Let $X\neq \emptyset$ be a set, let $\text{Part}(X)$ be the set of partitions of $X$ (where we require that $\emptyset \notin P$ whenever $P\in\text{Part}(X)$).
For $P, Q\in \text{Part}(X)$ we say that $P$ covers $X$ more efficiently than $Q$ if $$\text{card}(P\setminus Q) < \text{card}(Q\setminus P), $$ and we write $P<_{\text{eff}}Q$ for this.
Is $<_{\text{eff}}$ transitive?
I believe the answer is yes. Let $X\neq \emptyset$ and let $P,Q,R\in {\rm Part}(X)$.
First, note that if $|P|<|Q|$, then $P<_{\rm eff}Q$, since $|P-Q|\leq |P|<|Q|=|Q-P|$.
Now we are ready for the proof. Assume $P<_{\rm eff}Q<_{\rm eff}R$. Without loss of generality, we may remove $P\cap Q\cap R$ from all the sets, and reduce to the case that $P\cap Q\cap R=\emptyset$. By our previous paragraph, we also have $|P|\leq |Q|\leq |R|$. If inequality holds anywhere, then we have $P<_{\rm eff}R$ and we are done. Assume the contrary case holds, and so all three sets have equal cardinality. Note that $|P|$ must be infinite, else the computation from my comment above applies.
From the fact that $|P-Q|<|Q-P|$ and $|P|=|Q|$, we see that $|P\cap Q|=|P|=|Q|$. Now because $(P\cap Q)\cap R=\emptyset$, we have $|Q-R|\geq |P\cap Q|=|Q|=|R|\geq |R-Q|$, contradicting the fact that $Q<_{\rm eff}R$. Thus, this contrary case cannot actually happen after all.
