Divisor Divides Multiple/Proof 1

Theorem
Let $a, b$ be integers.

Let:
 * $a \mathrel \backslash b$

where $\backslash$ denotes divisibility.

Then:
 * $\forall c \in \Z: a \mathrel \backslash b c$

Proof
Let $a \mathrel \backslash b$.

From Integer Divides Zero:
 * $a \mathrel\backslash 0$

Thus $a$ is a common divisor of $b$ and $0$.

From Common Divisor Divides Integer Combination:
 * $\forall p, q \in \Z: a \mathrel \backslash \left({p \cdot b + q \cdot 0}\right)$

Putting $p = c$ and $q = 1$ (for example):
 * $a \mathrel \backslash \left({c b + 0}\right)$

Hence the result.