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)

-- SCC decomposition.

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

p1: app#(app(filter(),f),app(app(cons(),y),ys)) -> app#(app(app(filtersub(),app(f,y)),f),app(app(cons(),y),ys))
p2: app#(app(filter(),f),app(app(cons(),y),ys)) -> app#(app(filtersub(),app(f,y)),f)
p3: app#(app(filter(),f),app(app(cons(),y),ys)) -> app#(filtersub(),app(f,y))
p4: app#(app(filter(),f),app(app(cons(),y),ys)) -> app#(f,y)
p5: app#(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app#(app(cons(),y),app(app(filter(),f),ys))
p6: app#(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app#(app(filter(),f),ys)
p7: app#(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app#(filter(),f)
p8: app#(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app#(app(filter(),f),ys)
p9: app#(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app#(filter(),f)

and R consists of:

r1: app(app(filter(),f),nil()) -> nil()
r2: app(app(filter(),f),app(app(cons(),y),ys)) -> app(app(app(filtersub(),app(f,y)),f),app(app(cons(),y),ys))
r3: app(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app(app(cons(),y),app(app(filter(),f),ys))
r4: app(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app(app(filter(),f),ys)

The estimated dependency graph contains the following SCCs:

  {p1, p4, p6, p8}


-- Reduction pair.

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

p1: app#(app(filter(),f),app(app(cons(),y),ys)) -> app#(app(app(filtersub(),app(f,y)),f),app(app(cons(),y),ys))
p2: app#(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app#(app(filter(),f),ys)
p3: app#(app(filter(),f),app(app(cons(),y),ys)) -> app#(f,y)
p4: app#(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app#(app(filter(),f),ys)

and R consists of:

r1: app(app(filter(),f),nil()) -> nil()
r2: app(app(filter(),f),app(app(cons(),y),ys)) -> app(app(app(filtersub(),app(f,y)),f),app(app(cons(),y),ys))
r3: app(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app(app(cons(),y),app(app(filter(),f),ys))
r4: app(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app(app(filter(),f),ys)

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      app#_A(x1,x2) = x1 + ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,1,0,0)) x2
      app_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,1,0,0)) x1 + x2 + (1,1,1,1)
      filter_A() = (1,1,1,1)
      cons_A() = (1,1,1,0)
      filtersub_A() = (1,1,1,0)
      false_A() = (1,1,1,1)
      true_A() = (1,1,1,1)
      nil_A() = (1,1,1,1)

The next rules are strictly ordered:

  p2, p3, p4

We remove them from the problem.

-- SCC decomposition.

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

p1: app#(app(filter(),f),app(app(cons(),y),ys)) -> app#(app(app(filtersub(),app(f,y)),f),app(app(cons(),y),ys))

and R consists of:

r1: app(app(filter(),f),nil()) -> nil()
r2: app(app(filter(),f),app(app(cons(),y),ys)) -> app(app(app(filtersub(),app(f,y)),f),app(app(cons(),y),ys))
r3: app(app(app(filtersub(),true()),f),app(app(cons(),y),ys)) -> app(app(cons(),y),app(app(filter(),f),ys))
r4: app(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app(app(filter(),f),ys)

The estimated dependency graph contains the following SCCs:

  (no SCCs)