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))

-- 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)

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))

The estimated dependency graph contains the following SCCs:

  {p3, p4}


-- 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(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))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          app#_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (7,2)
          app_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (2,0)
          map_A() = (4,0)
          cons_A() = (1,1)
  
    precedence:
    
      app# = app = map = cons
    
    partial status:
  
      pi(app#) = []
      pi(app) = [2]
      pi(map) = []
      pi(cons) = []

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(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))

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#(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))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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