Let's say I have $(E, \leq_{E})$ and $(F, R)$ where $\leq_{E}$ is an order on $E$, and $R$ is some binary relation defined over $F$.
Now if I have a bijection $f : E \to F$ such that,
$$\forall x, y \in E,\: x \leq_{E} y \Leftrightarrow f(x) R f(y)$$
Is it enough to conclude that $R$ defines an order on $F$, as $f$ is an order isomorphism ?
I know it should work, but, can I use the word "order isomorphism" if I have not yet proven $R$ is an order on $F$ ?
It's more a question about an eventual lack or rigour.
I believe that if $E$ has some kind of order on it, then since $f$ is a bijection, then that same kind of order structure will be preserved entirely in $F$. But really, it shouldn't be too hard just to check that the order is preserved, so check it. Then there will be no scruples over calling $f$ an "order isomorphism".
You should verify to yourself as an exercise that the order structure is preserved in $F$. As for whether you include that proof in whatever you are writing up, that is a matter of who your audience is. If this is for a class and you are writing this up for a professor, just include it. If this were for a paper or something where the audience consists of experts in the field, then I would just assert that the order is preserved in $F$.