Chinese Remainder Theorem

Theorem
Let $a, b, r, s \in \Z$.

Let $r \perp s$ (i.e. let $r$ and $s$ be coprime).

Then:
 * $a \equiv b \left({\bmod\, rs}\right)$ iff $a \equiv b \left({\bmod\, r}\right)$ and $a \equiv b \left({\bmod\, s}\right)$

where $a \equiv b \left({\bmod\, r}\right)$ denotes that $a$ is congruent modulo $r$ to $b$.

Proof
Suppose $a \equiv b \left({\bmod\, rs}\right)$.

Then from Congruence by Divisor of Modulus it follows directly that $a \equiv b \left({\bmod\, r}\right)$ and $a \equiv b \left({\bmod\, s}\right)$.

Note that for this result it is not necessary for $r \perp s$.

Now suppose that $a \equiv b \left({\bmod\, r}\right)$ and $a \equiv b \left({\bmod\, s}\right)$.

We have $a \equiv b \left({\bmod\, r}\right) \implies \exists k \in \Z: a - b = kr$ by definition of congruence.

We also have that $k r \equiv 0 \left({\bmod\, s}\right)$ because $a \equiv b \left({\bmod\, s}\right)$.

As $r \perp s$, we have from Common Factor Cancelling in Congruence that $k \equiv 0 \left({\bmod\, s}\right)$.

So $\exists q \in \Z: a - b = qsr$.

Hence by definition of congruence $a \equiv b \left({\bmod\, rs}\right)$.