In ECC, how do I prove that point addition is commutative? I am studying elliptic curve cryptography and this question is related to the commutative property of point addition operation.

Point addition $P_3(x_3,y_3)$ of two points $P_1(x_1, y_1)$ and $P_2(x_2,y_2)$ is given by the following rules:
$$x_3 = (\lambda^2 - x_1 - x_2) \bmod n$$
and
$$y_3 = (\lambda (x_1 - x_3) - y_1) \bmod n$$
where 
$$\lambda = \frac{y_2-y_1}{x_2-x_1} \bmod n.$$

To prove that the point addition operation is commutative, I simply interchange $x_1$ by $x_2$, and $y_1$ by $y_2$, and what I get is:     

$$x_3' = (\lambda^2 - x_2 - x_1) \bmod n$$
and
$$y_3' = (\lambda (x_2 - x_3) - y_2) \bmod n.$$

Now it can be seen that $x_3 = x_3'$ but $y_3 \ne y_3'$; that is, I am not getting the same point $(x_3, y_3)$. But I have read everywhere that point addition is a commutative operation. 

Please help me understand the problem.
If I am not mistaken, $y_3 = y'_3$:

$y_3 = \frac{y_2 - y_1}{x_2 - x_1}(x_1 - x_3) - y_1 = \frac{y_2 x_1 - y_2 x_3 - y_1 x_1 + y_1 x_3 - y_1 x_2 + y_1 x_1}{x_2 - x_1} = \frac{y_2 x_1 - y_2 x_3 + y_1 x_3 - y_1 x_2}{x_2 - x_1}$

$y'_3= \frac{y_1-y_2}{x_1-x_2}(x_2-x_3) - y_2 = \frac{y_1 x_2 - y_1 x_3 - y_2 x_2 + y_2 x_3 - y_2 x_1 + y_2 x_2}{x_1-x_2} = \frac{y_2 x_1 - y_2 x_3 + y_1 x_3 - y_1 x_2}{x_2 - x_1}$