Assume to the contrary, that $m$ is not a prime number. Then $m = ab$ for some integers $a$ and $b$ with $1 < a < m$ and $1 < b < m$. Let $p$ be the smallest prime that divides $a$ and let $q$ be the smallest prime that divides $b$. WLOG we may assume that $p \le q$. We now consider two cases according to whether $p \in X$ or $p \notin X$.

* If $p \in X$, then either $p = q_i$ for some $i$ with $1 \le i \le s$ or $p = r_j$ for some $j$ with $1 \le j \le t$, but not both ($\{A, B\}$ is a partition). Suppose that $p = q_i$ where $1 \le i \le s$. Since $p \mid a$, then $p \mid m$. Also $p \mid q_1q_2\dots q_s$. Thus $p \mid (m - q_1q_2\dots q_s)$ and so $p \mid r_1r_2\dots r_t$. This implies that $p = r_j$ for some $j$ with $1 \le j \le t$. Contradiction.

* If $p \notin X$, then $q \ge p \ge p_{n+1}$ and so $m \ge pq \ge p_{n+1}^2$. Contradiction.

From the preceding we conclude that $m$ is prime, and we are done.