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

  r5, r24

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{23, x1 + 21, x2 + 10}
        filter_1_A(x1) = max{43, x1 + 18}
        filter_2#_A(x1,x2) = max{x1 + 24, x2 + 3}
        cons_2_A(x1,x2) = max{x1 + 17, x2 + 17}
        map_1_A(x1) = max{65, x1 + 2}
        map_2#_A(x1,x2) = max{66, x1 + 22, x2 - 6}
        filter2_3_A(x1,x2,x3) = max{x2 + 19, x3 + 19}
        filter2_4#_A(x1,x2,x3,x4) = max{x2 + 24, x3 - 1, x4 + 3}
        false_0_A = 20
        app_A(x1,x2) = max{64, x2 + 19}
        true_0_A = 20
        filter2_0_A = 3
        filter2_1_A(x1) = max{2, x1 + 1}
        filter_2_A(x1,x2) = 42
        nil_0_A = 41
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            app#_A(x1,x2) = x1 + x2
            filter_1_A(x1) = x1 + 3
            filter_2#_A(x1,x2) = x1 + x2 + 3
            cons_2_A(x1,x2) = x1 + x2 + 10
            map_1_A(x1) = x1 + 9
            map_2#_A(x1,x2) = x1 + x2
            filter2_3_A(x1,x2,x3) = x1 + x2 + x3 + 6
            filter2_4#_A(x1,x2,x3,x4) = x2 + x4 + 3
            false_0_A() = 1
            app_A(x1,x2) = x1 + x2 + 1
            true_0_A() = 1
            filter2_0_A() = 3
            filter2_1_A(x1) = x1 + 1
            filter_2_A(x1,x2) = x1 + x2 + 9
            nil_0_A() = 12
            app#!_A(x1,x2) = x1 + 20
            filter_1!_A(x1) = 14
            filter_2#!_A(x1,x2) = x2 + 1
            cons_2!_A(x1,x2) = x2 + 18
            map_1!_A(x1) = x1 + 19
            map_2#!_A(x1,x2) = 19
            filter2_3!_A(x1,x2,x3) = 16
            filter2_4#!_A(x1,x2,x3,x4) = x4 + 10
            false_0!_A() = 0
            app!_A(x1,x2) = x1 + 12
            true_0!_A() = 0
            filter2_0!_A() = 0
            filter2_1!_A(x1) = 14
            filter_2!_A(x1,x2) = x2 + 21
            nil_0!_A() = 32
  
    partial status:
  
      pi(app#) = [2]
      pi(filter_1) = [1]
      pi(filter_2#) = [2]
      pi(cons_2) = [1, 2]
      pi(map_1) = [1]
      pi(map_2#) = [1]
      pi(filter2_3) = [3]
      pi(filter2_4#) = []
      pi(false_0) = []
      pi(app) = [2]
      pi(true_0) = []
      pi(filter2_0) = []
      pi(filter2_1) = [1]
      pi(filter_2) = []
      pi(nil_0) = []

The following rules are removed:

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



-- SCC decomposition.

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 estimated dependency graph contains the following SCCs:

  {p1}


-- 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

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = 1
        cons_2_A(x1,x2) = max{11, x1 - 32}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            map_2#_A(x1,x2) = x2
            cons_2_A(x1,x2) = x1 + x2 + 5
            map_2#!_A(x1,x2) = x2 + 4
            cons_2!_A(x1,x2) = x1 + x2 + 5
  
    partial status:
  
      pi(map_2#) = []
      pi(cons_2) = []

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

  r24

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        times_2#_A(x1,x2) = 1
        s_1_A(x1) = 16
        filter_2_A(x1,x2) = 11
        nil_0_A = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            times_2#_A(x1,x2) = x2
            s_1_A(x1) = x1 + 4
            filter_2_A(x1,x2) = x2 + 14
            nil_0_A() = 8
            times_2#!_A(x1,x2) = x2 + 7
            s_1!_A(x1) = 6
            filter_2!_A(x1,x2) = 6
            nil_0!_A() = 0
  
    partial status:
  
      pi(times_2#) = []
      pi(s_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

  r24

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        plus_2#_A(x1,x2) = max{3, x1 - 13}
        s_1_A(x1) = max{12, x1 + 4}
        filter_2_A(x1,x2) = max{x1 + 6, x2 - 13}
        nil_0_A = 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            plus_2#_A(x1,x2) = x1 + x2
            s_1_A(x1) = x1 + 4
            filter_2_A(x1,x2) = x2 + 14
            nil_0_A() = 8
            plus_2#!_A(x1,x2) = x1 + x2
            s_1!_A(x1) = 7
            filter_2!_A(x1,x2) = x1 + x2 + 9
            nil_0!_A() = 0
  
    partial status:
  
      pi(plus_2#) = []
      pi(s_1) = [1]
      pi(filter_2) = []
      pi(nil_0) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.