$\Gamma$ is disconnected then $\Gamma$ is the disjoint union of complete graphs and hence contains cliques of size $\lambda + 2$. Therefore the clique adjacency bound is at least $\lambda + 2$. Now we assume that $\Gamma$ is connected.

By Lemma \[lem:NoOtherZeros\], the clique adjacency bound is less than $\lambda+2$ only if there exists some integer $b$ such that $C_\Gamma(b,\lambda + 2)$ is less than zero. To ease notation set $f(x) := C_\Gamma(x,\lambda + 2)$. Hence $$f(x) = x \left ( (v-\lambda-2)x + v + (2\lambda-2k+1)(\lambda+2) \right ).$$ It suffices to show that there does not exist any integer $b$ such that $f(b) < 0$.

Observe that the polynomial $f(x)$ is a quadratic polynomial in the variable $x$. Furthermore, the leading coefficient of $f(x)$ is $v-\lambda-2 \geqslant 0$, and $f(0) = 0$. Let $\xi$ be the other zero of $f(x)$. Now, $f(x)$ is negative if and only if $x$ is between $0$ and $\xi$. Hence, if $f(-1)$ and $f(1)$ are both nonnegative then there are no integers $b$ such that $f(b) < 0$. As in the proof of the previous result $f(-1)$ is nonnegative. Therefore Lemma \[lem:lam1le\] completes the proof for type-II strongly regular graphs.

The inequality $\lambda+1 \leqslant -k/s$ 