Rooted block exposure and attachment separation

The predecessor of $B$ is a cutvertex node because the incidence tree is bipartite; it is unique by the unique-path property. If a cutvertex $x$ in $B$ differs from $a$, its node is a child of $B$: its path to the root starts through $B$. Every other block containing $x$ is then a child of $x$, so its distance from the root is two more than that of $B$. Indeed, any other direction would give a second path to the root. Thus a block other than $B$ containing $x$ is processed strictly later. If $x$ is not a cutvertex, no other block contains it. Consequently no $x$ in $V(B)$ other than $a$ is exposed earlier, including by a block on another branch.

Let $x$ be such a vertex and suppose $xy$ is an edge with $y$ in an earlier block $D$. Choose a block $C$ containing $xy$. If $C=B$, then $y$ belongs to both $B$ and $D$; since $D$ is earlier, step 1.1 forces $y=a$. If $C$ differs from $B$, then $x$ is a cutvertex and $C$ is a child block of $x$, hence lies farther from the root than $B$. As $x$ and $y$ are distinct, the cutvertex node $y$, if it occurs in the incidence tree, cannot be the predecessor $x$ of $C$. If $D=C$, then $D$ is later than $B$, a contradiction. Otherwise $y$ is a cutvertex shared by $C$ and $D$; it is a child of $C$, and $D$ is a child of $y$. Thus $D$ is still farther from the root than $C$, contradicting its being earlier than $B$. The only possible earlier exposed neighbour is therefore $a$, as required.