Let $M$ be a monoid (e.g. $M = \Sigma^*$) and $L \subseteq M$.
Then $\mathsf{RAT}(M)$ is the set of rational sets of $M$ and the elements of $\mathsf{RAT}(M)$ are inductively defined as follows:
- $|L| < \infty \implies L \in \mathsf{RAT}(M)$
- $L_1, L_2 \in \mathsf{RAT}(M) \implies L_1 \cup L_2 \in \mathsf{RAT}(M)$
- $L_1, L_2 \in \mathsf{RAT}(M) \implies L_1 \cdot L_2 \in \mathsf{RAT}(M)$
- $L \in \mathsf{RAT}(M) \implies L^\ast = \bigcup_{i \geq 0} L^i \in \mathsf{RAT}(M)$ with $L^0=\{1_m\}$ and $L^{i+1} = L^iL$
And in my understanding, a rational expression is then an expression using only the operations in this definition, i.e. $$L_1 \cup L_2 \cup L_3L_4 \cup L_3^*$$.
On the other hand we have the definition of regular languages where $\mathsf{REG(M)}$ is the set of regular sets inductivly defined as follows:
- $\emptyset$ is a regular expression
- $\epsilon $ is a regular expression
- $a \in \Sigma$ is a regular expression
- if $\alpha$ and $\beta$ are regular expressions, then $\alpha\mid\beta$ is a regular expression
- $\alpha^*$ and $\alpha\beta$ are regular expressions
Futhermore we know that regular languages are closed under complement unlike rational languages.
Question: It is often stated that regular expressions are the same as rational expressions. But hence this defintion of a rational expression doesn't allow for example "the decision"-operator, ($a \mid b$), it seems that regular expressions cover more than rational ones. What am I missing out? Are they really the same and if not, where are they distinct?