So if i have two languages $L_1$ and $L_2$ are regular, how can I prove that $L_1 \cap L_2$ is regular with Myhill Nerode.
By Myhill Nerode, I know that $L_1$ and $L_2$ are both made up of the union of equivalence classes over some relations $\simeq_1$ and $\simeq_2$ such that the number of equivalence classes is finite in both cases. Additionally, I know that they're both right invariant relations. So I can (maybe?) say that
$$ L_1 = \bigcup\limits_{C_i \in \Sigma^* / \simeq_1} C_i $$
and for all $x, y, z \in L_1$ i know that for $x \neq y$ that $xz \simeq_1 yz$.
I also know the same for the other language
$$ L_2 = \bigcup\limits_{D_i \in \Sigma^* / \simeq_2} D_i $$
and for all $x, y, z \in L_2$ i know that for $x \neq y$ that $xz \simeq_2 yz$.
So now to prove $L_1$ and $L_2$ are equivalent I need show there is some equivalence relation $\simeq$ over $L_1 \cap L_2$ that has a finite number of equivalence classes and is right invariant. i.e.
$$ L_1 \cap L_2 = (\bigcup\limits_{C_i \in \Sigma^* / \simeq_1} C_i) \,\, \cap (\bigcup\limits_{D_i \in \Sigma^* / \simeq_2} D_i) $$
and for all $x, y, z \in \Sigma$ i know that for $x \neq y$, $xz \simeq yz$
For proving the number of equivalence classes is finite, i think I can just say that the maximum number of equivalence classes will just at most $mn$ given $| \Sigma^*/\simeq_1| = m$ and $| \Sigma^*/\simeq_2| = n$, but i am not sure if that is justification or not. Also another thing i am unclear on is whether an equivalence relation covers the entire language or not (and whether it needs to).
For right invariance, no idea on how to proceed on that front. If $a,b,c \in L_1 \cap L_2$, then maybe I can say that $ac \simeq_1 bc$ and $ac \simeq_2 bc$...but i am not sure how to prove that $ac \simeq bc$.
The Myhill–Nerode theorem is better stated in terms of residuals. Given a word $u$ and a language $L$ of $A^*$ let $$ u^{-1}L = \{v \in A^* \mid uv \in L \} $$ The Myhill–Nerode theorem states that a language is regular if and only if the set $\{u^{-1}L \mid u \in A^*\}$ is finite. Actually its size is the number of states of the minimal automaton of $L$.
Now let $L_1$ and $L_2$ be regular languages and let $n_1$, respectively $n_2$, be the number of states of their minimum DFA. I let you verify that for each word $u$ $$ u^{-1}(L_1 \cap L_2) = u^{-1}L_1 \cap u^{-1}L_2 $$ Now $u^{-1}L_1$ (resp. $u^{-1}L_2$) can take $n_1$ (resp. $n_2$) different values and thus $u^{-1}(L_1 \cap L_2)$ can take at most $n_1n_2$ different values. It follows that $u^{-1}L_1 \cap u^{-1}L_2$ is regular and the minimum DFA of $L_1 \cap L_2$ has at most $n_1n_2$ states.