This is proved below.

For graphs $X$ and $Y$, we use the notation $X+Y$ to denote a graph that is a disjoint union of two graphs isomorphic to $X$ and $Y$, respectively. Suppose $G$ and $H$ are connected graphs having the same vertex deck. Consider graphs $2G = G+G$ and $2H=H+H$.

\[lem-kelly2\] Let $F$ be a graph on fewer than $2v(G)$ vertices. If $F$ has a component isomorphic to $G$ (in which case we write $F=G+X$) then $\displaystyle\binom{2G}{G+X} = \displaystyle\binom{2H}{H+X}$. If $F$ has no component isomorphic to $G$ then $\displaystyle\binom{2G}{F} = \displaystyle\binom{2H}{F}$.

When $F=G+X$, $X$ must have fewer than $v(G)-1$ vertices. Since $G$ and $H$ have identical vertex decks, by Kelly’s Lemma  \[lem-kelly\], $\displaystyle\binom{G}{X} = \displaystyle\binom{H}{X}$. Therefore, $\displaystyle\binom{2G}{G+X} = \displaystyle\binom{2H}{H+X}$.

When $F$ does not have a component isomorphic to $G$, then if $F$ has a component on $v(G)$ vertices then $\displaystyle\binom{2G}{F} = \displaystyle\binom{2H}{F} = 0$. Therefore, assume that all components of $F$ have at most $v(G)-1$ vertices. Any realisation of $F$ as an induced subgraph of $2G$ is a disjoint union of graphs isomorphic to $X$ and $Y$ such that $X$ is an induced subgraph of one component of $2G$ and $Y$ is an induced subgraph of the other component 