YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [cons](x0, x1) = x0 + 7x1 + 1,
   
   [f](x0, x1) = x0 + 4x1 + 1,
   
   [empty] = 4
  orientation:
   f(x,empty()) = x + 17 >= x = x
   
   f(empty(),cons(a,k)) = 4a + 28k + 9 >= a + 11k + 2 = f(cons(a,k),k)
   
   f(cons(a,k),y) = a + 7k + 4y + 2 >= 4k + y + 1 = f(y,k)
  problem:
   
  Qed