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 monotone reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      f#_A(x1,x2) = x1 + ((1,1),(1,1)) x2
      empty_A() = (1,1)
      cons_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3

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