# Show that $L=\{a^nb^m : m \neq n\}$ is context free language using closure under union.

I'm trying to solve the following problem. I am asked to show that $$L = \{a^nb^m : m \neq n\}$$ is context free by expressing this language $$L$$ as the union of two other context-free languages.

However, I can't seem to figure out what two context-free languages can be combined via union?!

I know that if I let $$L_1=\{a^nb^m : n < m\}$$ and $$L_2 = \{a^nb^m : n > m\}$$, then $$L = L_1 \cup L_2$$, but how do I go about showing that $$L_1$$ and $$L_2$$ are context free? I can't seem to come up with any context free grammars with them. Can anyone give some advice or suggest a different two context free languages to form $$L$$ via union?

You can show that $$L_1$$ is a context free language because it is produced by the following context-free grammar with starting symbol $$S$$: $$S\longrightarrow Sb \text{ } | \text{ }Cb \\ C\longrightarrow \epsilon \text{ } | \text{ }aCb$$
A similar grammar will show that $$L_2$$ is a context free language.