YES
We show the termination of the TRS R:

  app(app(times(),x),|0|()) -> |0|()
  app(app(times(),x),app(s(),y)) -> app(app(plus(),app(app(times(),x),y)),x)
  app(app(plus(),x),|0|()) -> x
  app(app(plus(),|0|()),x) -> x
  app(app(plus(),x),app(s(),y)) -> app(s(),app(app(plus(),x),y))
  app(app(plus(),app(s(),x)),y) -> app(s(),app(app(plus(),x),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(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(filter_0(),y) -> filter_1(y)
  app(filter_1(x1),y) -> filter_2(x1,y)
  app(filter2_0(),y) -> filter2_1(y)
  app(filter2_1(x1),y) -> filter2_2(x1,y)
  app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
  app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
  app(map_0(),y) -> map_1(y)
  app(map_1(x1),y) -> map_2(x1,y)
  app(plus_0(),y) -> plus_1(y)
  app(plus_1(x1),y) -> plus_2(x1,y)
  app(s_0(),y) -> s_1(y)
  app(times_0(),y) -> times_1(y)
  app(times_1(x1),y) -> times_2(x1,y)
  times_2(x,|0|_0()) -> |0|_0()
  times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
  plus_2(x,|0|_0()) -> x
  plus_2(|0|_0(),x) -> x
  plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
  plus_2(s_1(x),y) -> s_1(plus_2(x,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#(filter_1(x1),y) -> filter_2#(x1,y)
p2: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p3: app#(map_1(x1),y) -> map_2#(x1,y)
p4: app#(plus_1(x1),y) -> plus_2#(x1,y)
p5: app#(times_1(x1),y) -> times_2#(x1,y)
p6: times_2#(x,s_1(y)) -> plus_2#(times_2(x,y),x)
p7: times_2#(x,s_1(y)) -> times_2#(x,y)
p8: plus_2#(x,s_1(y)) -> plus_2#(x,y)
p9: plus_2#(s_1(x),y) -> plus_2#(x,y)
p10: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p11: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p12: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p13: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p14: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p15: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p10, p11, p12, p13, p14, p15}
  {p7}
  {p8, p9}


-- Reduction pair.

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

p1: app#(filter_1(x1),y) -> filter_2#(x1,y)
p2: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p3: app#(map_1(x1),y) -> map_2#(x1,y)
p4: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p5: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p6: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y)
p7: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs)
p8: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p9: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r24

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{4, x1}
        filter_1_A(x1) = x1 + 25
        filter_2#_A(x1,x2) = x1 + 24
        cons_2_A(x1,x2) = max{3, x1 - 35, x2 - 35}
        map_1_A(x1) = max{7, x1 + 6}
        map_2#_A(x1,x2) = max{7, x1 + 5}
        filter2_3_A(x1,x2,x3) = max{x1 + 24, x2 + 25, x3 + 24}
        filter2_4#_A(x1,x2,x3,x4) = x2 + 24
        false_0_A = 16
        app_A(x1,x2) = max{21, x1 - 26}
        true_0_A = 1
        filter_2_A(x1,x2) = max{20, x1 - 44, x2 - 17}
        nil_0_A = 7
  
    base order 2:
  
      precedence:
      
        map_1 = map_2# = nil_0 > app# = filter2_3 = app = filter_2 > filter_1 = cons_2 = false_0 > true_0 > filter_2# = filter2_4#
      
  
    partial status:
  
      pi(app#) = [1]
      pi(filter_1) = [1]
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(map_1) = [1]
      pi(map_2#) = [1]
      pi(filter2_3) = []
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(app) = []
      pi(true_0) = []
      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(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: 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(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r3, r5, r6, r9, r11, r13, r14, r24

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 + 5
        cons_2_A(x1,x2) = max{14, x2 - 25}
        app#_A(x1,x2) = max{4, x1 - 1}
        filter2_3_A(x1,x2,x3) = max{x1 + 2, x2 + 7, x3 + 2}
        filter2_4#_A(x1,x2,x3,x4) = x2 + 5
        true_0_A = 1
        app_A(x1,x2) = max{13, x1 - 23, x2 + 8}
        false_0_A = 3
        map_1_A(x1) = max{9, x1 + 7}
        map_2#_A(x1,x2) = x1 + 5
        cons_0_A = 11
        cons_1_A(x1) = max{12, x1 - 1}
        filter_0_A = 9
        filter_1_A(x1) = max{10, x1 - 77}
        filter2_0_A = 10
        filter2_1_A(x1) = max{11, x1 - 1}
        filter2_2_A(x1,x2) = max{12, x1 - 25, x2 - 1}
        map_0_A = 8
        plus_0_A = 1
        plus_1_A(x1) = max{9, x1 - 1}
        s_0_A = 8
        s_1_A(x1) = max{9, x1 - 36}
        times_0_A = 10
        times_1_A(x1) = max{11, x1 - 1}
        filter_2_A(x1,x2) = max{11, x1 - 77, x2 - 15}
        nil_0_A = 10
  
    base order 2:
  
      precedence:
      
        map_2# > app# > filter_2# = filter2_4# = app > map_1 > cons_2 = filter2_3 = cons_0 = map_0 = times_1 = filter_2 > true_0 = false_0 = cons_1 = filter_1 = filter2_2 = plus_1 = s_1 = times_0 = nil_0 > filter_0 = filter2_1 = plus_0 = s_0 > filter2_0
      
  
    partial status:
  
      pi(filter_2#) = [1]
      pi(cons_2) = []
      pi(app#) = []
      pi(filter2_3) = [2, 3]
      pi(filter2_4#) = [2]
      pi(true_0) = []
      pi(app) = []
      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(plus_0) = []
      pi(plus_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(times_0) = []
      pi(times_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: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
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(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

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


-- 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(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r2, r3, r5, r6, r7, r9, r11, r13, r14, r24

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{14, x3 + 3, x4 + 12}
        false_0_A = 36
        filter_2#_A(x1,x2) = x2 + 11
        cons_2_A(x1,x2) = max{60, x1 + 4, x2 + 13}
        app_A(x1,x2) = max{34, x1 + 15, x2 + 15}
        true_0_A = 16
        cons_0_A = 47
        cons_1_A(x1) = max{61, x1 + 14}
        filter_0_A = 1
        filter_1_A(x1) = max{34, x1 - 29}
        filter2_0_A = 1
        filter2_1_A(x1) = max{17, x1 - 16}
        filter2_2_A(x1,x2) = max{33, x1 - 17, x2 - 16}
        filter2_3_A(x1,x2,x3) = max{46, x1 - 51, x2 - 30, x3 - 16}
        map_0_A = 63
        map_1_A(x1) = max{62, x1 - 30}
        plus_0_A = 1
        plus_1_A(x1) = max{17, x1 - 35}
        s_0_A = 1
        s_1_A(x1) = max{17, x1 - 36}
        times_0_A = 1
        times_1_A(x1) = max{21, x1 - 16}
        filter_2_A(x1,x2) = max{48, x2 - 25}
        nil_0_A = 1
  
    base order 2:
  
      precedence:
      
        cons_2 > app = map_0 > cons_1 = filter_1 = filter2_3 = map_1 = filter_2 > false_0 = cons_0 = filter_0 = filter2_2 = times_1 = nil_0 > filter2_4# = filter_2# = filter2_1 = plus_1 = s_1 = times_0 > true_0 = filter2_0 = plus_0 = s_0
      
  
    partial status:
  
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(filter_2#) = [2]
      pi(cons_2) = [1, 2]
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(cons_0) = []
      pi(cons_1) = [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(plus_0) = []
      pi(plus_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(times_0) = []
      pi(times_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)) -> filter2_4#(app(f,x),f,x,xs)
p2: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p2: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r2, r3, r5, r6, r7, r9, r11, r13, r14, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        filter_2#_A(x1,x2) = max{2, x2 + 1}
        cons_2_A(x1,x2) = max{14, x1 + 3, x2 + 3}
        filter2_4#_A(x1,x2,x3,x4) = max{x3 + 3, x4 + 3}
        app_A(x1,x2) = max{16, x1 - 5, x2 + 10}
        true_0_A = 11
        cons_0_A = 3
        cons_1_A(x1) = x1 + 9
        filter_0_A = 4
        filter_1_A(x1) = max{15, x1 - 1}
        filter2_0_A = 1
        filter2_1_A(x1) = max{2, x1 - 1}
        filter2_2_A(x1,x2) = max{3, x1 - 7, x2 - 1}
        filter2_3_A(x1,x2,x3) = max{4, x1 - 13, x2 - 7, x3 - 1}
        map_0_A = 1
        map_1_A(x1) = max{2, x1 - 1}
        plus_0_A = 3
        plus_1_A(x1) = max{4, x1 - 6}
        s_0_A = 4
        s_1_A(x1) = max{9, x1 - 24}
        times_0_A = 3
        times_1_A(x1) = max{4, x1 - 1}
        filter_2_A(x1,x2) = max{12, x1 - 25}
        nil_0_A = 4
  
    base order 2:
  
      precedence:
      
        app > filter_1 = filter2_3 > cons_2 = filter_0 = filter2_2 = times_1 = filter_2 > filter_2# = filter2_4# = cons_1 = filter2_1 = map_1 = plus_1 = s_1 = times_0 = nil_0 > true_0 = cons_0 = filter2_0 = map_0 = plus_0 = s_0
      
  
    partial status:
  
      pi(filter_2#) = [2]
      pi(cons_2) = [2]
      pi(filter2_4#) = []
      pi(app) = [2]
      pi(true_0) = []
      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(plus_0) = []
      pi(plus_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(times_0) = []
      pi(times_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: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: 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: app#(map_1(x1),y) -> map_2#(x1,y)
p2: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p3: map_2#(f,cons_2(x,xs)) -> app#(f,x)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(filter_0(),y) -> filter_1(y)
r4: app(filter_1(x1),y) -> filter_2(x1,y)
r5: app(filter2_0(),y) -> filter2_1(y)
r6: app(filter2_1(x1),y) -> filter2_2(x1,y)
r7: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y)
r8: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y)
r9: app(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r3, r5, r6, r7, r11, r13, r14, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{9, x1 - 8}
        map_1_A(x1) = x1 + 19
        map_2#_A(x1,x2) = x1 + 10
        cons_2_A(x1,x2) = max{3, x1 - 21, x2 - 21}
        app_A(x1,x2) = max{6, x1 - 36, x2 - 48}
        cons_0_A = 1
        cons_1_A(x1) = max{2, x1 - 49}
        filter_0_A = 4
        filter_1_A(x1) = max{5, x1 - 49}
        filter2_0_A = 2
        filter2_1_A(x1) = max{3, x1 - 49}
        filter2_2_A(x1,x2) = max{4, x1 - 86, x2 - 49}
        filter2_3_A(x1,x2,x3) = max{5, x1 - 123, x2 - 86, x3 - 49}
        plus_0_A = 1
        plus_1_A(x1) = max{2, x1 - 48}
        s_0_A = 2
        s_1_A(x1) = max{3, x1 - 87}
        times_0_A = 3
        times_1_A(x1) = max{4, x1 - 49}
        filter_2_A(x1,x2) = max{7, x1 - 23, x2 - 26}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        map_2# > map_1 = filter_2 = nil_0 > app# = cons_2 = app > cons_0 = filter2_3 > cons_1 = filter_1 = filter2_0 = filter2_2 = plus_1 = s_1 = times_1 > filter_0 = filter2_1 = plus_0 = s_0 = times_0
      
  
    partial status:
  
      pi(app#) = []
      pi(map_1) = [1]
      pi(map_2#) = [1]
      pi(cons_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(plus_0) = []
      pi(plus_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(times_0) = []
      pi(times_1) = []
      pi(filter_2) = []
      pi(nil_0) = []

The following rules are removed:

  p3



-- SCC decomposition.

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

p1: app#(map_1(x1),y) -> map_2#(x1,y)
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(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: 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(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r3, r5, r6, r7, r9, r11, r13, r14, r24

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{x1 + 5, x2 + 24}
        cons_2_A(x1,x2) = max{3, x1 - 19, x2 + 1}
        app_A(x1,x2) = max{18, x1 - 3, x2 - 23}
        cons_0_A = 1
        cons_1_A(x1) = max{2, x1 - 24}
        filter_0_A = 1
        filter_1_A(x1) = max{2, x1 - 24}
        filter2_0_A = 1
        filter2_1_A(x1) = max{0, x1 - 24}
        filter2_2_A(x1,x2) = max{0, x1 - 28, x2 - 24}
        filter2_3_A(x1,x2,x3) = max{0, x1 - 32, x2 - 28, x3 - 24}
        map_0_A = 1
        map_1_A(x1) = max{2, x1 - 24}
        plus_0_A = 1
        plus_1_A(x1) = max{2, x1 - 24}
        s_0_A = 1
        s_1_A(x1) = max{2, x1 - 28}
        times_0_A = 1
        times_1_A(x1) = max{2, x1 - 23}
        filter_2_A(x1,x2) = max{18, x1 - 3, x2 - 4}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        filter_2 > map_2# = app > cons_2 = filter_1 > cons_1 = filter_0 = filter2_3 = map_1 = times_1 = nil_0 > cons_0 = filter2_2 = map_0 = plus_1 = s_1 = times_0 > filter2_1 = plus_0 = s_0 > filter2_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(plus_0) = []
      pi(plus_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(times_0) = []
      pi(times_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: times_2#(x,s_1(y)) -> times_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(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r2, r3, r5, r6, r7, r9, r11, r14, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        times_2#_A(x1,x2) = max{x1 + 26, x2 + 25}
        s_1_A(x1) = x1 + 24
        app_A(x1,x2) = max{6, x1 - 7, x2 - 15}
        cons_1_A(x1) = max{2, x1 - 1}
        cons_2_A(x1,x2) = max{3, x1 - 9, x2 - 17}
        filter_0_A = 4
        filter_1_A(x1) = max{5, x1 - 16}
        filter2_0_A = 3
        filter2_1_A(x1) = max{4, x1 - 16}
        filter2_2_A(x1,x2) = max{5, x1 - 24, x2 - 16}
        filter2_3_A(x1,x2,x3) = max{6, x1 - 31, x2 - 23, x3 - 15}
        map_0_A = 5
        map_1_A(x1) = max{6, x1 - 15}
        plus_0_A = 1
        plus_1_A(x1) = max{2, x1 - 16}
        times_0_A = 3
        times_1_A(x1) = max{4, x1 - 15}
        filter_2_A(x1,x2) = max{15, x1 - 6, x2 - 5}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        filter_2 > app > cons_2 > times_2# = cons_1 = nil_0 > s_1 = filter_1 = filter2_3 = map_1 = plus_1 = times_1 > filter_0 = filter2_2 = map_0 = plus_0 = times_0 > filter2_1 > filter2_0
      
  
    partial status:
  
      pi(times_2#) = [2]
      pi(s_1) = [1]
      pi(app) = []
      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(plus_0) = []
      pi(plus_1) = []
      pi(times_0) = []
      pi(times_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: plus_2#(x,s_1(y)) -> plus_2#(x,y)
p2: plus_2#(s_1(x),y) -> plus_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(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r2, r3, r5, r6, r7, r9, r11, r14, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus_2#_A(x1,x2) = max{x1 + 2, x2 + 3}
        s_1_A(x1) = max{6, x1 + 2}
        app_A(x1,x2) = max{64, x1 - 8, x2 - 1}
        cons_0_A = 7
        cons_1_A(x1) = max{6, x1 - 2}
        cons_2_A(x1,x2) = max{30, x1 - 98, x2 - 33}
        filter_0_A = 6
        filter_1_A(x1) = max{7, x1 - 28}
        filter2_0_A = 4
        filter2_1_A(x1) = max{5, x1 - 2}
        filter2_2_A(x1,x2) = max{6, x1 - 11, x2 - 2}
        filter2_3_A(x1,x2,x3) = max{7, x1 - 19, x2 - 11, x3 - 1}
        map_0_A = 6
        map_1_A(x1) = max{7, x1 - 2}
        plus_0_A = 1
        plus_1_A(x1) = max{2, x1 - 2}
        times_0_A = 6
        times_1_A(x1) = max{7, x1 - 2}
        filter_2_A(x1,x2) = max{35, x1 - 37, x2 - 31}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        app > plus_2# = filter2_3 > s_1 = cons_2 = filter2_2 = map_1 = times_1 = filter_2 > cons_1 = filter_1 = filter2_1 = map_0 = plus_1 = times_0 > cons_0 = filter_0 = filter2_0 = plus_0 = nil_0
      
  
    partial status:
  
      pi(plus_2#) = [1, 2]
      pi(s_1) = [1]
      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(plus_0) = []
      pi(plus_1) = []
      pi(times_0) = []
      pi(times_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: plus_2#(s_1(x),y) -> plus_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(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: plus_2#(s_1(x),y) -> plus_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(map_0(),y) -> map_1(y)
r10: app(map_1(x1),y) -> map_2(x1,y)
r11: app(plus_0(),y) -> plus_1(y)
r12: app(plus_1(x1),y) -> plus_2(x1,y)
r13: app(s_0(),y) -> s_1(y)
r14: app(times_0(),y) -> times_1(y)
r15: app(times_1(x1),y) -> times_2(x1,y)
r16: times_2(x,|0|_0()) -> |0|_0()
r17: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r18: plus_2(x,|0|_0()) -> x
r19: plus_2(|0|_0(),x) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: plus_2(s_1(x),y) -> s_1(plus_2(x,y))
r22: map_2(f,nil_0()) -> nil_0()
r23: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r24: filter_2(f,nil_0()) -> nil_0()
r25: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r26: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r27: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r2, r3, r5, r6, r7, r9, r11, r14, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus_2#_A(x1,x2) = max{x1 + 11, x2 + 12}
        s_1_A(x1) = max{10, x1 + 1}
        app_A(x1,x2) = max{12, x1 - 13, x2 - 1}
        cons_0_A = 1
        cons_1_A(x1) = max{2, x1 - 30}
        cons_2_A(x1,x2) = max{3, x1 - 44, x2 - 50}
        filter_0_A = 8
        filter_1_A(x1) = max{9, x1 - 42}
        filter2_0_A = 10
        filter2_1_A(x1) = max{11, x1 - 2}
        filter2_2_A(x1,x2) = max{12, x1 - 15, x2 - 1}
        filter2_3_A(x1,x2,x3) = max{10, x1 - 28, x2 - 14, x3 - 1}
        map_0_A = 1
        map_1_A(x1) = max{2, x1 - 2}
        plus_0_A = 1
        plus_1_A(x1) = max{2, x1 - 2}
        times_0_A = 9
        times_1_A(x1) = max{10, x1 - 2}
        filter_2_A(x1,x2) = max{10, x1 - 56, x2 - 42}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        app > plus_2# = cons_2 = plus_0 > s_1 = cons_1 = filter_0 = filter2_3 = map_1 = plus_1 = times_1 = filter_2 = nil_0 > cons_0 = filter_1 = filter2_2 = map_0 = times_0 > filter2_1 > filter2_0
      
  
    partial status:
  
      pi(plus_2#) = [1]
      pi(s_1) = [1]
      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(plus_0) = []
      pi(plus_1) = []
      pi(times_0) = []
      pi(times_1) = []
      pi(filter_2) = []
      pi(nil_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.