It is well known that if $P$ is a proper notion of forcing and $\Vdash_P \dot{Q} \text{ is proper}$ then the iteration $P \ast \dot{Q}$ is proper. Is the converse also true, i.e

Suppose that we have a two step iteration, such that $P$ and $P \ast \dot{Q}$ are proper forcings. Can we conclude that $\Vdash_P \dot{Q} \text{ is proper}$?

My intuition would tell me no, we could probably add in the first step a new stationary set, which gets killed after the second forcing. We have to ensure however that while doing this the iterated forcing remains proper, i.e. preserves stationary subsets of $[\lambda]^\omega$ of $V$ (for $\lambda$ an arbitrary uncountable cardinal).

If the answer to my question is negative, could at least the following be true: Suppose that $P$ and $P\times Q$ are proper. Does it follow that $\Vdash_P Q \text{ is proper}$?