In the paper of Hsiao and Reyzin, Section $1.4$:
Note that to show that no general reduction from $P$ to $Q$ exists requires proving that $Q$ does not exist
Since the statement is about trying to rule out the implication $Q \rightarrow P$, showing that $Q$ does not exist will indeed do the opposite: it makes the implication true for the trivial reason that $(\mathsf{false} \rightarrow \mathsf{proposition})$ is $\mathsf{true}$ for any $\mathsf{proposition}$.
Could someone explain?
(Note that this attempted answer contains a significant amount of guessing and must not be considered reliable!)
If one assumes "there is a general reduction from $P$ to $Q$" to mean "existence of $Q$ implies existence of $P$" (which is up to interpretation since the authors do not clearly define that term, but it seems somewhat reasonable), then the statement in question is wrong: Under this assumption, it can be reformulated as
If the existence of $Q$ does not imply the existence of $P$, then $Q$ does not exist.
...whose contrapositive
If $Q$ exists, then the existence of $Q$ implies the existence of $P$.
is equivalent to "there is a reduction from $P$ to $Q$", and this is clearly untrue in general.
My best guess is, still under the assumption mentioned in the beginning, that this is a simple typo and they actually meant
Note that to show that no general reduction from $P$ to $Q$ exists requires proving that $P$ does not exist
...which is true (and still conveys the point they are trying to illustrate) since its contrapositive
If $P$ exists, then the existence of $Q$ implies the existence of $P$.
is obviously a tautology.
