YES

TRS:

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

max/plus interpretations on N:

  f_A(x1,x2) = max{1, 2 + x1, 3 + x2}
  f#_A(x1,x2) = max{1, 2 + x1, 3 + x2}
  empty_A = 0
  empty#_A = 0
  cons_A(x1,x2) = max{0, x1, 2 + x2}
  cons#_A(x1,x2) = max{0, x1, 2 + x2}

precedence:

  f > empty = cons