%or loop~$f$, then $\pi(f)$ would be removed when flattening~$G/e$, so $f$ never participates
%in a MST.
The right-hand side of the equality is a spanning tree of~$G$, let us denote it by~$T$ and
-the MST of $G/e$ by~$T'$. If $T$ were not minimum, there would exist a $T$-light edge~$f$ in~$G$.
-If the path $T[f]$ covered by~$f$ does not contain~$e$, then $\pi[T[f]]$ is a path covered by~$\pi(f)$
-in~$T'$. Otherwise $\pi(T[f]-e)$ is such a path. In both cases, $f$ is $T'$-light, which should
-contradict the minimality of~$T'$, but we need a multigraph version of Theorem \thmref{mstthm}
-which we did not prove.
-
-To avoid it, we consider the flattening~$F$ of~$G/e$ and apply the theorem to it.
-According to the Flattening lemma (\thmref{flattening}), $F$~has the same MST as~$G/e$
-and therefore it contains all edges of~$T'$. If $\pi(f)\not\in E(F)$, it
-has been removed in favor of some lighter edge~$f'$ and if $\pi(f)$ was
-$T$-light, then $f'$ is even so.
+the MST of $G/e$ by~$T'$. If $T$ were not minimum, there would exist a $T$-light edge~$f$ in~$G$
+(according to Theorem \thmref{mstthm}). If the path $T[f]$ covered by~$f$ does not contain~$e$,
+then $\pi[T[f]]$ is a path covered by~$\pi(f)$ in~$T'$. Otherwise $\pi(T[f]-e)$ is such a path.
+In both cases, $f$ is $T'$-light, which contradicts the minimality of~$T'$. (We do not have
+a multigraph version of the theorem, but this direction is a straightforward edge exchange,
+which of course works in multigraphs as well.)
\qed
\rem
-In the previous algorithm, the role of the mapping~$\pi^{-1}$ is played by the edge labels~$\ell$.
+In the previous algorithm, the role of the mapping~$\pi^{-1}$ is of course played by the edge labels~$\ell$.
\section{Minor-closed graph classes}