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:

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

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(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(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:

  {p1, p2, p3}


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

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

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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)

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:

  {p1, p2}


-- 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)

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:

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

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: app#(app(map(),f),app(app(cons(),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:

  {p1}


-- 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)

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:

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

The next rules are strictly ordered:

  p1

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