> [!NOTE] Theorem > For all $k,l\in \mathbb{N}^{+},$ $R(k,l)=R(l,k).$ ###### Proof: Sufficient to show that for all $k,l\in \mathbb{N},$ $R(k,l)\leq R(l,k)$ since we have then that $R(l,k)\leq R(k,l),$ since $k$ and $l$ are arbitrary, which then gives $R(k,l)=R(l,k).$ Let $G = (V, E)$ such that $|V | = R(l, k)$. Then we need to show that $G$ contains a $k$-clique or an independent set on $l$ vertices. Consider its complement $G' = (V, {V \choose 2} \setminus E)$. Since $G'$ has $R(l, k)$ vertices, it follows that $G'$ has an $l$-clique or an independent set on $k$ vertices. However, this implies that $G$ contains an independent set on $l$ vertices or a $k$-clique as required. This completes the proof of the theorem.