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(app(treemap(),f),app(app(node(),x),xs)) -> app(app(node(),app(f,x)),app(app(map(),app(treemap(),f)),xs))

-- 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#(app(treemap(),f),app(app(node(),x),xs)) -> app#(app(node(),app(f,x)),app(app(map(),app(treemap(),f)),xs))
p6: app#(app(treemap(),f),app(app(node(),x),xs)) -> app#(node(),app(f,x))
p7: app#(app(treemap(),f),app(app(node(),x),xs)) -> app#(f,x)
p8: app#(app(treemap(),f),app(app(node(),x),xs)) -> app#(app(map(),app(treemap(),f)),xs)
p9: app#(app(treemap(),f),app(app(node(),x),xs)) -> app#(map(),app(treemap(),f))

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(app(treemap(),f),app(app(node(),x),xs)) -> app(app(node(),app(f,x)),app(app(map(),app(treemap(),f)),xs))

The estimated dependency graph contains the following SCCs:

  {p3, p4, p7, p8}


-- 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#(app(treemap(),f),app(app(node(),x),xs)) -> app#(app(map(),app(treemap(),f)),xs)
p3: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(map(),f),xs)
p4: app#(app(treemap(),f),app(app(node(),x),xs)) -> app#(f,x)

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(app(treemap(),f),app(app(node(),x),xs)) -> app(app(node(),app(f,x)),app(app(map(),app(treemap(),f)),xs))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      app > map > treemap > app# > node > cons
    
    argument filter:
  
      pi(app#) = 2
      pi(app) = [1, 2]
      pi(map) = []
      pi(cons) = []
      pi(treemap) = []
      pi(node) = []

The next rules are strictly ordered:

  p1, p2, p3, p4

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