Extra-large comment
With reference to the linked post, let assume for the current discussion the theory of arithmetic with the well-known Peano axioms.
We can formalize the theory in predicate logic with some primitive symbols (expressing the corresponding notions): $0, S, +, \times$, where $s$ is the successor function.
With them we can define new relations, like $<$ and $\le$.
Finally, we introduce the following definition:
$\text {Div}(n,m) =_{\text{def}} \exists k (0 < k \land n=k \times m)$,
expressing the fact that "$m$ Divides $n$".
Now, using the $\mathsf {PA}$ axioms and already proved theorems, we can prove the following result: if $\text {Div}(n,6)$, then $\text {Div}(n,2)$, for a number $n$ whatever.
The converse obviously does not hold, because e.g. $\text {Div}(4,2)$ holds but $\text {Div}(4,6)$ does not.
In a more formal way, we have proved that "if $\text {Div}(n,2)$, then $\text {Div}(n,6)$" is a theorem of arithmetic, i.e. that:
$\mathsf {PA} \vdash \text {Div}(n,2) \to \text {Div}(n,6)$, where the symbol $\vdash$ expresses derivability.
Due to Completeness of predicate logic this is equivalent to: $\mathsf {PA} \vDash \text {Div}(n,2) \to \text {Div}(n,6)$, where the symbol $\vDash$ expresses consequence.
Due to this fact, we can read the above result in an ambiguous way using "implies":
in arithmetic, divisibility by $2$ implies divisibility by $6$,
and if we want to use a symbol for "implies", some author uses $\implies$.
But the use of the symbol does not add anything to the result.
See e.g.: Ethan Bloch, Proofs and Fundamentals: A First Course in Abstract Mathematics (Springer, 2nd ed., 2011), page 87:
In the final write-up of a proof, do not use logical symbols, such as $\land, \lor, \exists, \forall$ and $\Rightarrow$, as abbreviations for words. Unless you are writing about logic, where logical
symbols are necessary, the use of logical symbols makes proofs harder for others to read. Of course, you may use any symbols you want in your scratch work.
Bad: $\forall$ distinct real numbers $x \land y$, if $x < y \Rightarrow \exists$ rational $q$ such that $x<q<y$.
Good: For all distinct real numbers $x$ and $y$ if $x < y$ then there exists a rational number $q$ such that $x < q < y$.