# Is language decideable (subset)?

I'm working on a proof for following question

$$L=\{(R,S)\mid \text{R,S are regular expressions and } L(R)\subset L(S)\}$$. Show that this language is/isn't decidable.

A language is decidable iff we have a Turing machine that halts for all input strings. If $$L$$ is regular, we can construct a DFA and therefore a Turing machine. Now we just have to show that this DFA/TM accepts/declines every input.

At least that's what I was told. But isn't it sufficient to show that $$L$$ is regular? Since reg. languages have a DFA and im pretty sure a DFA is decidable (because it can be converted to a TM and is finite)?

Since $$L(R)\subset L(S) \Rightarrow L(R)\cap L(S) = L(R)$$ we get that $$L=L(R) \Rightarrow L$$ regular?

• – D.W. Apr 24 at 23:44
• @D.W. Okay, so the first 2 links confirm, that $L$ is in fact decidable. I don't really get the meaning from the 3rd link. – Quotenbanane Apr 25 at 0:05
• If you can answer your question now, may I invite you to write an answer below? – D.W. Apr 25 at 2:30
• @D.W. Unfortunately, I haven't found a solution yet. Hopefully, I'll find one in the next couple days. – Quotenbanane Apr 26 at 20:48