Smallest Positive Integer Combination is Greatest Common Divisor/Proof 2

Theorem
Let $a, b \in \Z_{>0}$ be positive integers.

Let $d \in \Z_{>0}$ be the smallest positive integer such that:
 * $d = a s + b t$

where $s, t \in \Z$.

Then:
 * $(1): \quad d \mathop \backslash a \land d \mathop \backslash b$
 * $(2): \quad c \mathop \backslash a \land c \mathop \backslash b \implies c \mathop \backslash d$

where $\backslash$ denotes divisibility.

That is, by GCD iff Divisible by Common Divisor, $d$ is the greatest common divisor of $a$ and $b$.

Proof
From Bézout's Lemma we have: Let $a, b \in \Z$ such that $a$ and $b$ are not both zero.

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

Then:
 * $\exists x, y \in \Z: a x + b y = \gcd \left\{{a, b}\right\}$

Furthermore, $\gcd \left\{{a, b}\right\}$ is the smallest positive integer combination of $a$ and $b$.

In this instance, $a, b \in \Z_{>0}$ and are therefore both non-zero.

The result then follows from GCD iff Divisible by Common Divisor:

$d = \gcd \left\{{a, b}\right\}$ iff:


 * $(1): \quad d \mathop \backslash a \land d \mathop \backslash b$
 * $(2): \quad c \mathop \backslash a \land c \mathop \backslash b \implies c \mathop \backslash d$