Proof: First, it should be pointed out that the form of Theorem 14 is different from the previous theorems; the condition here is both sufficient and necessary, making the conclusion stronger. We will first prove it for a complete residue system. We start by proving the sufficiency (which can be derived using Theorem 11 and Theorem 10, but we will provide a direct proof here). At this point, $s = m_{1}, t = m_{2}$. Therefore, there are $m_{1} m_{2}$ numbers $x_{ij}$. For any $1 \leqslant i_{1}, i_{2} \leqslant m_{1}$, $1 \leqslant j_{1}, j_{2} \leqslant m_{2}$, by the condition $\left(m_{1}, m_{2}\right) = 1$ and Property IX of $\S 1$, we have:
$$x_{i_{1} j_{1}} \equiv x_{i_{2} j_{2}} (\bmod m)$$

is equivalent to
$$x_{i_{1} j_{1}} \equiv x_{i_{2} j_{2}} (\bmod m_{1}), \quad x_{i_{1} j_{1}} \equiv x_{i_{2} j_{2}} (\bmod m_{2})$$

which is equivalent to
$$m_{2} x_{i_{1}}^{(1)} \equiv m_{2} x_{i_{2}}^{(1)} (\bmod m_{1}), \quad m_{1} x_{j_{1}}^{(2)} \equiv m_{1} x_{j_{2}}^{(2)} (\bmod m_{2})$$

By Property V of $\S 1$ and $\left(m_{1}, m_{2}\right) = 1$, this is equivalent to
$$x_{i_{1}}^{(1)} \equiv x_{i_{2}}^{(1)} (\bmod m_{1}), \quad x_{j_{1}}^{(2)} \equiv x_{j_{2}}^{(2)} (\bmod m_{2})$$

Since $x_{i}^{(1)}, x_{j}^{(2)}$ take values in the same complete residue systems modulo $m_{1}$ and $m_{2}$ respectively, it must be that $i_{1} = i_{2}, j_{1} = j_{2}$. This proves that these $m_{1} m_{2}$ numbers $x_{ij}$ are pairwise incongruent modulo $m$, i.e., they form a complete residue system modulo $m$.

Next, we prove the necessity. Since $x_{ij} (1 \leqslant i \leqslant s, 1 \leqslant j \leqslant t)$ is a complete residue system modulo $m$, we have $st = m = m_{1} m_{2}$. Fix $x_{1}^{(2)}$, and consider
$$x_{i1} = m_{2} x_{i}^{(1)} + m_{1} x_{1}^{(2)}, \quad 1 \leqslant i \leqslant s$$

which are pairwise incongruent modulo $m = m_{1} m_{2}$. Therefore, $m_{2} x_{i}^{(1)}$ are also pairwise incongruent modulo $m_{1} m_{2}$, i.e.,
$$m_{2} x_{i_{1}}^{(1)} \not\equiv m_{2} x_{i_{2}}^{(1)} (\bmod m_{1} m_{2}), \quad 1 \leqslant i_{1} \neq i_{2} \leqslant s$$

This is equivalent to
$$x_{i_{1}}^{(1)} \not\equiv x_{i_{2}}^{(1)} (\bmod m_{1})$$

which means these $s$ numbers $x_{i}^{(1)}$ are pairwise incongruent modulo $m_{1}$, so $s \leqslant m_{1}$. Similarly, we can prove that the $t$ numbers $x_{j}^{(2)}$ are pairwise incongruent modulo $m_{2}$, so $t \leqslant m_{2}$. From $st = m_{1} m_{2}$, we conclude $s = m_{1}, t = m_{2}$, which proves the necessity.

To prove the conclusion for a reduced residue system, we need to prove (why):
$$\left(x, m_{1} m_{2}\right) = 1$$

is equivalent to
$$\left(x^{(1)}, m_{1}\right) = \left(x^{(2)}, m_{2}\right) = 1$$

Since $\left(x, m_{1} m_{2}\right) = 1$ is equivalent to
$$\left(m_{2} x^{(1)} + m_{1} x^{(2)}, m_{1}\right) = \left(m_{2} x^{(1)} + m_{1} x^{(2)}, m_{2}\right) = 1$$

which is
$$\left(m_{2} x^{(1)}, m_{1}\right) = \left(m_{1} x^{(2)}, m_{2}\right) = 1$$

Since $\left(m_{1}, m_{2}\right) = 1$, by Theorem 5 of Chapter 1 $\S 4$, the above is equivalent to equation (19). Proof complete.