Proving that a language defined by a regular expression is equivalent to a right linear grammar After thinking for a bit, I am not able to prove a double inclusion proof for the following problem. It seems very interesting to me.
Consider the regular expression $r= ((1(00)^∗1 + 0)1)^∗$ and the right-linear grammar $G= (\{S,A\},\{0,1\},S,P)$, where $P$ consists of the following rules:

$S\rightarrow 1A|01S|\lambda$
$A\rightarrow 00A|11S$

Prove that $L(G)\subseteq L(r)$ and vice versa.
In general, how exactly do I prove that a regular grammar describes the same language as a regular expression?
Let's try to simplify your grammar. First, notice that $A$ generates $(00)^*11S$, and so we can get rid of $A$, obtaining
$$
S \to 1(00)^*11S \mid 01S \mid \lambda
$$
Similar reasoning shows that your grammar generates the language
$$
(1(00)^*11+01)^* = ((1(00)^*1+0)1)^*,
$$
which is identical to your regular expression.
While the reasoning above is informal, it can be made formal with some work.