For each of these implications, state whether it holds for arbitrary regular expressions R, S and justify your answer
If $R \equiv RR$, then $R \equiv \emptyset$ or $\epsilon \in L(R)$
Solution:
$R = RR$
$R + \emptyset \equiv RR + R \emptyset$
$R + \emptyset \equiv R(R + \emptyset) \to R \equiv \emptyset$
$R \epsilon \equiv RR \epsilon$
$R \epsilon \equiv R(R\epsilon)$
$\epsilon \equiv R \epsilon$
$\epsilon \equiv R \to \epsilon \in L(R)$
Pretty lost.. can someone help me?
Suppose that $R \equiv RR$. Now assume, towards a contradiction, that $R \not\equiv \emptyset$ and $\epsilon \notin L(R)$. Then there must exist some $x \in L(R)$ such that $x$ has minimal length, and we know that for every $s \in L(R)$, we have that $|s| > 0$.
Now since $R \equiv RR$, we know that $x \in L(RR)$, so there must exist some $y, z \in L(R)$ such that $x = yz$. But since $|x| = |y| + |z|$ and $|y|, |z| > 0$, we know that $|y| < |x|$ and $|z| < |x|$, contradicting the fact that $x$ was suppose to have the minimal length out of all the strings accepted by $R$.
We conclude that $R \equiv \emptyset$ or $\epsilon \in L(R)$, as desired. $~~\blacksquare$