Consider a homogeneous Poisson process with inter-arrival times $T_i$, which follows the exponential distribution with rate $\lambda$. Let $N(t)$ denote the number of arrivals by time $t$.
Suppose I start monitoring the arrivals from time $t$ onwards. Let $U_t$ denote the time I need to wait for an arrival. Now suppose I know $N(t)=1$, and I'd like to show that $$ P(U_t>x\mid N(t)=1)=e^{-\lambda x}. $$
Here's what I've tried: I expressed the event $\{N(t)=1\}$ as $\{T_1<t,T_1+T_2>t\}$ and the event $\{U_t>x\}$ as $\{T_1+T_2>t+x\}$. Then I did: $$ \frac{P(T_1+T_2>t+x,T_1+T_2>t,T_1<t)}{P(N(t)=1)}=\frac{P(T_1<t)-P(T_1+T_2<t+x)}{P(N(t)=1)}, $$ where the distributions of $T_1,T_1+T_2,N(t)$ are all known. But after calculations I got something far away from an exponential distribution. I think my calculations were okay. Then it must be something conceptually wrong I assume? Please help!
\begin{align} & \Pr( U_t >x \mid N(t) = 1) \\[10pt] = {} & \Pr(N(t+x) -N(t)=0 \mid N(t)=1) \\[10pt] = {} & \Pr(N(t+x)-N(t)=0) \\[3pt] & \text{ since } N(t) \text{ and } N(t+x)-N(t) \text{ are independent} \\[10pt] = {} & \frac{(\lambda x)^0 e^{-\lambda x}}{0!} = e^{-\lambda x}. \end{align}
You considered the event $$[T_1+T_2>t+x\,\,\, \&\,\,\, T_1+T_2 >t\,\,\, \&\,\,\, T_1< t].$$ Notice that that is the same as $$[T_1+T_2>t+x\,\,\, \&\,\,\, T_1<t],$$ so there is some redundancy there. You equated $\Pr(T_1+T_2>t+x\,\,\, \&\,\,\, T_1<t)$ with $\Pr(T_1<t)-\Pr(T_1+T_2<t+x).$ That would be correct if the event $[T_1+T_2<t+x]$ occurs only when the event $[T_1<t]$ occurs. But that is not correct.