Every hereditary graph class is determined by its unique minimal forbidden induced subgraphs

If $G\in \mathcal{C}$, then no induced subgraph of $G$ lies outside $\mathcal{C}$, so in particular $G$ contains no member of $\mathcal{B}(\mathcal{C})$.

Suppose $G\mathrm{\notin }\mathcal{C}$. Among vertex sets $W\subseteq V(G)$ for which $G[W]\mathrm{\notin }\mathcal{C}$, choose one of least cardinality; it exists because $W=V(G)$ is available.

Let $\mathcal{F}$ be any forbidden induced-subgraph basis for $\mathcal{C}$. Every $J\in \mathcal{F}$ lies outside $\mathcal{C}$: otherwise $J\in \mathcal{C}$ would contain itself as an induced copy of a member of $\mathcal{F}$, contradicting the defining equivalence for $\mathcal{F}$.

Every proper induced subgraph of $G[W]$ lies in $\mathcal{C}$ by minimality of $\mathrm{\mid }W\mathrm{\mid }$. Hence $G[W]\in \mathcal{B}(\mathcal{C})$.

Fix $H\in \mathcal{B}(\mathcal{C})$. Since $H\mathrm{\notin }\mathcal{C}$, it is not $\mathcal{F}$-free, so some $J\in \mathcal{F}$ occurs as an induced subgraph of $H$. If that copy were proper, then it would lie in $\mathcal{C}$ by the minimality of $H$; closure under isomorphism would give $J\in \mathcal{C}$, contradicting step 1.3. Thus the copy uses all vertices of $H$, and $J\cong H$.

Thus $G$ is not $\mathcal{B}(\mathcal{C})$-free. Together with step 1.1 this proves the equivalence.

Hence every forbidden basis for $\mathcal{C}$ contains, up to isomorphism, every member of $\mathcal{B}(\mathcal{C})$. Since $\mathcal{B}(\mathcal{C})$ is itself a basis by step 3.1, it is inclusion-minimal, and any inclusion-minimal forbidden basis has no additional members. This proves uniqueness up to isomorphism.