Let $T,C$ be two real valued independent random variables. I want to see the following equality, where $dt$ denotes interval $[t,t+\delta)$ for any $\delta>0$
$P(T\in dt|T\geq t)=P(T\in dt,T\leq C| min(T,C)\geq t)$.
$LHS=\frac{P(T\in dt,T\leq C,T\geq t\cap C\geq t)}{P(T\geq t\cap C\geq t)}$. Denominator can be factorized by $C,T$ independence. Thus it suffices to check numerator $P(T\in dt,T\geq t)=P(T\in dt,T\leq C,T\geq t\cap C\geq t|C\geq t)$. This will show equality.
$P(T\in dt,T\leq C,T\geq t\cap C\geq t|C\geq t)=P(T\in dt,t\leq T\leq C|C\geq t)$
$Q:$ How do I get rid of $T\leq C$ part in above inequality? $C,T$ independence only means for event $a,b$, $P(C\in a,T\in b)=P(C\in a,T\in b)$. $T\leq C$ does not cut either along $C$ or $T$ direction and it cuts in a slanted direction. Have I made a mistake somewhere? If this does not work, how should I argue about equality?