YES
We show the termination of the TRS R:

  app(app(app(consif(),true()),x),ys) -> app(app(cons(),x),ys)
  app(app(app(consif(),false()),x),ys) -> ys
  app(app(filter(),f),nil()) -> nil()
  app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(consif(),app(f,x)),x),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(consif_0(),y) -> consif_1(y)
  app(consif_1(x1),y) -> consif_2(x1,y)
  app(consif_2(x1,x2),y) -> consif_3(x1,x2,y)
  app(filter_0(),y) -> filter_1(y)
  app(filter_1(x1),y) -> filter_2(x1,y)
  consif_3(true_0(),x,ys) -> cons_2(x,ys)
  consif_3(false_0(),x,ys) -> ys
  filter_2(f,nil_0()) -> nil_0()
  filter_2(f,cons_2(x,xs)) -> consif_3(app(f,x),x,filter_2(f,xs))


-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: app#(consif_2(x1,x2),y) -> consif_3#(x1,x2,y)
p2: app#(filter_1(x1),y) -> filter_2#(x1,y)
p3: filter_2#(f,cons_2(x,xs)) -> consif_3#(app(f,x),x,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(consif_0(),y) -> consif_1(y)
r4: app(consif_1(x1),y) -> consif_2(x1,y)
r5: app(consif_2(x1,x2),y) -> consif_3(x1,x2,y)
r6: app(filter_0(),y) -> filter_1(y)
r7: app(filter_1(x1),y) -> filter_2(x1,y)
r8: consif_3(true_0(),x,ys) -> cons_2(x,ys)
r9: consif_3(false_0(),x,ys) -> ys
r10: filter_2(f,nil_0()) -> nil_0()
r11: filter_2(f,cons_2(x,xs)) -> consif_3(app(f,x),x,filter_2(f,xs))

The estimated dependency graph contains the following SCCs:

  {p2, 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(consif_0(),y) -> consif_1(y)
r4: app(consif_1(x1),y) -> consif_2(x1,y)
r5: app(consif_2(x1,x2),y) -> consif_3(x1,x2,y)
r6: app(filter_0(),y) -> filter_1(y)
r7: app(filter_1(x1),y) -> filter_2(x1,y)
r8: consif_3(true_0(),x,ys) -> cons_2(x,ys)
r9: consif_3(false_0(),x,ys) -> ys
r10: filter_2(f,nil_0()) -> nil_0()
r11: filter_2(f,cons_2(x,xs)) -> consif_3(app(f,x),x,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:
  
      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{0, x1 - 16, x2 - 4}
            filter_1_A(x1) = x1 + 15
            filter_2#_A(x1,x2) = max{0, x1 - 16, x2 - 4}
            cons_2_A(x1,x2) = max{x1 + 3, x2}
            app#!_A(x1,x2) = max{3, x1 - 10, x2 - 3}
            filter_1!_A(x1) = max{1, x1 - 2}
            filter_2#!_A(x1,x2) = max{4, x1 - 2, x2 - 5}
            cons_2!_A(x1,x2) = max{2, x2 - 5}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            app#_A(x1,x2) = max{0, x1 - 10, x2 - 4}
            filter_1_A(x1) = x1 + 9
            filter_2#_A(x1,x2) = max{0, x1 - 10, x2 - 4}
            cons_2_A(x1,x2) = max{3, x1 + 2, x2}
            app#!_A(x1,x2) = max{5, x1 - 2, x2 - 27}
            filter_1!_A(x1) = max{2, x1 - 54}
            filter_2#!_A(x1,x2) = max{x1 + 6, x2 - 28}
            cons_2!_A(x1,x2) = max{3, x1 - 28, x2 - 27}
        
  
    partial status:
  
      pi(app#) = []
      pi(filter_1) = [1]
      pi(filter_2#) = []
      pi(cons_2) = [2]

The following rules are removed:

  p1, p3



-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: 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(consif_0(),y) -> consif_1(y)
r4: app(consif_1(x1),y) -> consif_2(x1,y)
r5: app(consif_2(x1,x2),y) -> consif_3(x1,x2,y)
r6: app(filter_0(),y) -> filter_1(y)
r7: app(filter_1(x1),y) -> filter_2(x1,y)
r8: consif_3(true_0(),x,ys) -> cons_2(x,ys)
r9: consif_3(false_0(),x,ys) -> ys
r10: filter_2(f,nil_0()) -> nil_0()
r11: filter_2(f,cons_2(x,xs)) -> consif_3(app(f,x),x,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: 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(consif_0(),y) -> consif_1(y)
r4: app(consif_1(x1),y) -> consif_2(x1,y)
r5: app(consif_2(x1,x2),y) -> consif_3(x1,x2,y)
r6: app(filter_0(),y) -> filter_1(y)
r7: app(filter_1(x1),y) -> filter_2(x1,y)
r8: consif_3(true_0(),x,ys) -> cons_2(x,ys)
r9: consif_3(false_0(),x,ys) -> ys
r10: filter_2(f,nil_0()) -> nil_0()
r11: filter_2(f,cons_2(x,xs)) -> consif_3(app(f,x),x,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:
  
      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:
          
            filter_2#_A(x1,x2) = x2
            cons_2_A(x1,x2) = max{x1 - 3, x2 + 3}
            filter_2#!_A(x1,x2) = max{1, x2 - 2}
            cons_2!_A(x1,x2) = max{2, x1 - 10, x2 - 1}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            filter_2#_A(x1,x2) = 0
            cons_2_A(x1,x2) = max{4, x1 - 4, x2 - 6}
            filter_2#!_A(x1,x2) = max{1, x1 - 1, x2 - 5}
            cons_2!_A(x1,x2) = max{0, x1 - 11, x2 - 10}
        
  
    partial status:
  
      pi(filter_2#) = [2]
      pi(cons_2) = [2]

The following rules are removed:

  p1

Then no dependency pair remains.