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

  r1, r5, r6, r12, r14, r16, r19, r21, r25

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{6, x1 + 2}
        filter_1_A(x1) = x1 + 30
        filter_2#_A(x1,x2) = max{6, x1 + 2}
        cons_2_A(x1,x2) = max{7, x1 - 11, x2 - 11}
        map_1_A(x1) = x1 + 8
        map_2#_A(x1,x2) = x1 + 7
        filter2_3_A(x1,x2,x3) = max{x1 - 23, x2 + 3, x3 - 11}
        filter2_4#_A(x1,x2,x3,x4) = max{6, x2 + 2}
        false_0_A = 5
        app_A(x1,x2) = max{10, x1 - 31, x2 - 31}
        true_0_A = 5
        cons_0_A = 5
        cons_1_A(x1) = max{6, x1 - 31}
        filter2_0_A = 9
        filter2_1_A(x1) = max{10, x1 - 31}
        filter2_2_A(x1,x2) = max{2, x1 - 63, x2 - 32}
        minus_0_A = 30
        minus_1_A(x1) = max{7, x1 - 169}
        quot_0_A = 7
        quot_1_A(x1) = max{8, x1 - 102}
        s_0_A = 8
        s_1_A(x1) = max{9, x1 - 33}
        quot_2_A(x1,x2) = max{9, x1 - 134, x2 - 32}
        |0|_0_A = 8
        log_1_A(x1) = max{9, x1 - 68}
        filter_2_A(x1,x2) = max{8, x1 - 54, x2 - 20}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        filter2_0 = quot_0 = filter_2 > filter_1 = map_2# > cons_2 = filter2_3 = app > app# = filter_2# = map_1 = filter2_4# = cons_1 = filter2_2 = s_1 = nil_0 > false_0 = true_0 = cons_0 = filter2_1 = s_0 = log_1 > quot_2 > quot_1 = |0|_0 > minus_1 > minus_0
      
  
    partial status:
  
      pi(app#) = [1]
      pi(filter_1) = [1]
      pi(filter_2#) = [1]
      pi(cons_2) = []
      pi(map_1) = [1]
      pi(map_2#) = [1]
      pi(filter2_3) = [2]
      pi(filter2_4#) = [2]
      pi(false_0) = []
      pi(app) = []
      pi(true_0) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(minus_0) = []
      pi(minus_1) = []
      pi(quot_0) = []
      pi(quot_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(quot_2) = []
      pi(|0|_0) = []
      pi(log_1) = []
      pi(filter_2) = []
      pi(nil_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)) -> 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: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p5: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p6: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p7: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p8: 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 estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p2: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p3: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p4: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p5: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p6: app#(map_1(x1),y) -> map_2#(x1,y)
p7: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p8: 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(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

  r1, r2, r3, r5, r6, r10, r12, r14, r16, r19, r21, r25

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        filter_2#_A(x1,x2) = x1 + 10
        cons_2_A(x1,x2) = max{5, x1 - 13, x2 - 13}
        app#_A(x1,x2) = max{5, x1}
        filter2_3_A(x1,x2,x3) = max{x1 + 2, x2 + 24, x3 + 2}
        filter2_4#_A(x1,x2,x3,x4) = x2 + 10
        true_0_A = 4
        app_A(x1,x2) = max{9, x1 - 10, x2 + 3}
        false_0_A = 11
        map_1_A(x1) = max{5, x1 + 2}
        map_2#_A(x1,x2) = max{5, x1 + 1}
        cons_0_A = 3
        cons_1_A(x1) = max{4, x1 - 1}
        filter_0_A = 7
        filter_1_A(x1) = max{8, x1 - 27}
        filter2_0_A = 9
        filter2_1_A(x1) = max{2, x1 - 1}
        filter2_2_A(x1,x2) = max{8, x1 - 12, x2 - 1}
        map_0_A = 4
        minus_0_A = 3
        minus_1_A(x1) = max{4, x1 - 60}
        quot_0_A = 1
        quot_1_A(x1) = max{5, x1 - 1}
        s_0_A = 6
        s_1_A(x1) = max{7, x1 - 9}
        quot_2_A(x1,x2) = max{6, x1 - 61, x2 - 8}
        |0|_0_A = 1
        log_1_A(x1) = max{8, x1 - 42}
        filter_2_A(x1,x2) = max{12, x1 - 26, x2 - 12}
        nil_0_A = 5
  
    base order 2:
  
      precedence:
      
        app = cons_0 = filter2_0 > s_1 > cons_2 = filter2_3 = map_1 = s_0 = log_1 = filter_2 > map_2# = cons_1 = filter_1 = filter2_2 = map_0 = quot_2 = nil_0 > app# = false_0 = filter_0 = filter2_1 = quot_1 = |0|_0 > filter_2# = filter2_4# = minus_1 = quot_0 > true_0 = minus_0
      
  
    partial status:
  
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(app#) = [1]
      pi(filter2_3) = []
      pi(filter2_4#) = []
      pi(true_0) = []
      pi(app) = [2]
      pi(false_0) = []
      pi(map_1) = [1]
      pi(map_2#) = [1]
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(map_0) = []
      pi(minus_0) = []
      pi(minus_1) = []
      pi(quot_0) = []
      pi(quot_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(quot_2) = []
      pi(|0|_0) = []
      pi(log_1) = []
      pi(filter_2) = []
      pi(nil_0) = []

The following rules are removed:

  p1



-- SCC decomposition.

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

p1: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p2: filter2_4#(true_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#(false_0(),f,x,xs) -> filter_2#(f,xs)
p5: app#(map_1(x1),y) -> map_2#(x1,y)
p6: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p7: 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(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:

  {p5, p6, p7}
  {p2, p3, p4}


-- 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)

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

  r1, r2, r3, r5, r6, r12, r14, r16, r19, r21, r25

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = x1 + 5
        cons_2_A(x1,x2) = max{8, x1 - 56, x2 - 56}
        app#_A(x1,x2) = x1 + 4
        map_1_A(x1) = x1 + 6
        app_A(x1,x2) = max{45, x1 - 14, x2 - 9}
        cons_0_A = 13
        cons_1_A(x1) = max{7, x1 - 42}
        filter_0_A = 12
        filter_1_A(x1) = max{13, x1 - 10}
        filter2_0_A = 11
        filter2_1_A(x1) = max{12, x1 - 9}
        filter2_2_A(x1,x2) = max{13, x1 - 24, x2 - 10}
        minus_0_A = 12
        minus_1_A(x1) = max{13, x1 - 10}
        quot_0_A = 12
        quot_1_A(x1) = max{13, x1 - 10}
        s_0_A = 2
        s_1_A(x1) = max{3, x1 - 78}
        quot_2_A(x1,x2) = max{43, x1 - 52, x2 - 10}
        |0|_0_A = 1
        log_1_A(x1) = max{2, x1 - 39}
        filter_2_A(x1,x2) = max{12, x1 - 60, x2 - 46}
        nil_0_A = 8
  
    base order 2:
  
      precedence:
      
        cons_0 = quot_0 = s_1 = quot_2 = filter_2 = nil_0 > app# = app = s_0 > cons_2 = filter_1 = filter2_2 = minus_1 = quot_1 = log_1 > map_1 = cons_1 = filter_0 = filter2_1 = minus_0 = |0|_0 > map_2# = filter2_0
      
  
    partial status:
  
      pi(map_2#) = []
      pi(cons_2) = []
      pi(app#) = [1]
      pi(map_1) = []
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(minus_0) = []
      pi(minus_1) = []
      pi(quot_0) = []
      pi(quot_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(quot_2) = []
      pi(|0|_0) = []
      pi(log_1) = []
      pi(filter_2) = []
      pi(nil_0) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p2: 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(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:

  {p2}


-- 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(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

  r1, r3, r5, r6, r7, r10, r12, r14, r19, r21, r25

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = max{4, x2 + 2}
        cons_2_A(x1,x2) = max{6, x1 + 3, x2 + 3}
        app_A(x1,x2) = max{25, x1 - 8, x2 - 23}
        cons_0_A = 6
        cons_1_A(x1) = max{7, x1 - 24}
        filter_0_A = 1
        filter_1_A(x1) = max{0, x1 - 23}
        filter2_0_A = 1
        filter2_1_A(x1) = max{2, x1 - 24}
        filter2_2_A(x1,x2) = max{3, x1 - 32, x2 - 23}
        filter2_3_A(x1,x2,x3) = max{4, x1 - 40, x2 - 32, x3 - 24}
        map_0_A = 1
        map_1_A(x1) = max{2, x1 - 24}
        minus_0_A = 6
        minus_1_A(x1) = max{7, x1 - 24}
        quot_0_A = 1
        quot_1_A(x1) = max{2, x1 - 24}
        quot_2_A(x1,x2) = max{24, x1 - 43, x2 - 66}
        |0|_0_A = 21
        s_1_A(x1) = max{32, x1 + 10}
        log_1_A(x1) = max{23, x1 - 77}
        filter_2_A(x1,x2) = max{3, x2 - 76}
        nil_0_A = 1
  
    base order 2:
  
      precedence:
      
        quot_2 > s_1 > app > filter2_3 = minus_1 = quot_1 = log_1 > cons_2 = filter_1 = filter2_1 = filter2_2 = minus_0 = quot_0 = |0|_0 = filter_2 > map_2# = cons_1 = filter_0 = filter2_0 = map_1 = nil_0 > cons_0 = map_0
      
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [2]
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(filter2_3) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(minus_0) = []
      pi(minus_1) = []
      pi(quot_0) = []
      pi(quot_1) = []
      pi(quot_2) = []
      pi(|0|_0) = []
      pi(s_1) = [1]
      pi(log_1) = []
      pi(filter_2) = []
      pi(nil_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: filter2_4#(true_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#(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 set of usable rules consists of

  r2, r3, r4, r5, r6, r8, r10, r12, r14, r16, r19, r21, r25, r26, r27, r28

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        filter2_4#_A(x1,x2,x3,x4) = max{x3 + 4, x4 + 2}
        true_0_A = 6
        filter_2#_A(x1,x2) = max{1, x2 - 1}
        cons_2_A(x1,x2) = max{13, x1 + 6, x2 + 4}
        app_A(x1,x2) = max{14, x1 - 16, x2 + 7}
        false_0_A = 8
        cons_1_A(x1) = x1 + 23
        filter_0_A = 1
        filter_1_A(x1) = max{8, x1 - 1}
        filter_2_A(x1,x2) = max{9, x2}
        filter2_0_A = 30
        filter2_1_A(x1) = max{13, x1 - 1}
        filter2_2_A(x1,x2) = max{0, x1 - 18, x2 - 1}
        filter2_3_A(x1,x2,x3) = max{x2 + 12, x3 + 28}
        filter2_4_A(x1,x2,x3,x4) = max{13, x3 + 6, x4 + 4}
        map_0_A = 15
        map_1_A(x1) = max{0, x1 - 1}
        minus_0_A = 7
        minus_1_A(x1) = max{8, x1 - 106}
        quot_0_A = 17
        quot_1_A(x1) = max{14, x1 - 70}
        s_0_A = 15
        s_1_A(x1) = max{13, x1 - 17}
        quot_2_A(x1,x2) = max{15, x1 - 87, x2 - 16}
        |0|_0_A = 12
        log_1_A(x1) = max{16, x1 - 52}
        nil_0_A = 1
  
    base order 2:
  
      precedence:
      
        map_0 > map_1 > app = nil_0 > filter_2 = s_1 = |0|_0 > filter2_4 = log_1 > cons_2 = filter_1 = filter2_3 = minus_0 = quot_1 = s_0 = quot_2 > filter2_4# = cons_1 = filter_0 = filter2_2 = quot_0 > filter_2# = filter2_1 = minus_1 > true_0 = false_0 = filter2_0
      
  
    partial status:
  
      pi(filter2_4#) = []
      pi(true_0) = []
      pi(filter_2#) = []
      pi(cons_2) = [1, 2]
      pi(app) = [2]
      pi(false_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(filter_2) = [2]
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(filter2_3) = []
      pi(filter2_4) = [3, 4]
      pi(map_0) = []
      pi(map_1) = []
      pi(minus_0) = []
      pi(minus_1) = []
      pi(quot_0) = []
      pi(quot_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(quot_2) = []
      pi(|0|_0) = []
      pi(log_1) = []
      pi(nil_0) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

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

  (no SCCs)

-- 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

  r1, r2, r3, r5, r6, r7, r10, r12, r14, r17, r18, r19, r20, r21, r25

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        log_1#_A(x1) = max{41, x1 + 40}
        s_1_A(x1) = max{83, x1 + 42}
        quot_2_A(x1,x2) = max{38, x1}
        |0|_0_A = 166
        app_A(x1,x2) = max{39, x1 - 5, x2 - 13}
        cons_0_A = 3
        cons_1_A(x1) = max{4, x1 - 14}
        cons_2_A(x1,x2) = max{12, x1 - 27, x2 - 44}
        filter_0_A = 4
        filter_1_A(x1) = 39
        filter2_0_A = 1
        filter2_1_A(x1) = max{2, x1 - 13}
        filter2_2_A(x1,x2) = max{3, x1 - 18, x2 - 13}
        filter2_3_A(x1,x2,x3) = max{4, x1 - 24, x2 - 19, x3 - 14}
        map_0_A = 1
        map_1_A(x1) = max{2, x1 - 14}
        minus_0_A = 1
        minus_1_A(x1) = max{0, x1 - 14}
        quot_0_A = 43
        quot_1_A(x1) = max{37, x1 - 14}
        minus_2_A(x1,x2) = max{40, x1}
        log_1_A(x1) = max{251, x1 - 85}
        filter_2_A(x1,x2) = max{33, x1 - 19, x2 - 14}
        nil_0_A = 4
  
    base order 2:
  
      precedence:
      
        quot_2 > |0|_0 = minus_2 > log_1# = s_1 = filter_0 = filter2_3 = log_1 > filter_2 > app > filter_1 = filter2_2 = map_0 = minus_1 = quot_1 > cons_1 = cons_2 = filter2_1 = map_1 = minus_0 = quot_0 = nil_0 > cons_0 = filter2_0
      
  
    partial status:
  
      pi(log_1#) = [1]
      pi(s_1) = []
      pi(quot_2) = [1]
      pi(|0|_0) = []
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(cons_2) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(filter2_3) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(minus_0) = []
      pi(minus_1) = []
      pi(quot_0) = []
      pi(quot_1) = []
      pi(minus_2) = [1]
      pi(log_1) = []
      pi(filter_2) = []
      pi(nil_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: 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

  r1, r2, r3, r5, r6, r7, r10, r12, r14, r16, r17, r18, r19, r21, r25

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        quot_2#_A(x1,x2) = max{x1 + 16, x2 - 13}
        s_1_A(x1) = x1 + 28
        minus_2_A(x1,x2) = max{13, x1 + 5}
        app_A(x1,x2) = max{x1 - 30, x2 + 58}
        cons_0_A = 29
        cons_1_A(x1) = max{2, x1 - 51}
        cons_2_A(x1,x2) = max{3, x1 - 82, x2 - 90}
        filter_0_A = 8
        filter_1_A(x1) = max{9, x1 - 92}
        filter2_0_A = 29
        filter2_1_A(x1) = max{32, x1 - 1}
        filter2_2_A(x1,x2) = max{1, x1 - 31, x2 - 52}
        filter2_3_A(x1,x2,x3) = max{12, x1 - 61, x2 - 82, x3 - 80}
        map_0_A = 1
        map_1_A(x1) = max{2, x1 - 114}
        minus_0_A = 1
        minus_1_A(x1) = max{0, x1 - 1}
        quot_0_A = 11
        quot_1_A(x1) = max{12, x1 - 1}
        s_0_A = 11
        |0|_0_A = 1
        quot_2_A(x1,x2) = max{57, x1 - 32, x2 - 86}
        log_1_A(x1) = max{28, x1 - 131}
        filter_2_A(x1,x2) = max{10, x1 - 123, x2 - 83}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        cons_0 > app = cons_1 = cons_2 > filter2_3 = map_1 = quot_0 = quot_2 = filter_2 = nil_0 > quot_2# = s_1 = filter_1 = filter2_0 = filter2_2 = map_0 = quot_1 > minus_2 = filter_0 = filter2_1 = minus_1 = s_0 = log_1 > minus_0 = |0|_0
      
  
    partial status:
  
      pi(quot_2#) = [1]
      pi(s_1) = [1]
      pi(minus_2) = []
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(cons_2) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(filter2_3) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(minus_0) = []
      pi(minus_1) = []
      pi(quot_0) = []
      pi(quot_1) = []
      pi(s_0) = []
      pi(|0|_0) = []
      pi(quot_2) = []
      pi(log_1) = []
      pi(filter_2) = []
      pi(nil_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

  r1, r3, r5, r6, r7, r10, r12, r14, r19, r21, r25

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        minus_2#_A(x1,x2) = max{x1 - 1, x2 + 18}
        s_1_A(x1) = max{57, x1 + 18}
        app_A(x1,x2) = max{77, x1 - 15, x2 - 94}
        cons_0_A = 1
        cons_1_A(x1) = max{2, x1 - 95}
        filter_0_A = 13
        filter_1_A(x1) = max{14, x1 - 94}
        filter2_0_A = 1
        filter2_1_A(x1) = max{2, x1 - 95}
        filter2_2_A(x1,x2) = max{3, x1 - 111, x2 - 95}
        filter2_3_A(x1,x2,x3) = max{4, x1 - 127, x2 - 111, x3 - 95}
        map_0_A = 14
        map_1_A(x1) = max{16, x1 - 95}
        minus_0_A = 14
        minus_1_A(x1) = max{16, x1 - 95}
        quot_0_A = 14
        quot_1_A(x1) = max{16, x1 - 95}
        quot_2_A(x1,x2) = max{38, x1 - 128, x2 - 110}
        |0|_0_A = 37
        log_1_A(x1) = max{76, x1 - 164}
        filter_2_A(x1,x2) = max{145, x1 - 14, x2 - 7}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        minus_2# > s_1 = filter_2 > app = filter2_0 = nil_0 > cons_1 = filter_1 = filter2_3 = map_1 = quot_1 = log_1 > cons_0 = filter_0 = filter2_1 = filter2_2 = map_0 = quot_0 = quot_2 > |0|_0 > minus_1 > minus_0
      
  
    partial status:
  
      pi(minus_2#) = [2]
      pi(s_1) = [1]
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(filter2_0) = []
      pi(filter2_1) = []
      pi(filter2_2) = []
      pi(filter2_3) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(minus_0) = []
      pi(minus_1) = []
      pi(quot_0) = []
      pi(quot_1) = []
      pi(quot_2) = []
      pi(|0|_0) = []
      pi(log_1) = []
      pi(filter_2) = []
      pi(nil_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.