I am facing problem in understanding the last part of the proof of Theorem 6 (Chinese Remainder Theorem. I cannot understand why $(m_j,n_j)=1$. I do not understand the line "Solving $m_j x\equiv 1 (\mod n_j)$ using Theorem 5, we have a unique solution $x\equiv m_j (\mod n_j)$". What is $m_j'$? I also do not understand the rest of the proof. Please help.
The numbers $n_1,\ldots,n_k$ are relatively prime. Put $M=\prod_{i=1}^k n_i$ and $m_j=M/n_j$ for $1\leq j\leq k$. Then clearly $n_j$ and $m_j$ are relatively prime, i.e., $(m_j,n_j)=1$. Then by Bezout's theorem, $1 = x_jm_j+y_jn_j$ for some numbers $x_j,y_j$. Then $m_jx_j\equiv 1 \mod n_j$. These solutions $x_j$ (which are called $m'_j$) are then put together to give a solution $x_0$ for the system in question.