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, r5, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{2, x1 - 5, x2 - 1}
            filter_1_A(x1) = max{4, x1 + 3}
            filter_2#_A(x1,x2) = max{1, x1 - 2, x2 - 54}
            cons_2_A(x1,x2) = max{60, x1 + 53, x2 + 38}
            filtersub_2_A(x1,x2) = max{4, x1 + 3, x2 + 3}
            filtersub_3#_A(x1,x2,x3) = max{2, x2 - 2, x3 - 54}
            false_0_A = 60
            app_A(x1,x2) = max{60, x1 - 5, x2 + 3}
            true_0_A = 60
            cons_0_A = 4
            cons_1_A(x1) = max{4, x1 + 3}
            filtersub_0_A = 4
            filtersub_1_A(x1) = max{4, x1 + 3}
            filter_2_A(x1,x2) = max{5, x1 - 40, x2 + 1}
            nil_0_A = 6
            app#!_A(x1,x2) = max{32, x2 + 28}
            filter_1!_A(x1) = max{0, x1 - 2}
            filter_2#!_A(x1,x2) = max{27, x2 - 24}
            cons_2!_A(x1,x2) = max{30, x1 + 24, x2 + 10}
            filtersub_2!_A(x1,x2) = max{0, x1 - 3, x2 - 2}
            filtersub_3#!_A(x1,x2,x3) = max{31, x3 - 61}
            false_0!_A = 26
            app!_A(x1,x2) = max{37, x1 - 5, x2 - 1}
            true_0!_A = 24
            cons_0!_A = 38
            cons_1!_A(x1) = max{37, x1 - 1}
            filtersub_0!_A = 1
            filtersub_1!_A(x1) = max{0, x1 - 4}
            filter_2!_A(x1,x2) = max{25, x2 - 29}
            nil_0!_A = 1
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{1, x1 - 3, x2 - 4}
            filter_1_A(x1) = x1 + 2
            filter_2#_A(x1,x2) = max{1, x1 - 3, x2 - 11}
            cons_2_A(x1,x2) = max{10, x1 + 7, x2 + 7}
            filtersub_2_A(x1,x2) = max{x1 + 2, x2}
            filtersub_3#_A(x1,x2,x3) = max{1, x2 - 3, x3 - 11}
            false_0_A = 20
            app_A(x1,x2) = max{20, x1 - 3, x2 - 3}
            true_0_A = 20
            cons_0_A = 2
            cons_1_A(x1) = max{0, x1 - 3}
            filtersub_0_A = 0
            filtersub_1_A(x1) = max{0, x1 - 3}
            filter_2_A(x1,x2) = max{13, x1 - 10, x2 + 2}
            nil_0_A = 0
            app#!_A(x1,x2) = max{1, x1 - 5, x2 - 13}
            filter_1!_A(x1) = max{4, x1 - 3}
            filter_2#!_A(x1,x2) = max{3, x1 - 2, x2 - 12}
            cons_2!_A(x1,x2) = 0
            filtersub_2!_A(x1,x2) = max{0, x1 - 22, x2 - 4}
            filtersub_3#!_A(x1,x2,x3) = max{x2 + 4, x3 - 11}
            false_0!_A = 4
            app!_A(x1,x2) = max{5, x1 - 1, x2 - 3}
            true_0!_A = 7
            cons_0!_A = 3
            cons_1!_A(x1) = max{2, x1 - 4}
            filtersub_0!_A = 1
            filtersub_1!_A(x1) = max{0, x1 - 23}
            filter_2!_A(x1,x2) = max{6, x1 + 2, x2 - 1}
            nil_0!_A = 3
        
  
    partial status:
  
      pi(app#) = [2]
      pi(filter_1) = [1]
      pi(filter_2#) = [2]
      pi(cons_2) = [1, 2]
      pi(filtersub_2) = [2]
      pi(filtersub_3#) = [3]
      pi(false_0) = []
      pi(app) = [2]
      pi(true_0) = []
      pi(cons_0) = []
      pi(cons_1) = [1]
      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.