YES
We show the termination of the TRS R:

  app(app(minus(),x),|0|()) -> x
  app(app(minus(),app(s(),x)),app(s(),y)) -> app(app(minus(),x),y)
  app(app(quot(),|0|()),app(s(),y)) -> |0|()
  app(app(quot(),app(s(),x)),app(s(),y)) -> app(s(),app(app(quot(),app(app(minus(),x),y)),app(s(),y)))
  app(log(),app(s(),|0|())) -> |0|()
  app(log(),app(s(),app(s(),x))) -> app(s(),app(log(),app(s(),app(app(quot(),x),app(s(),app(s(),|0|()))))))
  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(log_0(),y) -> log_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(minus_1(x1),y) -> minus_2(x1,y)
  app(quot_0(),y) -> quot_1(y)
  app(quot_1(x1),y) -> quot_2(x1,y)
  app(s_0(),y) -> s_1(y)
  minus_2(x,|0|_0()) -> x
  minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
  quot_2(|0|_0(),s_1(y)) -> |0|_0()
  quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
  log_1(s_1(|0|_0())) -> |0|_0()
  log_1(s_1(s_1(x))) -> s_1(log_1(s_1(quot_2(x,s_1(s_1(|0|_0()))))))
  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#(log_0(),y) -> log_1#(y)
p4: app#(map_1(x1),y) -> map_2#(x1,y)
p5: app#(minus_1(x1),y) -> minus_2#(x1,y)
p6: app#(quot_1(x1),y) -> quot_2#(x1,y)
p7: minus_2#(s_1(x),s_1(y)) -> minus_2#(x,y)
p8: quot_2#(s_1(x),s_1(y)) -> quot_2#(minus_2(x,y),s_1(y))
p9: quot_2#(s_1(x),s_1(y)) -> minus_2#(x,y)
p10: log_1#(s_1(s_1(x))) -> log_1#(s_1(quot_2(x,s_1(s_1(|0|_0())))))
p11: log_1#(s_1(s_1(x))) -> quot_2#(x,s_1(s_1(|0|_0())))
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(log_0(),y) -> log_1(y)
r10: app(map_0(),y) -> map_1(y)
r11: app(map_1(x1),y) -> map_2(x1,y)
r12: app(minus_0(),y) -> minus_1(y)
r13: app(minus_1(x1),y) -> minus_2(x1,y)
r14: app(quot_0(),y) -> quot_1(y)
r15: app(quot_1(x1),y) -> quot_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: minus_2(x,|0|_0()) -> x
r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r19: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r20: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r21: log_1(s_1(|0|_0())) -> |0|_0()
r22: log_1(s_1(s_1(x))) -> s_1(log_1(s_1(quot_2(x,s_1(s_1(|0|_0()))))))
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, p4, p12, p13, p14, p15, p16, p17}
  {p10}
  {p8}
  {p7}


-- 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(log_0(),y) -> log_1(y)
r10: app(map_0(),y) -> map_1(y)
r11: app(map_1(x1),y) -> map_2(x1,y)
r12: app(minus_0(),y) -> minus_1(y)
r13: app(minus_1(x1),y) -> minus_2(x1,y)
r14: app(quot_0(),y) -> quot_1(y)
r15: app(quot_1(x1),y) -> quot_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: minus_2(x,|0|_0()) -> x
r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r19: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r20: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r21: log_1(s_1(|0|_0())) -> |0|_0()
r22: log_1(s_1(s_1(x))) -> s_1(log_1(s_1(quot_2(x,s_1(s_1(|0|_0()))))))
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

  r13, r17, r18, r19, 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:
          
            app#_A(x1,x2) = max{1, x1 - 12, x2 - 1}
            filter_1_A(x1) = max{9, x1 + 1}
            filter_2#_A(x1,x2) = max{1, x1 - 11, x2 - 1}
            cons_2_A(x1,x2) = max{x1, x2}
            map_1_A(x1) = max{14, x1}
            map_2#_A(x1,x2) = max{2, x1 - 12, x2 - 1}
            filter2_3_A(x1,x2,x3) = max{x2 + 1, x3 + 11}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 11, x3 - 1, x4 - 1}
            false_0_A = 0
            app_A(x1,x2) = max{8, x1 - 10}
            true_0_A = 3
            minus_1_A(x1) = x1 + 22
            minus_2_A(x1,x2) = max{12, x1}
            |0|_0_A = 10
            s_1_A(x1) = max{9, x1}
            quot_2_A(x1,x2) = max{12, x1 - 11}
            filter_2_A(x1,x2) = max{3, x1 - 15, x2 + 1}
            nil_0_A = 8
            app#!_A(x1,x2) = 0
            filter_1!_A(x1) = 2
            filter_2#!_A(x1,x2) = 0
            cons_2!_A(x1,x2) = x1 + 2
            map_1!_A(x1) = x1 + 1
            map_2#!_A(x1,x2) = 0
            filter2_3!_A(x1,x2,x3) = max{x1 - 2, x2 - 14, x3 + 1}
            filter2_4#!_A(x1,x2,x3,x4) = 0
            false_0!_A = 0
            app!_A(x1,x2) = max{12, x2 - 12}
            true_0!_A = 0
            minus_1!_A(x1) = 11
            minus_2!_A(x1,x2) = 11
            |0|_0!_A = 9
            s_1!_A(x1) = max{0, x1 - 12}
            quot_2!_A(x1,x2) = max{8, x2 + 1}
            filter_2!_A(x1,x2) = x1 + 13
            nil_0!_A = 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{1, x2 - 33}
            filter_1_A(x1) = 66
            filter_2#_A(x1,x2) = max{1, x2 - 61}
            cons_2_A(x1,x2) = max{60, x1 + 28, x2 + 19}
            map_1_A(x1) = x1 + 24
            map_2#_A(x1,x2) = max{1, x2 - 61}
            filter2_3_A(x1,x2,x3) = max{198, x1 - 64}
            filter2_4#_A(x1,x2,x3,x4) = max{1, x4 - 42}
            false_0_A = 63
            app_A(x1,x2) = max{x1 + 65, x2 + 30}
            true_0_A = 63
            minus_1_A(x1) = max{200, x1 + 31}
            minus_2_A(x1,x2) = max{265, x1, x2}
            |0|_0_A = 67
            s_1_A(x1) = max{68, x1}
            quot_2_A(x1,x2) = max{132, x1, x2 - 69}
            filter_2_A(x1,x2) = max{132, x1, x2 + 30}
            nil_0_A = 28
            app#!_A(x1,x2) = max{7, x2 + 4}
            filter_1!_A(x1) = max{1, x1 - 265}
            filter_2#!_A(x1,x2) = max{2, x2 - 13}
            cons_2!_A(x1,x2) = 16
            map_1!_A(x1) = max{0, x1 - 1}
            map_2#!_A(x1,x2) = max{5, x2 - 20}
            filter2_3!_A(x1,x2,x3) = max{1, x1 - 1, x2 - 2, x3 - 267}
            filter2_4#!_A(x1,x2,x3,x4) = max{5, x4 + 3}
            false_0!_A = 15
            app!_A(x1,x2) = max{16, x2 - 266}
            true_0!_A = 3
            minus_1!_A(x1) = max{0, x1 - 1}
            minus_2!_A(x1,x2) = 0
            |0|_0!_A = 1
            s_1!_A(x1) = max{1, x1 - 266}
            quot_2!_A(x1,x2) = max{17, x2 - 268}
            filter_2!_A(x1,x2) = max{14, x2 - 1}
            nil_0!_A = 27
        
  
    partial status:
  
      pi(app#) = []
      pi(filter_1) = []
      pi(filter_2#) = []
      pi(cons_2) = [2]
      pi(map_1) = []
      pi(map_2#) = []
      pi(filter2_3) = []
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(app) = [1]
      pi(true_0) = []
      pi(minus_1) = [1]
      pi(minus_2) = [1]
      pi(|0|_0) = []
      pi(s_1) = [1]
      pi(quot_2) = [2]
      pi(filter_2) = []
      pi(nil_0) = []

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: log_1#(s_1(s_1(x))) -> log_1#(s_1(quot_2(x,s_1(s_1(|0|_0())))))

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(log_0(),y) -> log_1(y)
r10: app(map_0(),y) -> map_1(y)
r11: app(map_1(x1),y) -> map_2(x1,y)
r12: app(minus_0(),y) -> minus_1(y)
r13: app(minus_1(x1),y) -> minus_2(x1,y)
r14: app(quot_0(),y) -> quot_1(y)
r15: app(quot_1(x1),y) -> quot_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: minus_2(x,|0|_0()) -> x
r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r19: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r20: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r21: log_1(s_1(|0|_0())) -> |0|_0()
r22: log_1(s_1(s_1(x))) -> s_1(log_1(s_1(quot_2(x,s_1(s_1(|0|_0()))))))
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

  r12, r17, r18, r19, r20

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:
          
            log_1#_A(x1) = max{0, x1 - 72}
            s_1_A(x1) = max{19, x1}
            quot_2_A(x1,x2) = max{71, x1 - 137, x2 - 136}
            |0|_0_A = 47
            app_A(x1,x2) = max{18, x1 - 92, x2 - 137}
            minus_0_A = 0
            minus_1_A(x1) = 0
            minus_2_A(x1,x2) = max{19, x1, x2 - 122}
            log_1#!_A(x1) = 140
            s_1!_A(x1) = max{114, x1 + 68}
            quot_2!_A(x1,x2) = max{x1 + 121, x2 + 94}
            |0|_0!_A = 168
            app!_A(x1,x2) = x1 + 46
            minus_0!_A = 7
            minus_1!_A(x1) = 6
            minus_2!_A(x1,x2) = max{67, x1 - 20}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            log_1#_A(x1) = 0
            s_1_A(x1) = x1 + 25
            quot_2_A(x1,x2) = x1 + 24
            |0|_0_A = 18
            app_A(x1,x2) = max{1, x1 - 72, x2 - 72}
            minus_0_A = 0
            minus_1_A(x1) = max{1, x1 - 72}
            minus_2_A(x1,x2) = x1
            log_1#!_A(x1) = max{1, x1 - 47}
            s_1!_A(x1) = x1 + 24
            quot_2!_A(x1,x2) = 23
            |0|_0!_A = 22
            app!_A(x1,x2) = max{65, x1 - 67, x2 + 25}
            minus_0!_A = 1
            minus_1!_A(x1) = max{0, x1 - 26}
            minus_2!_A(x1,x2) = 22
        
  
    partial status:
  
      pi(log_1#) = []
      pi(s_1) = [1]
      pi(quot_2) = [1]
      pi(|0|_0) = []
      pi(app) = []
      pi(minus_0) = []
      pi(minus_1) = []
      pi(minus_2) = [1]

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: quot_2#(s_1(x),s_1(y)) -> quot_2#(minus_2(x,y),s_1(y))

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(log_0(),y) -> log_1(y)
r10: app(map_0(),y) -> map_1(y)
r11: app(map_1(x1),y) -> map_2(x1,y)
r12: app(minus_0(),y) -> minus_1(y)
r13: app(minus_1(x1),y) -> minus_2(x1,y)
r14: app(quot_0(),y) -> quot_1(y)
r15: app(quot_1(x1),y) -> quot_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: minus_2(x,|0|_0()) -> x
r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r19: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r20: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r21: log_1(s_1(|0|_0())) -> |0|_0()
r22: log_1(s_1(s_1(x))) -> s_1(log_1(s_1(quot_2(x,s_1(s_1(|0|_0()))))))
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

  r17, 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:
          
            quot_2#_A(x1,x2) = max{0, x1 - 29, x2 - 33}
            s_1_A(x1) = max{28, x1 + 19}
            minus_2_A(x1,x2) = max{25, x1 + 18, x2 - 20}
            |0|_0_A = 52
            quot_2#!_A(x1,x2) = max{1, x1 - 26, x2 - 17}
            s_1!_A(x1) = max{0, x1 - 121}
            minus_2!_A(x1,x2) = max{0, x1 - 121}
            |0|_0!_A = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            quot_2#_A(x1,x2) = max{0, x1 - 20, x2 - 38}
            s_1_A(x1) = x1 + 19
            minus_2_A(x1,x2) = max{x1 + 2, x2 - 20}
            |0|_0_A = 9
            quot_2#!_A(x1,x2) = max{3, x1 - 13, x2 - 15}
            s_1!_A(x1) = max{7, x1 - 63}
            minus_2!_A(x1,x2) = max{5, x1 - 64}
            |0|_0!_A = 0
        
  
    partial status:
  
      pi(quot_2#) = [1]
      pi(s_1) = [1]
      pi(minus_2) = [1]
      pi(|0|_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: minus_2#(s_1(x),s_1(y)) -> minus_2#(x,y)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(log_0(),y) -> log_1(y)
r10: app(map_0(),y) -> map_1(y)
r11: app(map_1(x1),y) -> map_2(x1,y)
r12: app(minus_0(),y) -> minus_1(y)
r13: app(minus_1(x1),y) -> minus_2(x1,y)
r14: app(quot_0(),y) -> quot_1(y)
r15: app(quot_1(x1),y) -> quot_2(x1,y)
r16: app(s_0(),y) -> s_1(y)
r17: minus_2(x,|0|_0()) -> x
r18: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r19: quot_2(|0|_0(),s_1(y)) -> |0|_0()
r20: quot_2(s_1(x),s_1(y)) -> s_1(quot_2(minus_2(x,y),s_1(y)))
r21: log_1(s_1(|0|_0())) -> |0|_0()
r22: log_1(s_1(s_1(x))) -> s_1(log_1(s_1(quot_2(x,s_1(s_1(|0|_0()))))))
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:
          
            minus_2#_A(x1,x2) = max{0, x1 - 15, x2 - 15}
            s_1_A(x1) = max{64, x1 + 14}
            minus_2#!_A(x1,x2) = max{7, x1 - 137, x2 - 137}
            s_1!_A(x1) = max{6, x1 - 136}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus_2#_A(x1,x2) = max{0, x1 - 8, x2 - 8}
            s_1_A(x1) = max{49, x1 + 7}
            minus_2#!_A(x1,x2) = max{10, x1 - 50}
            s_1!_A(x1) = x1 + 9
        
  
    partial status:
  
      pi(minus_2#) = [1]
      pi(s_1) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.