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

  r1, r3

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          app#_A(x1,x2) = x2 + 5
          filter_1_A(x1) = x1 + 3
          filter_2#_A(x1,x2) = x2 + 4
          cons_2_A(x1,x2) = x1 + x2 + 6
          app_A(x1,x2) = 2
          cons_0_A() = 2
          cons_1_A(x1) = 1
          consif_0_A() = 2
          consif_1_A(x1) = 1
  
    base order 2:
  
      precedence:
      
        cons_2 > app# = filter_2# > filter_1 > app = cons_0 = consif_0 > cons_1 = consif_1
      
  
    partial status:
  
      pi(app#) = [2]
      pi(filter_1) = []
      pi(filter_2#) = [2]
      pi(cons_2) = [2]
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(consif_0) = []
      pi(consif_1) = []

The following rules are removed:

  p1



-- 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)
p2: 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 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

  r1, r3, r4, r6, r9

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          filter_2#_A(x1,x2) = x2
          cons_2_A(x1,x2) = x1 + x2 + 8
          app_A(x1,x2) = x2 + 3
          cons_0_A() = 2
          cons_1_A(x1) = 1
          consif_0_A() = 3
          consif_1_A(x1) = 2
          consif_2_A(x1,x2) = 1
          filter_0_A() = 1
          filter_1_A(x1) = 2
          consif_3_A(x1,x2,x3) = x1 + x3 + 4
          false_0_A() = 1
  
    base order 2:
  
      precedence:
      
        consif_3 > app = cons_0 = consif_0 = false_0 > filter_2# = cons_2 = consif_1 > cons_1 = consif_2 = filter_1 > filter_0
      
  
    partial status:
  
      pi(filter_2#) = []
      pi(cons_2) = [2]
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(consif_0) = []
      pi(consif_1) = []
      pi(consif_2) = []
      pi(filter_0) = []
      pi(filter_1) = []
      pi(consif_3) = []
      pi(false_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.