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The join ${\Gamma }_1\ast {\Gamma }_2$ is obtained from the disjoint union of graphs ${\Gamma }_1\bigsqcup {\Gamma }_2$ by adding every edge between a vertex of ${\Gamma }_1$ and a vertex of ${\Gamma }_2$.

Write $q=m/s$, an integer under the stated divisibility assumption. The degree relation gives $n=rm/s=qr$. Take the disjoint union of graphs of $q$ copies of the complete bipartite graph $K_{s,r}$, placing its $s$-vertex part in $U$ and its $r$-vertex part in $W$. The total part sizes are then $qs=m$ and $qr=n$. Each left vertex has vertex degree $r$, and each right vertex has vertex degree $s$, so the graph has exactly the required biregular graph parameters.

An independent set in one component can be any subset of its left part or any subset of its right part. It cannot contain a vertex from both parts, because all cross-pairs are edges. The two families share only the empty set, so the component has $2^s+2^r-1$ independent sets. Choices in different connected components of a graph are independent combinatorial choices, and their counts multiply. HenceThis attains equality in part (i). It also explains the entropy weight $2^r$ there: the empty selected left neighbourhood permits all $2^r$ subsets of the opposite part.

If $\Gamma ={\Gamma }_1\bigsqcup {\Gamma }_2$ is a nontrivial disjoint union of graphs, no defining relation mixes the two vertex sets, and henceOn the topological side, every clique lies in one component, soa one-point union of Salvetti complexes.

If $\Gamma ={\Gamma }_1\ast {\Gamma }_2$ is a nontrivial join of graphs, every generator from the first part commutes with every generator from the second. ThereforeEvery clique of the join is the union of a clique in each factor, which gives the cubical identity