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## 1 Answer

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A language is a set.

So, $L_{1} = \{a^{n}b^{n} : n \geq 0\}$

Which can be written like this: $\{\epsilon , ab, aabb, aaabbb, aaaabbbb, ... \}$.

And $L_{2} = \{a^{n}b^{m} : n \neq m\}$

Which is: $\{aab, abb, aaab, aaabb, abbb, aabbb, ... \}$.

When you union these two languages (or, sets), you get another set, which contains elements (or strings), which is in at least one of the two sets.

From this you can see, the union of these two languages will give:

$\{\epsilon , ab, aabb, aaabbb, aaaabbbb, ... , aab, abb, aaab, aaabb, abbb, aabbb, ... \}$

Which is nothing but the set of strings of the type $a^{*}b^{*}$.

Which is a regular language.

Hope it helps :)

answered by Boss (8.7k points)
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@Rishabh Gupta 2 Yes, it very much does! Thanks a lot.

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