vertex set $V$, such that $\langle C\rangle$ is a complete graph.

\[l:clique\] Let $G$ be any ${\Delta}$-regular graph with order $n$, ${\Delta}\ge3$ and $2{\Delta}\le n \le 2{\Delta}+1$. If $G$ contains two cliques of cardinality ${\Delta}$, then these cliques are disjoint. In particular, $G$ contains at most two cliques of cardinality ${\Delta}$.

Seeking for a contradiction, assume that there exist $S_1,S_2\subset V$ cliques of cardinality ${\Delta}$ with $S_1\cap S_2\neq\emptyset$. Denote by $r$ the number $r:= |S_1\cap S_2|$; then $1\le r \le {\Delta}-1$. Note that for any $v\in S_1\cap S_2$ we have ${\delta}_{S_1\cup S_2}(v)= |S_1|-1 + |S_2\setminus S_1|= {\Delta}-1+{\Delta}-r$, so, we obtain $r={\Delta}-1$. Then, we have $|S_1\cup S_2|={\Delta}+1$ and ${\Delta}-1\le|\overline{S_1\cup S_2}|\le {\Delta}$. Besides, we have $N(S_1\cup S_2,\overline{S_1\cup S_2}) = 2 = |(S_1\cup S_2)\setminus (S_1\cap S_2)|$ and, since $|\overline{S_1\cup S_2}|\le{\Delta}$, $N(\overline{S_1\cup S_2},S_1\cup S_2)\ge |\overline{S_1\cup S_2}|\ge {\Delta}-1$. Since $N(S_1\cup S_2,\overline{S_1\cup S_2}) = N(\overline{S_1\cup S_2},S_1\cup S_2)$, we obtain ${\Delta}=3$ and $n=6$; therefore, $G$ is a graph isomorphic to either $K_{3,3}$ or the Cartesian product $P_2\Box K_3$. Thus, we obtain that there are not two non-disjoint cliques in $G$ with cardinality ${\Delta}$. This finishes the proof since, by $n\le 2{\Delta}+1$, it is impossible to have three disjoint cliques of cardinality ${\Delta}$ contained in $G$.

\[r:exitClique\] If 