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:

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

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

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

  (no SCCs)