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(D(),app(minus(),x)) -> app(minus(),app(D(),x))
  app(D(),app(app(div(),x),y)) -> app(app(-(),app(app(div(),app(D(),x)),y)),app(app(div(),app(app(*(),x),app(D(),y))),app(app(pow(),y),|2|())))
  app(D(),app(ln(),x)) -> app(app(div(),app(D(),x)),x)
  app(D(),app(app(pow(),x),y)) -> app(app(+(),app(app(*(),app(app(*(),y),app(app(pow(),x),app(app(-(),y),|1|())))),app(D(),x))),app(app(*(),app(app(*(),app(app(pow(),x),y)),app(ln(),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(div_0(),y) -> div_1(y)
  app(div_1(x1),y) -> div_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(ln_0(),y) -> ln_1(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(pow_0(),y) -> pow_1(y)
  app(pow_1(x1),y) -> pow_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))
  D_1(minus_1(x)) -> minus_1(D_1(x))
  D_1(div_2(x,y)) -> -_2(div_2(D_1(x),y),div_2(*_2(x,D_1(y)),pow_2(y,|2|_0())))
  D_1(ln_1(x)) -> div_2(D_1(x),x)
  D_1(pow_2(x,y)) -> +_2(*_2(*_2(y,pow_2(x,-_2(y,|1|_0()))),D_1(x)),*_2(*_2(pow_2(x,y),ln_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: D_1#(minus_1(x)) -> D_1#(x)
p12: D_1#(div_2(x,y)) -> D_1#(x)
p13: D_1#(div_2(x,y)) -> D_1#(y)
p14: D_1#(ln_1(x)) -> D_1#(x)
p15: D_1#(pow_2(x,y)) -> D_1#(x)
p16: D_1#(pow_2(x,y)) -> D_1#(y)
p17: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p18: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p19: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p20: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p21: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p22: 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(div_0(),y) -> div_1(y)
r11: app(div_1(x1),y) -> div_2(x1,y)
r12: app(filter_0(),y) -> filter_1(y)
r13: app(filter_1(x1),y) -> filter_2(x1,y)
r14: app(filter2_0(),y) -> filter2_1(y)
r15: app(filter2_1(x1),y) -> filter2_2(x1,y)
r16: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r17: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r18: app(ln_0(),y) -> ln_1(y)
r19: app(map_0(),y) -> map_1(y)
r20: app(map_1(x1),y) -> map_2(x1,y)
r21: app(minus_0(),y) -> minus_1(y)
r22: app(pow_0(),y) -> pow_1(y)
r23: app(pow_1(x1),y) -> pow_2(x1,y)
r24: D_1(t_0()) -> |1|_0()
r25: D_1(constant_0()) -> |0|_0()
r26: D_1(+_2(x,y)) -> +_2(D_1(x),D_1(y))
r27: D_1(*_2(x,y)) -> +_2(*_2(y,D_1(x)),*_2(x,D_1(y)))
r28: D_1(-_2(x,y)) -> -_2(D_1(x),D_1(y))
r29: D_1(minus_1(x)) -> minus_1(D_1(x))
r30: D_1(div_2(x,y)) -> -_2(div_2(D_1(x),y),div_2(*_2(x,D_1(y)),pow_2(y,|2|_0())))
r31: D_1(ln_1(x)) -> div_2(D_1(x),x)
r32: D_1(pow_2(x,y)) -> +_2(*_2(*_2(y,pow_2(x,-_2(y,|1|_0()))),D_1(x)),*_2(*_2(pow_2(x,y),ln_1(x)),D_1(y)))
r33: map_2(f,nil_0()) -> nil_0()
r34: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r35: filter_2(f,nil_0()) -> nil_0()
r36: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r37: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r38: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p17, p18, p19, p20, p21, p22}
  {p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16}


-- 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(div_0(),y) -> div_1(y)
r11: app(div_1(x1),y) -> div_2(x1,y)
r12: app(filter_0(),y) -> filter_1(y)
r13: app(filter_1(x1),y) -> filter_2(x1,y)
r14: app(filter2_0(),y) -> filter2_1(y)
r15: app(filter2_1(x1),y) -> filter2_2(x1,y)
r16: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r17: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r18: app(ln_0(),y) -> ln_1(y)
r19: app(map_0(),y) -> map_1(y)
r20: app(map_1(x1),y) -> map_2(x1,y)
r21: app(minus_0(),y) -> minus_1(y)
r22: app(pow_0(),y) -> pow_1(y)
r23: app(pow_1(x1),y) -> pow_2(x1,y)
r24: D_1(t_0()) -> |1|_0()
r25: D_1(constant_0()) -> |0|_0()
r26: D_1(+_2(x,y)) -> +_2(D_1(x),D_1(y))
r27: D_1(*_2(x,y)) -> +_2(*_2(y,D_1(x)),*_2(x,D_1(y)))
r28: D_1(-_2(x,y)) -> -_2(D_1(x),D_1(y))
r29: D_1(minus_1(x)) -> minus_1(D_1(x))
r30: D_1(div_2(x,y)) -> -_2(div_2(D_1(x),y),div_2(*_2(x,D_1(y)),pow_2(y,|2|_0())))
r31: D_1(ln_1(x)) -> div_2(D_1(x),x)
r32: D_1(pow_2(x,y)) -> +_2(*_2(*_2(y,pow_2(x,-_2(y,|1|_0()))),D_1(x)),*_2(*_2(pow_2(x,y),ln_1(x)),D_1(y)))
r33: map_2(f,nil_0()) -> nil_0()
r34: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r35: filter_2(f,nil_0()) -> nil_0()
r36: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r37: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r38: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r6, r22, r25, r33, r35

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 - 2, x2 - 2}
            cons_2_A(x1,x2) = max{1, x1, x2}
            app#_A(x1,x2) = max{1, x1 - 126, x2 - 2}
            map_1_A(x1) = x1 + 124
            filter2_3_A(x1,x2,x3) = max{x1 - 15, x2 + 3, x3 + 124}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 123, x3 - 2, x4 - 2}
            false_0_A = 62
            filter_2#_A(x1,x2) = max{1, x1 - 123, x2 - 2}
            filter_1_A(x1) = max{125, x1 + 124}
            app_A(x1,x2) = max{x1 - 7, x2 + 58}
            true_0_A = 128
            -_1_A(x1) = x1 + 40
            -_2_A(x1,x2) = max{x1 + 33, x2 + 35}
            pow_0_A = 8
            pow_1_A(x1) = x1 + 8
            D_1_A(x1) = x1 + 46
            constant_0_A = 33
            |0|_0_A = 79
            map_2_A(x1,x2) = max{60, x1 - 1, x2 + 58}
            nil_0_A = 0
            filter_2_A(x1,x2) = x2
            map_2#!_A(x1,x2) = 0
            cons_2!_A(x1,x2) = max{127, x1 + 6, x2 + 59}
            app#!_A(x1,x2) = x2 + 5
            map_1!_A(x1) = 0
            filter2_3!_A(x1,x2,x3) = max{1, x1 - 11, x2 - 122, x3 - 1}
            filter2_4#!_A(x1,x2,x3,x4) = 2
            false_0!_A = 1
            filter_2#!_A(x1,x2) = 2
            filter_1!_A(x1) = max{3, x1 - 1}
            app!_A(x1,x2) = max{x1 - 2, x2 + 125}
            true_0!_A = 128
            -_1!_A(x1) = 4
            -_2!_A(x1,x2) = 3
            pow_0!_A = 122
            pow_1!_A(x1) = x1 + 124
            D_1!_A(x1) = max{0, x1 - 31}
            constant_0!_A = 2
            |0|_0!_A = 1
            map_2!_A(x1,x2) = max{x1 + 121, x2 - 2}
            nil_0!_A = 121
            filter_2!_A(x1,x2) = max{62, x1, x2 + 59}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            map_2#_A(x1,x2) = 1
            cons_2_A(x1,x2) = max{49, x1 + 9}
            app#_A(x1,x2) = 1
            map_1_A(x1) = max{16, x1}
            filter2_3_A(x1,x2,x3) = 1
            filter2_4#_A(x1,x2,x3,x4) = 1
            false_0_A = 39
            filter_2#_A(x1,x2) = 1
            filter_1_A(x1) = x1 + 9
            app_A(x1,x2) = 37
            true_0_A = 39
            -_1_A(x1) = max{10, x1 + 9}
            -_2_A(x1,x2) = 3
            pow_0_A = 57
            pow_1_A(x1) = 1
            D_1_A(x1) = 7
            constant_0_A = 4
            |0|_0_A = 1
            map_2_A(x1,x2) = x2
            nil_0_A = 1
            filter_2_A(x1,x2) = x2
            map_2#!_A(x1,x2) = 6
            cons_2!_A(x1,x2) = 0
            app#!_A(x1,x2) = max{0, x1 - 6, x2 - 1}
            map_1!_A(x1) = max{5, x1 + 1}
            filter2_3!_A(x1,x2,x3) = max{50, x1 - 38, x2 - 1, x3 - 1}
            filter2_4#!_A(x1,x2,x3,x4) = max{x3 + 4, x4 + 3}
            false_0!_A = 3
            filter_2#!_A(x1,x2) = max{2, x2 - 4}
            filter_1!_A(x1) = 47
            app!_A(x1,x2) = max{47, x1 - 9, x2 - 40}
            true_0!_A = 3
            -_1!_A(x1) = max{48, x1 - 41}
            -_2!_A(x1,x2) = x1 + 50
            pow_0!_A = 1
            pow_1!_A(x1) = max{48, x1 - 40}
            D_1!_A(x1) = x1 + 62
            constant_0!_A = 1
            |0|_0!_A = 0
            map_2!_A(x1,x2) = max{x1 + 46, x2 - 2}
            nil_0!_A = 45
            filter_2!_A(x1,x2) = 0
        
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [1, 2]
      pi(app#) = [2]
      pi(map_1) = [1]
      pi(filter2_3) = [3]
      pi(filter2_4#) = [4]
      pi(false_0) = []
      pi(filter_2#) = [2]
      pi(filter_1) = []
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(-_1) = [1]
      pi(-_2) = []
      pi(pow_0) = []
      pi(pow_1) = [1]
      pi(D_1) = []
      pi(constant_0) = []
      pi(|0|_0) = []
      pi(map_2) = [1, 2]
      pi(nil_0) = []
      pi(filter_2) = [2]

The following rules are removed:

  p1, p2, p3, p4, p5, p6, p7, p8, p9

Then no dependency pair remains.

-- 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#(pow_2(x,y)) -> D_1#(y)
p3: D_1#(pow_2(x,y)) -> D_1#(x)
p4: D_1#(ln_1(x)) -> D_1#(x)
p5: D_1#(div_2(x,y)) -> D_1#(y)
p6: D_1#(div_2(x,y)) -> D_1#(x)
p7: D_1#(minus_1(x)) -> 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: D_1#(*_2(x,y)) -> D_1#(x)
p12: 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(div_0(),y) -> div_1(y)
r11: app(div_1(x1),y) -> div_2(x1,y)
r12: app(filter_0(),y) -> filter_1(y)
r13: app(filter_1(x1),y) -> filter_2(x1,y)
r14: app(filter2_0(),y) -> filter2_1(y)
r15: app(filter2_1(x1),y) -> filter2_2(x1,y)
r16: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r17: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r18: app(ln_0(),y) -> ln_1(y)
r19: app(map_0(),y) -> map_1(y)
r20: app(map_1(x1),y) -> map_2(x1,y)
r21: app(minus_0(),y) -> minus_1(y)
r22: app(pow_0(),y) -> pow_1(y)
r23: app(pow_1(x1),y) -> pow_2(x1,y)
r24: D_1(t_0()) -> |1|_0()
r25: D_1(constant_0()) -> |0|_0()
r26: D_1(+_2(x,y)) -> +_2(D_1(x),D_1(y))
r27: D_1(*_2(x,y)) -> +_2(*_2(y,D_1(x)),*_2(x,D_1(y)))
r28: D_1(-_2(x,y)) -> -_2(D_1(x),D_1(y))
r29: D_1(minus_1(x)) -> minus_1(D_1(x))
r30: D_1(div_2(x,y)) -> -_2(div_2(D_1(x),y),div_2(*_2(x,D_1(y)),pow_2(y,|2|_0())))
r31: D_1(ln_1(x)) -> div_2(D_1(x),x)
r32: D_1(pow_2(x,y)) -> +_2(*_2(*_2(y,pow_2(x,-_2(y,|1|_0()))),D_1(x)),*_2(*_2(pow_2(x,y),ln_1(x)),D_1(y)))
r33: map_2(f,nil_0()) -> nil_0()
r34: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r35: filter_2(f,nil_0()) -> nil_0()
r36: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r37: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r38: 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) = max{0, x1 - 44}
            +_2_A(x1,x2) = max{23, x1 + 22, x2 + 22}
            pow_2_A(x1,x2) = max{x1 + 43, x2 + 22}
            ln_1_A(x1) = max{23, x1 + 22}
            div_2_A(x1,x2) = max{x1 + 23, x2 + 27}
            minus_1_A(x1) = x1 + 22
            -_2_A(x1,x2) = max{42, x1 + 22, x2 + 22}
            *_2_A(x1,x2) = max{26, x1 + 22, x2 + 22}
            D_1#!_A(x1) = 0
            +_2!_A(x1,x2) = max{15, x1 - 242, x2 - 239}
            pow_2!_A(x1,x2) = max{11, x2 - 93}
            ln_1!_A(x1) = 14
            div_2!_A(x1,x2) = max{9, x1 - 211, x2 - 186}
            minus_1!_A(x1) = max{8, x1 - 29}
            -_2!_A(x1,x2) = max{10, x1 - 212, x2 - 163}
            *_2!_A(x1,x2) = max{13, x1 - 212, x2 - 117}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            D_1#_A(x1) = max{0, x1 - 120}
            +_2_A(x1,x2) = max{x1 + 119, x2 + 44}
            pow_2_A(x1,x2) = max{x1 + 35, x2 + 1}
            ln_1_A(x1) = max{62, x1 + 1}
            div_2_A(x1,x2) = max{x1 + 2, x2 + 8}
            minus_1_A(x1) = x1 + 46
            -_2_A(x1,x2) = max{x1 + 60, x2 + 76}
            *_2_A(x1,x2) = max{x1 + 41, x2 + 10}
            D_1#!_A(x1) = max{8, x1 - 110}
            +_2!_A(x1,x2) = max{12, x1 - 352, x2 - 529}
            pow_2!_A(x1,x2) = max{7, x1 - 473, x2 - 240}
            ln_1!_A(x1) = max{5, x1 - 445}
            div_2!_A(x1,x2) = max{0, x1 - 314, x2 - 435}
            minus_1!_A(x1) = 0
            -_2!_A(x1,x2) = max{0, x2 - 44}
            *_2!_A(x1,x2) = max{9, x1 - 313, x2 - 109}
        
  
    partial status:
  
      pi(D_1#) = []
      pi(+_2) = []
      pi(pow_2) = [1, 2]
      pi(ln_1) = [1]
      pi(div_2) = []
      pi(minus_1) = []
      pi(-_2) = [1, 2]
      pi(*_2) = [2]

The following rules are removed:

  p1, p12



-- SCC decomposition.

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

p1: D_1#(pow_2(x,y)) -> D_1#(y)
p2: D_1#(pow_2(x,y)) -> D_1#(x)
p3: D_1#(ln_1(x)) -> D_1#(x)
p4: D_1#(div_2(x,y)) -> D_1#(y)
p5: D_1#(div_2(x,y)) -> D_1#(x)
p6: D_1#(minus_1(x)) -> D_1#(x)
p7: D_1#(-_2(x,y)) -> D_1#(y)
p8: D_1#(-_2(x,y)) -> D_1#(x)
p9: D_1#(*_2(x,y)) -> D_1#(y)
p10: D_1#(*_2(x,y)) -> D_1#(x)

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(div_0(),y) -> div_1(y)
r11: app(div_1(x1),y) -> div_2(x1,y)
r12: app(filter_0(),y) -> filter_1(y)
r13: app(filter_1(x1),y) -> filter_2(x1,y)
r14: app(filter2_0(),y) -> filter2_1(y)
r15: app(filter2_1(x1),y) -> filter2_2(x1,y)
r16: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r17: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r18: app(ln_0(),y) -> ln_1(y)
r19: app(map_0(),y) -> map_1(y)
r20: app(map_1(x1),y) -> map_2(x1,y)
r21: app(minus_0(),y) -> minus_1(y)
r22: app(pow_0(),y) -> pow_1(y)
r23: app(pow_1(x1),y) -> pow_2(x1,y)
r24: D_1(t_0()) -> |1|_0()
r25: D_1(constant_0()) -> |0|_0()
r26: D_1(+_2(x,y)) -> +_2(D_1(x),D_1(y))
r27: D_1(*_2(x,y)) -> +_2(*_2(y,D_1(x)),*_2(x,D_1(y)))
r28: D_1(-_2(x,y)) -> -_2(D_1(x),D_1(y))
r29: D_1(minus_1(x)) -> minus_1(D_1(x))
r30: D_1(div_2(x,y)) -> -_2(div_2(D_1(x),y),div_2(*_2(x,D_1(y)),pow_2(y,|2|_0())))
r31: D_1(ln_1(x)) -> div_2(D_1(x),x)
r32: D_1(pow_2(x,y)) -> +_2(*_2(*_2(y,pow_2(x,-_2(y,|1|_0()))),D_1(x)),*_2(*_2(pow_2(x,y),ln_1(x)),D_1(y)))
r33: map_2(f,nil_0()) -> nil_0()
r34: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r35: filter_2(f,nil_0()) -> nil_0()
r36: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r37: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r38: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10}


-- Reduction pair.

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

p1: D_1#(pow_2(x,y)) -> D_1#(y)
p2: D_1#(*_2(x,y)) -> D_1#(x)
p3: D_1#(*_2(x,y)) -> D_1#(y)
p4: D_1#(-_2(x,y)) -> D_1#(x)
p5: D_1#(-_2(x,y)) -> D_1#(y)
p6: D_1#(minus_1(x)) -> D_1#(x)
p7: D_1#(div_2(x,y)) -> D_1#(x)
p8: D_1#(div_2(x,y)) -> D_1#(y)
p9: D_1#(ln_1(x)) -> D_1#(x)
p10: D_1#(pow_2(x,y)) -> D_1#(x)

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(div_0(),y) -> div_1(y)
r11: app(div_1(x1),y) -> div_2(x1,y)
r12: app(filter_0(),y) -> filter_1(y)
r13: app(filter_1(x1),y) -> filter_2(x1,y)
r14: app(filter2_0(),y) -> filter2_1(y)
r15: app(filter2_1(x1),y) -> filter2_2(x1,y)
r16: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r17: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r18: app(ln_0(),y) -> ln_1(y)
r19: app(map_0(),y) -> map_1(y)
r20: app(map_1(x1),y) -> map_2(x1,y)
r21: app(minus_0(),y) -> minus_1(y)
r22: app(pow_0(),y) -> pow_1(y)
r23: app(pow_1(x1),y) -> pow_2(x1,y)
r24: D_1(t_0()) -> |1|_0()
r25: D_1(constant_0()) -> |0|_0()
r26: D_1(+_2(x,y)) -> +_2(D_1(x),D_1(y))
r27: D_1(*_2(x,y)) -> +_2(*_2(y,D_1(x)),*_2(x,D_1(y)))
r28: D_1(-_2(x,y)) -> -_2(D_1(x),D_1(y))
r29: D_1(minus_1(x)) -> minus_1(D_1(x))
r30: D_1(div_2(x,y)) -> -_2(div_2(D_1(x),y),div_2(*_2(x,D_1(y)),pow_2(y,|2|_0())))
r31: D_1(ln_1(x)) -> div_2(D_1(x),x)
r32: D_1(pow_2(x,y)) -> +_2(*_2(*_2(y,pow_2(x,-_2(y,|1|_0()))),D_1(x)),*_2(*_2(pow_2(x,y),ln_1(x)),D_1(y)))
r33: map_2(f,nil_0()) -> nil_0()
r34: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r35: filter_2(f,nil_0()) -> nil_0()
r36: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r37: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r38: 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) = max{0, x1 - 162}
            pow_2_A(x1,x2) = max{81, x1 + 68, x2 + 36}
            *_2_A(x1,x2) = max{83, x1 + 1, x2 + 18}
            -_2_A(x1,x2) = max{19, x1 + 17, x2 + 13}
            minus_1_A(x1) = x1 + 161
            div_2_A(x1,x2) = max{x1 + 91, x2 + 18}
            ln_1_A(x1) = x1 + 57
            D_1#!_A(x1) = max{6, x1 - 769}
            pow_2!_A(x1,x2) = max{4, x2 - 764}
            *_2!_A(x1,x2) = max{5, x1 - 799, x2 - 770}
            -_2!_A(x1,x2) = max{3, x1 - 765, x2 - 710}
            minus_1!_A(x1) = x1 + 3
            div_2!_A(x1,x2) = max{7, x1 - 390, x2 - 535}
            ln_1!_A(x1) = max{0, x1 - 783}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            D_1#_A(x1) = max{0, x1 - 35}
            pow_2_A(x1,x2) = max{46, x1 + 1, x2 + 12}
            *_2_A(x1,x2) = max{62, x1 + 10, x2 + 2}
            -_2_A(x1,x2) = max{x1 + 34, x2 + 1}
            minus_1_A(x1) = max{121, x1 + 1}
            div_2_A(x1,x2) = max{21, x1 + 11, x2 + 1}
            ln_1_A(x1) = max{21, x1 + 10}
            D_1#!_A(x1) = max{11, x1 - 22}
            pow_2!_A(x1,x2) = max{0, x2 - 554}
            *_2!_A(x1,x2) = max{8, x1 - 607, x2 - 280}
            -_2!_A(x1,x2) = max{10, x1 - 555, x2 - 290}
            minus_1!_A(x1) = max{1, x1 - 415}
            div_2!_A(x1,x2) = max{9, x1 - 279, x2 - 401}
            ln_1!_A(x1) = max{3, x1 - 402}
        
  
    partial status:
  
      pi(D_1#) = [1]
      pi(pow_2) = [1, 2]
      pi(*_2) = [1, 2]
      pi(-_2) = [1, 2]
      pi(minus_1) = [1]
      pi(div_2) = [1, 2]
      pi(ln_1) = [1]

The following rules are removed:

  p1, p2, p3, p4, p5, p6, p7, p8, p9, p10

Then no dependency pair remains.