Smallest Positive Integer Combination is Greatest Common Divisor/Proof 1

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
Let $D$ be the subset of $\Z_{>0}$ defined as:
 * $D = \left\{{a s + b t: s, t \in \Z, a s + b t > 0}\right\}$

Setting $s = 1$ and $t = 0$ it is clear that $a = \left({a \times 1 + b \times 0}\right) \in D$.

So $D \ne \varnothing$.

By Integers Bounded Below has Smallest Element, $D$ has a smallest element $d$, say.

Thus $d = a s + b t$ for some $s, t \in \Z$.

Proof of $(1)$
From the Division Theorem we can write $a = q d + r$ for some $q, r$ with $0 \le r < d$.

If $r \ne 0$ we have:

So by definition of $D$, it is clear that $r \in D$.

But $r < d$ which contradicts the stipulation that $d$ is the smallest element of $D$.

Thus $r = 0$ and so $a = q d$.

That is $d \mathop \backslash a$.

By the same argument, it follows that $d \mathop \backslash b$ also.

Proof of $(2)$
Let $c \mathop \backslash a$ and $c \mathop \backslash b$.

Then:
 * $\exists u, v \in \Z: a = c u, b = c v$

Thus:

demonstrating that $c \mathop \backslash d$.