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

  r3, r5, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        app#_A(x1,x2) = max{5, x1 - 16, x2 + 1}
        filter_1_A(x1) = x1 + 33
        filter_2#_A(x1,x2) = max{x1 + 16, x2 - 7}
        cons_2_A(x1,x2) = max{x1 + 8, x2 + 4}
        filtersub_2_A(x1,x2) = x2 + 33
        filtersub_3#_A(x1,x2,x3) = max{x2 + 16, x3 - 10}
        false_0_A = 33
        app_A(x1,x2) = max{36, x1 + 35, x2 + 35}
        true_0_A = 36
        filter_0_A = 1
        filtersub_0_A = 1
        filtersub_1_A(x1) = 0
        filter_2_A(x1,x2) = max{34, x1 - 5, x2 + 5}
        nil_0_A = 35
  
    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) = 0
            filter_1_A(x1) = 0
            filter_2#_A(x1,x2) = 0
            cons_2_A(x1,x2) = x1 + x2 + 3
            filtersub_2_A(x1,x2) = x1
            filtersub_3#_A(x1,x2,x3) = 0
            false_0_A() = 1
            app_A(x1,x2) = 1
            true_0_A() = 1
            filter_0_A() = 1
            filtersub_0_A() = 0
            filtersub_1_A(x1) = 0
            filter_2_A(x1,x2) = x1 + 3
            nil_0_A() = 0
            app#!_A(x1,x2) = 3
            filter_1!_A(x1) = 6
            filter_2#!_A(x1,x2) = x1 + x2 + 2
            cons_2!_A(x1,x2) = 1
            filtersub_2!_A(x1,x2) = 8
            filtersub_3#!_A(x1,x2,x3) = x2 + x3 + 1
            false_0!_A() = 0
            app!_A(x1,x2) = x2 + 6
            true_0!_A() = 0
            filter_0!_A() = 7
            filtersub_0!_A() = 8
            filtersub_1!_A(x1) = 7
            filter_2!_A(x1,x2) = 7
            nil_0!_A() = 0
  
    partial status:
  
      pi(app#) = []
      pi(filter_1) = []
      pi(filter_2#) = [1]
      pi(cons_2) = []
      pi(filtersub_2) = []
      pi(filtersub_3#) = []
      pi(false_0) = []
      pi(app) = [1, 2]
      pi(true_0) = []
      pi(filter_0) = []
      pi(filtersub_0) = []
      pi(filtersub_1) = []
      pi(filter_2) = [2]
      pi(nil_0) = []

The following rules are removed:

  p1, p2, p3, p4, p5, p6

Then no dependency pair remains.