18: rev_1(cons_2(x,l)) -> cons_2(rev1_2(x,l),rev2_2(x,l))
r19: rev1_2(|0|_0(),nil_0()) -> |0|_0()
r20: rev1_2(s_1(x),nil_0()) -> s_1(x)
r21: rev1_2(x,cons_2(y,l)) -> rev1_2(y,l)
r22: rev2_2(x,nil_0()) -> nil_0()
r23: rev2_2(x,cons_2(y,l)) -> rev_1(cons_2(x,rev2_2(y,l)))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r19, r20, r24, r26

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{1, x1 - 92, x2 - 20}
            filter_1_A(x1) = x1 + 50
            filter_2#_A(x1,x2) = max{1, x1 - 42, x2 - 20}
            cons_2_A(x1,x2) = max{19, x1, x2}
            map_1_A(x1) = max{87, x1 + 82}
            map_2#_A(x1,x2) = max{1, x1 - 10, x2 - 20}
            filter2_3_A(x1,x2,x3) = max{x2 + 52, x3 + 91}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 42, x3 - 20, x4 - 20}
            false_0_A = 161
            app_A(x1,x2) = max{161, x1 - 88, x2 + 140}
            true_0_A = 161
            rev1_2_A(x1,x2) = max{16, x2 - 175}
            |0|_0_A = 16
            nil_0_A = 174
            s_1_A(x1) = 16
            map_2_A(x1,x2) = max{173, x1 - 5, x2 + 141}
            filter_2_A(x1,x2) = max{24, x2 + 4}
            app#!_A(x1,x2) = max{0, x1 - 51}
            filter_1!_A(x1) = max{49, x1 + 1}
            filter_2#!_A(x1,x2) = max{0, x1 - 2}
            cons_2!_A(x1,x2) = max{29, x1 + 15, x2 + 3}
            map_1!_A(x1) = 179
            map_2#!_A(x1,x2) = max{30, x1 - 51}
            filter2_3!_A(x1,x2,x3) = max{177, x2 + 49, x3 + 1}
            filter2_4#!_A(x1,x2,x3,x4) = max{0, x2 - 2}
            false_0!_A = 49
            app!_A(x1,x2) = max{178, x1 - 90, x2 + 28}
            true_0!_A = 30
            rev1_2!_A(x1,x2) = 29
            |0|_0!_A = 30
            nil_0!_A = 178
            s_1!_A(x1) = 178
            map_2!_A(x1,x2) = max{177, x1 - 1, x2 + 156}
            filter_2!_A(x1,x2) = max{48, x2 + 28}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{1, x1 - 14, x2 - 27}
            filter_1_A(x1) = max{3, x1 + 2}
            filter_2#_A(x1,x2) = max{1, x1 - 14, x2 - 27}
            cons_2_A(x1,x2) = max{26, x1, x2}
            map_1_A(x1) = max{3, x1}
            map_2#_A(x1,x2) = max{1, x1 - 14, x2 - 27}
            filter2_3_A(x1,x2,x3) = max{13, x1 - 21, x2, x3 + 11}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 14, x3 - 27, x4 - 27}
            false_0_A = 25
            app_A(x1,x2) = 2
            true_0_A = 25
            rev1_2_A(x1,x2) = max{53, x2 - 107}
            |0|_0_A = 53
            nil_0_A = 53
            s_1_A(x1) = 3
            map_2_A(x1,x2) = max{52, x1 - 12, x2 + 25}
            filter_2_A(x1,x2) = max{x1 - 32, x2 + 1}
            app#!_A(x1,x2) = 28
            filter_1!_A(x1) = max{30, x1 - 1}
            filter_2#!_A(x1,x2) = 28
            cons_2!_A(x1,x2) = 4
            map_1!_A(x1) = 10
            map_2#!_A(x1,x2) = max{29, x1 - 1, x2 + 1}
            filter2_3!_A(x1,x2,x3) = max{29, x1 - 14, x2 + 7, x3 + 5}
            filter2_4#!_A(x1,x2,x3,x4) = 28
            false_0!_A = 7
            app!_A(x1,x2) = max{30, x1 - 4, x2 + 9}
            true_0!_A = 29
            rev1_2!_A(x1,x2) = 4
            |0|_0!_A = 5
            nil_0!_A = 6
            s_1!_A(x1) = 5
            map_2!_A(x1,x2) = max{31, x1 - 1, x2 - 54}
            filter_2!_A(x1,x2) = 6
        
  
    partial status:
  
      pi(app#) = []
      pi(filter_1) = []
      pi(filter_2#) = []
      pi(cons_2) = [2]
      pi(map_1) = []
      pi(map_2#) = []
      pi(filter2_3) = [3]
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(app) = [2]
      pi(true_0) = []
      pi(rev1_2) = [2]
      pi(|0|_0) = []
      pi(nil_0) = []
      pi(s_1) = []
      pi(map_2) = [2]
      pi(filter_2) = []

The following rules are removed:

  p3, p5, p6



-- 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: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p5: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p6: 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(rev_0(),y) -> rev_1(y)
r12: app(rev1_0(),y) -> rev1_1(y)
r13: app(rev1_1(x1),y) -> rev1_2(x1,y)
r14: app(rev2_0(),y) -> rev2_1(y)
r15: app(rev2_1(x1),y) -> rev2_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: rev_1(nil_0()) -> nil_0()
r18: rev_1(cons_2(x,l)) -> cons_2(rev1_2(x,l),rev2_2(x,l))
r19: rev1_2(|0|_0(),nil_0()) -> |0|_0()
r20: rev1_2(s_1(x),nil_0()) -> s_1(x)
r21: rev1_2(x,cons_2(y,l)) -> rev1_2(y,l)
r22: rev2_2(x,nil_0()) -> nil_0()
r23: rev2_2(x,cons_2(y,l)) -> rev_1(cons_2(x,rev2_2(y,l)))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p4, p5, p6}
  {p3}


-- 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: 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(rev_0(),y) -> rev_1(y)
r12: app(rev1_0(),y) -> rev1_1(y)
r13: app(rev1_1(x1),y) -> rev1_2(x1,y)
r14: app(rev2_0(),y) -> rev2_1(y)
r15: app(rev2_1(x1),y) -> rev2_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: rev_1(nil_0()) -> nil_0()
r18: rev_1(cons_2(x,l)) -> cons_2(rev1_2(x,l),rev2_2(x,l))
r19: rev1_2(|0|_0(),nil_0()) -> |0|_0()
r20: rev1_2(s_1(x),nil_0()) -> s_1(x)
r21: rev1_2(x,cons_2(y,l)) -> rev1_2(y,l)
r22: rev2_2(x,nil_0()) -> nil_0()
r23: rev2_2(x,cons_2(y,l)) -> rev_1(cons_2(x,rev2_2(y,l)))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 22, x2 - 43}
            filter_1_A(x1) = x1 + 9
            filter_2#_A(x1,x2) = max{0, x1 - 22, x2 - 43}
            cons_2_A(x1,x2) = max{42, x1 + 9, x2}
            filter2_4#_A(x1,x2,x3,x4) = max{0, x1 - 65, x2 - 22, x3 - 34, x4 - 43}
            app_A(x1,x2) = max{44, x1 - 16, x2 + 31}
            true_0_A = 31
            false_0_A = 0
            cons_0_A = 9
            cons_1_A(x1) = max{32, x1 + 25}
            filter_0_A = 9
            filter_2_A(x1,x2) = max{44, x1 - 23, x2 + 1}
            filter2_0_A = 9
            filter2_1_A(x1) = x1 + 31
            filter2_2_A(x1,x2) = max{x1 + 15, x2 + 9}
            filter2_3_A(x1,x2,x3) = max{x1 - 1, x2 - 7, x3 + 25}
            filter2_4_A(x1,x2,x3,x4) = max{44, x1 - 45, x2 - 23, x3 + 9, x4 + 1}
            map_0_A = 9
            map_1_A(x1) = x1 + 9
            map_2_A(x1,x2) = max{41, x1 - 7, x2 + 31}
            rev_0_A = 9
            rev_1_A(x1) = max{42, x1 - 44}
            rev1_0_A = 11
            rev1_1_A(x1) = x1 + 9
            rev1_2_A(x1,x2) = max{32, x2 - 144}
            rev2_0_A = 9
            rev2_1_A(x1) = x1 + 9
            rev2_2_A(x1,x2) = max{42, x1 - 35, x2 - 45}
            s_0_A = 9
            s_1_A(x1) = 32
            nil_0_A = 0
            |0|_0_A = 8
            app#!_A(x1,x2) = max{75, x1 + 68, x2 + 6}
            filter_1!_A(x1) = max{1, x1 - 49}
            filter_2#!_A(x1,x2) = max{76, x1 + 74, x2 - 3}
            cons_2!_A(x1,x2) = max{x1 + 7, x2 - 93}
            filter2_4#!_A(x1,x2,x3,x4) = max{76, x1 - 138, x2 + 74, x4 - 3}
            app!_A(x1,x2) = max{x1 - 10, x2 + 76}
            true_0!_A = 1
            false_0!_A = 9
            cons_0!_A = 1
            cons_1!_A(x1) = max{8, x1 - 1}
            filter_0!_A = 2
            filter_2!_A(x1,x2) = max{8, x1 - 48, x2 - 1}
            filter2_0!_A = 1
            filter2_1!_A(x1) = max{2, x1 - 35}
            filter2_2!_A(x1,x2) = max{3, x1 - 34, x2 - 49}
            filter2_3!_A(x1,x2,x3) = max{4, x1 - 33, x2 - 48, x3 - 1}
            filter2_4!_A(x1,x2,x3,x4) = max{x1 - 32, x2 - 48, x3 + 8, x4 - 1}
            map_0!_A = 1
            map_1!_A(x1) = max{37, x1 - 1}
            map_2!_A(x1,x2) = max{38, x1 - 2, x2 + 30}
            rev_0!_A = 1
            rev_1!_A(x1) = max{40, x1 - 136}
            rev1_0!_A = 1
            rev1_1!_A(x1) = max{0, x1 - 1}
            rev1_2!_A(x1,x2) = max{3, x2 - 137}
            rev2_0!_A = 75
            rev2_1!_A(x1) = max{74, x1 - 3}
            rev2_2!_A(x1,x2) = max{75, x1 - 2, x2 - 50}
            s_0!_A = 1
            s_1!_A(x1) = 2
            nil_0!_A = 1
            |0|_0!_A = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 15, x2 - 47}
            filter_1_A(x1) = max{14, x1 + 2}
            filter_2#_A(x1,x2) = max{0, x1 - 13, x2 - 47}
            cons_2_A(x1,x2) = max{46, x1 + 4, x2}
            filter2_4#_A(x1,x2,x3,x4) = max{0, x1 - 65, x2 - 13, x3 - 43, x4 - 47}
            app_A(x1,x2) = max{64, x1 - 18, x2 + 22}
            true_0_A = 0
            false_0_A = 54
            cons_0_A = 0
            cons_1_A(x1) = x1 + 22
            filter_0_A = 14
            filter_2_A(x1,x2) = max{48, x1 - 23, x2 + 1}
            filter2_0_A = 14
            filter2_1_A(x1) = max{14, x1 + 13}
            filter2_2_A(x1,x2) = max{14, x1 - 5, x2 + 13}
            filter2_3_A(x1,x2,x3) = max{x1 - 39, x2 - 5, x3 + 22}
            filter2_4_A(x1,x2,x3,x4) = max{48, x1 - 65, x2 - 23, x3 + 4, x4 + 1}
            map_0_A = 16
            map_1_A(x1) = max{16, x1 + 13}
            map_2_A(x1,x2) = max{28, x1 - 14, x2 + 22}
            rev_0_A = 14
            rev_1_A(x1) = max{1, x1}
            rev1_0_A = 17
            rev1_1_A(x1) = 17
            rev1_2_A(x1,x2) = max{23, x2 - 47}
            rev2_0_A = 14
            rev2_1_A(x1) = x1 + 22
            rev2_2_A(x1,x2) = max{44, x1 + 4, x2}
            s_0_A = 14
            s_1_A(x1) = 23
            nil_0_A = 2
            |0|_0_A = 23
            app#!_A(x1,x2) = max{0, x1 - 13, x2 - 43}
            filter_1!_A(x1) = max{45, x1 - 1}
            filter_2#!_A(x1,x2) = max{0, x1 - 12, x2 - 47}
            cons_2!_A(x1,x2) = max{x1 - 24, x2 + 12}
            filter2_4#!_A(x1,x2,x3,x4) = max{0, x1 - 66, x2 - 12, x3 - 44, x4 - 47}
            app!_A(x1,x2) = max{82, x1 - 15}
            true_0!_A = 61
            false_0!_A = 45
            cons_0!_A = 84
            cons_1!_A(x1) = max{83, x1 - 25}
            filter_0!_A = 46
            filter_2!_A(x1,x2) = max{44, x1 - 12, x2 - 3}
            filter2_0!_A = 84
            filter2_1!_A(x1) = max{83, x1 - 2}
            filter2_2!_A(x1,x2) = max{84, x1 - 1, x2 + 1}
            filter2_3!_A(x1,x2,x3) = max{85, x1 - 53, x2 - 1, x3 - 2}
            filter2_4!_A(x1,x2,x3,x4) = max{44, x1 - 53, x2 - 12, x3 - 1, x4 - 3}
            map_0!_A = 1
            map_1!_A(x1) = max{83, x1 - 2}
            map_2!_A(x1,x2) = max{84, x1 - 1, x2 + 35}
            rev_0!_A = 1
            rev_1!_A(x1) = max{84, x1 - 71}
            rev1_0!_A = 2
            rev1_1!_A(x1) = max{1, x1 - 42}
            rev1_2!_A(x1,x2) = max{85, x1 - 43, x2 - 47}
            rev2_0!_A = 2
            rev2_1!_A(x1) = max{1, x1 - 24}
            rev2_2!_A(x1,x2) = max{83, x1 - 23, x2 - 72}
            s_0!_A = 38
            s_1!_A(x1) = max{37, x1 - 21}
            nil_0!_A = 38
            |0|_0!_A = 86
        
  
    partial status:
  
      pi(app#) = []
      pi(filter_1) = [1]
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(filter2_4#) = []
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(false_0) = []
      pi(cons_0) = []
      pi(cons_1) = [1]
      pi(filter_0) = []
      pi(filter_2) = [1]
      pi(filter2_0) = []
      pi(filter2_1) = [1]
      pi(filter2_2) = [1, 2]
      pi(filter2_3) = [2, 3]
      pi(filter2_4) = [2]
      pi(map_0) = []
      pi(map_1) = []
      pi(map_2) = []
      pi(rev_0) = []
      pi(rev_1) = []
      pi(rev1_0) = []
      pi(rev1_1) = [1]
      pi(rev1_2) = []
      pi(rev2_0) = []
      pi(rev2_1) = [1]
      pi(rev2_2) = [1, 2]
      pi(s_0) = []
      pi(s_1) = []
      pi(nil_0) = []
      pi(|0|_0) = []

The following rules are removed:

  p1



-- 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#(true_0(),f,x,xs) -> filter_2#(f,xs)
p3: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p4: 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(rev_0(),y) -> rev_1(y)
r12: app(rev1_0(),y) -> rev1_1(y)
r13: app(rev1_1(x1),y) -> rev1_2(x1,y)
r14: app(rev2_0(),y) -> rev2_1(y)
r15: app(rev2_1(x1),y) -> rev2_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: rev_1(nil_0()) -> nil_0()
r18: rev_1(cons_2(x,l)) -> cons_2(rev1_2(x,l),rev2_2(x,l))
r19: rev1_2(|0|_0(),nil_0()) -> |0|_0()
r20: rev1_2(s_1(x),nil_0()) -> s_1(x)
r21: rev1_2(x,cons_2(y,l)) -> rev1_2(y,l)
r22: rev2_2(x,nil_0()) -> nil_0()
r23: rev2_2(x,cons_2(y,l)) -> rev_1(cons_2(x,rev2_2(y,l)))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p4}


-- 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(rev_0(),y) -> rev_1(y)
r12: app(rev1_0(),y) -> rev1_1(y)
r13: app(rev1_1(x1),y) -> rev1_2(x1,y)
r14: app(rev2_0(),y) -> rev2_1(y)
r15: app(rev2_1(x1),y) -> rev2_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: rev_1(nil_0()) -> nil_0()
r18: rev_1(cons_2(x,l)) -> cons_2(rev1_2(x,l),rev2_2(x,l))
r19: rev1_2(|0|_0(),nil_0()) -> |0|_0()
r20: rev1_2(s_1(x),nil_0()) -> s_1(x)
r21: rev1_2(x,cons_2(y,l)) -> rev1_2(y,l)
r22: rev2_2(x,nil_0()) -> nil_0()
r23: rev2_2(x,cons_2(y,l)) -> rev_1(cons_2(x,rev2_2(y,l)))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            filter_2#_A(x1,x2) = max{0, x1 - 3, x2 - 12}
            cons_2_A(x1,x2) = max{9, x1 - 34, x2}
            filter2_4#_A(x1,x2,x3,x4) = max{0, x1 - 58, x2 - 3, x3 - 46, x4 - 12}
            app_A(x1,x2) = max{56, x1 - 9, x2 + 8}
            false_0_A = 57
            true_0_A = 43
            cons_0_A = 6
            cons_1_A(x1) = max{6, x1 - 25}
            filter_0_A = 8
            filter_1_A(x1) = x1 + 6
            filter_2_A(x1,x2) = max{14, x1 - 12, x2 + 2}
            filter2_0_A = 6
            filter2_1_A(x1) = x1 + 6
            filter2_2_A(x1,x2) = max{x1 - 3, x2 + 6}
            filter2_3_A(x1,x2,x3) = max{x1 - 12, x2 - 3, x3 + 8}
            filter2_4_A(x1,x2,x3,x4) = max{14, x1 - 44, x2 - 12, x3 - 32, x4 + 2}
            map_0_A = 6
            map_1_A(x1) = x1 + 6
            map_2_A(x1,x2) = max{22, x1 - 3, x2 + 8}
            rev_0_A = 8
            rev_1_A(x1) = max{23, x1 - 29}
            rev1_0_A = 6
            rev1_1_A(x1) = x1 + 6
            rev1_2_A(x1,x2) = max{32, x2 - 24}
            rev2_0_A = 6
            rev2_1_A(x1) = x1 + 6
            rev2_2_A(x1,x2) = max{23, x1 - 63, x2 - 30}
            s_0_A = 6
            s_1_A(x1) = 29
            nil_0_A = 23
            |0|_0_A = 29
            filter_2#!_A(x1,x2) = max{12, x1 - 8, x2 - 1}
            cons_2!_A(x1,x2) = 11
            filter2_4#!_A(x1,x2,x3,x4) = max{12, x1 - 44, x2 - 8, x3 - 35, x4 - 1}
            app!_A(x1,x2) = max{x1 - 7, x2 + 17}
            false_0!_A = 17
            true_0!_A = 1
            cons_0!_A = 11
            cons_1!_A(x1) = max{10, x1 - 32}
            filter_0!_A = 1
            filter_1!_A(x1) = max{0, x1 - 1}
            filter_2!_A(x1,x2) = 16
            filter2_0!_A = 13
            filter2_1!_A(x1) = max{14, x1 - 9}
            filter2_2!_A(x1,x2) = max{16, x1 - 10, x2 - 3}
            filter2_3!_A(x1,x2,x3) = max{0, x1 - 11, x2 - 2, x3 - 1}
            filter2_4!_A(x1,x2,x3,x4) = 16
            map_0!_A = 1
            map_1!_A(x1) = max{16, x1 - 7}
            map_2!_A(x1,x2) = max{11, x1 - 6, x2 + 1}
            rev_0!_A = 1
            rev_1!_A(x1) = max{12, x1 - 25}
            rev1_0!_A = 17
            rev1_1!_A(x1) = x1 + 16
            rev1_2!_A(x1,x2) = 10
            rev2_0!_A = 1
            rev2_1!_A(x1) = max{14, x1 - 58}
            rev2_2!_A(x1,x2) = max{13, x1 - 58, x2 - 24}
            s_0!_A = 0
            s_1!_A(x1) = 0
            nil_0!_A = 2
            |0|_0!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            filter_2#_A(x1,x2) = max{1, x1 - 23}
            cons_2_A(x1,x2) = 13
            filter2_4#_A(x1,x2,x3,x4) = max{1, x1 - 48, x2 - 23}
            app_A(x1,x2) = 26
            false_0_A = 47
            true_0_A = 0
            cons_0_A = 2
            cons_1_A(x1) = 2
            filter_0_A = 4
            filter_1_A(x1) = 4
            filter_2_A(x1,x2) = 13
            filter2_0_A = 2
            filter2_1_A(x1) = 2
            filter2_2_A(x1,x2) = 2
            filter2_3_A(x1,x2,x3) = 2
            filter2_4_A(x1,x2,x3,x4) = 13
            map_0_A = 2
            map_1_A(x1) = 4
            map_2_A(x1,x2) = 18
            rev_0_A = 2
            rev_1_A(x1) = 13
            rev1_0_A = 2
            rev1_1_A(x1) = 4
            rev1_2_A(x1,x2) = 26
            rev2_0_A = 2
            rev2_1_A(x1) = 1
            rev2_2_A(x1,x2) = 26
            s_0_A = 2
            s_1_A(x1) = 1
            nil_0_A = 13
            |0|_0_A = 1
            filter_2#!_A(x1,x2) = 5
            cons_2!_A(x1,x2) = max{4, x2 - 14}
            filter2_4#!_A(x1,x2,x3,x4) = max{5, x1 - 41}
            app!_A(x1,x2) = max{6, x1 - 3, x2 + 1}
            false_0!_A = 9
            true_0!_A = 9
            cons_0!_A = 3
            cons_1!_A(x1) = max{2, x1 - 1}
            filter_0!_A = 1
            filter_1!_A(x1) = 7
            filter_2!_A(x1,x2) = 8
            filter2_0!_A = 3
            filter2_1!_A(x1) = max{2, x1 - 41}
            filter2_2!_A(x1,x2) = max{7, x1 - 40, x2 - 1}
            filter2_3!_A(x1,x2,x3) = max{0, x1 - 39, x2 - 2, x3 - 1}
            filter2_4!_A(x1,x2,x3,x4) = max{5, x1 - 38}
            map_0!_A = 1
            map_1!_A(x1) = max{0, x1 - 1}
            map_2!_A(x1,x2) = max{5, x1 - 2}
            rev_0!_A = 1
            rev_1!_A(x1) = max{0, x1 - 14}
            rev1_0!_A = 8
            rev1_1!_A(x1) = x1 + 38
            rev1_2!_A(x1,x2) = 3
            rev2_0!_A = 1
            rev2_1!_A(x1) = 2
            rev2_2!_A(x1,x2) = max{0, x1 - 2}
            s_0!_A = 1
            s_1!_A(x1) = 2
            nil_0!_A = 5
            |0|_0!_A = 1
        
  
    partial status:
  
      pi(filter_2#) = [1]
      pi(cons_2) = []
      pi(filter2_4#) = [2]
      pi(app) = []
      pi(false_0) = []
      pi(true_0) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(filter_2) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(filter2_3) = []
      pi(filter2_4) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(map_2) = []
      pi(rev_0) = []
      pi(rev_1) = []
      pi(rev1_0) = []
      pi(rev1_1) = []
      pi(rev1_2) = []
      pi(rev2_0) = []
      pi(rev2_1) = []
      pi(rev2_2) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(nil_0) = []
      pi(|0|_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)
p2: 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(rev_0(),y) -> rev_1(y)
r12: app(rev1_0(),y) -> rev1_1(y)
r13: app(rev1_1(x1),y) -> rev1_2(x1,y)
r14: app(rev2_0(),y) -> rev2_1(y)
r15: app(rev2_1(x1),y) -> rev2_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: rev_1(nil_0()) -> nil_0()
r18: rev_1(cons_2(x,l)) -> cons_2(rev1_2(x,l),rev2_2(x,l))
r19: rev1_2(|0|_0(),nil_0()) -> |0|_0()
r20: rev1_2(s_1(x),nil_0()) -> s_1(x)
r21: rev1_2(x,cons_2(y,l)) -> rev1_2(y,l)
r22: rev2_2(x,nil_0()) -> nil_0()
r23: rev2_2(x,cons_2(y,l)) -> rev_1(cons_2(x,rev2_2(y,l)))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: 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#(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(rev_0(),y) -> rev_1(y)
r12: app(rev1_0(),y) -> rev1_1(y)
r13: app(rev1_1(x1),y) -> rev1_2(x1,y)
r14: app(rev2_0(),y) -> rev2_1(y)
r15: app(rev2_1(x1),y) -> rev2_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: rev_1(nil_0()) -> nil_0()
r18: rev_1(cons_2(x,l)) -> cons_2(rev1_2(x,l),rev2_2(x,l))
r19: rev1_2(|0|_0(),nil_0()) -> |0|_0()
r20: rev1_2(s_1(x),nil_0()) -> s_1(x)
r21: rev1_2(x,cons_2(y,l)) -> rev1_2(y,l)
r22: rev2_2(x,nil_0()) -> nil_0()
r23: rev2_2(x,cons_2(y,l)) -> rev_1(cons_2(x,rev2_2(y,l)))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r17, r19, r24, r26

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            filter_2#_A(x1,x2) = max{1, x1 - 1, x2 - 4}
            cons_2_A(x1,x2) = max{x1 - 27, x2 + 3}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 1, x3 - 31, x4 - 1}
            app_A(x1,x2) = max{47, x1 - 7}
            true_0_A = 47
            rev_1_A(x1) = max{67, x1 - 51}
            nil_0_A = 50
            rev1_2_A(x1,x2) = 26
            |0|_0_A = 26
            map_2_A(x1,x2) = x2 + 18
            filter_2_A(x1,x2) = max{49, x2 + 15}
            filter_2#!_A(x1,x2) = x2 + 30
            cons_2!_A(x1,x2) = max{72, x1 + 25, x2 + 34}
            filter2_4#!_A(x1,x2,x3,x4) = max{33, x3 + 2, x4 + 31}
            app!_A(x1,x2) = max{x1 + 37, x2 + 109}
            true_0!_A = 38
            rev_1!_A(x1) = 110
            nil_0!_A = 37
            rev1_2!_A(x1,x2) = x2 + 30
            |0|_0!_A = 36
            map_2!_A(x1,x2) = max{x1 + 36, x2 + 69}
            filter_2!_A(x1,x2) = max{x1 + 27, x2 + 66}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            filter_2#_A(x1,x2) = max{1, x1 - 1, x2 - 5}
            cons_2_A(x1,x2) = max{4, x1 - 29, x2}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 1, x3 - 34, x4 - 5}
            app_A(x1,x2) = max{x1 - 58, x2 + 31}
            true_0_A = 31
            rev_1_A(x1) = max{32, x1 - 56}
            nil_0_A = 15
            rev1_2_A(x1,x2) = max{28, x2 + 3}
            |0|_0_A = 28
            map_2_A(x1,x2) = max{x1 - 17, x2 + 32}
            filter_2_A(x1,x2) = max{14, x1 - 94, x2 + 6}
            filter_2#!_A(x1,x2) = 0
            cons_2!_A(x1,x2) = 0
            filter2_4#!_A(x1,x2,x3,x4) = 0
            app!_A(x1,x2) = max{x1 - 25, x2 + 30}
            true_0!_A = 1
            rev_1!_A(x1) = max{11, x1 - 62}
            nil_0!_A = 9
            rev1_2!_A(x1,x2) = 30
            |0|_0!_A = 1
            map_2!_A(x1,x2) = max{8, x1 - 24, x2 - 5}
            filter_2!_A(x1,x2) = 0
        
  
    partial status:
  
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(filter2_4#) = []
      pi(app) = [2]
      pi(true_0) = []
      pi(rev_1) = [1]
      pi(nil_0) = []
      pi(rev1_2) = []
      pi(|0|_0) = []
      pi(map_2) = []
      pi(filter_2) = []

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(rev_0(),y) -> rev_1(y)
r12: app(rev1_0(),y) -> rev1_1(y)
r13: app(rev1_1(x1),y) -> rev1_2(x1,y)
r14: app(rev2_0(),y) -> rev2_1(y)
r15: app(rev2_1(x1),y) -> rev2_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: rev_1(nil_0()) -> nil_0()
r18: rev_1(cons_2(x,l)) -> cons_2(rev1_2(x,l),rev2_2(x,l))
r19: rev1_2(|0|_0(),nil_0()) -> |0|_0()
r20: rev1_2(s_1(x),nil_0()) -> s_1(x)
r21: rev1_2(x,cons_2(y,l)) -> rev1_2(y,l)
r22: rev2_2(x,nil_0()) -> nil_0()
r23: rev2_2(x,cons_2(y,l)) -> rev_1(cons_2(x,rev2_2(y,l)))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: 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: 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(rev_0(),y) -> rev_1(y)
r12: app(rev1_0(),y) -> rev1_1(y)
r13: app(rev1_1(x1),y) -> rev1_2(x1,y)
r14: app(rev2_0(),y) -> rev2_1(y)
r15: app(rev2_1(x1),y) -> rev2_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: rev_1(nil_0()) -> nil_0()
r18: rev_1(cons_2(x,l)) -> cons_2(rev1_2(x,l),rev2_2(x,l))
r19: rev1_2(|0|_0(),nil_0()) -> |0|_0()
r20: rev1_2(s_1(x),nil_0()) -> s_1(x)
r21: rev1_2(x,cons_2(y,l)) -> rev1_2(y,l)
r22: rev2_2(x,nil_0()) -> nil_0()
r23: rev2_2(x,cons_2(y,l)) -> rev_1(cons_2(x,rev2_2(y,l)))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: 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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = 0
            cons_2_A(x1,x2) = max{7, x1 - 21, x2 - 7}
            map_2#!_A(x1,x2) = max{1, x2 - 58}
            cons_2!_A(x1,x2) = max{3, x1 - 54, x2 - 57}
        
        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{4, x1 - 9, x2 - 4}
            map_2#!_A(x1,x2) = max{7, x1 - 1, x2 - 5}
            cons_2!_A(x1,x2) = max{6, x1 - 72, x2 - 43}
        
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: rev2_2#(x,cons_2(y,l)) -> rev_1#(cons_2(x,rev2_2(y,l)))
p2: rev_1#(cons_2(x,l)) -> rev2_2#(x,l)
p3: rev2_2#(x,cons_2(y,l)) -> rev2_2#(y,l)

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(rev_0(),y) -> rev_1(y)
r12: app(rev1_0(),y) -> rev1_1(y)
r13: app(rev1_1(x1),y) -> rev1_2(x1,y)
r14: app(rev2_0(),y) -> rev2_1(y)
r15: app(rev2_1(x1),y) -> rev2_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: rev_1(nil_0()) -> nil_0()
r18: rev_1(cons_2(x,l)) -> cons_2(rev1_2(x,l),rev2_2(x,l))
r19: rev1_2(|0|_0(),nil_0()) -> |0|_0()
r20: rev1_2(s_1(x),nil_0()) -> s_1(x)
r21: rev1_2(x,cons_2(y,l)) -> rev1_2(y,l)
r22: rev2_2(x,nil_0()) -> nil_0()
r23: rev2_2(x,cons_2(y,l)) -> rev_1(cons_2(x,rev2_2(y,l)))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r17, r18, r19, r20, r21, r22, r23

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            rev2_2#_A(x1,x2) = max{76, x2 + 1}
            cons_2_A(x1,x2) = max{74, x2 + 7}
            rev_1#_A(x1) = max{76, x1 + 1}
            rev2_2_A(x1,x2) = max{67, x2}
            rev_1_A(x1) = x1
            nil_0_A = 78
            rev1_2_A(x1,x2) = 3
            |0|_0_A = 0
            s_1_A(x1) = 0
            rev2_2#!_A(x1,x2) = max{1, x2 - 4}
            cons_2!_A(x1,x2) = max{10, x1 - 4}
            rev_1#!_A(x1) = max{70, x1 - 5}
            rev2_2!_A(x1,x2) = max{83, x1}
            rev_1!_A(x1) = 0
            nil_0!_A = 80
            rev1_2!_A(x1,x2) = max{0, x2 - 75}
            |0|_0!_A = 3
            s_1!_A(x1) = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            rev2_2#_A(x1,x2) = max{15, x2 + 1}
            cons_2_A(x1,x2) = max{13, x2 + 8}
            rev_1#_A(x1) = max{15, x1}
            rev2_2_A(x1,x2) = max{5, x2}
            rev_1_A(x1) = max{12, x1}
            nil_0_A = 5
            rev1_2_A(x1,x2) = max{0, x2 - 39}
            |0|_0_A = 0
            s_1_A(x1) = 0
            rev2_2#!_A(x1,x2) = max{7, x2 + 1}
            cons_2!_A(x1,x2) = max{x1 + 24, x2 + 16}
            rev_1#!_A(x1) = max{8, x1 + 1}
            rev2_2!_A(x1,x2) = max{x1 + 25, x2 + 9}
            rev_1!_A(x1) = 23
            nil_0!_A = 25
            rev1_2!_A(x1,x2) = x2
            |0|_0!_A = 1
            s_1!_A(x1) = max{35, x1 - 46}
        
  
    partial status:
  
      pi(rev2_2#) = []
      pi(cons_2) = []
      pi(rev_1#) = []
      pi(rev2_2) = [1]
      pi(rev_1) = [1]
      pi(nil_0) = []
      pi(rev1_2) = []
      pi(|0|_0) = []
      pi(s_1) = []

The following rules are removed:

  p2, p3



-- SCC decomposition.

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

p1: rev2_2#(x,cons_2(y,l)) -> rev_1#(cons_2(x,rev2_2(y,l)))

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(rev_0(),y) -> rev_1(y)
r12: app(rev1_0(),y) -> rev1_1(y)
r13: app(rev1_1(x1),y) -> rev1_2(x1,y)
r14: app(rev2_0(),y) -> rev2_1(y)
r15: app(rev2_1(x1),y) -> rev2_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: rev_1(nil_0()) -> nil_0()
r18: rev_1(cons_2(x,l)) -> cons_2(rev1_2(x,l),rev2_2(x,l))
r19: rev1_2(|0|_0(),nil_0()) -> |0|_0()
r20: rev1_2(s_1(x),nil_0()) -> s_1(x)
r21: rev1_2(x,cons_2(y,l)) -> rev1_2(y,l)
r22: rev2_2(x,nil_0()) -> nil_0()
r23: rev2_2(x,cons_2(y,l)) -> rev_1(cons_2(x,rev2_2(y,l)))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: 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: rev1_2#(x,cons_2(y,l)) -> rev1_2#(y,l)

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(rev_0(),y) -> rev_1(y)
r12: app(rev1_0(),y) -> rev1_1(y)
r13: app(rev1_1(x1),y) -> rev1_2(x1,y)
r14: app(rev2_0(),y) -> rev2_1(y)
r15: app(rev2_1(x1),y) -> rev2_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: rev_1(nil_0()) -> nil_0()
r18: rev_1(cons_2(x,l)) -> cons_2(rev1_2(x,l),rev2_2(x,l))
r19: rev1_2(|0|_0(),nil_0()) -> |0|_0()
r20: rev1_2(s_1(x),nil_0()) -> s_1(x)
r21: rev1_2(x,cons_2(y,l)) -> rev1_2(y,l)
r22: rev2_2(x,nil_0()) -> nil_0()
r23: rev2_2(x,cons_2(y,l)) -> rev_1(cons_2(x,rev2_2(y,l)))
r24: map_2(f,nil_0()) -> nil_0()
r25: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r26: filter_2(f,nil_0()) -> nil_0()
r27: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r28: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r29: 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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            rev1_2#_A(x1,x2) = 0
            cons_2_A(x1,x2) = max{4, x1 - 4, x2 - 4}
            rev1_2#!_A(x1,x2) = max{3, x1 - 36, x2 - 43}
            cons_2!_A(x1,x2) = max{3, x1 - 36, x2 - 42}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            rev1_2#_A(x1,x2) = 0
            cons_2_A(x1,x2) = max{15, x1 - 21, x2 - 16}
            rev1_2#!_A(x1,x2) = max{4, x1 - 73, x2 - 53}
            cons_2!_A(x1,x2) = max{4, x1 - 72, x2 - 16}
        
  
    partial status:
  
      pi(rev1_2#) = [2]
      pi(cons_2) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.