When proving that conditions $A$ and $B$ are equivalent, it is often an arbitrary choice whether to first prove $A\implies B$ or $B\implies A$. Are there examples where the second implication uses the first in a nontrivial way, and becomes significantly more difficult without invoking the former?



Asymmetric $A \iff B$ proofs
The proof of the Shephard-Todd theorem proceeds exactly that way. The theorem can be stated as:


For a finite subgroup $G\subset GL(n,\mathbb C)$ the following are equivalent:


(A) $G$ is generated by complex reflections (i.e., elements $g$ with $\text{rank}(g-\mathbf1\_n)=1$).


(B) The ring of invariants $\mathbb C[x\_1,\ldots,x\_n]^G$ is a polynomial ring.


First, Shephard-Todd prove $A\Rightarrow B$ by classifying all complex reflection groups and then proceed by inspection (Chevalley found later a conceptual proof of this direction).


To prove $B\Rightarrow A$ let $H\subseteq G$ be the subgroup generated by all complex reflections. From $A\Rightarrow B$ one gets that $\mathbb C[x\_1,\ldots,x\_n]^H=\mathbb C[y\_1,\ldots,y\_n]=:P$ is a polynomial ring. By assumption $\mathbb C[x\_1,\ldots,x\_n]^G=\mathbb C[z\_1,\ldots,z\_n]=:Q$. The ring extension $P|Q$ is, by construction, unramified in codimension one. On the other hand it is ramified along the zero set of the Jacobian $\det(\partial z\_i/\partial y\_j)$. So $P=Q$ and therefore $H=G$.


