A quotient block of connected or anticonnected blocks is again connected or anticonnected

Assume first that every original block contained in $D$ is connected. Fix one such block $A_0\subseteq D$. Let $B\subseteq D$ be any other member block. By [L1], choose a mixed chain $A_0=B_0,B_1,\dots ,B_m=B$ inside $D$. We prove by induction on $m$ that $G[B_0\cup \cdots \cup B_m]$ is connected. The case $m=0$ is immediate because $B_0=A_0$ is connected. If $m>0$, then the induction hypothesis gives connectedness of $G[B_0\cup \cdots \cup B_{m-1}]$, the block $B_m$ is connected by assumption, and [L2] gives a cross-edge between $B_{m-1}$ and $B_m$ because that pair is mixed. Hence the union up to $B_m$ is connected. Since $B$ was arbitrary, every member block of $D$ lies in the same connected component of $G[D]$, and therefore $G[D]$ is connected.

Now assume every original block contained in $D$ is anticonnected. By [L3] and [L4], each member block induces a connected subgraph of $\bar{G}$. If two member blocks are consecutive on a mixed chain in $D$, then [L2] gives a cross-nonedge between them in $G$, hence a cross-edge in $\bar{G}$. Repeating the argument of step 1.1 inside $\bar{G}$ shows that $\bar{G}[D]$ is connected. By [L3], this means that $G[D]$ is anticonnected.

Steps 1.1 and 2.1 prove the connected and anticonnected conclusions.