0$ we have that $s \ne -1$ and hence $\Gamma$ is connected. Then, using Proposition \[pro:comp\], we see that the complement of $\Gamma$ also satisfies the hypothesis of Corollary \[cor:beatDel01\] if $$0 < {\operatorname{frac} \left ( (v-k-1)/(r+1) \right )} < 1-(s^2 +s)/\mu.$$ In particular, for the corollary to hold for both $\Gamma$ and its complement, we must have both $(r^2 + r)/(v - 2k + \lambda) < 1$ and $(s^2 + s)/\mu < 1$. But we find that $(r^2 + r)/(v - 2k + \lambda) < 1$ if and only if $(s^2 + s)/\mu > 1$. One can see this by using the equalities in Proposition \[pro:SRG\] (see Appendix \[sec:comp\]) to obtain the equality $$\mu(v-2k+\lambda) = (r^2+r)(s^2+s).$$

How sharp is the clique adjacency bound? {#sec:a_limitation_of_the_bound}
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In this section we show that the clique adjacency bound is sharp for strongly regular graphs in certain instances. We also comment on the sharpness of the clique adjacency bound for general strongly regular graphs.

\[thm:limitation\] Let $\Gamma$ be a strongly regular graph with parameters $(v,k,\lambda,\mu)$ and eigenvalues $k > r \geqslant s$. Suppose that $\lambda + 1 \leqslant -k/s$. Then the clique adjacency bound is equal to $\lambda + 2$.

As the second 