YES
We show the termination of the TRS R:

  app(app(app(if(),true()),x),y) -> x
  app(app(app(if(),false()),x),y) -> y
  app(app(filter(),f),nil()) -> nil()
  app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(if(),app(f,x)),app(app(cons(),x),app(app(filter(),f),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(if_0(),y) -> if_1(y)
  app(if_1(x1),y) -> if_2(x1,y)
  app(if_2(x1,x2),y) -> if_3(x1,x2,y)
  if_3(true_0(),x,y) -> x
  if_3(false_0(),x,y) -> y
  filter_2(f,nil_0()) -> nil_0()
  filter_2(f,cons_2(x,xs)) -> if_3(app(f,x),cons_2(x,filter_2(f,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#(if_2(x1,x2),y) -> if_3#(x1,x2,y)
p3: filter_2#(f,cons_2(x,xs)) -> if_3#(app(f,x),cons_2(x,filter_2(f,xs)),filter_2(f,xs))
p4: filter_2#(f,cons_2(x,xs)) -> app#(f,x)
p5: filter_2#(f,cons_2(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(if_0(),y) -> if_1(y)
r6: app(if_1(x1),y) -> if_2(x1,y)
r7: app(if_2(x1,x2),y) -> if_3(x1,x2,y)
r8: if_3(true_0(),x,y) -> x
r9: if_3(false_0(),x,y) -> y
r10: filter_2(f,nil_0()) -> nil_0()
r11: filter_2(f,cons_2(x,xs)) -> if_3(app(f,x),cons_2(x,filter_2(f,xs)),filter_2(f,xs))

The estimated dependency graph contains the following SCCs:

  {p1, p4, p5}


-- 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)) -> filter_2#(f,xs)
p3: filter_2#(f,cons_2(x,xs)) -> app#(f,x)

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(if_0(),y) -> if_1(y)
r6: app(if_1(x1),y) -> if_2(x1,y)
r7: app(if_2(x1,x2),y) -> if_3(x1,x2,y)
r8: if_3(true_0(),x,y) -> x
r9: if_3(false_0(),x,y) -> y
r10: filter_2(f,nil_0()) -> nil_0()
r11: filter_2(f,cons_2(x,xs)) -> if_3(app(f,x),cons_2(x,filter_2(f,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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          app#_A(x1,x2) = x2 + 4
          filter_1_A(x1) = x1 + 2
          filter_2#_A(x1,x2) = x2 + 3
          cons_2_A(x1,x2) = x1 + x2 + 5
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          app#_A(x1,x2) = max{0, x1 - 5, x2 - 1}
          filter_1_A(x1) = max{4, x1}
          filter_2#_A(x1,x2) = max{0, x1 - 5, x2 - 12}
          cons_2_A(x1,x2) = max{x1 + 11, x2 + 8}
          app#!_A(x1,x2) = max{3, x1 - 5, x2 - 8}
          filter_1!_A(x1) = 6
          filter_2#!_A(x1,x2) = max{1, x2 - 9}
          cons_2!_A(x1,x2) = max{x1 + 4, x2 - 8}
  
    partial status:
  
      pi(app#) = []
      pi(filter_1) = [1]
      pi(filter_2#) = []
      pi(cons_2) = [2]

The following rules are removed:

  p1, p2, p3

Then no dependency pair remains.