YES
We show the termination of the TRS R:

  app(D(),t()) -> |1|()
  app(D(),constant()) -> |0|()
  app(D(),app(app(+(),x),y)) -> app(app(+(),app(D(),x)),app(D(),y))
  app(D(),app(app(*(),x),y)) -> app(app(+(),app(app(*(),y),app(D(),x))),app(app(*(),x),app(D(),y)))
  app(D(),app(app(-(),x),y)) -> app(app(-(),app(D(),x)),app(D(),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(*_0(),y) -> *_1(y)
  app(*_1(x1),y) -> *_2(x1,y)
  app(+_0(),y) -> +_1(y)
  app(+_1(x1),y) -> +_2(x1,y)
  app(-_0(),y) -> -_1(y)
  app(-_1(x1),y) -> -_2(x1,y)
  app(D_0(),y) -> D_1(y)
  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)
  D_1(t_0()) -> |1|_0()
  D_1(constant_0()) -> |0|_0()
  D_1(+_2(x,y)) -> +_2(D_1(x),D_1(y))
  D_1(*_2(x,y)) -> +_2(*_2(y,D_1(x)),*_2(x,D_1(y)))
  D_1(-_2(x,y)) -> -_2(D_1(x),D_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#(D_0(),y) -> D_1#(y)
p2: app#(filter_1(x1),y) -> filter_2#(x1,y)
p3: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p4: app#(map_1(x1),y) -> map_2#(x1,y)
p5: D_1#(+_2(x,y)) -> D_1#(x)
p6: D_1#(+_2(x,y)) -> D_1#(y)
p7: D_1#(*_2(x,y)) -> D_1#(x)
p8: D_1#(*_2(x,y)) -> D_1#(y)
p9: D_1#(-_2(x,y)) -> D_1#(x)
p10: D_1#(-_2(x,y)) -> D_1#(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(*_0(),y) -> *_1(y)
r2: app(*_1(x1),y) -> *_2(x1,y)
r3: app(+_0(),y) -> +_1(y)
r4: app(+_1(x1),y) -> +_2(x1,y)
r5: app(-_0(),y) -> -_1(y)
r6: app(-_1(x1),y) -> -_2(x1,y)
r7: app(D_0(),y) -> D_1(y)
r8: app(cons_0(),y) -> cons_1(y)
r9: app(cons_1(x1),y) -> cons_2(x1,y)
r10: app(filter_0(),y) -> filter_1(y)
r11: app(filter_1(x1),y) -> filter_2(x1,y)
r12: app(filter2_0(),y) -> filter2_1(y)
r13: app(filter2_1(x1),y) -> filter2_2(x1,y)
r14: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r15: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r16: app(map_0(),y) -> map_1(y)
r17: app(map_1(x1),y) -> map_2(x1,y)
r18: D_1(t_0()) -> |1|_0()
r19: D_1(constant_0()) -> |0|_0()
r20: D_1(+_2(x,y)) -> +_2(D_1(x),D_1(y))
r21: D_1(*_2(x,y)) -> +_2(*_2(y,D_1(x)),*_2(x,D_1(y)))
r22: D_1(-_2(x,y)) -> -_2(D_1(x),D_1(y))
r23: map_2(f,nil_0()) -> nil_0()
r24: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r25: filter_2(f,nil_0()) -> nil_0()
r26: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r27: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r28: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p11, p12, p13, p14, p15, p16}
  {p5, p6, p7, p8, p9, p10}


-- Reduction pair.

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

p1: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p2: app#(map_1(x1),y) -> map_2#(x1,y)
p3: map_2#(f,cons_2(x,xs)) -> map_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)
p6: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p7: app#(filter_1(x1),y) -> filter_2#(x1,y)
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(*_0(),y) -> *_1(y)
r2: app(*_1(x1),y) -> *_2(x1,y)
r3: app(+_0(),y) -> +_1(y)
r4: app(+_1(x1),y) -> +_2(x1,y)
r5: app(-_0(),y) -> -_1(y)
r6: app(-_1(x1),y) -> -_2(x1,y)
r7: app(D_0(),y) -> D_1(y)
r8: app(cons_0(),y) -> cons_1(y)
r9: app(cons_1(x1),y) -> cons_2(x1,y)
r10: app(filter_0(),y) -> filter_1(y)
r11: app(filter_1(x1),y) -> filter_2(x1,y)
r12: app(filter2_0(),y) -> filter2_1(y)
r13: app(filter2_1(x1),y) -> filter2_2(x1,y)
r14: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r15: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r16: app(map_0(),y) -> map_1(y)
r17: app(map_1(x1),y) -> map_2(x1,y)
r18: D_1(t_0()) -> |1|_0()
r19: D_1(constant_0()) -> |0|_0()
r20: D_1(+_2(x,y)) -> +_2(D_1(x),D_1(y))
r21: D_1(*_2(x,y)) -> +_2(*_2(y,D_1(x)),*_2(x,D_1(y)))
r22: D_1(-_2(x,y)) -> -_2(D_1(x),D_1(y))
r23: map_2(f,nil_0()) -> nil_0()
r24: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r25: filter_2(f,nil_0()) -> nil_0()
r26: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r27: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r28: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r5, r6, r8, r9, r10, r12, r13, r14, r16, r18, r19, r23, r25

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) = max{1, x1 - 27}
            cons_2_A(x1,x2) = max{27, x2}
            app#_A(x1,x2) = max{1, x1 - 29}
            map_1_A(x1) = max{10, x1 + 2}
            filter2_3_A(x1,x2,x3) = max{x1 - 26, x2 + 8, x3 + 29}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 23}
            false_0_A = 57
            filter_2#_A(x1,x2) = max{1, x1 - 23}
            filter_1_A(x1) = max{15, x1 + 6}
            app_A(x1,x2) = max{57, x1 - 16, x2 + 29}
            true_0_A = 57
            -_0_A = 7
            -_1_A(x1) = max{30, x1 + 2}
            -_2_A(x1,x2) = max{30, x1 - 34, x2 - 34}
            cons_0_A = 2
            cons_1_A(x1) = max{2, x1 - 27}
            filter_0_A = 14
            filter2_0_A = 2
            filter2_1_A(x1) = x1 + 6
            filter2_2_A(x1,x2) = max{x1 - 10, x2 + 24}
            map_0_A = 2
            D_1_A(x1) = 33
            t_0_A = 0
            |1|_0_A = 1
            constant_0_A = 32
            |0|_0_A = 33
            map_2_A(x1,x2) = max{57, x1 - 14, x2 + 29}
            nil_0_A = 1
            filter_2_A(x1,x2) = 30
            map_2#!_A(x1,x2) = max{2, x1 - 4}
            cons_2!_A(x1,x2) = 0
            app#!_A(x1,x2) = max{1, x1 - 7}
            map_1!_A(x1) = max{3, x1 - 3}
            filter2_3!_A(x1,x2,x3) = max{0, x1 - 29, x2 - 3, x3 - 1}
            filter2_4#!_A(x1,x2,x3,x4) = max{2, x2 - 2}
            false_0!_A = 12
            filter_2#!_A(x1,x2) = max{2, x1 - 2}
            filter_1!_A(x1) = max{10, x1 - 1}
            app!_A(x1,x2) = max{9, x1 - 3, x2 + 5}
            true_0!_A = 0
            -_0!_A = 2
            -_1!_A(x1) = max{3, x1 - 33}
            -_2!_A(x1,x2) = max{4, x1 - 32, x2 - 66}
            cons_0!_A = 1
            cons_1!_A(x1) = max{2, x1 - 1}
            filter_0!_A = 11
            filter2_0!_A = 1
            filter2_1!_A(x1) = max{6, x1 + 4}
            filter2_2!_A(x1,x2) = max{2, x2 - 4}
            map_0!_A = 4
            D_1!_A(x1) = 34
            t_0!_A = 2
            |1|_0!_A = 1
            constant_0!_A = 2
            |0|_0!_A = 1
            map_2!_A(x1,x2) = max{8, x1 - 2, x2 - 28}
            nil_0!_A = 9
            filter_2!_A(x1,x2) = max{11, x1 + 1, x2 - 28}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = max{25, x1 - 36, x2 - 34}
            cons_2_A(x1,x2) = max{x1 + 33, x2}
            app#_A(x1,x2) = max{25, x1 - 69, x2 - 1}
            map_1_A(x1) = max{34, x1 + 33}
            filter2_3_A(x1,x2,x3) = max{x1 - 37, x2 + 33, x3 + 68}
            filter2_4#_A(x1,x2,x3,x4) = max{25, x2 - 69, x3 - 1, x4 - 9}
            false_0_A = 101
            filter_2#_A(x1,x2) = max{25, x1 - 69, x2 - 9}
            filter_1_A(x1) = max{34, x1 + 33}
            app_A(x1,x2) = max{x1 - 35, x2 + 130}
            true_0_A = 106
            -_0_A = 34
            -_1_A(x1) = max{34, x1 + 33}
            -_2_A(x1,x2) = max{107, x1 - 6, x2 - 6}
            cons_0_A = 34
            cons_1_A(x1) = x1 + 68
            filter_0_A = 34
            filter2_0_A = 34
            filter2_1_A(x1) = max{34, x1 + 33}
            filter2_2_A(x1,x2) = max{x1 - 2, x2 + 68}
            map_0_A = 34
            D_1_A(x1) = max{5, x1 - 119}
            t_0_A = 0
            |1|_0_A = 0
            constant_0_A = 0
            |0|_0_A = 0
            map_2_A(x1,x2) = max{x1 - 2, x2 + 130}
            nil_0_A = 214
            filter_2_A(x1,x2) = max{131, x1 - 2, x2 + 1}
            map_2#!_A(x1,x2) = max{0, x1 - 220}
            cons_2!_A(x1,x2) = max{1, x1 - 107, x2 - 215}
            app#!_A(x1,x2) = max{2, x2 - 108}
            map_1!_A(x1) = max{27, x1 - 219}
            filter2_3!_A(x1,x2,x3) = max{2, x1 - 34, x2 - 1, x3 - 1}
            filter2_4#!_A(x1,x2,x3,x4) = max{0, x4 - 3}
            false_0!_A = 1
            filter_2#!_A(x1,x2) = max{0, x2 - 3}
            filter_1!_A(x1) = max{32, x1 - 1}
            app!_A(x1,x2) = 31
            true_0!_A = 31
            -_0!_A = 7
            -_1!_A(x1) = max{6, x1 - 6}
            -_2!_A(x1,x2) = max{7, x1 - 6, x2 - 110}
            cons_0!_A = 33
            cons_1!_A(x1) = max{32, x1 - 108}
            filter_0!_A = 1
            filter2_0!_A = 1
            filter2_1!_A(x1) = max{32, x1 - 35}
            filter2_2!_A(x1,x2) = max{1, x1 - 35}
            map_0!_A = 28
            D_1!_A(x1) = max{8, x1 - 109}
            t_0!_A = 1
            |1|_0!_A = 0
            constant_0!_A = 0
            |0|_0!_A = 1
            map_2!_A(x1,x2) = max{28, x1 - 218, x2 - 215}
            nil_0!_A = 29
            filter_2!_A(x1,x2) = 30
        
  
    partial status:
  
      pi(map_2#) = [1]
      pi(cons_2) = []
      pi(app#) = [1]
      pi(map_1) = [1]
      pi(filter2_3) = [2, 3]
      pi(filter2_4#) = [2]
      pi(false_0) = []
      pi(filter_2#) = [1]
      pi(filter_1) = [1]
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(-_0) = []
      pi(-_1) = [1]
      pi(-_2) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = [2]
      pi(map_0) = []
      pi(D_1) = []
      pi(t_0) = []
      pi(|1|_0) = []
      pi(constant_0) = []
      pi(|0|_0) = []
      pi(map_2) = [1, 2]
      pi(nil_0) = []
      pi(filter_2) = [1]

The following rules are removed:

  p1, p2, p4, p6, p7



-- 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(*_0(),y) -> *_1(y)
r2: app(*_1(x1),y) -> *_2(x1,y)
r3: app(+_0(),y) -> +_1(y)
r4: app(+_1(x1),y) -> +_2(x1,y)
r5: app(-_0(),y) -> -_1(y)
r6: app(-_1(x1),y) -> -_2(x1,y)
r7: app(D_0(),y) -> D_1(y)
r8: app(cons_0(),y) -> cons_1(y)
r9: app(cons_1(x1),y) -> cons_2(x1,y)
r10: app(filter_0(),y) -> filter_1(y)
r11: app(filter_1(x1),y) -> filter_2(x1,y)
r12: app(filter2_0(),y) -> filter2_1(y)
r13: app(filter2_1(x1),y) -> filter2_2(x1,y)
r14: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r15: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r16: app(map_0(),y) -> map_1(y)
r17: app(map_1(x1),y) -> map_2(x1,y)
r18: D_1(t_0()) -> |1|_0()
r19: D_1(constant_0()) -> |0|_0()
r20: D_1(+_2(x,y)) -> +_2(D_1(x),D_1(y))
r21: D_1(*_2(x,y)) -> +_2(*_2(y,D_1(x)),*_2(x,D_1(y)))
r22: D_1(-_2(x,y)) -> -_2(D_1(x),D_1(y))
r23: map_2(f,nil_0()) -> nil_0()
r24: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r25: filter_2(f,nil_0()) -> nil_0()
r26: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r27: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r28: 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(*_0(),y) -> *_1(y)
r2: app(*_1(x1),y) -> *_2(x1,y)
r3: app(+_0(),y) -> +_1(y)
r4: app(+_1(x1),y) -> +_2(x1,y)
r5: app(-_0(),y) -> -_1(y)
r6: app(-_1(x1),y) -> -_2(x1,y)
r7: app(D_0(),y) -> D_1(y)
r8: app(cons_0(),y) -> cons_1(y)
r9: app(cons_1(x1),y) -> cons_2(x1,y)
r10: app(filter_0(),y) -> filter_1(y)
r11: app(filter_1(x1),y) -> filter_2(x1,y)
r12: app(filter2_0(),y) -> filter2_1(y)
r13: app(filter2_1(x1),y) -> filter2_2(x1,y)
r14: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r15: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r16: app(map_0(),y) -> map_1(y)
r17: app(map_1(x1),y) -> map_2(x1,y)
r18: D_1(t_0()) -> |1|_0()
r19: D_1(constant_0()) -> |0|_0()
r20: D_1(+_2(x,y)) -> +_2(D_1(x),D_1(y))
r21: D_1(*_2(x,y)) -> +_2(*_2(y,D_1(x)),*_2(x,D_1(y)))
r22: D_1(-_2(x,y)) -> -_2(D_1(x),D_1(y))
r23: map_2(f,nil_0()) -> nil_0()
r24: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r25: filter_2(f,nil_0()) -> nil_0()
r26: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r27: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r28: 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{3, x1 - 9, x2 - 4}
            map_2#!_A(x1,x2) = max{0, x2 - 5}
            cons_2!_A(x1,x2) = max{9, x1 - 4, x2 - 4}
        
        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{60, x1 - 92, x2 - 72}
            map_2#!_A(x1,x2) = max{6, x1 - 1, x2 - 61}
            cons_2!_A(x1,x2) = max{5, x1 - 152, x2 - 70}
        
  
    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: 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(*_0(),y) -> *_1(y)
r2: app(*_1(x1),y) -> *_2(x1,y)
r3: app(+_0(),y) -> +_1(y)
r4: app(+_1(x1),y) -> +_2(x1,y)
r5: app(-_0(),y) -> -_1(y)
r6: app(-_1(x1),y) -> -_2(x1,y)
r7: app(D_0(),y) -> D_1(y)
r8: app(cons_0(),y) -> cons_1(y)
r9: app(cons_1(x1),y) -> cons_2(x1,y)
r10: app(filter_0(),y) -> filter_1(y)
r11: app(filter_1(x1),y) -> filter_2(x1,y)
r12: app(filter2_0(),y) -> filter2_1(y)
r13: app(filter2_1(x1),y) -> filter2_2(x1,y)
r14: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r15: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r16: app(map_0(),y) -> map_1(y)
r17: app(map_1(x1),y) -> map_2(x1,y)
r18: D_1(t_0()) -> |1|_0()
r19: D_1(constant_0()) -> |0|_0()
r20: D_1(+_2(x,y)) -> +_2(D_1(x),D_1(y))
r21: D_1(*_2(x,y)) -> +_2(*_2(y,D_1(x)),*_2(x,D_1(y)))
r22: D_1(-_2(x,y)) -> -_2(D_1(x),D_1(y))
r23: map_2(f,nil_0()) -> nil_0()
r24: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r25: filter_2(f,nil_0()) -> nil_0()
r26: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r27: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r28: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r3, r5, r10, r16, r18

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:
          
            filter2_4#_A(x1,x2,x3,x4) = max{2, x2 - 79, x3 - 135, x4 - 67}
            false_0_A = 67
            filter_2#_A(x1,x2) = max{2, x1 - 79, x2 - 67}
            cons_2_A(x1,x2) = max{68, x1 - 68, x2}
            app_A(x1,x2) = max{67, x1 - 12, x2 + 28}
            true_0_A = 67
            +_0_A = 311
            +_1_A(x1) = x1 + 11
            -_0_A = 11
            -_1_A(x1) = 67
            filter_0_A = 2
            filter_1_A(x1) = max{29, x1 - 14}
            map_0_A = 1
            map_1_A(x1) = 1
            D_1_A(x1) = max{74, x1 - 105}
            t_0_A = 0
            |1|_0_A = 0
            filter2_4#!_A(x1,x2,x3,x4) = 0
            false_0!_A = 140
            filter_2#!_A(x1,x2) = 0
            cons_2!_A(x1,x2) = max{278, x2 + 42}
            app!_A(x1,x2) = max{x1, x2 + 311}
            true_0!_A = 0
            +_0!_A = 309
            +_1!_A(x1) = 310
            -_0!_A = 2
            -_1!_A(x1) = max{1, x1 - 69}
            filter_0!_A = 2
            filter_1!_A(x1) = max{1, x1 - 16}
            map_0!_A = 3
            map_1!_A(x1) = max{2, x1 - 1}
            D_1!_A(x1) = max{114, x1 - 69}
            t_0!_A = 1
            |1|_0!_A = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            filter2_4#_A(x1,x2,x3,x4) = 1
            false_0_A = 25
            filter_2#_A(x1,x2) = 1
            cons_2_A(x1,x2) = max{47, x1 - 24, x2}
            app_A(x1,x2) = max{23, x1 - 21, x2 + 22}
            true_0_A = 0
            +_0_A = 3
            +_1_A(x1) = max{23, x1 + 20}
            -_0_A = 3
            -_1_A(x1) = max{3, x1 - 69}
            filter_0_A = 3
            filter_1_A(x1) = max{3, x1 - 70}
            map_0_A = 3
            map_1_A(x1) = x1 + 22
            D_1_A(x1) = max{26, x1 - 27}
            t_0_A = 0
            |1|_0_A = 0
            filter2_4#!_A(x1,x2,x3,x4) = 0
            false_0!_A = 21
            filter_2#!_A(x1,x2) = 0
            cons_2!_A(x1,x2) = max{0, x2 - 73}
            app!_A(x1,x2) = max{21, x1 - 4}
            true_0!_A = 0
            +_0!_A = 21
            +_1!_A(x1) = max{22, x1 - 1}
            -_0!_A = 21
            -_1!_A(x1) = max{22, x1 - 75}
            filter_0!_A = 20
            filter_1!_A(x1) = max{19, x1 - 74}
            map_0!_A = 1
            map_1!_A(x1) = max{22, x1 - 1}
            D_1!_A(x1) = max{13, x1 - 73}
            t_0!_A = 1
            |1|_0!_A = 0
        
  
    partial status:
  
      pi(filter2_4#) = [4]
      pi(false_0) = []
      pi(filter_2#) = [2]
      pi(cons_2) = [2]
      pi(app) = [2]
      pi(true_0) = []
      pi(+_0) = []
      pi(+_1) = []
      pi(-_0) = []
      pi(-_1) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(D_1) = []
      pi(t_0) = []
      pi(|1|_0) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

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

and R consists of:

r1: app(*_0(),y) -> *_1(y)
r2: app(*_1(x1),y) -> *_2(x1,y)
r3: app(+_0(),y) -> +_1(y)
r4: app(+_1(x1),y) -> +_2(x1,y)
r5: app(-_0(),y) -> -_1(y)
r6: app(-_1(x1),y) -> -_2(x1,y)
r7: app(D_0(),y) -> D_1(y)
r8: app(cons_0(),y) -> cons_1(y)
r9: app(cons_1(x1),y) -> cons_2(x1,y)
r10: app(filter_0(),y) -> filter_1(y)
r11: app(filter_1(x1),y) -> filter_2(x1,y)
r12: app(filter2_0(),y) -> filter2_1(y)
r13: app(filter2_1(x1),y) -> filter2_2(x1,y)
r14: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r15: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r16: app(map_0(),y) -> map_1(y)
r17: app(map_1(x1),y) -> map_2(x1,y)
r18: D_1(t_0()) -> |1|_0()
r19: D_1(constant_0()) -> |0|_0()
r20: D_1(+_2(x,y)) -> +_2(D_1(x),D_1(y))
r21: D_1(*_2(x,y)) -> +_2(*_2(y,D_1(x)),*_2(x,D_1(y)))
r22: D_1(-_2(x,y)) -> -_2(D_1(x),D_1(y))
r23: map_2(f,nil_0()) -> nil_0()
r24: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r25: filter_2(f,nil_0()) -> nil_0()
r26: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r27: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r28: 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: D_1#(+_2(x,y)) -> D_1#(x)
p2: D_1#(-_2(x,y)) -> D_1#(y)
p3: D_1#(-_2(x,y)) -> D_1#(x)
p4: D_1#(*_2(x,y)) -> D_1#(y)
p5: D_1#(*_2(x,y)) -> D_1#(x)
p6: D_1#(+_2(x,y)) -> D_1#(y)

and R consists of:

r1: app(*_0(),y) -> *_1(y)
r2: app(*_1(x1),y) -> *_2(x1,y)
r3: app(+_0(),y) -> +_1(y)
r4: app(+_1(x1),y) -> +_2(x1,y)
r5: app(-_0(),y) -> -_1(y)
r6: app(-_1(x1),y) -> -_2(x1,y)
r7: app(D_0(),y) -> D_1(y)
r8: app(cons_0(),y) -> cons_1(y)
r9: app(cons_1(x1),y) -> cons_2(x1,y)
r10: app(filter_0(),y) -> filter_1(y)
r11: app(filter_1(x1),y) -> filter_2(x1,y)
r12: app(filter2_0(),y) -> filter2_1(y)
r13: app(filter2_1(x1),y) -> filter2_2(x1,y)
r14: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r15: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r16: app(map_0(),y) -> map_1(y)
r17: app(map_1(x1),y) -> map_2(x1,y)
r18: D_1(t_0()) -> |1|_0()
r19: D_1(constant_0()) -> |0|_0()
r20: D_1(+_2(x,y)) -> +_2(D_1(x),D_1(y))
r21: D_1(*_2(x,y)) -> +_2(*_2(y,D_1(x)),*_2(x,D_1(y)))
r22: D_1(-_2(x,y)) -> -_2(D_1(x),D_1(y))
r23: map_2(f,nil_0()) -> nil_0()
r24: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r25: filter_2(f,nil_0()) -> nil_0()
r26: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r27: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r28: 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:
          
            D_1#_A(x1) = 0
            +_2_A(x1,x2) = max{7, x1 - 7, x2 - 7}
            -_2_A(x1,x2) = max{7, x1 - 7, x2 - 7}
            *_2_A(x1,x2) = max{3, x1 + 1, x2}
            D_1#!_A(x1) = max{1, x1 - 25}
            +_2!_A(x1,x2) = max{17, x1 - 16, x2 - 24}
            -_2!_A(x1,x2) = max{18, x1 - 25, x2 - 20}
            *_2!_A(x1,x2) = max{0, x2 - 9}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            D_1#_A(x1) = 0
            +_2_A(x1,x2) = max{4, x1 - 6, x2 - 6}
            -_2_A(x1,x2) = max{7, x1 - 3, x2 - 3}
            *_2_A(x1,x2) = max{x1 + 5, x2 + 1}
            D_1#!_A(x1) = max{3, x1 - 35}
            +_2!_A(x1,x2) = max{4, x1 - 40, x2 - 27}
            -_2!_A(x1,x2) = max{7, x1 - 35, x2 - 32}
            *_2!_A(x1,x2) = max{2, x1 - 49, x2 - 36}
        
  
    partial status:
  
      pi(D_1#) = [1]
      pi(+_2) = [1, 2]
      pi(-_2) = [1, 2]
      pi(*_2) = [1, 2]

The following rules are removed:

  p1, p2, p3, p4, p5, p6

Then no dependency pair remains.