YES

We show the termination of the TRS R:

  app(app(append(),nil()),ys) -> ys
  app(app(append(),app(app(cons(),x),xs)),ys) -> app(app(cons(),x),app(app(append(),xs),ys))
  app(app(flatwith(),f),app(leaf(),x)) -> app(app(cons(),app(f,x)),nil())
  app(app(flatwith(),f),app(node(),xs)) -> app(app(flatwithsub(),f),xs)
  app(app(flatwithsub(),f),nil()) -> nil()
  app(app(flatwithsub(),f),app(app(cons(),x),xs)) -> app(app(append(),app(app(flatwith(),f),x)),app(app(flatwithsub(),f),xs))

-- SCC decomposition.

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

p1: app#(app(append(),app(app(cons(),x),xs)),ys) -> app#(app(cons(),x),app(app(append(),xs),ys))
p2: app#(app(append(),app(app(cons(),x),xs)),ys) -> app#(app(append(),xs),ys)
p3: app#(app(append(),app(app(cons(),x),xs)),ys) -> app#(append(),xs)
p4: app#(app(flatwith(),f),app(leaf(),x)) -> app#(app(cons(),app(f,x)),nil())
p5: app#(app(flatwith(),f),app(leaf(),x)) -> app#(cons(),app(f,x))
p6: app#(app(flatwith(),f),app(leaf(),x)) -> app#(f,x)
p7: app#(app(flatwith(),f),app(node(),xs)) -> app#(app(flatwithsub(),f),xs)
p8: app#(app(flatwith(),f),app(node(),xs)) -> app#(flatwithsub(),f)
p9: app#(app(flatwithsub(),f),app(app(cons(),x),xs)) -> app#(app(append(),app(app(flatwith(),f),x)),app(app(flatwithsub(),f),xs))
p10: app#(app(flatwithsub(),f),app(app(cons(),x),xs)) -> app#(append(),app(app(flatwith(),f),x))
p11: app#(app(flatwithsub(),f),app(app(cons(),x),xs)) -> app#(app(flatwith(),f),x)
p12: app#(app(flatwithsub(),f),app(app(cons(),x),xs)) -> app#(flatwith(),f)
p13: app#(app(flatwithsub(),f),app(app(cons(),x),xs)) -> app#(app(flatwithsub(),f),xs)

and R consists of:

r1: app(app(append(),nil()),ys) -> ys
r2: app(app(append(),app(app(cons(),x),xs)),ys) -> app(app(cons(),x),app(app(append(),xs),ys))
r3: app(app(flatwith(),f),app(leaf(),x)) -> app(app(cons(),app(f,x)),nil())
r4: app(app(flatwith(),f),app(node(),xs)) -> app(app(flatwithsub(),f),xs)
r5: app(app(flatwithsub(),f),nil()) -> nil()
r6: app(app(flatwithsub(),f),app(app(cons(),x),xs)) -> app(app(append(),app(app(flatwith(),f),x)),app(app(flatwithsub(),f),xs))

The estimated dependency graph contains the following SCCs:

  {p6, p7, p11, p13}
  {p2}


-- Reduction pair.

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

p1: app#(app(flatwithsub(),f),app(app(cons(),x),xs)) -> app#(app(flatwithsub(),f),xs)
p2: app#(app(flatwithsub(),f),app(app(cons(),x),xs)) -> app#(app(flatwith(),f),x)
p3: app#(app(flatwith(),f),app(node(),xs)) -> app#(app(flatwithsub(),f),xs)
p4: app#(app(flatwith(),f),app(leaf(),x)) -> app#(f,x)

and R consists of:

r1: app(app(append(),nil()),ys) -> ys
r2: app(app(append(),app(app(cons(),x),xs)),ys) -> app(app(cons(),x),app(app(append(),xs),ys))
r3: app(app(flatwith(),f),app(leaf(),x)) -> app(app(cons(),app(f,x)),nil())
r4: app(app(flatwith(),f),app(node(),xs)) -> app(app(flatwithsub(),f),xs)
r5: app(app(flatwithsub(),f),nil()) -> nil()
r6: app(app(flatwithsub(),f),app(app(cons(),x),xs)) -> app(app(append(),app(app(flatwith(),f),x)),app(app(flatwithsub(),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:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{5, x1 - 1, x2}
          app_A(x1,x2) = max{x1, x2 + 5}
          flatwithsub_A = 1
          cons_A = 5
          flatwith_A = 3
          node_A = 5
          leaf_A = 0
    
      precedence:
      
        app# = app = flatwithsub = cons = flatwith = node = leaf
      
      partial status:
    
        pi(app#) = [2]
        pi(app) = [1, 2]
        pi(flatwithsub) = []
        pi(cons) = []
        pi(flatwith) = []
        pi(node) = []
        pi(leaf) = []
    
    2. 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 + 9, x2 + 16}
          flatwithsub_A = 6
          cons_A = 5
          flatwith_A = 13
          node_A = 7
          leaf_A = 0
    
      precedence:
      
        app# = app = flatwithsub = cons = flatwith = node = leaf
      
      partial status:
    
        pi(app#) = []
        pi(app) = [2]
        pi(flatwithsub) = []
        pi(cons) = []
        pi(flatwith) = []
        pi(node) = []
        pi(leaf) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4

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

-- Reduction pair.

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

p1: app#(app(append(),app(app(cons(),x),xs)),ys) -> app#(app(append(),xs),ys)

and R consists of:

r1: app(app(append(),nil()),ys) -> ys
r2: app(app(append(),app(app(cons(),x),xs)),ys) -> app(app(cons(),x),app(app(append(),xs),ys))
r3: app(app(flatwith(),f),app(leaf(),x)) -> app(app(cons(),app(f,x)),nil())
r4: app(app(flatwith(),f),app(node(),xs)) -> app(app(flatwithsub(),f),xs)
r5: app(app(flatwithsub(),f),nil()) -> nil()
r6: app(app(flatwithsub(),f),app(app(cons(),x),xs)) -> app(app(append(),app(app(flatwith(),f),x)),app(app(flatwithsub(),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:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{x1 + 4, x2 + 5}
          app_A(x1,x2) = max{x1 + 2, x2 + 4}
          append_A = 2
          cons_A = 3
    
      precedence:
      
        app = append > app# = cons
      
      partial status:
    
        pi(app#) = [1, 2]
        pi(app) = [1, 2]
        pi(append) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          app#_A(x1,x2) = max{x1 - 15, x2 + 3}
          app_A(x1,x2) = max{x1 + 4, x2 + 7}
          append_A = 2
          cons_A = 3
    
      precedence:
      
        app# = app = append = cons
      
      partial status:
    
        pi(app#) = []
        pi(app) = [1, 2]
        pi(append) = []
        pi(cons) = []
    

The next rules are strictly ordered:

  p1

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