YES
We show the termination of the TRS R:

  app(app(times(),x),app(app(plus(),y),app(s(),z))) -> app(app(plus(),app(app(times(),x),app(app(plus(),y),app(app(times(),app(s(),z)),|0|())))),app(app(times(),x),app(s(),z)))
  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(),x),app(s(),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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
  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(x,s_1(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,plus_2(y,s_1(z))) -> plus_2#(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
p7: times_2#(x,plus_2(y,s_1(z))) -> times_2#(x,plus_2(y,times_2(s_1(z),|0|_0())))
p8: times_2#(x,plus_2(y,s_1(z))) -> plus_2#(y,times_2(s_1(z),|0|_0()))
p9: times_2#(x,plus_2(y,s_1(z))) -> times_2#(s_1(z),|0|_0())
p10: times_2#(x,plus_2(y,s_1(z))) -> times_2#(x,s_1(z))
p11: times_2#(x,s_1(y)) -> plus_2#(times_2(x,y),x)
p12: times_2#(x,s_1(y)) -> times_2#(x,y)
p13: plus_2#(x,s_1(y)) -> plus_2#(x,y)
p14: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p15: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p16: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p17: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p18: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p19: 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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p14, p15, p16, p17, p18, p19}
  {p7, p10, p12}
  {p13}


-- 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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: 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, r17, r23

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 + 2}
        filter_1_A(x1) = x1 + 6
        filter_2#_A(x1,x2) = x1 + 5
        cons_2_A(x1,x2) = max{3, x1 + 2, x2 - 33}
        map_1_A(x1) = x1 + 6
        map_2#_A(x1,x2) = max{6, x1 + 5}
        filter2_3_A(x1,x2,x3) = max{x1 + 10, x2 + 6, x3 + 4}
        filter2_4#_A(x1,x2,x3,x4) = x2 + 5
        false_0_A = 1
        app_A(x1,x2) = max{x1 - 9, x2 + 30}
        true_0_A = 31
        cons_0_A = 1
        cons_1_A(x1) = max{2, x1}
        filter_0_A = 1
        filter2_0_A = 1
        filter2_1_A(x1) = x1 + 29
        filter2_2_A(x1,x2) = max{x1 + 19, x2}
        map_0_A = 1
        plus_0_A = 7
        plus_1_A(x1) = x1 + 8
        s_0_A = 8
        s_1_A(x1) = max{0, x1 - 11}
        times_0_A = 1
        times_1_A(x1) = max{8, x1}
        times_2_A(x1,x2) = max{14, x1 - 10, x2}
        |0|_0_A = 0
        filter_2_A(x1,x2) = max{32, x1 - 9, x2 - 38}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        app = filter_2 > app# = filter_1 = map_1 = filter2_3 = false_0 = times_2 = nil_0 > filter_2# = cons_2 = filter2_4# = filter_0 = filter2_2 = map_0 = plus_1 = s_1 = times_1 > map_2# = true_0 = cons_1 = filter2_1 = plus_0 = s_0 = times_0 = |0|_0 > cons_0 = filter2_0
      
  
    partial status:
  
      pi(app#) = [1]
      pi(filter_1) = [1]
      pi(filter_2#) = []
      pi(cons_2) = [1]
      pi(map_1) = [1]
      pi(map_2#) = []
      pi(filter2_3) = [3]
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(app) = [2]
      pi(true_0) = []
      pi(cons_0) = []
      pi(cons_1) = [1]
      pi(filter_0) = []
      pi(filter2_0) = []
      pi(filter2_1) = [1]
      pi(filter2_2) = [2]
      pi(map_0) = []
      pi(plus_0) = []
      pi(plus_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(times_0) = []
      pi(times_1) = [1]
      pi(times_2) = [2]
      pi(|0|_0) = []
      pi(filter_2) = []
      pi(nil_0) = []

The following rules are removed:

  p6



-- SCC decomposition.

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: 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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: 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: app#(filter_1(x1),y) -> filter_2#(x1,y)
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)
p4: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
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)
p8: 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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r5, r6, r7, r11, r13, r14, r17, r23

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = x1 + 5
        filter_1_A(x1) = max{8, x1 + 6}
        filter_2#_A(x1,x2) = x1 + 6
        cons_2_A(x1,x2) = max{2, x1 - 17, x2 - 17}
        filter2_4#_A(x1,x2,x3,x4) = x2 + 6
        app_A(x1,x2) = max{10, x1 - 11, x2 - 11}
        true_0_A = 1
        map_1_A(x1) = x1 + 9
        map_2#_A(x1,x2) = x1 + 8
        false_0_A = 5
        cons_0_A = 12
        cons_1_A(x1) = max{10, x1 - 11}
        filter2_0_A = 6
        filter2_1_A(x1) = max{7, x1 - 11}
        filter2_2_A(x1,x2) = max{8, x1 - 23, x2 - 12}
        filter2_3_A(x1,x2,x3) = max{9, x1 - 35, x2 - 24, x3 - 12}
        plus_0_A = 12
        plus_1_A(x1) = max{10, x1 - 11}
        s_0_A = 3
        s_1_A(x1) = max{4, x1 - 36}
        times_0_A = 1
        times_1_A(x1) = max{2, x1 - 12}
        times_2_A(x1,x2) = max{3, x1 - 82, x2 - 24}
        |0|_0_A = 1
        filter_2_A(x1,x2) = max{9, x1 + 7, x2 - 10}
        nil_0_A = 8
  
    base order 2:
  
      precedence:
      
        app = cons_0 = plus_0 > map_1 = filter2_3 = times_2 = filter_2 > filter_1 = false_0 = filter2_2 = times_1 = nil_0 > filter_2# = filter2_4# = filter2_1 = plus_1 = s_1 = times_0 > cons_2 = filter2_0 = s_0 = |0|_0 > app# = true_0 = cons_1 > map_2#
      
  
    partial status:
  
      pi(app#) = [1]
      pi(filter_1) = []
      pi(filter_2#) = []
      pi(cons_2) = []
      pi(filter2_4#) = []
      pi(app) = []
      pi(true_0) = []
      pi(map_1) = [1]
      pi(map_2#) = []
      pi(false_0) = []
      pi(cons_0) = []
      pi(cons_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(times_2) = []
      pi(|0|_0) = []
      pi(filter_2) = [1]
      pi(nil_0) = []

The following rules are removed:

  p4



-- SCC decomposition.

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)) -> filter2_4#(app(f,x),f,x,xs)
p3: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs)
p4: app#(map_1(x1),y) -> map_2#(x1,y)
p5: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p6: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p7: 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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The estimated dependency graph contains the following SCCs:

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


-- 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(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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r1, r3, r5, r6, r7, r14, r23

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 + 38
        cons_2_A(x1,x2) = max{35, x1 - 67, x2 - 67}
        app#_A(x1,x2) = x1 + 37
        map_1_A(x1) = max{41, x1 + 39}
        app_A(x1,x2) = max{25, x1 - 40, x2 - 103}
        cons_0_A = 39
        cons_1_A(x1) = max{24, x1 - 104}
        filter_0_A = 23
        filter_1_A(x1) = max{24, x1 - 104}
        filter2_0_A = 39
        filter2_1_A(x1) = max{22, x1 - 104}
        filter2_2_A(x1,x2) = max{23, x1 - 144, x2 - 104}
        filter2_3_A(x1,x2,x3) = max{24, x1 - 185, x2 - 145, x3 - 104}
        times_0_A = 22
        times_1_A(x1) = max{23, x1 - 104}
        filter_2_A(x1,x2) = max{34, x1 - 39, x2 - 34}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        app# > cons_2 = times_0 > times_1 = filter_2 = nil_0 > app > map_1 = cons_1 = filter_1 = filter2_3 > map_2# = cons_0 = filter_0 = filter2_1 = filter2_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(filter2_3) = []
      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: 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(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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: 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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: 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, r17, r23

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = x2 + 46
        cons_2_A(x1,x2) = max{48, x1 - 44, x2 + 47}
        app_A(x1,x2) = max{21, x1 - 3, x2 - 21}
        cons_0_A = 1
        cons_1_A(x1) = max{2, x1 - 41}
        filter_0_A = 1
        filter_1_A(x1) = max{2, x1 - 74}
        filter2_0_A = 1
        filter2_1_A(x1) = max{2, x1 - 22}
        filter2_2_A(x1,x2) = max{1, x1 - 26, x2 - 22}
        filter2_3_A(x1,x2,x3) = max{2, x1 - 30, x2 - 26, x3 - 22}
        map_0_A = 1
        map_1_A(x1) = 0
        plus_0_A = 2
        plus_1_A(x1) = max{4, x1 - 22}
        s_0_A = 2
        s_1_A(x1) = max{20, x1 - 60}
        times_0_A = 1
        times_1_A(x1) = max{2, x1 - 23}
        times_2_A(x1,x2) = max{8, x1 - 22, x2 - 11}
        |0|_0_A = 6
        filter_2_A(x1,x2) = max{49, x1 - 73, x2 - 50}
        nil_0_A = 48
  
    base order 2:
  
      precedence:
      
        filter_2 > cons_2 = app = nil_0 > map_2# = cons_1 = filter_1 = filter2_3 = map_1 = times_2 > cons_0 = filter_0 = filter2_2 = map_0 = plus_1 = s_1 = times_1 > filter2_1 = plus_0 = s_0 = times_0 = |0|_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(times_2) = []
      pi(|0|_0) = []
      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: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs)
p2: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs)

The set of usable rules consists of

  r2, r5, r6, r7, r11, r13, r17, r23

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{1, x2 - 1}
        cons_2_A(x1,x2) = max{x1 + 16, x2 + 4}
        filter2_4#_A(x1,x2,x3,x4) = max{x3 + 14, x4 + 2}
        app_A(x1,x2) = max{17, x1 - 20, x2 + 5}
        false_0_A = 6
        true_0_A = 8
        cons_1_A(x1) = x1 + 37
        filter2_0_A = 5
        filter2_1_A(x1) = max{6, x1 - 1}
        filter2_2_A(x1,x2) = max{7, x1 - 22, x2 - 1}
        filter2_3_A(x1,x2,x3) = max{6, x1 - 43, x2 - 22, x3 - 1}
        plus_0_A = 5
        plus_1_A(x1) = max{6, x1 - 1}
        s_0_A = 7
        s_1_A(x1) = max{6, x1 - 18}
        times_2_A(x1,x2) = max{21, x1 + 8, x2 - 11}
        |0|_0_A = 5
        filter_2_A(x1,x2) = max{22, x1 - 5, x2 - 10}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        filter_2 > filter2_4# > cons_2 = nil_0 > true_0 = filter2_3 > filter_2# = app > false_0 = s_0 > cons_1 = filter2_0 = filter2_2 = plus_1 = s_1 = times_2 > filter2_1 = plus_0 = |0|_0
      
  
    partial status:
  
      pi(filter_2#) = []
      pi(cons_2) = [2]
      pi(filter2_4#) = [3]
      pi(app) = []
      pi(false_0) = []
      pi(true_0) = []
      pi(cons_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_2) = []
      pi(|0|_0) = []
      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: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,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(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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: 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#(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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: 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, r13, r14, r17, r23

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{x1 + 41, x2 + 4}
        filter2_4#_A(x1,x2,x3,x4) = max{x3 + 40, x4 + 3}
        app_A(x1,x2) = max{43, x1 - 45, x2 + 19}
        false_0_A = 45
        cons_1_A(x1) = x1 + 87
        filter_0_A = 44
        filter_1_A(x1) = max{43, x1 - 45}
        filter2_0_A = 44
        filter2_1_A(x1) = max{17, x1 - 1}
        filter2_2_A(x1,x2) = max{18, x1 - 47, x2 - 1}
        filter2_3_A(x1,x2,x3) = max{42, x1 - 92, x2 - 46, x3 - 1}
        map_0_A = 44
        map_1_A(x1) = max{18, x1 - 1}
        plus_0_A = 4
        plus_1_A(x1) = max{5, x1 - 68}
        s_0_A = 1
        s_1_A(x1) = max{7, x1 - 66}
        times_0_A = 41
        times_1_A(x1) = max{42, x1 + 18}
        times_2_A(x1,x2) = max{41, x1 - 27, x2 - 35}
        |0|_0_A = 5
        filter_2_A(x1,x2) = max{46, x1 - 44, x2 - 21}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        filter_0 = filter2_0 = map_0 = times_0 = filter_2 > cons_2 = app = nil_0 > filter2_4# = cons_1 = filter_1 = filter2_3 = map_1 = times_2 > filter_2# = filter2_2 = s_1 = times_1 > false_0 = filter2_1 = plus_1 = s_0 = |0|_0 > plus_0
      
  
    partial status:
  
      pi(filter_2#) = [2]
      pi(cons_2) = [2]
      pi(filter2_4#) = []
      pi(app) = []
      pi(false_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) = [1]
      pi(times_2) = []
      pi(|0|_0) = []
      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)) -> filter2_4#(app(f,x),f,x,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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: 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: times_2#(x,s_1(y)) -> times_2#(x,y)
p2: times_2#(x,plus_2(y,s_1(z))) -> times_2#(x,s_1(z))
p3: times_2#(x,plus_2(y,s_1(z))) -> times_2#(x,plus_2(y,times_2(s_1(z),|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(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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: 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, r14, r17, r19, r20, r23

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{6, x2}
        s_1_A(x1) = max{55, x1}
        plus_2_A(x1,x2) = max{x1 + 2, x2 + 5}
        times_2_A(x1,x2) = max{7, x1 - 81, x2 - 45}
        |0|_0_A = 3
        app_A(x1,x2) = max{8, x1 - 22, x2 - 44}
        cons_0_A = 1
        cons_1_A(x1) = max{2, x1 - 45}
        filter_0_A = 6
        filter_1_A(x1) = max{7, x1 - 45}
        filter2_0_A = 7
        filter2_1_A(x1) = max{8, x1 - 44}
        filter2_2_A(x1,x2) = max{8, x1 - 66, x2 - 44}
        filter2_3_A(x1,x2,x3) = max{7, x1 - 88, x2 - 66, x3 - 44}
        map_0_A = 6
        map_1_A(x1) = max{7, x1 - 45}
        plus_0_A = 2
        plus_1_A(x1) = max{3, x1 - 45}
        times_0_A = 5
        times_1_A(x1) = max{6, x1 - 45}
        filter_2_A(x1,x2) = max{9, x1 - 21, x2 - 24}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        filter_2 = nil_0 > app = cons_0 = map_0 > plus_2 = times_2 = cons_1 = filter_1 = filter2_0 = filter2_3 = map_1 > times_2# = s_1 = filter_0 = filter2_2 = plus_1 = times_1 > |0|_0 = filter2_1 = plus_0 = times_0
      
  
    partial status:
  
      pi(times_2#) = [2]
      pi(s_1) = [1]
      pi(plus_2) = [2]
      pi(times_2) = []
      pi(|0|_0) = []
      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(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: times_2#(x,s_1(y)) -> times_2#(x,y)
p2: times_2#(x,plus_2(y,s_1(z))) -> times_2#(x,s_1(z))

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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: 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: times_2#(x,s_1(y)) -> times_2#(x,y)
p2: times_2#(x,plus_2(y,s_1(z))) -> times_2#(x,s_1(z))

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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: 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, r17, r23

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        times_2#_A(x1,x2) = x2 + 53
        s_1_A(x1) = max{598, x1 + 98}
        plus_2_A(x1,x2) = max{749, x1 - 51, x2 + 99}
        app_A(x1,x2) = max{47, x1 - 5, x2 - 1}
        cons_0_A = 4
        cons_1_A(x1) = max{3, x1 - 2}
        cons_2_A(x1,x2) = max{22, x1 - 72, x2 - 24}
        filter_0_A = 1
        filter_1_A(x1) = max{2, x1 - 20}
        filter2_0_A = 1
        filter2_1_A(x1) = max{2, x1 - 2}
        filter2_2_A(x1,x2) = max{3, x1 - 8, x2 - 2}
        filter2_3_A(x1,x2,x3) = max{4, x1 - 14, x2 - 8, x3 - 2}
        map_0_A = 3
        map_1_A(x1) = max{4, x1 - 2}
        plus_0_A = 3
        plus_1_A(x1) = max{4, x1 - 52}
        times_0_A = 3
        times_1_A(x1) = max{4, x1 - 850}
        times_2_A(x1,x2) = max{x1 - 849, x2 + 100}
        |0|_0_A = 298
        filter_2_A(x1,x2) = max{26, x1 - 26, x2 - 23}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        plus_2 = times_2 > s_1 = plus_1 > times_2# = app = |0|_0 > filter2_3 = times_1 > cons_1 = cons_2 = filter2_2 = map_1 = plus_0 = times_0 = filter_2 > cons_0 = filter_1 = filter2_1 = map_0 = nil_0 > filter_0 = filter2_0
      
  
    partial status:
  
      pi(times_2#) = [2]
      pi(s_1) = [1]
      pi(plus_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(plus_0) = []
      pi(plus_1) = []
      pi(times_0) = []
      pi(times_1) = []
      pi(times_2) = []
      pi(|0|_0) = []
      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: 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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: 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: 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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: 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, r14, r23

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{3, x2}
        s_1_A(x1) = x1 + 3
        app_A(x1,x2) = max{12, x1 - 23, x2 - 32}
        cons_0_A = 22
        cons_1_A(x1) = max{1, x1 - 33}
        filter_0_A = 11
        filter_1_A(x1) = max{12, x1 - 33}
        filter2_0_A = 9
        filter2_1_A(x1) = max{10, x1 - 33}
        filter2_2_A(x1,x2) = max{11, x1 - 57, x2 - 33}
        filter2_3_A(x1,x2,x3) = max{12, x1 - 80, x2 - 56, x3 - 32}
        map_0_A = 10
        map_1_A(x1) = max{11, x1 - 33}
        plus_0_A = 9
        plus_1_A(x1) = max{10, x1 - 33}
        times_0_A = 1
        times_1_A(x1) = max{2, x1 - 33}
        filter_2_A(x1,x2) = max{16, x1 - 57, x2 - 19}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        s_1 > times_2# = app = filter_0 > map_1 = plus_1 = times_1 > cons_1 = filter2_2 = filter2_3 = map_0 = plus_0 = times_0 = filter_2 > cons_0 = filter_1 = filter2_1 = nil_0 > filter2_0
      
  
    partial status:
  
      pi(times_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(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)

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,plus_2(y,s_1(z))) -> plus_2(times_2(x,plus_2(y,times_2(s_1(z),|0|_0()))),times_2(x,s_1(z)))
r17: times_2(x,|0|_0()) -> |0|_0()
r18: times_2(x,s_1(y)) -> plus_2(times_2(x,y),x)
r19: plus_2(x,|0|_0()) -> x
r20: plus_2(x,s_1(y)) -> s_1(plus_2(x,y))
r21: map_2(f,nil_0()) -> nil_0()
r22: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r23: filter_2(f,nil_0()) -> nil_0()
r24: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs)
r25: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs))
r26: 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, r14, r17, r23

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 + 3, x2 + 2}
        s_1_A(x1) = max{5, x1 + 1}
        app_A(x1,x2) = max{37, x1 - 38, x2 - 15}
        cons_0_A = 37
        cons_1_A(x1) = max{1, x1 - 16}
        filter_0_A = 37
        filter_1_A(x1) = max{1, x1 - 173}
        filter2_0_A = 1
        filter2_1_A(x1) = max{2, x1 - 16}
        filter2_2_A(x1,x2) = max{3, x1 - 55, x2 - 15}
        filter2_3_A(x1,x2,x3) = max{4, x1 - 94, x2 - 59, x3 - 58}
        map_0_A = 36
        map_1_A(x1) = max{37, x1 - 25}
        plus_0_A = 1
        plus_1_A(x1) = max{2, x1 - 16}
        times_0_A = 1
        times_1_A(x1) = max{2, x1 - 16}
        times_2_A(x1,x2) = max{36, x1 - 90, x2 - 29}
        |0|_0_A = 2
        filter_2_A(x1,x2) = max{2, x1 - 172, x2 - 7}
        nil_0_A = 0
  
    base order 2:
  
      precedence:
      
        plus_2# = times_2 > s_1 = filter_2 > app = map_0 = |0|_0 = nil_0 > cons_1 = filter_1 = filter2_2 = filter2_3 = map_1 = plus_1 = times_1 > cons_0 = filter_0 = filter2_1 = plus_0 = times_0 > filter2_0
      
  
    partial status:
  
      pi(plus_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(plus_0) = []
      pi(plus_1) = []
      pi(times_0) = []
      pi(times_1) = []
      pi(times_2) = []
      pi(|0|_0) = []
      pi(filter_2) = []
      pi(nil_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.