## formal languages: if \$ L1⊆L2 \$ and \$ L1 not∈RE \$, it is possible that \$ L2∈RE \$

Yes $$L1⊆L2$$and $$L1 not∈RE$$, It is possible that $$L2∈RE$$ ?

It's also hard for me to find languages ​​that aren't RE at all, I've heard of arithmetic hierarchy, but we didn't really learn it in class, so I guess the solution doesn't necessarily have to be correlated with arithmetic hierarchy.

I know that he $$A_ {TM} = {(〈M〉, w): M accepts w }$$ where M is a Turing machine
so $$overline {A_ {TM}} notin RE$$, but it doesn't seem to help.

## semi decidability: there are two languages ​​\$ L_1 \$ and \$ L_2 \$ so that \$ L_1 \$ and \$ L_2 \$ are recursive, but \$ L_1L_2 \$ is not recursive

True or false statements.

There are two languages $$L_1$$ Y $$L_2$$ such that $$L_1$$ Y $$L_2$$ are recurvise but $$L_1L_2$$ It is not recursive.

For me, the statements are true because I can describe a non-recursive one (I understood recursively enumerable) procedure of two recurring languages.

Test:

Suppose that $$L_1$$ Y $$L_2$$ they are recursive languages ​​Then there is a Turing M1 machine that accepts any input in $$L_1$$ and reject any entry that is not in $$L_1$$. Similarly, there is a Turing M2 machine that accepts any input in $$L_2$$ and reject any entry that is not in $$L_2$$. To determine if $$w ∈ L_1L_2$$, we guess in a non-determining way where to divide the chain into $$w_1w_2$$and run M1 in $$w_1$$ and M2 in $$w_2$$. So, if w is in the language, M1 and M2 will eventually accept.

## Does it involve \$ L_1L_2 = L_2L_1 \$ \$ L_1 = L_2?

Leave $$L_1, L_2 subseteq Sigma ^ *$$ be two languages, where $$Sigma$$ It is a finite alphabet.

Make $$L_1L_2 = L_2L_1$$ to imply $$L_1 = L_2$$?

What if $$L_1$$ Y $$L_2$$ Are they regular languages?

Can you give counterexamples?

## demonstrating L1 * ∪ L2 * ⊆ (L1∪L2) *

x∈ L1 * ∪ L2 * ⇔ x∈ L1 * ∨ x ∈L2 * ⇔ x ∈ (L1) * ∨ x∈ (L2) * ⇔ x ∈L1 * ∪ L2 * ⇔ x∈ (L1∪L2) *

Is it enough to try it this way?