The induced copies of $H_1$ in $G$ are counted by summing, over the induced embeddings of $H_1-v$, the number of vertices that extend them at $v$

If $\psi \in \mathrm{\Phi }$ then its restriction ${\psi }^{\mathrm{\prime }}$ to $V(H_1)\setminus \{v\}$ is injective, and for distinct $x,y\in V(H_1)\setminus \{v\}$ the condition $xy\in E(H_1-v)$ is the condition $xy\in E(H_1)$, which holds exactly when $\psi (x)\psi (y)\in E(G)$; so ${\psi }^{\mathrm{\prime }}\in \mathrm{\Psi }$, and it is the only member of $\mathrm{\Psi }$ that $\psi$ restricts to.

Let $R\subseteq \mathrm{\Psi }\times \mathrm{\Phi }$ consist of the pairs $(\phi ,\psi )$ whose second entry restricts to the first. Both $\mathrm{\Phi }$ and $\mathrm{\Psi }$ are sets of functions between finite sets, hence finite.

The column fibre of $R$ at $\psi \in \mathrm{\Phi }$ has exactly one element by step 1.1, so ${\sum }_{\psi \in \mathrm{\Phi }}\mathrm{\mid }{R}^{\psi }\mathrm{\mid }=\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }$.

The row fibre of $R$ at $\phi \in \mathrm{\Psi }$ is carried bijectively onto $X_{\phi }$ by $\psi \mapsto \psi (v)$: the map is injective because $\psi$ is determined by $\phi$ together with $\psi (v)$, and its image is exactly $X_{\phi }$, because a vertex $u$ arises as some $\psi (v)$ precisely when extending $\phi$ by $v\mapsto u$ gives an induced embedding of $H_1$, and injectivity of that extension is exactly the requirement $u\mathrm{\notin }\phi [V(H_1)\setminus \{v\}]$.

Double counting $R$ therefore gives ${\sum }_{\phi \in \mathrm{\Psi }}\mathrm{\mid }{X}_{\phi }\mathrm{\mid }=\mathrm{\mid }R\mathrm{\mid }=\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }={\mathrm{ind}}_{H_1}(G)$.

Every member of $\mathrm{\Psi }$ is a function from $V(H_1)\setminus \{v\}$, a set of $h_1-1$ elements, to $V(G)$, so $\mathrm{\Psi }$ is a subset of a set of size $n^{h_1-1}$ and $\mathrm{\mid }\mathrm{\Psi }\mathrm{\mid }\le n^{h_1-1}$.