Suppose I have two independent alarm clocks which I set right before I go to bed. Their ring times are exponentially distributed with rates $\lambda_1$ and $\lambda_2$. Whenever alarm 1 goes off I immediately reset both alarms, but if alarm 2 goes off I actually get up.
How long do I stay in bed? Am I right in saying that alarm clock one is redundant with respect to bed staying time?
Yes, you are right. This comes down to the fact that exponential random variables are memoryless. The idea is that a "reset" exponential has the same distribution as a conditioned one. Here's the math:
Let $X$ be exponential with rate $\lambda$, and let $b > a$. Then $$\mathbb{P}(X > b | X > a) = \frac{\mathbb{P}(X > b)}{\mathbb{P}(X > a)} = \frac{e^{-\lambda b}}{ e^{-\lambda a}} = e^{-\lambda(b-a)} = \mathbb{P}(X > b - a).$$
In other words, if we know that alarm 2 hasn't gone off after $a$ minutes, then the probability it won't go off after another $b - a$ minutes is the same as the probability a reset clock won't go off for $b - a$ minutes. This proves that resetting doesn't make a difference for the distribution of alarm 2.