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))

-- SCC decomposition.

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

p1: 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))
p2: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(app(if(),app(f,x)),app(app(cons(),x),app(app(filter(),f),xs)))
p3: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(if(),app(f,x))
p4: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(f,x)
p5: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(app(cons(),x),app(app(filter(),f),xs))
p6: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(app(filter(),f),xs)

and R consists of:

r1: app(app(app(if(),true()),x),y) -> x
r2: app(app(app(if(),false()),x),y) -> y
r3: app(app(filter(),f),nil()) -> nil()
r4: 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))

The estimated dependency graph contains the following SCCs:

  {p4, p6}


-- Reduction pair.

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

p1: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(f,x)
p2: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(app(filter(),f),xs)

and R consists of:

r1: app(app(app(if(),true()),x),y) -> x
r2: app(app(app(if(),false()),x),y) -> y
r3: app(app(filter(),f),nil()) -> nil()
r4: 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))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        app#_A(x1,x2) = max{x1 + 8, x2 - 4}
        app_A(x1,x2) = max{x1 + 6, x2 + 3}
        filter_A = 2
        cons_A = 9
  
    precedence:
    
      app# = app = filter = cons
    
    partial status:
  
      pi(app#) = []
      pi(app) = [1, 2]
      pi(filter) = []
      pi(cons) = []

The next rules are strictly ordered:

  p1

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(),x),xs)) -> app#(app(filter(),f),xs)

and R consists of:

r1: app(app(app(if(),true()),x),y) -> x
r2: app(app(app(if(),false()),x),y) -> y
r3: app(app(filter(),f),nil()) -> nil()
r4: 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))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(app(filter(),f),xs)

and R consists of:

r1: app(app(app(if(),true()),x),y) -> x
r2: app(app(app(if(),false()),x),y) -> y
r3: app(app(filter(),f),nil()) -> nil()
r4: 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))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        app#_A(x1,x2) = max{0, x2 - 5}
        app_A(x1,x2) = max{x1 + 7, x2 + 4}
        filter_A = 3
        cons_A = 4
  
    precedence:
    
      app# = app = filter = cons
    
    partial status:
  
      pi(app#) = []
      pi(app) = [1, 2]
      pi(filter) = []
      pi(cons) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.