YES

We show the termination of the TRS R:

  app(app(map(),f),nil()) -> nil()
  app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
  app(flatten(),app(app(node(),x),xs)) -> app(app(cons(),x),app(concat(),app(app(map(),flatten()),xs)))
  app(concat(),nil()) -> nil()
  app(concat(),app(app(cons(),x),xs)) -> app(app(append(),x),app(concat(),xs))
  app(app(append(),nil()),xs) -> xs
  app(app(append(),app(app(cons(),x),xs)),ys) -> app(app(cons(),x),app(app(append(),xs),ys))

-- SCC decomposition.

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

p1: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(cons(),app(f,x)),app(app(map(),f),xs))
p2: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(cons(),app(f,x))
p3: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(f,x)
p4: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(map(),f),xs)
p5: app#(flatten(),app(app(node(),x),xs)) -> app#(app(cons(),x),app(concat(),app(app(map(),flatten()),xs)))
p6: app#(flatten(),app(app(node(),x),xs)) -> app#(cons(),x)
p7: app#(flatten(),app(app(node(),x),xs)) -> app#(concat(),app(app(map(),flatten()),xs))
p8: app#(flatten(),app(app(node(),x),xs)) -> app#(app(map(),flatten()),xs)
p9: app#(flatten(),app(app(node(),x),xs)) -> app#(map(),flatten())
p10: app#(concat(),app(app(cons(),x),xs)) -> app#(app(append(),x),app(concat(),xs))
p11: app#(concat(),app(app(cons(),x),xs)) -> app#(append(),x)
p12: app#(concat(),app(app(cons(),x),xs)) -> app#(concat(),xs)
p13: app#(app(append(),app(app(cons(),x),xs)),ys) -> app#(app(cons(),x),app(app(append(),xs),ys))
p14: app#(app(append(),app(app(cons(),x),xs)),ys) -> app#(app(append(),xs),ys)
p15: app#(app(append(),app(app(cons(),x),xs)),ys) -> app#(append(),xs)

and R consists of:

r1: app(app(map(),f),nil()) -> nil()
r2: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r3: app(flatten(),app(app(node(),x),xs)) -> app(app(cons(),x),app(concat(),app(app(map(),flatten()),xs)))
r4: app(concat(),nil()) -> nil()
r5: app(concat(),app(app(cons(),x),xs)) -> app(app(append(),x),app(concat(),xs))
r6: app(app(append(),nil()),xs) -> xs
r7: app(app(append(),app(app(cons(),x),xs)),ys) -> app(app(cons(),x),app(app(append(),xs),ys))

The estimated dependency graph contains the following SCCs:

  {p3, p4, p8}
  {p12}
  {p14}


-- Reduction pair.

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

p1: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(f,x)
p2: app#(flatten(),app(app(node(),x),xs)) -> app#(app(map(),flatten()),xs)
p3: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(map(),f),xs)

and R consists of:

r1: app(app(map(),f),nil()) -> nil()
r2: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r3: app(flatten(),app(app(node(),x),xs)) -> app(app(cons(),x),app(concat(),app(app(map(),flatten()),xs)))
r4: app(concat(),nil()) -> nil()
r5: app(concat(),app(app(cons(),x),xs)) -> app(app(append(),x),app(concat(),xs))
r6: app(app(append(),nil()),xs) -> xs
r7: app(app(append(),app(app(cons(),x),xs)),ys) -> app(app(cons(),x),app(app(append(),xs),ys))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        app > map > flatten > app# > node > cons
      
      argument filter:
    
        pi(app#) = [2]
        pi(app) = [1, 2]
        pi(map) = []
        pi(cons) = []
        pi(flatten) = []
        pi(node) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        app > map > flatten > app# > node > cons
      
      argument filter:
    
        pi(app#) = 2
        pi(app) = [1, 2]
        pi(map) = []
        pi(cons) = []
        pi(flatten) = []
        pi(node) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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#(concat(),app(app(cons(),x),xs)) -> app#(concat(),xs)

and R consists of:

r1: app(app(map(),f),nil()) -> nil()
r2: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r3: app(flatten(),app(app(node(),x),xs)) -> app(app(cons(),x),app(concat(),app(app(map(),flatten()),xs)))
r4: app(concat(),nil()) -> nil()
r5: app(concat(),app(app(cons(),x),xs)) -> app(app(append(),x),app(concat(),xs))
r6: app(app(append(),nil()),xs) -> xs
r7: app(app(append(),app(app(cons(),x),xs)),ys) -> app(app(cons(),x),app(app(append(),xs),ys))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        app# > concat > cons > app
      
      argument filter:
    
        pi(app#) = [1, 2]
        pi(concat) = []
        pi(app) = [1, 2]
        pi(cons) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        concat > app# > cons > app
      
      argument filter:
    
        pi(app#) = 2
        pi(concat) = []
        pi(app) = [1, 2]
        pi(cons) = []
    

The next rules are strictly ordered:

  p1

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(map(),f),nil()) -> nil()
r2: app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
r3: app(flatten(),app(app(node(),x),xs)) -> app(app(cons(),x),app(concat(),app(app(map(),flatten()),xs)))
r4: app(concat(),nil()) -> nil()
r5: app(concat(),app(app(cons(),x),xs)) -> app(app(append(),x),app(concat(),xs))
r6: app(app(append(),nil()),xs) -> xs
r7: app(app(append(),app(app(cons(),x),xs)),ys) -> app(app(cons(),x),app(app(append(),xs),ys))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        cons > append > app > app#
      
      argument filter:
    
        pi(app#) = [1]
        pi(app) = [2]
        pi(append) = []
        pi(cons) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        cons > append > app > app#
      
      argument filter:
    
        pi(app#) = []
        pi(app) = []
        pi(append) = []
        pi(cons) = []
    

The next rules are strictly ordered:

  p1

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