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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = x1 + (5,2)
            app_A(x1,x2) = ((1,0),(1,1)) x2 + (2,1)
            filter_A() = (4,4)
            cons_A() = (1,5)
    
      precedence:
      
        filter = cons > app# = app
      
      partial status:
    
        pi(app#) = [1]
        pi(app) = [2]
        pi(filter) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = ((1,0),(1,1)) x1 + (3,4)
            app_A(x1,x2) = x2 + (4,3)
            filter_A() = (1,1)
            cons_A() = (2,2)
    
      precedence:
      
        app# > cons > filter > app
      
      partial status:
    
        pi(app#) = [1]
        pi(app) = []
        pi(filter) = []
        pi(cons) = []
    

The next rules are strictly ordered:

  p2

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(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#(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 set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (1,1)
            app_A(x1,x2) = ((1,0),(1,0)) x2 + (3,2)
            filter_A() = (2,1)
            cons_A() = (1,4)
    
      precedence:
      
        app# = app = filter = cons
      
      partial status:
    
        pi(app#) = [1, 2]
        pi(app) = []
        pi(filter) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = (0,0)
            app_A(x1,x2) = (1,1)
            filter_A() = (0,0)
            cons_A() = (2,2)
    
      precedence:
      
        cons > app# = app = filter
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(filter) = []
        pi(cons) = []
    

The next rules are strictly ordered:

  p1

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