Common Divisor Divides GCD

Theorem
Let $$a, b \in \Z$$ such that not both of $$a$$ and $$b$$ are zero.

Let $$c$$ be any common divisor of $$a$$ and $$b$$.

That is, let $$c \in \Z: c \backslash a, c \backslash b$$.

Then:
 * $$c \backslash \gcd \left\{{a, b}\right\}$$

where $$\gcd \left\{{a, b}\right\}$$ is the greatest common divisor of $$a$$ and $$b$$.

Proof
Let $$d = \gcd \left\{{a, b}\right\}$$.

Then $$d \backslash a$$ and $$d \backslash b$$ by definition.

Then from Bézout's Identity, $$\exists u, v \in \Z: d = u a + v b$$.

From Common Divisor Divides Integer Combination, $$c \backslash a \and c \backslash b \implies c \backslash u a + v b$$ for all $$u, v \in \Z$$.

Thus $$c \backslash d$$.