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))

-- SCC decomposition.

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

p1: app#(app(app(consif(),true()),x),ys) -> app#(app(cons(),x),ys)
p2: app#(app(app(consif(),true()),x),ys) -> app#(cons(),x)
p3: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(app(app(consif(),app(f,x)),x),app(app(filter(),f),xs))
p4: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(app(consif(),app(f,x)),x)
p5: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(consif(),app(f,x))
p6: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(f,x)
p7: app#(app(filter(),f),app(app(cons(),x),xs)) -> app#(app(filter(),f),xs)

and R consists of:

r1: app(app(app(consif(),true()),x),ys) -> app(app(cons(),x),ys)
r2: app(app(app(consif(),false()),x),ys) -> ys
r3: app(app(filter(),f),nil()) -> nil()
r4: app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(consif(),app(f,x)),x),app(app(filter(),f),xs))

The estimated dependency graph contains the following SCCs:

  {p6, p7}


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

and R consists of:

r1: app(app(app(consif(),true()),x),ys) -> app(app(cons(),x),ys)
r2: app(app(app(consif(),false()),x),ys) -> ys
r3: app(app(filter(),f),nil()) -> nil()
r4: app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(consif(),app(f,x)),x),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 - 8}
        app_A(x1,x2) = max{x1 + 5, x2 + 2}
        filter_A = 1
        cons_A = 0
  
    precedence:
    
      app# = app = filter = cons
    
    partial status:
  
      pi(app#) = []
      pi(app) = [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#(f,x)

and R consists of:

r1: app(app(app(consif(),true()),x),ys) -> app(app(cons(),x),ys)
r2: app(app(app(consif(),false()),x),ys) -> ys
r3: app(app(filter(),f),nil()) -> nil()
r4: app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(consif(),app(f,x)),x),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#(f,x)

and R consists of:

r1: app(app(app(consif(),true()),x),ys) -> app(app(cons(),x),ys)
r2: app(app(app(consif(),false()),x),ys) -> ys
r3: app(app(filter(),f),nil()) -> nil()
r4: app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(consif(),app(f,x)),x),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) = x1 + 1
        app_A(x1,x2) = max{5, x1 - 3, x2 + 2}
        filter_A = 3
        cons_A = 8
  
    precedence:
    
      app# = app = filter = cons
    
    partial status:
  
      pi(app#) = [1]
      pi(app) = [2]
      pi(filter) = []
      pi(cons) = []

The next rules are strictly ordered:

  p1

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