equivalence of regular expression: $L(\alpha\emptyset) = L(\emptyset)$

Question

I need some help writing the proof for $L(\alpha\emptyset) = L(\emptyset)$ where $\alpha$ ist a regular epression and $L\widehat{=}$Language.
What I have done so far:

$L(\alpha\emptyset)=L(\alpha)L(\emptyset)=\{w\in\Sigma^*\mid \exists l\in L(\alpha)\wedge\exists m\in L(\emptyset):w=lm\}\overset{\emptyset \text{ has no elements}}{=} \{w\in\Sigma^*\mid \exists l\in L(\alpha)\wedge \color{red}{FALSE ....}\color{black}\}=\{w\in\Sigma^*\mid \color{red}{...FALSE ....}\color{black}\}= ...= L(\emptyset)$

I have no idea how to handle the FALSE within a discription of a set.

Thanks!

Answer 1

Consider the description of a subset $S$ of some universe $X$ of the form $$ S = \{x \in X \mid \varphi(x) \} $$ where $\varphi$ is some logical formula. What happens if $\varphi$ is $FALSE$? Then $S$ is the empty set. For instance $$ \{ n \in {\Bbb Z} \mid n^2 = -1 \} = \emptyset. $$