Suppose there are sets $ S\subseteq Q, T\subseteq Q$ such that $ T=\epsilon (T),S=\epsilon (S)$ .
Prove $ \epsilon(S\cap T)\subseteq S \cap T$
Definition of $ \epsilon$ – closure for epsilon NFA is:
$ \epsilon : 2^Q \rightarrow 2^Q $
a) $ S \subseteq \epsilon (S)$ Base case
b) If $ q \in \epsilon (S)$ then $ \delta(q,\epsilon )\subseteq \epsilon (S)$ Recursive case
c) and nothing else is in $ \epsilon (S)$
And also, S is a set of all states in epsilon-NFA.
My proof: Since $ S=\epsilon (S)$ and $ T=\epsilon (T)$ , we know that the states in S and T have no $ \epsilon$ -transitions coming out of those states. Therefore states in $ S\cap T$ also have no $ \epsilon$ -transitions coming out of their states. Therefore $ \epsilon(S\cap T) = S \cap T$ .
Is this a correct reasoning?