∅-free regular expressions? This is a question involving regular expressions for regular languages.

I am currently stuck trying to prove that the operand ∅ is not necessary unless the language is the empty set. That is, a regular expression is either equivalent to ∅ or is a ∅-free regular language. (A language is ∅ free if it contains no occurences of ∅)

This seems to be an induction proof, but I'm not really sure how to prove this. It seems like you're possibly trying to show that if a set is ∅ free, then doing anything involving ∅ will either result in the language ∅ or a ∅ free language.

So if r is ∅ free, then (r+∅) = r, r∅=∅ etc

I'm not really sure if this is really complete. I'm not overly comfortable with regular languages yet, so any help is much appreciated.
Preforming the proof by induction on the structure of the tree is good way to go about this. I like to think about how I would write a program to remove them because this directly leads to a constructive proof.

Say our regular expressions have the following form

$$
\begin{align}
regex ::=&amp; \,\, regex \,\, regex \\
        |&amp; \,\, regex^* \\
        |&amp; \,\, regex^+ \\
        |&amp; \,\, regex\,? \\
        |&amp; \,\, regex + regex \\
        |&amp; \,\, char \\
        |&amp; \,\, \varnothing
\end{align}
$$

The bases cases are the cases in which the regex is either a character or $\varnothing$

To give an inductive step say we have a regular expression of the form $regex + regex$. We can then recursively convert the two sub parts to regular expressions which either are the empty set or contain no empty set. So we have 4 cases

case 1: Both expressions contain no $\varnothing$ in which case we are done

case 2/3/4: One of the sides (possibly both) is $\varnothing$ so we remove that side. One needs to prove that this is valid but it isn't hard.

all the cases are going to precede as such with different proofs that the empty set can be eliminated.