I'm trying to understand a Chinese Remainder Theorem proof, with the base case $\displaystyle n=2$, and I'm hoping someone can help me in my jam.

Statement

Suppose $\displaystyle gcd(m_1,m_2)=1 $ Then the system of congruences

$\displaystyle x\equiv{a} (mod$ $\displaystyle m_1)$

$\displaystyle x\equiv{b} (mod$ $\displaystyle m_2)$

has a unique solution $\displaystyle (mod$ $\displaystyle mn)$

Proof

Under conditions

$\displaystyle x-a=m_1R$ and $\displaystyle x-b=m_2T$. So

$\displaystyle x=a+m_1R=b+m_2T$. So

$\displaystyle m_1R-m_2T=b-a

$

This has a solution $\displaystyle R_0 $ and $\displaystyle T_0 $ and the full set of solutions set given by

$\displaystyle R=R_0+m_2k $ and $\displaystyle T=T_0+m_1k$

So $\displaystyle x=a+m_1R_0+m_1m_2k=b+m_2T_0+m_1m_2k$

This apparently immediately demonstrates that this x solves our system of equations and also shows the solution is unique (modulo mn).

That final deduction confuses me. Can someone explain how $\displaystyle x=a+m_1R_0+m_1m_2k=b+m_2T_0+m_1m_2k$ demonstrates that this x solves the system and is unique $\displaystyle (mod$ $\displaystyle mn)$