On Proving That a Primitive Does Not Exist 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.