Duality exchanges B and C

${\mathrm{\Phi }}^{\vee }$ is a reduced crystallographic root system: it is finite, contains no zero vector, and spans $E$ because the ${\alpha }_i^{\vee }$ are positive multiples of the vector-space basis $\mathrm{\Delta }$. If ${\beta }^{\vee }=c{\alpha }^{\vee }$, then $\beta$ is parallel to $\alpha$, so reducedness of $\mathrm{\Phi }$ gives $\beta =\pm \alpha$ and hence ${\beta }^{\vee }=\pm {\alpha }^{\vee }$; thus the dual is reduced. Moreover $2({\beta }^{\vee },{\alpha }^{\vee })\mathrm{/}({\alpha }^{\vee },{\alpha }^{\vee })=2(\alpha ,\beta )\mathrm{/}(\beta ,\beta )\in \mathbb{Z}$, and direct substitution gives $s_{{\alpha }^{\vee }}({\beta }^{\vee })=(s_{\alpha }\beta {)}^{\vee }$, so integrality and reflection stability hold.

It remains to justify that ${\mathrm{\Delta }}^{\vee }=\{{\alpha }_i^{\vee }\}$ is a base, rather than merely a vector-space basis. Choose a regular vector $v$ whose positive system has base $\mathrm{\Delta }$. Since every ${\beta }^{\vee }$ is a positive scalar multiple of $\beta$, the same $v$ is regular for ${\mathrm{\Phi }}^{\vee }$ and makes ${\beta }^{\vee }$ positive exactly when $\beta$ is positive. Let ${\omega }_1,\dots ,{\omega }_r$ be the inner-product dual basis to ${\alpha }_1,\dots ,{\alpha }_r$, and for each $i$ put $x_i={\sum }_{j\mathrm{\ne }i}{\omega }_j$. If $\beta ={\sum }_j{n}_j{\alpha }_j$ is positive, [L2] gives $n_j\ge 0$, and $(x_i,\beta )={\sum }_{j\mathrm{\ne }i}{n}_j\ge 0$; equality holds only when $\beta$ lies on the positive ray of ${\alpha }_i$, hence only when $\beta ={\alpha }_i$ by reducedness. The same vanishing criterion holds for ${\beta }^{\vee }$ because it is a positive multiple of $\beta$. If ${\alpha }_i^{\vee }$ were a sum of two positive dual roots, pairing with $x_i$ would force both summands to equal ${\alpha }_i^{\vee }$, an impossibility. Thus every ${\alpha }_i^{\vee }$ is simple in the dual positive system. By [L2] the complete set of dual simple roots is a basis and has $r=\dim E$ elements; it therefore equals the $r$-element linearly independent set ${\mathrm{\Delta }}^{\vee }$. Its Cartan matrix has entries $a_{ij}^{\vee }=2({\alpha }_j^{\vee },{\alpha }_i^{\vee })\mathrm{/}({\alpha }_i^{\vee },{\alpha }_i^{\vee })=a_{ji}$, so it is $A^T$. The Dynkin diagram consequently reverses every arrow and keeps each edge multiplicity, since the multiplicity is $a_{ij}{a}_{ji}=a_{ji}^{\vee }{a}_{ij}^{\vee }$.

Inspecting the classified diagrams: the simply-laced types $A_n,D_n,E_6,E_7,E_8$ have symmetric Cartan matrices, so they are self-dual; the triple-edge diagram $G_2$ and the double-edge path $F_4$ are each isomorphic to their arrow-reversed diagrams (interchanging the two $G_2$ vertices, and reversing the $F_4$ path), so those types are self-dual up to isomorphism with long and short roots exchanged; and for $n\ge 3$ the transpose of the $B_n$ matrix is the $C_n$ matrix and conversely, while $B_2$ and $C_2$ have isomorphic diagrams and $B_1=C_1=A_1$.

Combining steps 1.1-3.1 gives the assertions: duality is an involution on reduced crystallographic root systems, transforms the Cartan matrix by transposition and the diagram by arrow reversal, and therefore exchanges $B_n$ with $C_n$ and fixes every other classified type up to isomorphism.