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 combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = ((0,0),(1,0)) x2 + (3,3)
            app_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (7,23)
            map_A() = (2,24)
            cons_A() = (8,25)
            treemap_A() = (4,1)
            node_A() = (5,2)
    
      precedence:
      
        app# = app = map = cons = treemap = node
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(map) = []
        pi(cons) = []
        pi(treemap) = []
        pi(node) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = (1,1)
            app_A(x1,x2) = (5,5)
            map_A() = (2,2)
            cons_A() = (3,3)
            treemap_A() = (3,3)
            node_A() = (4,4)
    
      precedence:
      
        treemap = node > cons > app# = app = map
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(map) = []
        pi(cons) = []
        pi(treemap) = []
        pi(node) = []
    

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(treemap(),f),app(app(node(),x),xs)) -> app#(app(map(),app(treemap(),f)),xs)
p2: app#(app(map(),f),app(app(cons(),x),xs)) -> app#(app(map(),f),xs)
p3: 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 estimated dependency graph contains the following SCCs:

  {p3}
  {p2}


-- Reduction pair.

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

p1: 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 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 + (4,4)
            app_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (2,2)
            treemap_A() = (5,1)
            node_A() = (1,1)
    
      precedence:
      
        app# = app = treemap = node
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(treemap) = []
        pi(node) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = (1,2)
            app_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (1,3)
            treemap_A() = (2,1)
            node_A() = (2,3)
    
      precedence:
      
        app# = app > treemap = node
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(treemap) = []
        pi(node) = []
    

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(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(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 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)) x2 + (5,2)
            app_A(x1,x2) = x2 + (1,3)
            map_A() = (4,5)
            cons_A() = (2,4)
    
      precedence:
      
        app# = app = map = cons
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(map) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            app#_A(x1,x2) = ((0,0),(1,0)) x2 + (4,4)
            app_A(x1,x2) = (2,2)
            map_A() = (1,1)
            cons_A() = (3,3)
    
      precedence:
      
        app = map > cons > app#
      
      partial status:
    
        pi(app#) = []
        pi(app) = []
        pi(map) = []
        pi(cons) = []
    

The next rules are strictly ordered:

  p1

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