Proof that congruence is compatible with multiplication

Proof

1. Let $n$ be a positive integer, and let $a, b, c, d$ be integers with $a \equiv b \pmod n$ and $c \equiv d \pmod n$.
2. By the definition of congruence modulo n, $n$ divides $a - b$ and $n$ divides $c - d$. So there are integers $s$ and $t$ with $a - b = n s$ and $c - d = n t$.
3. Expanding gives $a (c - d) + d (a - b) = a c - a d + a d - b d = a c - b d$.
4. By steps 2 and 3, $a c - b d = a n t + d n s = n (a t + d s)$.
5. Since $a t + d s$ is an integer, $n$ divides $a c - b d$.
6. Hence $a c \equiv b d \pmod n$.