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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = ((1,0),(1,1)) x1 + (3,0)
            app_A(x1,x2) = ((1,0),(1,1)) x2 + (4,0)
            filter_A() = (8,2)
            cons_A() = (2,1)
            filtersub_A() = (9,1)
            false_A() = (9,3)
            true_A() = (9,5)
            nil_A() = (1,1)
    
      precedence:
      
        app = filter > app# = cons = filtersub = true > false = nil
      
      partial status:
    
        pi(app#) = [1]
        pi(app) = [2]
        pi(filter) = []
        pi(cons) = []
        pi(filtersub) = []
        pi(false) = []
        pi(true) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = ((0,0),(1,0)) x1 + (2,6)
            app_A(x1,x2) = ((1,0),(0,0)) x2 + (4,4)
            filter_A() = (6,2)
            cons_A() = (3,5)
            filtersub_A() = (1,1)
            false_A() = (7,3)
            true_A() = (7,3)
            nil_A() = (5,1)
    
      precedence:
      
        true > filter > nil > false > app# = cons > app = filtersub
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(filter) = []
        pi(cons) = []
        pi(filtersub) = []
        pi(false) = []
        pi(true) = []
        pi(nil) = []
    

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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (9,1)
            app_A(x1,x2) = x2 + (4,4)
            filter_A() = (11,2)
            cons_A() = (1,1)
            filtersub_A() = (2,3)
            true_A() = (3,6)
            false_A() = (8,0)
            nil_A() = (1,0)
    
      precedence:
      
        app# = app = filter = cons = filtersub = true = false = nil
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(filter) = []
        pi(cons) = []
        pi(filtersub) = []
        pi(true) = []
        pi(false) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = x1 + ((0,0),(1,0)) x2
            app_A(x1,x2) = ((1,0),(0,0)) x2 + (1,2)
            filter_A() = (4,4)
            cons_A() = (4,4)
            filtersub_A() = (4,4)
            true_A() = (0,3)
            false_A() = (2,1)
            nil_A() = (0,3)
    
      precedence:
      
        false > true > nil > app# = app = filter = cons = filtersub
      
      partial status:
    
        pi(app#) = [1]
        pi(app) = []
        pi(filter) = []
        pi(cons) = []
        pi(filtersub) = []
        pi(true) = []
        pi(false) = []
        pi(nil) = []
    

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(),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}


-- 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)

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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = x1 + x2 + (6,0)
            app_A(x1,x2) = x2
            filter_A() = (4,3)
            cons_A() = (1,1)
            filtersub_A() = (2,2)
            true_A() = (5,0)
            nil_A() = (1,0)
            false_A() = (1,0)
    
      precedence:
      
        app# = app = filter = cons = filtersub = true = nil = false
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(filter) = []
        pi(cons) = []
        pi(filtersub) = []
        pi(true) = []
        pi(nil) = []
        pi(false) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (3,8)
            app_A(x1,x2) = x2 + (0,4)
            filter_A() = (1,2)
            cons_A() = (1,2)
            filtersub_A() = (1,2)
            true_A() = (2,3)
            nil_A() = (0,1)
            false_A() = (2,3)
    
      precedence:
      
        filtersub > true > nil > app# = app > filter > cons > false
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(filter) = []
        pi(cons) = []
        pi(filtersub) = []
        pi(true) = []
        pi(nil) = []
        pi(false) = []
    

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(),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)