YES
We show the termination of the TRS R:

  app(app(filter(),f),nil()) -> nil()
  app(app(filter(),f),app(app(cons(),y),ys)) -> app(app(app(filtersub(),app(f,y)),f),app(app(cons(),y),ys))
  app(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app(app(cons(),y),app(app(filter(),f),ys))
  app(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app(app(filter(),f),ys)

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(filtersub_0(),y) -> filtersub_1(y)
  app(filtersub_1(x1),y) -> filtersub_2(x1,y)
  app(filtersub_2(x1,x2),y) -> filtersub_3(x1,x2,y)
  filter_2(f,nil_0()) -> nil_0()
  filter_2(f,cons_2(y,ys)) -> filtersub_3(app(f,y),f,cons_2(y,ys))
  filtersub_3(true_0(),f,cons_2(y,ys)) -> cons_2(y,filter_2(f,ys))
  filtersub_3(false_0(),f,cons_2(y,ys)) -> filter_2(f,ys)


-- 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#(filtersub_2(x1,x2),y) -> filtersub_3#(x1,x2,y)
p3: filter_2#(f,cons_2(y,ys)) -> filtersub_3#(app(f,y),f,cons_2(y,ys))
p4: filter_2#(f,cons_2(y,ys)) -> app#(f,y)
p5: filtersub_3#(true_0(),f,cons_2(y,ys)) -> filter_2#(f,ys)
p6: filtersub_3#(false_0(),f,cons_2(y,ys)) -> filter_2#(f,ys)

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(filtersub_0(),y) -> filtersub_1(y)
r6: app(filtersub_1(x1),y) -> filtersub_2(x1,y)
r7: app(filtersub_2(x1,x2),y) -> filtersub_3(x1,x2,y)
r8: filter_2(f,nil_0()) -> nil_0()
r9: filter_2(f,cons_2(y,ys)) -> filtersub_3(app(f,y),f,cons_2(y,ys))
r10: filtersub_3(true_0(),f,cons_2(y,ys)) -> cons_2(y,filter_2(f,ys))
r11: filtersub_3(false_0(),f,cons_2(y,ys)) -> filter_2(f,ys)

The estimated dependency graph contains the following SCCs:

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


-- 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(y,ys)) -> app#(f,y)
p3: app#(filtersub_2(x1,x2),y) -> filtersub_3#(x1,x2,y)
p4: filtersub_3#(false_0(),f,cons_2(y,ys)) -> filter_2#(f,ys)
p5: filter_2#(f,cons_2(y,ys)) -> filtersub_3#(app(f,y),f,cons_2(y,ys))
p6: filtersub_3#(true_0(),f,cons_2(y,ys)) -> filter_2#(f,ys)

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(filtersub_0(),y) -> filtersub_1(y)
r6: app(filtersub_1(x1),y) -> filtersub_2(x1,y)
r7: app(filtersub_2(x1,x2),y) -> filtersub_3(x1,x2,y)
r8: filter_2(f,nil_0()) -> nil_0()
r9: filter_2(f,cons_2(y,ys)) -> filtersub_3(app(f,y),f,cons_2(y,ys))
r10: filtersub_3(true_0(),f,cons_2(y,ys)) -> cons_2(y,filter_2(f,ys))
r11: filtersub_3(false_0(),f,cons_2(y,ys)) -> filter_2(f,ys)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          app#_A(x1,x2) = x1
          filter_1_A(x1) = x1 + 5
          filter_2#_A(x1,x2) = x1 + 1
          cons_2_A(x1,x2) = 2
          filtersub_2_A(x1,x2) = x1 + x2 + 5
          filtersub_3#_A(x1,x2,x3) = x2 + 1
          false_0_A() = 2
          app_A(x1,x2) = x1 + x2 + 5
          true_0_A() = 3
          cons_0_A() = 1
          cons_1_A(x1) = x1 + 3
          filter_0_A() = 1
          filter_2_A(x1,x2) = x1 + 4
          filtersub_0_A() = 1
          filtersub_1_A(x1) = x1 + 2
          filtersub_3_A(x1,x2,x3) = x2 + 4
          nil_0_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 31, x2 - 30}
          filter_1_A(x1) = x1 + 30
          filter_2#_A(x1,x2) = max{0, x1 - 31, x2 - 37}
          cons_2_A(x1,x2) = max{13, x1 + 7, x2 + 11}
          filtersub_2_A(x1,x2) = max{x1 - 44, x2 + 28}
          filtersub_3#_A(x1,x2,x3) = max{0, x1 - 75, x2 - 31, x3 - 43}
          false_0_A = 61
          app_A(x1,x2) = max{52, x1 - 32, x2 + 45}
          true_0_A = 54
          cons_0_A = 18
          cons_1_A(x1) = x1 + 45
          filter_0_A = 18
          filter_2_A(x1,x2) = x2 + 20
          filtersub_0_A = 18
          filtersub_1_A(x1) = x1 + 18
          filtersub_3_A(x1,x2,x3) = x3 + 20
          nil_0_A = 23
          app#!_A(x1,x2) = max{x1 + 44, x2 - 6}
          filter_1!_A(x1) = max{9, x1 - 1}
          filter_2#!_A(x1,x2) = max{9, x2}
          cons_2!_A(x1,x2) = max{27, x1 - 9, x2 - 33}
          filtersub_2!_A(x1,x2) = max{0, x1 - 63, x2 - 1}
          filtersub_3#!_A(x1,x2,x3) = max{10, x3 - 6}
          false_0!_A = 1
          app!_A(x1,x2) = max{14, x1 - 19, x2 + 6}
          true_0!_A = 1
          cons_0!_A = 1
          cons_1!_A(x1) = max{2, x1 - 10}
          filter_0!_A = 1
          filter_2!_A(x1,x2) = max{x1 + 10, x2 - 24}
          filtersub_0!_A = 1
          filtersub_1!_A(x1) = max{0, x1 - 64}
          filtersub_3!_A(x1,x2,x3) = max{x1 - 62, x2 + 10, x3 - 25}
          nil_0!_A = 10
  
    partial status:
  
      pi(app#) = [1]
      pi(filter_1) = [1]
      pi(filter_2#) = [1]
      pi(cons_2) = []
      pi(filtersub_2) = [2]
      pi(filtersub_3#) = []
      pi(false_0) = []
      pi(app) = []
      pi(true_0) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(filter_0) = []
      pi(filter_2) = [1]
      pi(filtersub_0) = []
      pi(filtersub_1) = [1]
      pi(filtersub_3) = [2]
      pi(nil_0) = []

The following rules are removed:

  p1, p2, p3, p4, p5, p6

Then no dependency pair remains.