Proof.
Assume \(a \equiv b \pmod n\) and \(c \equiv d \pmod n\text{.}\) This means \(a = b + kn\) and \(c = d + jn\) for some integers \(k\) and \(j\text{.}\) Consider \(a-c\text{.}\) We have:
\begin{equation*}
a-c = b+kn - (d+jn) = b-d + (k-j)n\text{.}
\end{equation*}
In other words, \(a-c\) is \(b-d\) more than some multiple of \(n\text{,}\) so \(a-c \equiv b-d \pmod n\text{.}\)