How to prove $P(X_2 < X_3\mid X_1 = \min(X_1, X_2, X_3)) = P(X_2 < X_3)$? When $X$ is an exponential random varaible, the memoryless property is stated as
$$P(X>s+t\mid X>s)=P(X>t)$$
But, I am not sure how to prove 
$$P(X_2 < X_3\mid X_1 = \min(X_1, X_2, X_3)) = P(X_2 < X_3)$$

$X_1$~ $exp$ ($λ1$),  $X_2$~ $exp$ ($λ2$), $X_3$~ $exp$ ($λ3$),  and $X_1$,$X_2$,$X_3$ are independent variables.
Here is my proof:
$$P(X_2 < X_3\mid X_1 = \min(X_1, X_2, X_3)) = P(X_1< X_2 < X_3)/ P(X_1 = \min(X_1, X_2, X_3))$$
And, prove $$P(X_1< X_2 < X_3)/ P(X_1 = \min(X_1, X_2, X_3)) = P(X_2 < X_3)$$
{I have known how to calculate $P(X_1< X_2 < X_3)$ and $P(X_1 = \min(X_1, X_2, X_3))$ }
But, this procedure is too complicated , so I am wondering if there is much easier way too prove 
that.

Does this work?  First, prove that for any $s$, we have 
$$P(X_2<X_3 \mid s < X_2, s < X_3) = P(X_2<X_3)$$ using memorylessness.  Then note that 
$$P(X_2<X_3 \mid X_1 < X_2, X_1 < X_3) = \\ \int_x P(X_2<X_3 \mid X_1 =x, X_1 < X_2, X_1 < X_3)p(X_1=x \mid X_1 < X_2, X_1 < X_3)dx = \\ \int_x P(X_2<X_3 \mid x < X_2, x < X_3)p(X_1=x \mid X_1 < X_2, X_1 < X_3)dx = \\P(X_2<X_3) \int_x p(X_1=x \mid X_1 < X_2, X_1 < X_3)dx = P(X_2<X_3)$$
Here, lowercase $p$ denotes the conditional density function.  (If this works then it seems it's not necessary for $X_1$ to follow any particular distribution.)
