Independence between maximum and minimum of exponential $X$ and $Y$ are two independent variables with an exponential distribution with parameters $\lambda$ and $\mu$; $A = \min(X,Y)$, $B = \max(X,Y)$ and $C = B-A$. I want to prove that $A$ and $C$ are independent. 
I have provided two different demonstrations, but they are both wrong and i don't know why:
1° Demonstration:
$$ \begin{eqnarray}
\mathbb{P}\left(C<t \Big| A=X\right) &=& \mathbb{P}\left(X+Y-2A<t \Big| Y>X\right) \\ &=& 1-\mathbb{P}\left(Y>2A-X+t \Big| Y>X\right) \\ &=& 1-\mathbb{P}\left(Y>2A-2X-t\right)
\end{eqnarray}
$$ because exponential is memoryless. So:
$$ = \mathbb{P}\left(2X+Y-2A<t\right) = \mathbb{P}\left(C+Y < t\right)$$
So $C$ and $A$ are not independent...Something's wrong!
2° Demonstration:
$$\begin{eqnarray}
\mathbb{P}\left(C<t \big| A=X\right) &=&\mathbb{P}\left(X+Y-2A<t \big| A=X\right) \\ &=& \mathbb{P}\left(X+Y-2A<t \big| A=X, B=Y\right) \\ &=& \ldots
\end{eqnarray}$$ 
by conditioning
$$ \ldots = \mathbb{P}\left(A+B-2A < t\right) = \mathbb{P}\left(B-A < t\right) = \mathbb{P}\left(C<t\right)$$
This seems correct, but that would mean that this property is true for every $X$, $Y$. But this is not true. So where's the mistake?
Thanks
 A: The errors has been pointed by by joriki's aswer. I don't think that starting with $P(C<t∣A=X)$ is the best way to prove independence, you want to show that $P(C<t)$ is the same as $P(C<t∣A=a)$ for any value of $a$.
Then $P(C<t∣A=a) = P(B-A<t\mid A=a) =P(B<t+a\mid A=a) $
But the last term does not depends on $a$ (and this proves independence).
Because (here we used memorylessness) :
$$\begin{align}
P(B<b\mid A=a) =& \,P(B<b \mid A=a  \cap A=X) P(A=X)+\\
&  \,+P(B<b \mid A=a \cap A=Y) P(A=Y)\\
=& \,
P(Y< b-a)  P(A=X) + P(X<b-a)  P(A=Y)
\end{align}$$
So $P(B<t+a\mid A=a)=P(Y<t) P(A=X) + P(X<t) P(A=Y) $ which does not depend on $a$ - note that we don't need to compute $P(A=X)$.
A: In both cases, the step interrupted by text is wrong.
In the first one, it should be $-t$ on the right-hand side of the inequality.
In the second one, I don't know what you mean by "by conditioning"; it seems you used the equations in the condition to rewrite the inequality, which is OK, and then simply dropped the condition, for which I see no justification.
