r(e) \cap r(f)$ for some $z\neq u,v$. Then $z\in e\cap f$, a contradiction. Thus each vertex has at most one incident edge and $\phi(M)$ is a matching. Note that $\phi(M)\notin {\mathcal{M}}(G)$ since $uy\in \phi(M)$ for some $y\in N_G(u,{\overline}{v})$ and no such edge is in $E(G)$. So $\phi(M) \in {\mathcal{M}}(H)\setminus {\mathcal{M}}(G)$.\
To finish the proof of the lemma we need only show that $\phi$ is an injection. We’ll show that $\phi$ has a left inverse defined similarly to $\phi$. Consider $r': E(G) \to E(H)$ defined by $$r'(e)= \begin{cases}
vy &\text{ if } e = uy \text{ for any } y \in N_G(u,{\overline}{v})\\
uz &\text{ if } e = vz\\
e &\text{otherwise}
\end{cases}.$$ Define $\phi': {\mathcal{M}}(G)\setminus {\mathcal{M}}(H) \to {\mathcal{M}}(H)\setminus {\mathcal{M}}(G)$ by $\phi'(M) = \{r'(e): e\in M\}$. It is straightforward to check that $\phi'(\phi(M))= M$. Thus $\phi$ has a left inverse and so $\phi$ is injective.

By Corollary \[bipartitecompress\] and Lemma \[differencegraph\], a bipartite graph minimizing the total number of matchings can be found among the threshold bipartite graphs.

Say $P$ is an *out-corner* of a Young diagram $B$ if $P \in B$ and there is no box in the diagram to its right or beneath it. Say $Q$ is an *in-corner* if 