Is this a valid proof that $\{ax + by|x,y \in \Bbb Z\}= \{n\times \gcd(a, b) |n\in \Bbb Z\}$? I'm trying to prove that $\{ax + by|x,y \in \Bbb Z\}= \{n\times \gcd(a, b) |n\in \Bbb Z\}$, but I'm unsure on the . The main proposition I'm using to solve this is that $\exists x,y, ax+by = \gcd(a, b)$. However, I'm having trouble with the exact mathematical presentation of the proof. I don't have a formal education in proofs, and so that's why I ask; is this a valid proof of this proposition?
$$\{n\times \gcd(a, b) |n\in \Bbb Z\}$$
$$= \{n\times (as + bt) |n,s,t\in \Bbb Z\}$$
$$= \{a(ns) + b(nt) |n,s,t\in \Bbb Z\}$$
And, because we know that $ns=x\in \Bbb Z$ and $nt=y \in \Bbb Z$, we can conclude that:
$$= \{ax + by |x,y\in \Bbb Z\}$$
The main issue I'm having is with the first step: is that equality even valid? It seems that 
$= \{n\times (as + bt) |n,s,t\in \Bbb Z\}$ has more elements than $\{n\times \gcd(a, b) |n\in \Bbb Z\}$.

Theorem $\rm \ \ \ d\mid a,b,\ \ d = ax+by\ \Rightarrow\ a\Bbb Z + b\Bbb Z = d\Bbb Z$
Proof $\rm \qquad\ a\Bbb Z + b\Bbb Z \subseteq d\Bbb Z\ \ \ by\ \ \ a=dj,\, b = dk\ \Rightarrow\ a\Bbb Z + b\Bbb Z = dj\Bbb Z + dk\Bbb Z \subseteq d\Bbb Z$
Conversely $\rm\,\ d\Bbb Z\subseteq a\Bbb Z + b\Bbb Z \ \ \ by\ \ \ dn = anx+bny \in a\Bbb Z + b\Bbb Z\quad $ QED
