## Prove $\epsilon(S\cap T)\subseteq S \cap T$

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?