Proof

Suppose \(a \equiv b \pmod{n}\) and \(c \equiv d \pmod{n}\text{.}\) That means \(a = b + kn\) and \(c = d + jn\) for integers \(k\) and \(j\text{.}\) Add these equations:

But \(kn + jn = (k+j)n\text{,}\) which is just a multiple of \(n\text{.}\) So \(a+c = b+d + (j+k)n\text{,}\) or in other words, \(a+c \equiv b+d \pmod{n}\)