<p>I'm going by the question in the title, i.e., that $L_1\not\subset L_2$.
In the body of the question you instead wrote $L_1\subsetneq L_2$ which would in contrast mean that $L_1$ is a strict subset of $L_2$.</p>

<p>Let $L_1$ be any non-regular language. Let $x\in L_1$ be some element of $L_1$. Then set $L_2 = \Sigma\setminus\{x\}$.</p>

<ol>
<li>$L_2$ is clearly regular, since there's a trivial NFA for it that accepts any word except $x$.</li>
<li>$L_1$ is by definition not regular.</li>
<li>$L_1\setminus L_2=\{x\}\neq\emptyset$ and therefore $L_1 \not\subset L_2$</li>
<li>$L_1\cup L_2 = \Sigma$ which is clearly a trivial regular language.</li>
</ol>

CC BY-SA 3.0
2018-02-13T05:22:12.420Z
2018-02-13T05:22:12.420Z
Maeher
<p>Does there exist languages $L1$, $L2$ where $L1$ is non regular, $L2$ is regular $L1\not\subset L2$ and $L1 \cup L2$ is regular?</p>

CC BY-SA 3.0
2018-02-13T01:15:07.243Z
why the downvote?
Torsten Gang
<regular-language>
Is there a non regular and regular language where the non regular is not a subset of the regular and the union is regular?
Torsten Gang
# Is there a non regular and regular language where the non regular is not a subset of the regular and the union is regular?

(Asked by: username_0 on 2018-02-12)

Does there exist languages $L1$, $L2$ where $L1$ is non regular, $L2$ is regular $L1\not\subset L2$ and $L1 \cup L2$ is regular?

**Tags:** regular-language
> **Comments:**
> **username_0**: why the downvote?  

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## 1 answers

### Answer by username_1 on 2018-02-13. Score: 0 | Accepted answer

I'm going by the question in the title, i.e., that $L\_1\not\subset L\_2$.
In the body of the question you instead wrote $L\_1\subsetneq L\_2$ which would in contrast mean that $L\_1$ is a strict subset of $L\_2$.

Let $L\_1$ be any non-regular language. Let $x\in L\_1$ be some element of $L\_1$. Then set $L\_2 = \Sigma\setminus\{x\}$.

1. $L\_2$ is clearly regular, since there's a trivial NFA for it that accepts any word except $x$.
2. $L\_1$ is by definition not regular.
3. $L\_1\setminus L\_2=\{x\}\neq\emptyset$ and therefore $L\_1 \not\subset L\_2$
4. $L\_1\cup L\_2 = \Sigma$ which is clearly a trivial regular language.


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