YES

We show the termination of the TRS R:

  f(x,empty()) -> x
  f(empty(),cons(a,k)) -> f(cons(a,k),k)
  f(cons(a,k),y) -> f(y,k)

-- SCC decomposition.

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

p1: f#(empty(),cons(a,k)) -> f#(cons(a,k),k)
p2: f#(cons(a,k),y) -> f#(y,k)

and R consists of:

r1: f(x,empty()) -> x
r2: f(empty(),cons(a,k)) -> f(cons(a,k),k)
r3: f(cons(a,k),y) -> f(y,k)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: f#(empty(),cons(a,k)) -> f#(cons(a,k),k)
p2: f#(cons(a,k),y) -> f#(y,k)

and R consists of:

r1: f(x,empty()) -> x
r2: f(empty(),cons(a,k)) -> f(cons(a,k),k)
r3: f(cons(a,k),y) -> f(y,k)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{x1 + 5, x2 + 7}
          empty_A = 0
          cons_A(x1,x2) = max{x1, x2 + 3}
    
      precedence:
      
        f# = empty = cons
      
      partial status:
    
        pi(f#) = [1, 2]
        pi(empty) = []
        pi(cons) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{x1 + 3, x2 + 4}
          empty_A = 1
          cons_A(x1,x2) = max{x1 - 5, x2 + 2}
    
      precedence:
      
        f# = empty = cons
      
      partial status:
    
        pi(f#) = [1]
        pi(empty) = []
        pi(cons) = [2]
    

The next rules are strictly ordered:

  p1, p2

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