validation of a pumping lemma proof for regular languages I have the following regular expression:

Of course I could think of a word like $w=a^{m+2}b^{m+2}c^{2m+3}$ and continue with the proof BUT
I was just wondering, because $L$ is made up of a union of two expressions, is it valid to split $L$ into
$ L_1=\{a^ib^jc^{2j-1} | i&lt;=j \land i,j&gt;0\}$
$L_2=\{a^ib^jc^{2j-1}| i&gt;=2j-1 \land i,j&gt;0\}$
and show for  each $L_1$ and $L_2$ that the pumping lemma does not work on them (so for $L_1$ I just show that for each $k$, $i$ is not smaller or equal to $j$, and similarly for $L_2$.
By that I show that $L_1$ and $L_2$ are not regular which means that the union of the two will also be not regular. Is this correct?
thank you.
the union of 2 non-regular languages can be regular.
The trivial example is any non-regular language L and its complement. The union of those two is the set of all strings, which is a regular language.
There for the conclusion $L1, L2 \notin Regular \implies L1\cup L2 \notin Regular$ does not follow.