Prove that at this time, $x_{ij}$ has a total of $m=m_{1} m_{2}$ numbers, so it is only necessary to prove that they are pairwise distinct modulo $m$.
If $x_{i_{1}}^{(1)}+m_{1} x_{j_{1}}^{(2)} \equiv x_{i_{1} j_{1}} \equiv x_{i_{2} j_{2}} \equiv x_{i_{2}}^{(1)}+m_{1} x_{j_{2}}^{(2)}\left(\bmod m_{1} m_{2}\right)$,
then it must be that $x_{i_{1}}^{(1)} \equiv x_{i_{2}}^{(1)}\left(\bmod m_{1}\right)$,
thus it must be that $i_{1}=i_{2}, x_{i_{1}}^{(1)}=x_{i_{2}}^{(1)}$ (because $x_{i}^{(1)}$ takes values in the same complete residue system modulo $m$).
Furthermore, we get $m_{1} x_{j_{1}}^{(2)} \equiv m_{1} x_{j_{2}}^{(2)}\left(\bmod m_{1} m_{2}\right)$,
which means $x_{j_{2}}^{(2)} \equiv x_{1_{1}}^{(2)}\left(\bmod m_{2}\right)$.
Similarly, we have $j_{1}=j_{2}, x_{j_{1}}^{(2)}=x_{j_{2}}^{(2)}$, and the theorem is proved.