It is provable by contrapositive, as below (after the post):
If $((r \mid s),$ & $(r \nmid t))$, then $s\nmid t$,
i.e., $\exists k \in \mathbb{Z}, s=kr, \forall m \in \mathbb{Z}, t\ne ms \implies \forall m \in \mathbb{Z}, \exists k \in \mathbb{Z}, t\ne mkr$.
I mean that what should be the correct existential clauses to be used in conclusion.
It seems more correct to ignore $k$ in conclusion, i.e., $\exists k \in \mathbb{Z}, s=kr, \forall m \in \mathbb{Z}, t\ne ms \implies \forall m \in \mathbb{Z}, t\ne mr$.
The meaning should be that for all values of $m$ the relation is not satisfied, & $k$ is immaterial here.
Need prove that $r\mid s \wedge r\nmid t \nrightarrow s\nmid t$.
Assume that the implication has the form $H\nrightarrow C$, with $H$ being compound, i.e. it is composed of $P\wedge Q$, with $P: r\mid s, Q:r\nmid t$.
Further, assume $P$ to be fixed(tautology), then $Q: r\nmid t \implies C: s\nmid t$.
If use $P$ as tautology, need arrive at $\bot$ form by getting : $P \wedge \lnot P$.
Request vetting the above line, for proving by the contra-diction based approach.
Can use the contrapositive approach to replace $Q\implies C$, by $\lnot C \implies \lnot Q$, with assumption that only divisibility relation in the propositions is reversed under negation.
So, the contrapositive leads to : $P \wedge \lnot C \implies \lnot Q$, this leads to : $r\mid s \wedge s\mid t \implies r\mid t$, which is true by the transitive property of the divisibility rule, i.e. $r\mid s \wedge s\mid t \implies r\mid t$.
