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:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      app#_A(x1,x2) = max{0, x1 - 1}
      app_A(x1,x2) = x2 + 27
      filter_A = 5
      cons_A = 17
      filtersub_A = 0
      false_A = 0
      true_A = 0
      nil_A = 34
    
    2. max/plus interpretations on natural numbers:
    
      app#_A(x1,x2) = 0
      app_A(x1,x2) = 1
      filter_A = 0
      cons_A = 0
      filtersub_A = 0
      false_A = 0
      true_A = 0
      nil_A = 0
    

The next rules are strictly ordered:

  p3

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))
p2: app#(app(app(filtersub(),false()),f),app(app(cons(),y),ys)) -> app#(app(filter(),f),ys)
p3: 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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


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

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      app#_A(x1,x2) = max{0, x2 - 5}
      app_A(x1,x2) = x2 + 10
      filter_A = 7
      cons_A = 11
      filtersub_A = 14
      true_A = 0
      false_A = 12
      nil_A = 9
    
    2. max/plus interpretations on natural numbers:
    
      app#_A(x1,x2) = 0
      app_A(x1,x2) = 1
      filter_A = 0
      cons_A = 0
      filtersub_A = 0
      true_A = 0
      false_A = 0
      nil_A = 0
    

The next rules are strictly ordered:

  p2, p3

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)