In the Euclidean algorithm, the basic assumption is the conditional statement that any common divisor $c$ of $a,b$ will divide any linear combination too,
i.e., if $\exists c,a,b \in \mathbb{Z}, c\mid a,b \implies \forall u,v \in \mathbb{Z}, c \mid au + bv.$
I request vetting my reasoning that why it is not a bi-conditional ($\iff$):
If assume: $\exists c,a,b \in \mathbb{Z}, c\mid a,b \iff \forall u,v \in \mathbb{Z}, c \mid au + bv$, and take the 'only-if' part then it means that : $\forall u,v \in \mathbb{Z}, c \mid au + bv \implies \exists c,a,b \in \mathbb{Z}, c\mid a,b $
Then it is clear that need prove assume for all possible values of $u,v$, and hence ...
I am unable to pursue further, or apply any standard approach, as contradiction based one.
To state the theorem precisely: Let $a, b \in \mathbb{Z}$ be fixed, and suppose $c$ is such that $c \mid a, b$. Then for every $u, v \in \mathbb{Z}$, we have $c \mid au+bv$.
The converse is: Let $a, b \in \mathbb{Z}$ be fixed, and suppose $c$ is such that for every $u, v \in \mathbb{Z}$, we have $c \mid au+bv$. Then $c \mid a,b$.
This converse is trivially true: let $v=0, u=1$ to obtain $c \mid a$, and let $u=0, v=1$ to obtain $c \mid b$.